2CH-EP [474.21 KiB]

2CH-EP [474.21 KiB]
2
2. SCHRÖDINGER EQUATION I: INTRODUCTORY PROBLEMS
2.1 Introduction
A tension created by efforts and results described in the last chapter resulted during
the first quarter of twentieth century in a formulation of a novel approach to precesses of the
microscopic world. This new approach is called wave mechanics or quantum mechanics i.e. the theory of the microscopic processes which has replaced and surpassed in all aspects
the old Bohr semiclassical theory.
Quantum mechanics was first formulated by W. Heisenberg† , who was using a matrix mechanics, and in an equivalent form by E. Schrödinger.♯ In Schrödinger formulation,
in so called wave mechanics one finds a wave function ψ(r, t) which is a solution of the
wave equation, Schrödinger equation. In that form it is easier to learn quantum mechanics
so the words wave mechanics and quantum mechanics have acquired the same meaning especially since their physical background is the same and moreover, Schrödinger has shown
that these two versions are completely equivalent. Therefore we wil use here the name
quantum mechanics.
†
Werner Karl Heisenberg (1901.-1976.)
♯
Erwin Schrödinger (1887.-1961.)
Str. 2 - 2
2. Schrödinger Eqution I: Introductory Problems
Classical mechanics represents a bridge towards formulation of quantum mechanics. In an elementary approach to the classical mechanics one considers classical laws of
motion - Newton’s laws and using them one solves equations of motion by which acquires
a deterministic description of a path of a particle on which forces are acting. In a more
general approach, in so called analytic mehanics according to Lagrange and Euler† and
Hamilton and Jacobi† a system is described by kinetic and potential energy using generalized coordinates and momenta. In Hamilton’s approach a total energy is introduced as
a sum of the potential and kinetic energy. This function is called Hamilton’s function or
Hamiltonian of a system, H. Therefore, Hamiltonian is (in 1 dimension)
H(p, x, t) = Ek (p) + Ep (x, t) =
p2
+ Ep (x, t),
2m
(2-1)
where Ep is a potential energy (often written as V (x) or V (r)), and in a usual way we
have introduced a momentum of a particle p with a mass m
p = mv.
(2-2)
If energy of the system is conserved, then Hamiltonian does not depend on time. A
force for a conservative system can obtained from the potential energy
F =−
dV (x)
dx
or
F = −∇V (x).
(2-3)
Potential energy Ep = V (x) is, as one can see from the above equations, determined by
forces which act on a particle with mass m, and in general it depends on coordinates,
velocities and time. The fact that we can formulate a potential energy of the system
means that we are dealing with a conservative system.
From the above expression we can see that one can add to the potential energy
some constant and the force, derived from the energy, will not change. Therefore we are
able to define a potential energy freely, on a most convenient way.
Using the so called canonical equations in analytic mechanics we can derive equations of motion (the second Newton law) just from Hamiltonian (2-1). This simply means
†
Joseph-Louis Lagrange (1736-1813), Leonhard Euler (1707-1783)
†
Sir William Rowan Hamilton (1805-1865), Karl Gustav Jacob Jacobi (1804-1851)
D. Horvat: Physics of Materials: Concepts
Str. 2 - 3
that we are not required to know forces (acting on a particle) to write down equations of
motion but only the total energy. We will use this form of classical mechanics to make a
transition, in the most natural way, from the classical to quantum mechanics.
Therefore, in quantum mechanics we will have to formulate a total energy of a
system (Hamiltonian) and according to quantum mechanical postulates to every physical
quantity we will introduce a corresponding operator. Formulation of a Hamiltonian is
thoroughly classical procedure, so one has to know the classical mechanics very well in
order to make the transition classical→quantum.
2.2 Schrödinger equation
Here we will consider a construction of a wave equation. Its role is to provide a wave
function with a standard notation ψ(r, t) as a solution of the equation. The function has
to describe an electron (or any other microscopic particle or system) in accordance with
de Broglie’s relation between a wave length and a momentum of a particle, in accordance
with Planck’s relation i.e. relation between energy and frequency, in accordance with the
classical mechanics, in accordance with the principle of correspondence (see below) and in
accordance with the classical properties of Hamiltonian (the total energy).
Equation which solution will be a wave which will describe an electron has to obey
the following conditions:
Equation of motion, a solution of which will describe an electron i.e. a wave equation
will have to satisfy the following requirements:
(1) The equation has to be a (partial) differential equation of the first order in time so
that its solution - the wave function ψ(x, t) - could be thoroughly determined by
one initial condition ψ(r, 0).
(2) The wave equation has to be linear in the wave function ψ(r, t), so that the principle
of superposition is valid. In that way the effects of interference will be guaranteed,
and these effects would provide figures equivalent to those from the wave (or physical) optics (and these are a consequence of Maxwell equations being linear). Due
Str. 2 - 4
2. Schrödinger Eqution I: Introductory Problems
to the same reasons all constants in the wave equation must not contain physical quantities which are characteristic for a dynamics of the particle (like energy,
momentum, angular momentum, etc.).
(3) The wave equation has to be a homogenous one in order to have probability conserved in time (see below).
To fully appreciate this requirement we have to know and understand postulates of quantum mechanics which are given later. Nevertheless, if we had an equation of the form
∂
Ψ(r, t) = DΨ(r, t) + q
∂t
where D is some differential operator, and q is a nonhomogeneous
term, we would have
d
dt
Z
3
2
d x |Ψ(x, t)| =
=
=
Z
Z
Z
d3 x (Ψ̇∗ Ψ + Ψ∗ Ψ̇)
3
∗
∗
d x [(DΨ) Ψ + Ψ (DΨ)] +
dS · j +
Z
Z
d3 x(qΨ∗ + Ψq ∗ )
d3 x 2ℜe(qΨ∗ ).
The first term equals zero using the Gauss theorem (see also below), and the second term is different from zero and all these
bring in a nonconservation of probability. Namely, the expression
R
d3 x|Ψ(x, t)|2 is in the quantum mechanics interpreted as probabil-
ity, and non-vanishing of time derivative of this term simply means
its non-conservation.
(4) Plain waves have to be solution of the wave equation, i.e. a free particle has Hamiltonian of the form (notice that for a free particle only the kinetic energy term
H = p2 /2m = E is present) and the plain wave is
Ψ(x, t) = C exp[i(k · x − ωt)]
hi
i
p · x − Et
= C exp
h̄
hi
p2 i
p·x−
t .
= C exp
h̄
2m
(2-4)
D. Horvat: Physics of Materials: Concepts
Str. 2 - 5
In the above expression are ”hidden” de Broglie and Planck relations.
de Broglie
Planck
h
2π h
=
·
= kh̄
λ
λ 2π
hν
E = hν =
· 2π = h̄ω,
2π
p=
where
1
h
,
k= p
2π
h̄
According to the requirement (1)
h̄ ≡
k = |k| =
i
∂
Ψ(x, t) = − EΨ(x, t)
∂t
h̄
i p2
=−
Ψ(x, t)
h̄ 2m
∂
ih̄ Ψ(x, t) = EΨ(x, t).
∂t
(2-5)
2π
.
λ
(2-6)
or
(2-7)
The derivative with respect to x will bring down px component of the momentum
vector p, and ∇ will bring down p:
∇Ψ(x, t) =
i
p Ψ(x, t)
h̄
(2-8)
Another operation of the operator ∇ will give
∇2 Ψ(x, t) = −
1 2
p Ψ(x, t)
h̄2
h̄2 2
p2
−
∇ Ψ(x, t) =
Ψ(x, t).
2m
2m
or
(2-9)
From the above expressions we can formally relate kinetic energy and the operator
nabla squared, as well as the total energy with time derivative
h̄2 2
h̄2
p2
−
∇ ≡−
∆ −→ Ek =
2m
2m
2m
∂
ih̄ −→ E.
∂t
(2-10)
From the equality of the right hand sides of the expressions (2-7) and (2-9) the left
hand sides are equal also so we get
h̄2
∂
−
∆Ψ(x, t) = ih̄ Ψ(x, t)
2m
∂t
(2-11)
Str. 2 - 6
2. Schrödinger Eqution I: Introductory Problems
and we obtain the time dependent Schrödinger equation for a free particle (Ep ≡
V (x) = 0)!
E. Schrödinger has assumed in 1926.† that a behaviour of an ”electron wave” could
be obtained by solving a wave equation of the form
∂Ψ(r, t)
h̄2 ∂ 2 Ψ(r, t) ∂ 2 Ψ(r, t) ∂ 2 Ψ(r, t) + V (r, t)Ψ(r, t) = ih̄
+
+
−
2
2
2
2me
∂x
∂y
∂z
∂t
(2-12)
2
−
h̄
∂Ψ(r, t)
∆Ψ(r, t) + V (r, t)Ψ(r, t) = ih̄
2me
∂t
where me is the electron mass, Ψ is a wave function of the electron, and V (r, t) is its
potential energy. The first equation in (2-12) is the time dependent Schrödinger equation
written in Cartesian coordinates, and the second is Schrödinger equation , written in a
general form.
Kinetic energy Ek = p2 /2m is, according to (2-10) in Cartesian coordinates (x, z, y)
given as
2
h̄2
∂
h̄2
h̄2
∂2
∂2
Ek = −
∇∇ = −
∆=−
+ 2+ 2 ,
(2-13)
2m
2m
2m ∂x2
∂y
∂z
and in spherical coordinates for instance, (r, θ, φ), it is given as
h̄2 1
∂
1 ∂2
1 ∂
h̄2 1 ∂
2 ∂
r
−
sin θ
+
Ek = −
2m r 2 ∂r
∂r
2m r 2 sin θ ∂θ
∂θ
sin2 θ ∂φ2
h̄2 1 ∂
h̄2 1
2 ∂
=−
r
−
∆θ,φ
2m r 2 ∂r
∂r
2m r 2
1
h̄2 1 ∂
2 ∂
r
+
L2
=−
2
2
2m r ∂r
∂r
2mr
(2-14)
where the angular momentum operator L is given with the equation from the classical
mechanics L = r × p, and it is calculated by using relationships between Cartesian and
spherical coordinates (as given in Appendix 2) If we recall that the moment of inertia I of
a point particle of mass m is given by I = mr 2 , then the last term in the above equation
becomes
L2
Ek/rot =
(2-15)
2I
†
E. Schrödinger, Annalen der Physik, vol.79 (1926)
D. Horvat: Physics of Materials: Concepts
Str. 2 - 7
and this is precisely the expression for kinetic energy of rotation.
The above equation (2-12) we can get from the ”derived” equation for a free particle
(2-11) if we take Ek = H − Ep = H − V , and write
p2
h̄2
Ek Ψ(x, t) =
Ψ(x, t) = −
∆Ψ(x, t) = (E − Ep )Ψ(x, t)
2m
2m
∂
= (ih̄ − Ep )Ψ(x, t)
∂t
(2-16)
Here again we have to point out that we have not derived the Schrödinger equation (in the same way as E. Schrödinger has not derived it) but we have ”constructed”
it knowing the target form. The purpose of the whole procedure is to comprehend the
physical background and physical assumptions built in it. This equation has written down
E. Schrödinger in 1926. and he represents the equation of wave mechanics which has described the wave properties of electrons. Somewhat before Schrödinger, W. Heisenberg has
written (in 1925.) another version of the equation which has described (wave) properties
of electrons using a matrix algebra, and the matrix mechanics was born. But Schödinger
has pretty soon shown that his and Heisenberg’s version are equivalent so a common name
for both theories has become quantum mechanics. One must admit that Schrödinger’s
wave mechanics is easier to handle, especially with respect to different practical calculations so today under the name quantum mechanics one (mostly) refers to Schrödinger
version, written above! Schrödinger has applied his theory to a hydrogen atom and using
some crucial results from Bohr’s model, knowing that at least he had to derive Balmer’s
formula, has was able to completely solve the problem.
We can now try to solve Schrödinger equation for different one-dimensional V (x) or
three dimensional potentials V (r) (or, more precisely potential energies). Later on we will,
using postulate of quantum mechanics, give to quantum mechanics more stringent frame
by giving the meaning and interpretation to the wave function as well as mathematical
aspects upon which quantum mechanical formalism is based. As always in such situations
in which we want to solve a differential equation we have to specify boundary conditions.
Here, however, we have to discuss interpretation of solutions ψ(t, x) to the above equation
since we can see immediately that ”something is wrong with it” from the physical point
of view: the equation (2-12) is complex, and such is its solution ψ(t, x) (or ψ(t, r)) as
well. This means that it makes no sense to specify a position of an electron by writing
Str. 2 - 8
2. Schrödinger Eqution I: Introductory Problems
√
x = 3 i m (where i = −1), or to specify a time for an electron to cross certain distance
and put t = 2 is. Obviously the wave function (of an electron) ψ(t, x) does not have a
direct physical interpretation. The problem was solved by M. Born† . who has published
in 1926. in the scientific journal Zeitschrift für Physik two papers about interpretation of
a wave function: he has introduced so called ”probabilistic interpretation” or ”statistical
interpretation” of the wave function, which means that the wave function ψ(t, r), being in
general a complex function of time and coordinates, has no direct physical meaning. The
only physical meaning has its absolute square |ψ(t, r)|2 which gives a probability of finding
a quantum particle. So the expression
|ψ(t, r)|2 dV
(2-17)
gives probability that a quantum particle, described by the wave function ψ(t, r) be found
within a volume dV at time t. In one dimension, the expression
|ψ(t, x)|2 dx
(2-18)
gives probability that a quantum particle, described by a wave function ψ(t, x) will be found
on the x axis between x and x + dx, in a moment t. Therefore, |ψ(t, x)|2 is probability
density. When we ”sum” over all the range of a space, i.e. when we integrate over the
whole available space we have to get the total probability (of finding a particle) meaning
that this integral has to be equal to one (because the total probability of all possible
outcomes of an event must be equal one). Therefore
+∞
Z
|ψ(t, x)|2 dx = 1
−∞
Z
or
(2-19)
|ψ(t, r)|2 dV = 1.
the whole volume V
We say that the wave function is normalized. Such probabilistic or, popular, Copenhagen
interpretation of a wave function (and of quantum mechanics as a whole) has immediately
induced huge number of discussions among scientists (and philosophers as well). Even today decades after its introduction, there are still different ideas how to extract something
†
Max Born (1882.-1970.) a German physicist who was active in Göttingen and Frankfurt
on Main, and left Germany during the nazi rule and spent some time in Scotland to return to
Germany in 1953.
D. Horvat: Physics of Materials: Concepts
Str. 2 - 9
more than ”just the probability” from the quantum mechanical formalism. It is the well
known dispute between A. Einstein (as a critic of quantum mechanics) and N. Bohr (the
defender), and as history shows, their dialog always ended in favor of Bohr, no matter
how Einstein tried to dismantle quantum mechanics and its statistical interpretation. Regardless of some shortcomings and some paradoxes which follow the quantum mechanical
formalism and interpretations, quantum mechanics provides a description of microscopic
world of atoms, processes in atoms, molecules, their interactions, spectra, conductors,
semiconductors, magnetism, nuclear structure and processes, structure and evolution of
stars etc. Extension of quantum mechanics towards relativistic mechanics and to relativistic theory of quantum fields i.e. to quantum field theory has enabled us to formulate the
theory and to do practical calculations of fundamental processes at the level of elementary
particles. Also, the quantum theory has produced methods which gives high precision results which are unique and unsurpassed in any other human activity in history of sciences.
So, although there are some controversies concerning quantum mechanics, it is the best
theory we have and it is important to learn some of its results which are closely connected
with experimental data which are in agreement with the theory. Certainly, to fully appreciate the results we should learn the quantum theory in much more details which is beyond
our present goal, but the examples which we will study provide enough interesting aspects
and we will be able to appreciate the power of its arguments.
2.3 Quantum mechanics postulates
The wave equation, Schrödinger equation is a partial differential equations of the
second order. In order to solve it one has to specify boundary conditions. Apart from this
mathematical requirement, one has to assign a meaning to solutions to the equation and
the problem is subtle since the solutions are in general complex functions of coordinates
and time. Also one has to learn how to obtain results for different physical quantities
which are involved in a particular problem and which serve to specify physical content of
the quantum system under consideration. These problems will be resolved by following
quantum mechanical postulates.
The quantum mechanical postulates cannot be derived from classical mechanics.
They stem from general principles and they are in fact given as posteriori instructions, after
physicists have learned how to formulate and how to calculate and extract useful results
from quantum mechanical calculations. Therefore, the postulates serve to systemize and
Str. 2 - 10
2. Schrödinger Eqution I: Introductory Problems
to organize quantum mechanical knowledge and application. Parallel to the postulates we
will give their comments which are their explanations and extensions and the comments
deepen and broaden the postulate contents. So the postulates and their corresponding
comments are to be read and studied together. In literature one can find different number
of postulates, their comments, and their scope. Here we are trying to minimize their
number and their scope to keep this framework as transparent as possible but still trying
to thoroughly cover all aspects of the content of quantum mechanics. Some of the more
important mathematical aspects of quantum mechanics will be introduced in following
chapters.
Hence, postulates of quantum mechanics provide:
(1) rules to find a wave function ψ(x, y, z, t),
(2) interpretation of the wave function,
(3) ways to introduce physical quantities - observables in quantum mechanical formalism,
(4) rules to calculate expectation values (mean values) of observable quantities using
wave functions.
POSTULATE 1
To every physical state of a quantum mechanical system there belongs a wave function
ψ(x, t) (in one dimension) or ψ(r, t) (in three dimensions) which contains all data about
the state of the physical system, and it completely determines the quantum state. The
wave function belongs to an (infinite dimensional) Hilbert space H(∞) of functions and it
is square integrable complex function of all its arguments.
POSTULATE 2
To every physical quantity there corresponds an operator. Operators of physical quantities are linear and Hermitian. Physical equations in which these operators enter become in
general differential equations, in most cases eigenvalue problems which involve the operators and their eigenfunctions. Eigenvalues of Hermitian operators which belong to physical
quantities are real and they are results of measurements of these quantities. Immediately
D. Horvat: Physics of Materials: Concepts
Str. 2 - 11
after a measurement is performed, a system will be in a state which is described by the
eigenfunction which corresponds to the measured eigenvalue.
POSTULATE 3
Mean value of expectation value of a physical quantity A in some state of a quantum system
ψ is obtained by calculating the mean value, or expectation value of the corresponding
Hermitian operator Â, i.e.
Z
(2-20)
< A >ψ = Aψ = ψ ∗ Âψ dV
where the integration is performed over the whole space in which the quantum system is
defined.
POSTULATE 4
Time development of a physical system is described by a wave function ψ(x, t) (in one
dimension) and it is a solution of the Schrödinger equation
ih̄
∂
ψ(x, t) = Ĥ(x, t)ψ(x, t)
∂t
(2-21)
where H(x, t) is the total energy of - hamiltonian of the system, and Ĥ is the corresponding
Hermitian operator.
Comments of the postulates 1 - 4
First Postulate deals with wave functions, and here we specify some more facts and
requirements set upon them: a wave function ψ(t, x) and its first derivative dφ
dx have to
be continuous, finite and single valued u the whole domain. From these requirements a
quantization in different quantum system will follow. Furthermore, the function ψ(x, t)
has to be square integrable, i.e. (in one dimension)
+∞
Z
ψ ∗ (x, t)ψ(x, t) dx = 1
(2-22)
−∞
This requirement means that the above integral has to be finite, and that we have put it
equal to one means that this is connected with the interpretation of the wave function.
Namely, M. Born has, as already pointed out, introduced the probabilistic interpretation of
a wave function. According to this interpretation, if ψ(r, t) is a solution of the Schrödinger
Str. 2 - 12
2. Schrödinger Eqution I: Introductory Problems
equation for a particle, then a probability that this particle will be found in a moment t
and at a position r surrounded by a small volume d3 r equals
dP = |ψ(r, t)|2 d3 r
(2-23)
where |ψ|2 = ψ ∗ ψ is a probability density. In one dimension this equals
dP = |ψ(x, t)|2 dx
(2-24)
and this corresponds to probability to find a particle in an interval [x, x + dx]. The total
probability to find the particle anywhere in the domain equals to one, i.e.
Z
|ψ(r, t)|2 d3 r = 1.
(2-25)
Wave functions are, therefore, vectors of a unit length and these vectors belong to a
(Hilbert) vector space. They are eigenfunctions of Hermitian operators so they are mutually orthogonal with respect to a scalar product defined as
Z
ψn∗ ψm d3 r
=
1, for m = n
0, for m 6= n.
(2-26)
By this property, these functions become basis vectors so any wave function ψ could be
represented in terms of them
X
ci ψi
(2-27)
ψ=
i
Again, ψi are eigenfunctions of an operator (see below) a coefficients ci are in general
complex numbers.
Second Postulate gives a mapping between physical quantities and operators. Physical
quantities are measurable quantities in classical and quantum physics. Operators are
mathematical objects which act on ”something” staying (in most cases) on their right
hand side, like for instance ”taking the square root”
p
p
( )(”f unkcija”) = ”f unkcija”
(2-28)
or the operator of time derivative
d
(2t2 + 1) = 4t.
dt
(2-29)
D. Horvat: Physics of Materials: Concepts
Str. 2 - 13
In quantum mechanics we are mostly interested in operators which act on a function and
as a result, the function is simply multiplied by a ”number”, i.e.
Ôφ = Oφ φ
operator × eig. function = eig. value × eig. function
(2-30)
where we have the eigenfunction φ of an operator Ô and eigenvalues Oφ of the operator Ô for this particular eigenfunction φ. Operators corresponding to physical quantities
are Hermitian operators, (more about Hermitian operators will be given in next Chapter)
and for us at the moment the most important property is that they have real eigenvalues wich correspond to measurable physical quantities. Also, important here is that the
eigenfunctions are mutually orthogonal. Let the functions ψn and ψm are eigenfunctions
of a Hermitian operator (for instance of Hamiltonian for the hydrogen atom with corresponding eigenvalues i.e. quantized energies En and Em ), then there is valid the relation
of orthonormalization (2-26). We have shown before that these functions become basis
functions in terms of which any physical state of a quantum system could be expressed as
a linear combination.
The classical expression for total energy H (Hamiltonian) is in one dimension given
as
H=
1 2
p + V (x) = E
2m x
(2-31)
and it becomes a wave equation by the following replacements of dynamical variables
(x, px ) by operators
x→x
px → −ih̄
f (x) → f (x)
E → ih̄
h̄2 2
Ek → −
∇
2m
h̄2 2
H→−
∇ + V (r).
2m
∂
∂x
∂
∂t
(2-32)
L → −ih̄r × ∇
In that way the equation (2-31) becomes
h̄2 ∂ 2 ψ(x, t)
∂ψ(x, t)
−
+ V (x)ψ(x, t) = ih̄
2
2m ∂x
∂t
(2-33)
Str. 2 - 14
2. Schrödinger Eqution I: Introductory Problems
i.e. we have obtained the time dependent Schrödinger equation . Therefore, to every
observable, i.e. a measurable quantity (like E, x or p) there corresponds a (Hermitian) operator . Possible results of a measurement of an observable are eigenvalues of corresponding
operators.
Third Postulate
provides a way how to calculate mean values and fluctuations about
these means values. If wave functions which are given in (2-20) are eigenfunctions of the
operator Â,
Âψ = aψ
(2-34)
then the mean value is precisely equal to eigenvalue (due to orthonormalization of eigenfunctions) < Â >= a. Let a physical state be described by a wave function ψ which
is not an eigenfunction of Â.
The eigenfunction of the operator  belong to a set
S = {φi |i = 1, . . . , n}, and
Âφi = ai φi
and
Âψi = χ
(2-35)
where χ is some function, and the eigenvalue ai of the operator  belongs to a spectrum
A = {ai |i = 1, . . . , n}. We are interested to calculate possible values of the physical
quantity A in the state ψ. The corresponding operator to quantity A is Â. It is not possible
to calculate directly with the function ψ. However, we will expand or represent the state
ψ in terms of eigenfunctions of the physical quantity A, since these (eigen)functions are
basis vectors
ψ=
X
ci φi .
(2-36)
i
For wave functions φi there is the orthonormal relation (2-26). Also
ψ∗ =
X
c∗j φ∗i
(2-37)
j
and since the wave function ψ is normalized as well
Z
ψ ∗ ψ dV = 1
Z
φ∗j φi
we get
V
1=
XX
i
j
c∗j ci
dV =
XX
i
j
c∗j ci δij
=
X
i
(2-38)
2
|ci | .
D. Horvat: Physics of Materials: Concepts
Str. 2 - 15
On the state ψ we will apply the above expression (2-20)
Z
Z
XX
XX
∗
∗
∗
ci cj φ∗i aj φj dV
ci cj φi Âφj dV =
< Â > =
i
=
j
i
=
V
XX
j
X
j
c∗i cj aj
i
Z
φ∗i φj dV =
i
V
j
XX
V
c∗i cj aj δij
(2-39)
j
2
|cj | aj .
So the mean value of the physical quantity is
< A >= a1 |c1 |2 + a2 |c2 |2 + · · · + an |cn |2
(2-40)
and the meaning of this expression is the following: if we measure a physical quantity A,
a probability to measure a value ak of the quantity equals |ck |2 . So quantum mechanics
provides us with the possibility of calculating different values of physical quantities and a
probability that the physical quantity has a certain value. Total probability to measure
any value from the spectrum A equals to sum of all values of |ci |2 and this is, evidently,
equal to one, as shown above.
By using the above formulae (2-20) we can calculate fluctuations of physical quantities. We know that standard deviation is defined as
(∆A)2 = (A2 ) − (A)2
(2-41)
where the mean value of (A2 ) we can calculate using the above formula for a state ψ
Z
2
2
(A )ψ =< (Â) >= ψ ∗ (Â)2 ψ dV.
(2-42)
V
This calculation leads to uncertainty relations where we have to calculate a product of
deviations of two physical variables ∆A and ∆B which are related in a special way, namely
by their non vanishing commutator. Recall that the commutator of two objects P and Q
is defined as: [P, Q] = P Q − QP .
Fourth Postulate If Hamiltonian does not depend on time (as is the case for closed systems), then we can formally integrate the equation 2-21and we get
t
ψ(x, t) = e−i h̄ Ĥ ψ(x, 0) ≡ U (t)ψ(x, 0)
(2-43)
Str. 2 - 16
2. Schrödinger Eqution I: Introductory Problems
where a unitary operator U (t) is introduced and it is called time evolution operator . Unitary operators have a special role in quantum computing because the reversibility of quantum algorithms is achieved by use of the unitary operators. Reversibility is not present in
the classical computing.
2.3.1 Probability conservation
Consider now Schrödinger equation and its complex conjugate
h̄2 2 ∗
∂ψ ∗
=−
∇ ψ + U ψ∗
−ih̄
∂t
2m
(2-44)
We write
∂ψ ∂ψ ∗ ∂|ψ|2
ψ
= ψ∗
+
∂t
∂t
∂t
(2-45)
i ih̄ h
i
ih̄ h ∗ 2
2 ∗
∗
∗
=
ψ ∇ ψ − (∇ ψ )ψ
∇ ψ ∇ψ − (∇ψ )ψ ,
2m
2m
where we have assumed that potential energy is real. If we introduce, according to the
above expression probability density ρ ≡ |ψ|2 and probability current density
j=−
we have (in one dimension)
ih̄ ∗
ψ ∇ψ − (∇ψ ∗ )ψ
2m
∂ρ ∂jx
+
= 0 or
∂t
∂x
∂
∂
ih̄
∂ ∗
∗
2
ψ (t, x) ψ(t, x) −
|ψ(t, x)| + −
ψ (t, x) ψ(t, x) = 0.
∂t
2m
∂x
∂x
(2-46)
(2-47)
For a plane wave ψ = A exp(ikx), so ρ = |A|2 and j = |A|2 (h̄k)/m = |A|2 v. Using Gauss
law we get
Z
Z
d
2 3
|ψ| d r = −
j · n̂ dS
(2-48)
dt
where dS is surface element and n̂ is a unit vector on dS. If the wave function has a
behavior like |ψ| ∼ r −3/2 for r → ∞, then the right hand side equals zero and
Z
∂
|ψ|2 d3 r = 0
(2-49)
∂t
i.e. probability is conserved.
D. Horvat: Physics of Materials: Concepts
Str. 2 - 17
2.3.2 Solution of Schrödinger equation for a free particle
From the ”construction” of Schrödinger equation it is evident that the plane wave, which
serves here to represent a free particle, is a solution of the equation. Let us try to obtain
the formal solution. Time dependent equation for a free particle (electron for instance)
was derived before (in one dimension)
−
∂
h̄2 ∂ 2
ψ(t, x) = ih̄ ψ(t, x).
2
2m ∂x
∂t
(2-50)
One of the most important methods in solving partial differential equations is the method
of separation of variables. Assume a form of solution in the following form (in a separated
way)
ψ(t, x) = χ(t)φ(x)
(2-51)
and after plugging this function into the above equation we get
h̄2
χ(t)φ′′ (x) = ih̄φ(x)χ̇(t),
−
2m
(2-52)
where prime or a dot represent space or time derivative. Dividing the equation by χφ we
get
h̄2 1 ′′
1
−
φ (x) = ih̄
χ̇(t).
(2-53)
2m φ(x)
χ(t)
The left hand side is a function of x only, and the right hand side is a function of t only
and since they are equal the only way it to be true is that they are equal to a constant
−
1
h̄2 1 ′′
φ (x) = G = ih̄
χ̇(t).
2m φ(x)
χ(t)
(2-54)
The physical dimension of the constant is
[h̄]2 1
(ML2 T−1 )2 1
=
· 2 = ML2 T−2 ,
2
[m] L
M
L
(2-55)
(where L−2 comes from space derivative) and both sides have dimension of energy, so
instead of G we will use the symbol E. Therefore
φ′′ (x) = −
′′
2
φ (x) + k φ(x) = 0
2m
Eφ(x)
h̄2 k=
or
q
2
2mE/h̄
(2-56)
,
Str. 2 - 18
2. Schrödinger Eqution I: Introductory Problems
and we have obtained the equation for a simple harmonic oscillator the solution of which
is
φ(x) = Aeikx + Be−ikx .
(2-57)
Now we proceed to solve the time dependent part of the wave function. From the above
equation
ih̄
χ̇(t)
=E
χ(t)
(2-58)
we get
dχ(t)
E
= −i dt giving
χ(t)
h̄
E
ln χ(t) = −i t + ln C, i.e.
h̄
−iωt
χ(t) = Ce
(uz ω = E/h̄).
(2-59)
The complete solution for the wave function of a free particle is
ψ(t, x) = C1 ei(kx−ωt) + C2 e−i(kx+ωt) ,
(2-60)
and this is a wellknown expression for the wave theory: it is a progressive (plane) wave
which propagates along the x−axis in “+” or “–” direction. Probability density |ψ(t, x)|2
is time independent.
We have to point out here that in fact the description of a particle by means of a
plane wave is not physically acceptable, because a particle is localized (being in a
small, bounded part of the space) and not spread along the whole x axis. However,
by using the Fourier transform
1
Ψ(t, x) = √
2π
Z+∞
g(k)ψ(t, x) dk
−∞
with a careful choice of the spectral function g(k), (for instance g(k) ∼ exp(−αk 2 ),
hence Gaussian form of a function) we know how to build spatially bounded wave
function - wave packet Ψ(t, x), which is certainly a solution of the Schrödinger
equation since Schrödinger equation is linear (partial) differential equation so a
linear combination of solutions is again a solution. In that way, using the wave
packet, we represent a particle by solutions of the wave equation (see below). In
(2-61)
D. Horvat: Physics of Materials: Concepts
Str. 2 - 19
many practical calculations we will very often use, however, the plane wave representation because it is much simpler and the results in many cases are completely
acceptable.
From the above considerations we can see that using the method of separation of
variables we can always integrate Schrödinger equation either for a free particle or with a
potential if the potential does not depend on time, and this is very often the case. Therefore
we can write time independent Schrödinger equation u three dimensions
h̄2
−
∆ + V (r) ψ(r) = Eψ(r),
2m
(2-62)
a time dependent solution is given as
ψ(t, r) = ψ(r)e−iEt/h̄ .
(2-63)
The equation (2-62) is an eigenvalue equation. It is important to check if the operator on
the left hand side is Hermitian. In that case it will provide real eigenvalues i.e. energies E.
It will be shown that the operator H is Hermitian, so its eigenvalues are real, measurable
energies, what is a necessary condition for a description of a particle.
2.3.3 Time independent Schrödinger equation
We can solve the time dependent Schrödinger equation partially by using the method of
separation of variables. We assume that a solution of a (partial differential equation)
could be written as a product of a function φ(t) which describes its time dependence and
a function ψ(x) which gives the x−dependence
Ψ(x, t) = ψ(x)φ(t)
(2-64)
Plugging (2-64) in Schrödinger equation we get
h
−
i
h̄2
∆ψ(x) + V (x)ψ(x) φ(t) = ψ(x)ih̄φ̇(t)
2m
(2-65)
Str. 2 - 20
2. Schrödinger Eqution I: Introductory Problems
and dividing by ψ(x)φ(t) we obtain
h̄2
1
−
∆ψ + V ψ =
ψ
2m
|
{z
}
func. only of x
φ̇
ih̄
φ
|{z}
(2-66)
func. only of t
The left hand side (LHS) is, according to former considerations, a function of x only, the
right hand side (RHS) is a function of t only, so they have to be equal to some constant
which, from dimensional analysis has the dimension of energy. So, we write first for the
RHS
φ̇
E = ih̄
(2-67)
φ
so
E
(−i t)
h̄ .
(2-68)
φ(t) = Ce
In that way we get the time independent Schrödinger equation
−
h̄2
∆ψ(x) + V (x)ψ(x) = Eψ(x).
2m
A solution of the time dependent Schrödinger equation is
E
−i t
h̄
Ψ(x, t) = ψ(x) · e
(2-69)
(2-70)
where we have absorbed a constant C from (2-68) in a function ψ(x).
We can now introduce Hamiltonian
H = E = Ek + V
(2-71)
ĤΨ = EΨ
(2-72)
and we can write
or
i
h̄2 2
∂
ĤΨ(r, t) = −
∇ + V (r, t) Ψ(r, t) = ih̄ Ψ(r, t),
(2-73)
2m
∂t
and this is Schrödinger equation in three dimensions for a general potential V (r, t). Also
h
−
h̄2 d2 ψ(x)
+ V (x)ψ(x) = Eψ(x)
2m dx2
(2-74)
D. Horvat: Physics of Materials: Concepts
Str. 2 - 21
is Schrödinger equation in one dimension.
We have seen (see (2-14)) that in some other coordinate system (spherical, cylindrical, parabolic, etc.) the operator ∇2 has a different from from the one given in (2-74),
and geometry of the problem, i.e. a form of the potential (potential energy) will dictate
or require a coordinate choice.
2.3.4 Standard deviation
We have now a possibility to calculate mean values of observables using Postulate 3 together
with interpretation of a wave function (Postulate 1 i its comment). Also we can calculate
deviations of physical quantities from their mean values and a measure of this is mean
square deviation or standard deviation.
Let P (x) = ψ ∗ ψ be probability density in one dimension. The mean value of a
function ψ(x) (Postulate 3) is
+∞
Z
< f (x) >=
ψ ∗ (x)f (x)ψ(x) dx,
(2-75)
−∞
and the mean value of the square of the function is
+∞
Z
< (f (x)) >=
ψ ∗ (x)(f (x))2ψ(x) dx,
2
(2-76)
−∞
Standard deviation σ is defined in this way
σ 2 =< (f (x)− < f (x) >)2 >
+∞
Z
=
ψ ∗ (x)(f (x)− < f (x) >)2 ψ(x) dx.
(2-77)
−∞
Standard deviation or mean square deviation measures dispersion or uncertainty of a function f (x) from or around its mean value. The above expression could be written in the
Str. 2 - 22
2. Schrödinger Eqution I: Introductory Problems
following way
+∞
+∞
Z
Z
ψ ∗ (x)f (x)ψ(x) dx +
σ2 =
ψ ∗ (x)(f (x))2ψ(x) dx −2 < f (x) >
−∞
−∞
|
{z
}
=<f 2 >
+∞
Z
2
ψ ∗ (x)ψ(x) dx
+ < f (x) >
|
{z
=<f >
}
(2-78)
−∞
|
so
{z
}
=1
σ 2 =< f 2 > − < f >2 .
(2-79)
We can introduce a standard notation ∆f for σ, sa we write
∆f =
p
< f 2 > − < f >2 .
(2-80)
In that way we can write
∆x =
p
< x2 > − < x >2
(2-81)
Therefore when we solve a quantum mechanical problem by solving Schrödinger
equation we can, using the above expression, calculate dispersion of some physical quantity
which has no precisely determined value in this state.
2.4 Free particle and wave packet
We are going to consider some particular cases of quantum mechanical systems
where we will apply Postulates 1 to 4.
For the case when V (r) = 0, i.e. when there is no force acting on a particle so
we talk about a free particle, we have Schrödinger equation , after separation of time
coordinate
∆ψ(r) +
2m
Eψ(r) = 0
h̄2
(2-82)
D. Horvat: Physics of Materials: Concepts
and its solution, with k =
q
Str. 2 - 23
2mE/h̄2 is
ψ ∝ eikr
(2-83)
or, with the time dependence
Ψ(r, t) ∝ ei(±kr−ωt)
Ψ1 (r, t) = Aei(kr−ωt)
or
i Ψ2 (r, t) = Bei(−kr+ωt)
(2-84)
Schrödinger equation is a linear equation so when Ψ1 and Ψ2 are (separately) solutions of
Schrödinger equation , then their linear combination is a solution as well Ψ = αΨ1 + βΨ2 ,
where α and β are complex numbers in general. Therefore Ψ is a superposition of states of
solutions of Schrödinger equation and this is necessary to have, according to requirements
of wave optics (or general wave theory) a path difference which will produce a phase
difference and at the end the interference pattern. However, plane waves given in (284) have a homogenous probability density |Ψ(x, t)|2 = |A|2 , (or |B|2 ) and this leads to
spatially spread-out particle what is not in accordance of a notion of a particle having
localized position in the space.
De Broglie assumed that a plane wave (2-84) could be combined (superposed) in a
more general function of the form
Z
Ψ(r, t) =
d3 k φ(k)ei(kr−ωt)
Z
(2-85)
d3 p
i(kr−ωt)
=
φ(k)e
(2πh̄)3
where φ(k) is weight or spectral function, which we can choose in such a way to obtain a
wave packet, i.e. localized signal which we have encountered before while studying waves.
Angular frequency ω is in general a function of a wave number k: ω = ω(k) and p = h̄k.
The above expression is a Fourier integral by means of which a function Ψ(x, t) is
presented. This same integral, with an appropriate spectral function φ(k) represents and
”light wave packet” or light pulse. However, for arguments of the wave function in the
same time de Broglie and Planck relations have to hold, and this is not the case for light.
For a free particle the total energy will be just a kinetic energy.
We will choose a spectral function which will provide a localized form for Ψ(x, t).
Among different possibilities we will choose the Gaussian form for the spectral function
Str. 2 - 24
2. Schrödinger Eqution I: Introductory Problems
φ( p)
p
p
0
Figure 2.1 - Gaussian function as the spectral function for a wave packet with a maximum at p0 .
2
φ(p) ∼ e−p (with k = p/h̄) according to Fig. (2.1). With different constants due to
dimensional requirements we can write for a spectral function with a maximum at p0
φ(p) = A exp [−(p − p0 )2 D2 /h̄2 ].
(2-86)
In one dimensional case (which will be the only we are going to study) we get
A
Ψ(x, t) =
2πh̄
+∞
Z
dp exp − a(p − b/a)2 + b2 /a − c
−∞
=
A
2πh̄
r
(2-87)
2
b
π
exp
−c ,
a
a
where we have introduced some additional notation
D2
t
a= 2 +i
2mh̄
h̄
D2 p0
x
b=
2 +i
2h̄
h̄
2 2
D p0
c=
h̄2
and we have used a simple integral formula for Gaussian functions
+∞
r
Z
π
−αx2
.
I0 =
dx e
=
α
(2-88)
(2-89)
−∞
With this wave function - wave packet, we calculate now probability density |Ψ(x, t)|2 , and
from that the constant A will be calculated. Hence
b2
A 2 π
exp 2ℜe
−c .
(2-90)
|Ψ(x, t)|2 =
2πh̄ |a|
a
D. Horvat: Physics of Materials: Concepts
Str. 2 - 25
The real part of the exponent is
ℜe
D2 p0 x
x2 b2 D2 p0 2
=
+
i
·
−
a
h̄2
h̄3
4h̄2
D2 /h̄2 − it/2mh̄
·
(D2 /h̄2 )2 + (t/2mh̄)2
(2-91)
so
2ℜe(b2 /a − c)
D6 p20 /h̄6 − x2 D2 /4h̄4 + D2 p0 xt/2mh̄4 − D6 p20 /h̄6 − (t/2mh̄)2 · D2 p20 /h̄2
=2·
(2-92)
(D2 /h̄2 )2 + (t/2mh̄)2
2
2 2
2
x − 2xvt + v t
(x − vt)
=−
=
,
2
2
2
2
2D 1 + (h̄t/2mD )
2D 1 + (h̄t/2mD2 )2
where we have introduced a group velocity v ≡ vg ≡ p0 /m a quantity which we have met
in wave theory also. If we introduce
∆(t) ≡
we can write finally
2ℜe
Since
h̄t
,
2mD2
(2-93)
(x − vt)2
b2
−c =− 2
.
a
2D [1 + ∆(t)2 ]
(2-94)
1
1
=
2
2
2
|a|
[(D /h̄ ) + (t/2mh̄)2 ]1/2
(2-95)
h̄2
1
,
= 2·p
D
1 + ∆(t)2
the density (2-90) we can write in the following way
|Ψ(x, t)|2 =
h
(x − vt)2 i
1
A2
p
·
exp
−
.
·
4πD2
2D2 [1 + ∆(t)2 ]
1 + ∆(t)2
(2-96)
From the requirement of a wave function normalization - here the wave packet
Z
|Ψ(x, t)|2 dx = 1 we get
A = (8πD2 )1/4 ,
|Ψ(x, t)|2 =
D
p
so
h
2
(x − vt)
1
· exp −
2D2 [1 + ∆(t)2 ]
2π(1 + ∆(t)2 )
i
,
(2-97)
Str. 2 - 26
2. Schrödinger Eqution I: Introductory Problems
and ∆(t) is defined in (2-93). This wave packet will have a maximum when
∂ kr − ω(k)t = 0 (j = 1, 2, 3)
∂kj
k=k0
(2-98)
and maximum is for r = vg · t where we have above introduced group velocity vg
vg/j =
and with k =
p
dω ∂k
∂ω
=
∂kj
dk ∂kj
(2-99)
k12 + k22 + k32 and
kj
dω k
∂k
= , we have vg =
·
∂kj
k
dk k
or
vg =
dω
,
dk
(2-100)
what is in the above case equal vg = p0 /m. Group velocity is in the same time a velocity of
a particle which is represented by the wave packet. The expression for the group velocity
vg we can compare with a phase velocity, i.e. with velocity of a phase φ of a plane wave
φ = kr − ωt
or
vf = λν =
λ(2π ν)
λ
ω
=
ω= ,
2π
2π
k
(2-101)
(2-102)
and with Planck and de Broglie relations this becomes
h̄ p2
p
ω
=
.
vf = = ·
k
p 2mh̄
2m
(2-103)
We can see that the group velocity is equal twice the phase velocity for the case of a wave
packet. Group velocity has important physical meaning: it is the maximal velocity by
which a signal could be transmitted and this is in the same time velocity of a (classical)
representee by the wave packet.
Standard deviation of a coordinate x and momentum p for a wave packet
According to before discussed standard deviation we can now calculate the standard deviation of a coordinate and momentum and we can expect that the standard deviations will
be different from zero since we are dealing with a free particle.
D. Horvat: Physics of Materials: Concepts
Str. 2 - 27
The mean value of the coordinate x is
+∞
+∞
Z
Z
<x>=
Ψ∗ (x, t)xΨ(x, t) dx =
|Ψ(x, t)|2 x dx
−∞
=
−∞
+∞
Z
|Ψ(x, t)|2 (x − vt + vt) dx
−∞
(2-104)
+∞
+∞
Z
Z
∗
=
Ψ (x, t) x Ψ(x, t) dx + vt
|Ψ(x, t)|2 dx
−∞
−∞
|
= I1 + vt.
{z
=1
}
In the first integral we change variables (and the new variable is in fact the argument of
the wave packet - see below)
x − vt = y
dx = dy
(2-105)
so, according to (2-56b) |Ψ|2 ∼ exp[−(x − vt)2 /a] → exp[−y 2 /a], and this leads to an
integral of the form
+∞
Z
2
I1 ∼
e−y y dy
−∞
and this equals to zero since we are integrating an odd function over the symmetric domain.
Therefore, the mean value of the coordinate x is
Also
< x >= vt.
(2-106)
(∆x)2 =< (x− < x >)2 >=< (x − vt)2 >
+∞
Z
=
|Ψ(x, t|2 (x − vt)2 dx
(2-107)
−∞
+∞
Z
2
e−αy y 2 dy
= |B|
2
−∞
where
2
|B| =
h
1
p
2π(1 + ∆(t)2 )
1
α= 2
.
D [1 + ∆(t)2 ]
D
i2
and
(2-108)
Str. 2 - 28
2. Schrödinger Eqution I: Introductory Problems
The above integral we can calculate if we notice that by differentiation of the basic Gaussian integral (2-89) we get
∂
I0 =
∂α
Z+∞
∂ −αx2
dx
e
∂α
−∞
=−
Z+∞
dx(−x2 )e−αx
=
2
dx x2 e−αx =
Z+∞
2
−∞
√
√
∂
π
1 π
· 1/2 = −
∂α α
2 α3/2
−∞
Z+∞
2
dx(x )e
−αx2
=
√
π
2α3/2
−∞
In order to calculate the mean value of a momentum we have to calculate the following
expression
+∞
Z
Ψ∗ (x, t)(p̂)Ψ(x, t) dx
<p>=
−∞
+∞
Z
∂
=
Ψ∗ (x, t)(−ih̄
)Ψ(x, t) dx,
∂x
(2-109)
−∞
what is now easy. We get the following result
< p >= p0 .
(2-110)
In the same wave we could calculate < p2 >, so the final result is
∆p =
h̄
.
2D
(2-111)
The product of standard deviations (uncertainties) ∆x and ∆p is
∆x · ∆p =
h̄ p
1 + ∆2
2
or,
and if we throw away the time dependent term ∆ (that way we decrease the RHS) we get
inequality
> h̄/2.
∆x · ∆p ∼
(2-112)
D. Horvat: Physics of Materials: Concepts
Str. 2 - 29
Later on we will see that this relation plays a fundamental role in interpretation of measurements in quantum mechanics.
For a wave packet we can write ∆x = vg · ∆t and since E = p2 /2m and ∆E =
∆p · (p/m) = vg · ∆p we get
∆E · ∆t ≥ h̄
(2-113)
and these are uncertainty relations for conjugate variables E and t, for a wave packet.
The uncertainty relations or uncertainty principle set up a fundamental boundary
for (im)possibility of simultaneous determination, up to arbitrary precision, of two conjugate variables. The variables and corresponding operators - ”physical quantity”→”operator”
are related through Postulate 2.
In a similar way, in the case of hydrogen atom, the relations are given for the angular
momentum operator L for which, according to Postulate 2 there is a correspondence Lz →
h̄∂/∂φ where the angle φ is an angle of rotation around z axis or one of the three spherical
coordinates (r, θ, φ). So the uncertainty relations between the conjugate variables the
angular momentum operator L̂y and the angle φ are: (∆Lz ) · (∆φ) ≥ h̄ .
2.5 One dimensional infinite potential well
V(x)
Ψ(x)
x
a
Figure 2.2 One dimensional infinite potential well.
Str. 2 - 30
2. Schrödinger Eqution I: Introductory Problems
We will study here a very important and illustrative example for quantum mechanical calculations as well as applications of quantum postulates. The problem under study
is one dimensional well with infinite walls
0, for 0 < x < a
V (x) =
∞, for x < 0 i x > a.
(2-114)
q
Schrödinger equation has the form (with k = 2mE/h̄2 )
d2 ψ(x)
+ k 2 ψ(x) = 0
dx2
(2-115)
and its general solution is
ψ(x) = C1 eikx + C2 e−ikx
or
ψ(x) = A cos kx + B sin kx.
(2-116)
So according to Postulate 1 we have found a wave function of the given quantum system
by solving Schrödinger equation (Postulate 4). The quantum system is in fact a particle
of the mass m confined by the above potential.
The particle cannot penetrate in the region where potential is infinite so the boundary conditions imposed on its wave function are ψ(0) = 0 i and ψ(a) = 0 where a is the
well width. For x = 0 we get
ψ(0) = 0 = A cos 0 + B sin 0
⇒
A=0
(2-117)
and for x = a is
ψ(a) = 0 = B sin ka
i.e. from the definition of k
r
⇒
ka = nπ
(n = 1, 2, 3, . . . )
2mEn
· a = nπ
h̄
(2-118)
(2-119)
we get the quantized energy
n2 h̄2
En =
2ma2
where n is the quantum number .
(n = 1, 2, 3, . . . ),
(2-120)
D. Horvat: Physics of Materials: Concepts
Str. 2 - 31
We should notice (and remember) that the boundary condition imposed on the
wave function (Postulate 1 and its comment) has resulted in quantization of energy. For
n = 1 we have ground state energy which is the energy of the lowest state and sometimes
this energy is called zero point energy. Normalization condition (Postulate 1) provides the
value of the constant B:
Za
0
so B =
|ψ(x)|2 dx = 1 = |B|2
sin2
0
1
nπx
dx = |B|2 a
a
2
(2-121)
p
2/a and the eigenfunction of Hermitian operator H (Hamiltonian) is
ψn (x) =
and
Za
r
2
nπ
sin
x;
a
a
Ψn (x, t) =
r
En
t
2
nπ −i
h̄
sin
xe
a
a
n2 h̄2 ∂
Ψn (x, t).
HΨn (x, t) = ih̄ Ψn (x, t) = En Ψn (x, t) =
∂t
2ma2
(2-122)
(2-123)
Wave functions are mutually orthogonal (for different quantum numbers).
Orthogonality was already defined in (2-26) with respect to (some)
scalar product. Here on the interval [0, a] the role of the scalar
product of two (eigen)functions ψn and ψm has the integral
Za
ψn∗ (x)ψm (x) dx
= δn,m =
0
0,
for
1,
for
m 6= n
m = n.
More precisely this integral provides the orthonormality condition
for wave functions and it is valid for all bound state wave functions
in the corresponding domains.
By normalization and orthogonality, set of functions {ψ(x, t)} becomes a basis of the
space of physical states so any physical state for a particle in the one dimensional infinite
well can be represented as a linear combination of functions {Ψn (x, t)}.
Therefore, our problem is solved: we have calculated wave functions (2-122) and
(quantized) energies (2-120) so now we can proceed to calculate measurable physical quantities.
Str. 2 - 32
2. Schrödinger Eqution I: Introductory Problems
Example Consider a problem of a particle of mass m in one dimensional infinite square
well with the wave function:
φ(x, 0) =
(2b/a)x,
2b(1 − x/a),
for 0 ≤ x ≤ a/2
(2-124)
for a/2 ≤ x ≤ a.
Find a time dependent solution φ(x, t) and mean value of its energy hEi.
Solution Normalization of the wave function (Postulate 1) gives the value of b:
Za/2
0
4b2 2
x dx +
a2
Za
2x x2
4b (1 −
+ 2 ) dx = 1
a
a
2
⇒
b=
a/2
r
3
a
(2-125)
General time dependent solution could be given in terms of functions {Ψ(x, t)} (2-122)
φ(x, t) =
∞
X
n=1
En
∞
−i
t X
h̄
cn ψn (x) e
=
cn Ψn (x, t)
(2-126)
n=1
where cn are constants in the expansion
φ(x, 0) =
∞
X
cn ψn (x).
(2-127)
n=1
Since the functions ψn (x) are mutually orthogonal we can easily calculate constants in the
expansion (2-127)
cn =
Za
0
r
r h Za/2
Za
2
x
nπx
nπx i
2b
∗
dx +
2b(1 − ) sin
dx
sin
ψ (x)φ(x, 0) dx =
a
a
a
a
a
0
a/2
2 4ab nπ (n = 1, 2, 3, . . . )
sin
a n2 π 2
2
√ 4b = 2a 2 2 (−1)(n−1)/2 (n = 1, 3, 5, . . . ).
n π
=
Therefore
φ(x, t) =
X
n=1,3,5,...
(2-128)
4√6 (−1)(n−1)/2
π2
n2
Ψn (x, t).
(2-129)
D. Horvat: Physics of Materials: Concepts
Str. 2 - 33
The mean value of energy (Postulate 3) is the mean value of corresponding (Hermitian)
operator H (Hamiltonian)
hEi = E =
Za
φ∗ (x, t)Ĥφ(x, t) dx =
XX
n
0
c∗n cm Em
m
= (due to orthogonality, see (2-26)) =
ψn (x)∗ ψm (x) dx
0
X
n
so
Za
(2-130)
|cn |2 En
48h̄2 π 2 12
48h̄2 1 = 2 · E1 .
= 2 2
hEi =
π 2 ma2 n2
π ma 8
π
n=1,3,5,...
X
(2-131)
In the above example we have seen how we can represent a general wave function
which describes a state of a quantum mechanical system by using eigenfunctions of Hamiltonian and how we can calculate mean value of the system. Let us go back to the problem
of a particle in the one dimensional infinite square well.
Heisenberg uncertainty relations for the infinite square well
Using the wave functions {Ψn (x, t)} (2-122) we can calculate dispersion (standard deviation) of a coordinate or of a momentum in the same way as we did for a free particle (wave
packet). We will find useful two following integrals
Z
1
1
2
2 2
x sin ax dx = 2 a x − ax sin(2ax) − cos 2ax
4a
2
Z
x3
1 1
2
2
2 2
x sin ax dx =
+ 3
sin(2ax) − ax cos 2ax − a x sin(2ax) .
6
4a 2
Using the first one we can easily calculate mean value of x coordinate in n state
< x >n =
a
,
2
(2-132)
and the result is, as we could expect, independent of the quantum number n. Using the
second integral we calculate mean value of the square of the x coordinate
3
a2
2
1− 2 2 .
(2-133)
< x >n =
3
2n π
It is interesting to check a limit of the above expression for n → ∞
< x2 >(QM)
n→∞ =
a2
.
3
(2-134)
Str. 2 - 34
2. Schrödinger Eqution I: Introductory Problems
The mean value calculation in classical mechanics is
< x2 >(CM)
1
=
a
Za
x2 dx =
a2
3
0
2
(2-135)
=< x >(QM)
n→∞ .
This calculation shows that quantum mechanics (QM) gives results of classical mechanics
(CM) for large quantum numbers (n → ∞).
According to formulae introduced before we calculate the standard deviation ∆x of
the coordinate x
(∆x)2 = h(x − hxi)2 i = hx2 i − h2xhxii + hhxi2 i = hx2 i − hxi2 ,
(2-136)
and in the n−th state this becomes
p
< x2 >n − < x >2
1/2
a
6
= √
.
1− 2 2
n π
2 3
(∆x)n =
(2-137)
For the mean value of p in the n state we get
hpin = pn =
Za
ψn∗ (x)p̂ψn (x) dx =
0
Za
ψn∗ (x)(−ih̄
d
ψn (x) dx = 0
dx
(2-138)
0
because the derivative of the sinus function we get cosine which is orthogonal to the sinus
function (as we know from Fourier analysis).
Mean value of the square of momentum could be inferred from the mean value of
energy because the total energy just equals to kinetic energy so
n2 h̄2 π 2
< p >n = 2mEn =
.
a2
2
(2-139)
The deviation or uncertainty of p in n-th state is (∆p)n and it equals
(∆p)n =
p
nπh̄
.
hp2 i − hpi2 =
a
(2-140)
D. Horvat: Physics of Materials: Concepts
The product of uncertainties for the coordinate and momentum is
1/2
6
h̄
nπ
1− 2 2
· .
(∆x)n · (∆p)n = √
n π
2
3
Str. 2 - 35
(2-141)
For n = 1 we have the numerical factor equals 1.1357 in front of h̄/2 and if we remove it
the RHS decreases and we get an inequality
h̄
> .
∆x · ∆p ∼
2
(2-142)
and this is Heisenberg relation (principle) of uncertainty for conjugate variables x and p.
The result is the same as for the wave packet (2-112).
Probability distribution for a particle in the infinite square well in one dimension
We will calculate probability of finding a particle confined in the infinite square
potential well defined in interval [0, L] to be found within the interval [0, L/4] for (a)
ground state and (b) first excited state.
ψ ψ
ψ ψ
Figure 2.3 - Graphs to accompany the calculation
of probability distribution in one dimensional potential well
The wave function for a particle in the quantum state n is
r
2
nπx −iEt/h̄
ψn (x, t) =
sin
e
,
L
L
and the probability of finding the particle is for (a)
2
W1 =
L
L/4
Z
1
1
πx
dx = −
.
sin2
L
4 2π
0
Str. 2 - 36
2. Schrödinger Eqution I: Introductory Problems
and for (b)
2
W2 =
L
L/4
Z
2πx
1
sin2
dx = .
L
4
0
2.6 Potential step
ψ
Figure 2.4 - Incoming stream of free particles on
a potential step of height V0 and its reflection
The wave function in one dimension for a free particle (V (x) = 0) is given here as
a plane wave
ψ(x) = C1 eikx + C2 e−ikx
(2-143)
and it is conveniently written as
ψ (+) (x) = C1 eikx
ψ (−) (x) = C2 e−ikx .
(2-144)
Let us calculate how is the operator p̂x acting on a wave function ψ (+) (x)
p̂x ψ (+) (x) = (−ih̄
d (+)
)ψ (x) = (h̄k)ψ (+) (x).
dx
(2-145)
D. Horvat: Physics of Materials: Concepts
Str. 2 - 37
We can see that the function ψ (+) is an eigenfunction of this operator and that the eigenvalue of the operator is the value of the momentum according to de Broglie relation. We
can say, therefore, that the wave function ψ (+) describes the situation when a free particle
moves in the positive direction of the x axis. In the same wave we would come to conclusion that the wave function ψ (−) describes a particle moving in the negative direction of
the x axis.
Consider now the situation when a beam of free particles moves from left to right,
so when the situation is described by ψ (+) and ψ (−) . Hence, we are studying impact of
free particles on the potential of the form (according to Fig. 2.4)
V (x) =
(
0,
for
x<0
V0
for
x > 0.
(2-146)
We have two regions, I and II, where solutions of Schrödinger equation should be found.
We can write
ψ(x) = ψI (x) · Θ(−x) + ψII (x) · Θ(x),
(2-147)
where Θ(x) is Heaviside function (see Appendix B). In the region I, i.e. for x < 0 we have:
−
where kI =
q
h̄2 ′′
ψ (x) = EψI (x)
2m I
or
ψ ′′ (x)I + k12 ψI (x) = 0
(2-148)
2mE/h̄2 , and in the region II, i.e. for x > 0 is
where now kII
′′
2
ψII
(x) + kII
ψII (x) = 0
(2-149)
q
= 2m(E − V0 )/h̄2 . The solution in the region I is
(+)
(−)
ψI (x) = AeikI x + Be−ikI x ≡ ψI (x) + ψI (x).
(2-150)
This is a linear combination of wave functions discussed above. Probability density current
(its x component) for the moving from left to right is
(+)
jI (x) = −
i
d (+)
d (+)∗
h̄kI
ih̄ h (+) ∗
(+)
ψI
(x) ψI (x) − ( ψI (x))ψI (x) = |A|2
2m
dx
dx
m
and for right to left is
(−)
jI
(x) = −|B|2
h̄kI
.
m
(2-151)
(2-152)
Str. 2 - 38
2. Schrödinger Eqution I: Introductory Problems
Relative probability density for first and for the second probability density current is
(−)
ρI
(+)
ρI
(−)
|ψ |2
|B|2
,
= I(+) 2 =
|A|2
|ψ |I
(2-153)
In order to discuss a probability of finding transmitted and reflected particles one introduces coefficients of reflection R and transmission T which are defined like corresponding
ratios of probability densities
(+)
T =
jII
(+)
jI
(−)
R=
jI
(+)
jI
(2-154)
A form of solution in the region II depends upon the relative ratio of energy E and hight
of the potential step V0 .
Consider first the case when E > V0 . The solution in the region II is
ψII (x) = CeikII x + De−ikII x .
(2-155)
In this region the is no obstacle from which a wave could be reflected we can immediately
set D = 0 so the solution for this region is
ψII (x) = CeikII x .
(2-156)
The corresponding probability density current is
jII = |C|2
h̄kII
.
m
(2-157)
According to Postulate 1 and its comment a wave function and its first derivative have to
be continuous (we have ”to sew” solutions and their derivatives) simply meaning that on a
boundary of regions solutions have to be equal (continuity) and their first derivative have
to be equal as well (smoothness). So
′
ψI (0) = ψII (0) i ψI′ (0) = ψII
(0)
(2-158)
and we get two equations for the unknown constants
A+B =C
ikI A − ikI B = ikII C.
(2-159)
D. Horvat: Physics of Materials: Concepts
Str. 2 - 39
After we substitute B we get the ration
2kI
C
=
,
A
kI + kII
(2-160)
from which, using C = A + B we get another important ratio
B
kI − kII
=
.
A
kI + kII
(2-161)
From these ratios we proceed to calculate coefficients of transmission and reflection according to (2-154). We get
(kI − kII )2
R=
(kI + kII )2
T =
(2-162)
4kI kII
and
(kI + kII )2
R + T = 1.
A paradox is evident: although the energy E of incoming particles is higher than the height
of the potential step V0 , there is a partial reflection besides the (partial) transmission.
When, however, the energy E approaches V0 there is no transmission i.e. T → 0 and all
particles get reflected, R → 1.
When is E > V0 for the ration of probability densities is
P =
2 − α − 2(1 − α)1/2
|B|2
=
|A|2
2 − α + 2(1 − α)1/2
(2-163)
where α = V0 /E. If α ≤ 1/2, we can expand the square root (1 − α)1/2 ≃ 1 − α/2 − α2 /8 −
α3 /48 − . . . so the above ration becomes
P =
α2 /4
α2
→
.
4 − 2α − α2 /4
8(2 − α)
(2-164)
Therefore, we can see explicitly that, although the energy E is bigger then the height V0 of
the potential step, some of the particles get reflected. There is no reflection when α = 0,
a this is the case only for E ≃ ∞. Also if, for example E = 2V0 , i.e. α = 0.5, we get that
P = 0.021, meaning that 2% of incoming particles will be reflected!
If E < V0 then the expression under the square root in the definition of kII is
negative. We write it as kII = ik2 and the solution in the region II is
(−)
(+)
ψ̃II (x) = C̃eikII x + D̃e−ikII x = C̃e−k2 x + D̃ek2 x = ψII + ψII (x).
(2-165)
Str. 2 - 40
2. Schrödinger Eqution I: Introductory Problems
Using Postulate 1 and its comment, we require again continuity and smoothness meaning
(+)
that D̃ = 0 because ψII (x → ∞) → ∞. Boundary conditions (2-158) are now
A + B = C̃
i
ikI A − ikI B = −k2 C̃
(2-166)
and they give us the ratios
C̃
2ikI
=
A
ikI − k2
i
B
2ikI + k2
=
,
A
ikI − k2
and the probability current density is
∂
h̄i
∗ ∂
∗
ψ̃II (x)
ψ̃II (x) − ψ̃II (x) ψ̃II (x) = 0.
j̃II = −
m
∂x
∂x
(2-167)
(2-168)
According to this the transmission coefficient T equals to zero. The above ration (2-167)
gives the coefficient of reflection
2
B R = = 1.
A
For the ration of probability densities we get now
|C̃|2
4kI2
E
4
=
=4
= .
2
2
2
|A|
kI + k2
V0
α
(2-169)
Hence, for the finite barrier there is nonzero probability of finding particles in the region
x > 0 i.e. in the classically forbidden region. We can calculate the so called penetration
depth which shows at which depth ”in the wall” in the classically forbidden region the
amplitude of the wave function gets reduced by the factor e. So for the wave function in
the region II we have
ψ̃II (x) = C̃e−k2 x
and at the penetration depth xp
ψ̃II (xp ) =
so
xp =
C̃ −k2 xp
e
e
1
k2
(2-170)
(2-171)
For the case V0 > E we can write the wave function in both regions in the following form
using only amplitude A
ψ(x) =
ikI x
A
e
· (1 + ik2 /kI ) + (1 − ik2 /kI )e−ikI x · Θ(−x)
1 + ik2 /kI
k2 x
+ 2e
·Θ(x)}
(2-172)
D. Horvat: Physics of Materials: Concepts
where A is the amplitude of the incoming wave, and k2
q
2mE/h̄2 .
Str. 2 - 41
q
2m(V0 − E)/h̄2 i kI =
=
2.7 Tunnel effect
The result from the last chapter that there is a finite probability of finding a particle
within the potential step for E < V0 leads to so called tunnel effect or tunnelling meaning
that a quantum mechanical particle can pass through a region which is according to the
principles of classical mechanics forbidden, or as we say ”classically forbidden”.
ψ
ψ
ψ
Figure 2.5 - Potential barrier of height V0 and thickness a. The case E < V0 is given,
where E is the energy of incoming particles.
Consider a one dimensional potential, according to the Fig.2.5
V (x) =
(
0,
for
x < 0 and x > a
V0
for
0 < x < a.
(2-173)
and let E < V0 . With this potential we have defined three regions. In the region I and
III we have solutions
ψI (x) = Aeik1 x + Be−ik1 x
ψIII (x) = C3 eik1 x
(2-174)
Str. 2 - 42
2. Schrödinger Eqution I: Introductory Problems
and in the region II is
ψII (x) = C2 ek2 x + D2 e−k2 x
q
q
2
where k1 = 2mE/h̄ and k2 = 2m(V0 − E)/h̄2 , and k2 is a real number.
(2-175)
Continuity of the wave function at x = 0 and x = a (Postulate 2) gives
ψI (0) = ψII (0),
′
ψI′ (0) = ψII
(0)
ψII (a) = ψIII (a),
′
′
ψII
(a) = ψIII
(a)
(2-176)
and they relate the unknown constants
A+B =C +D
ik1 A − ik1 B = k2 C − k2 D
Cek2 a + De−k2 a = Eeik1 a
(2-177)
k2 Cek2 a − k2 De−k2 a = ik1 Eeik1 a
When we multiply the first equation by ik1 and add it to the second we get the relation
between coefficients A, C and D
2ik1 A = (ik1 + k2 )C + (ik1 − k2 )D,
(2-178)
and by multiplying the third equation by k2 and adding it or subtracting it from the fourth
we get two relations
k2 + ik1 ik1 a−k2
e
and
2k2
k2 − ik1 ik1 a+k2
e
.
D=E·
2k2
C=E·
(2-179)
Substituting these relation in (2-178) we get an equation connecting A and E, and the
relation will give the coefficient of transmission T . Hence,
E
4ik1 k2
=
· e−ik1 a .
2
−k
A
(ik1 + k2 ) e 2 − (ik1 − k2 )2 ek2
(2-180)
From a formula for hyperbolic function e±x = ch x ± sh x we can express the following
functions
2
E
· e−ik1 a
(2-181)
=
A
k22 − k12
sh k2 a
2ch k2 a + i
k1 k2
D. Horvat: Physics of Materials: Concepts
Str. 2 - 43
and multiplying this expression with its complex conjugate we get the expression for the
transmission coefficient
2
E 4
T = =
2
2
A
k2 − k12
2
4 · ch k2 a +
sh2 k2 a
k1 k2
(2-182)
1
.
=
(k12 + k22 )2 2
sh k2 a
1+
4k12 k22
If we introduce, as we did before, the value for a ratio α = V0 /E, we can write, with the
approximation
ek2 a − e−k2 a
1
sh k2 a =
≃ ek2 a
2
2
the expression for the transmission coefficient
T =
or, finally
1
2
α
· sh2 k2 a
1+
4(α − 1)
≃
E
T = 16 ·
V0
1
2
α
· e2k2 a
1+
16(α − 1)
E
1−
V0
−2a
e
r
≃
16(α − 1) −2k2 a
·e
α2
2m
(V0 − E)
h̄2
.
(2-183)
(2-184)
We can immediately see that T 6= 0, i.e. there is the tunnelling of a particle through the
barrier, and T decreases exponentially with barrier thickness a.
V(x)
V(x)
O
a
b
O
a
b
Fig. 2.6 - Potential V (x) and the classical turning points a and b.
Str. 2 - 44
2. Schrödinger Eqution I: Introductory Problems
For a general form of a barrier in one dimension V = V (x), Schrödinger equation
is
d2 ψ(x) 2m
+ 2 [E − V (x)]ψ(x) = 0.
(2-185)
dx2
h̄
In the case of a classical particle in a potential V (x), points which correspond to places
where the classical particle could approach having a energy E are called classical turning
points. At these points the classical particle is being reflected from the potential (according
to Fig. 2.6), irrespective of the particle being scattered by the potential (left panel) or
being confined by the potential V (x) (right panel).
∆
Fig. 2.7 - The barrier is divided in a series of thin rectangles.
Turning points are here x1 and x2 .
If we a general barrier, according to Fig. 2.7 (where the turning points are x1,2 )
divide in thin ”ribbons” of thickness ∆xi , then for a transition probability through one of
the barriers we can write
q
−2 2m[V (x) − E]/h̄2 ∆xi
(2-186)
Ti = e
and the probability for all the barriers is the ”AND” probability which is given as a product
of single probabilities Ti
q
X
p
2
2m/h̄
∆x
V (xi ) − E
−2
i
Y
i
(2-187)
Ti = e
T =
i
D. Horvat: Physics of Materials: Concepts
Str. 2 - 45
and for ∆xi → 0 the sum turns into an integral
−2
T =e
r
2m
·
h̄2
Zx2 p
x1
[V (x) − E)] dx
.
(2-188)
This formula could be obtained from so called the WKB approximation† so the formula is
often called the transmission coefficient in the WKB approximation. Points x1,2 are to be
calculated as solutions of the equations
V (x1 ) = E
and
V (x2 ) = E.
(2-189)
2.5.1 Applications of the tunnel effect
2.5.1.1 Field emission
Eγ
W
EF
Fig. 2.8 - Schematic presentation of electrons in a metal. Electrons are bounded in
a potential well of the total depth W + EF , and their expulsion from the metal is
possible by using a photon of an energy Eγ . W is called the work function according
to the earlier description of the photoelectric effect. The highest energy of bounded
electrons is EF called Fermi energy.
†
Gregor Wentzel (1898.-1978.), Hans Kramers (1894.-1952.), Leon Brillouin (1889.-1969.)
Str. 2 - 46
2. Schrödinger Eqution I: Introductory Problems
When an atom is in its lowest energy state called a ground state and when a lowest
temperature is reached, electrons are positioned in energy levels according to Pauli principle
according to which in one quantum state there can be only one electron (or, more generally,
only one fermion). If two electrons have all quantum numbers equal except for the spin
quantum number, then in states determined by these quantum numbers there can be only
two electrons, because they differ only in their spin quantum number (which can have
values ”up” - ”down”, or ”↑”-”↓”). With such ”ordering” of electrons in the energy levels,
from lowest to highest we come to Fermi energy EF which represents the highest (allowed)
energy of an atom in its ground state at temperature T = 0. When the temperature is
increased then electrons get excited and they start to populate higher energy states, higher
then EF , but it can be shown that these energies are not much higher then EF , so in a
qualitative considerations we can always take that in the ground state electrons fill all
energy states, from the lowest to the highest and this highest energy could be take to be
EF , as is shown in Fig. 2.8.
V(x)
V0
V(x)=− e|E|x
W
T
EF
O x0
x2
x
Fig. 2.9 - Field emission is process when electrons are enabled by using am external
field giving a potential energy V (x) = −e|E|x to tunnel through the barrier. Due to
the external field the potential V (x) gets deformed and the dashed line represents the
final potential energy produced by the mirror (induced) charge on the surface of the
metal.
Field emission represents the emission of electrons from a cold cathode. (Emission
of electrons due to thermal effects is called thermionic emission.) Electrons are bound in
a metal by a potential which is for the work function higher then the maximal energy
i.e. Fermi energy EF which have bound electrons in the metal. External field Ez deforms
the shape of the potential well which now becomes thin enough at the level of Fermi
energy, according to Fig. 2.9. In that way the conditions for tunneling are created, the
tunneling through classically forbidden region. Similar situation is given in Fig. 2.10where
D. Horvat: Physics of Materials: Concepts
E
Str. 2 - 47
z
zm
C
φ=− Ez z
T
E3
E2
E
E1
Fig. 2.10 - Field emission where is the Coulomb potential (C) deformed by the
external field so that the potential is given as φ(z) = −Ez z so the effective potential
(E) is obtained (curved dashed line). The electron tunneling is now possible for
electrons with energy E3 .
M
V
q
-q
x
x
Fig. 2.11 - Mirror image is formed in the metal (M) and its charge is opposite (−q)
from the charge Q which has induced it. The charge q is in vacuum (V) and the
distance between charges is 2x.
the attractive Coulomb potential (C) is changed by the application of the external field
(dashed line in Fig.) so the final potential energy presented by the dashed curve is obtained.
However, since we are dealing with a conductor from which electrons could be
ejected, there is also an induced charge of opposite sign from the charge which is a source
of the external electric field. This problem could be solved in the classical electrodynamics,
using the method of mirror images, according to Fig. 2.11. Therefore, the initial potential
energy V0 changes first to V (x) = V0 − e|E|x and after that due to the image there is
Str. 2 - 48
2. Schrödinger Eqution I: Introductory Problems
another term
e2
.
V (x) = V0 − e|E|x −
4πε0 4x2
(2-190)
This curve (on Fig. 2.9) has a maximum on x0 , when dV (x)/dx = 0.
2.5.1.2 Contact potential
V
M1
V
M2
V
W1
W2
EF 1
E F2
∆W
(a)
V
V
W2
W1
E F2
EF 1
(b)
Fig. 2.12 - On the panel (a) we have two by vacuum (V) separated metals M1 and
M2 . Their Fermi energies EFi are different as well as their work functions Wi . When
the metals are close enough, electrons by tunneling move from M1 to M2 and form
the potential ∆W which is called the contact potential .
A contact potential is formed between two metals which have different Fermi energies according to Fig. 2.12. When the metals are brought into contact potential barrier
formed between the metals is such that electrons with higher Fermi energy can tunnel
through the barrier as long as their Fermi energies become equal. The metal which has received electrons become electro negative so a potential difference is formed. This potential
difference is called the ”contact potential”.
D. Horvat: Physics of Materials: Concepts
Str. 2 - 49
2.5.1.3 α decay
C
Z .2e 2
4πε0 r
r0
O
k
-V0
Fig. 2.13 - Tunneling in nuclear physics. A particle is confined u a potential well
of width r0 and by tunneling it exits with energy Ek which is lower then the hight
of the Coulomb barrier. The Coulomb barrier is given by the expression EC =
2Ze2 /(4πε0 r).
Uranium isotope
238 U
238 U
decays through the α− decay
→ α[Eα = 0.67 pJ = 4.2 MeV] + 234 X
(2-191)
Measurements of the potential barrier show that its height is 1.4 pJ=8.78 MeV.
In order to leave the confinement region α−particle should have, according to classical
approach, the energy equals to the barrier height (8.78 MeV). With this energy the α
particle would hit a detector as well. However, according to the above calculation their
energy is only about 4.2 MeV, and they leave the nucleus and reach a detector. This
simply means that α particles tunnel through the barrier and it is a consequence of the
Coulomb attraction in its attractive part and the Coulomb repulsion in its repulsive part.
Str. 2 - 50
2. Schrödinger Eqution I: Introductory Problems
Let x1 = a and x2 = b = 0 be classical turning points (see Fig. 213). The point a could
be obtained from
Z1 Z2 e2
E=−
4πε0 a
a
V (x) = − · E.
x
⇒
Z1 Z2 e2
= −aE
4πε0
so
(2-192)
The formula for T (2-188) we can write in the following way
T = e−2γ
(2-193)
and we calculate the exponent γ
γ=
√
E
r
2m
h̄
Z0 r
a
a
− 1 dx =
x
r
2mE
·a·
h̄2
Z0 r
1
− 1 du
u
1
(2-194)
2
=
Z1 Z2 e π
.
4πε0 h̄v
where a speed of the α particle is obtained from its kinetic energy Ek = mv 2 /2 from which
an assessment for the value of v is calculated v ≃ 7 × 106 m/s. Also the formula for the
algebraic integrals was used
Z1
0
xα (1 − x)β dx =
Γ(α + 1)Γ(β + 1)
= B[α + 1, β + 1]
Γ(α + β + 2)
with values of Γ−function
√
1√
1
3
= π Γ
=
Γ
π
2
2
2
Γ(n + 1) = n!
The above result (2-194) is known as Gamov factor , by which it is easy to calculate the
coefficient of transmission T .
2.6 Quantum mechanical harmonic oscillator
2.6.1 Introduction. Classical harmonic oscillator
D. Horvat: Physics of Materials: Concepts
Str. 2 - 51
Very often we have in different physical situations that a particle (or a body) is in
stable equilibrium and a force is present which returns a particle to the equilibrium position
when a disturbance moves it from the equilibrium. If the force is linearly proportional to
a displacement from the equilibrium position and a particle has an energy then it will
oscillate around the equilibrium. This kind of force we call the harmonic force, and the
oscillating system which is moving due to the harmonic force is called harmonic oscillator .
The examples of a harmonic oscillator is a body on a spring, simple (or mathematical) pendulum for small amplitudes, atoms in a crystal structure, vibration of atoms in a
molecule etc. Potential function U (x) of a particle is a function which has a minimum at a
point x = x0 (which we can shift to x = 0 by change of coordinates). If we make a Taylor
(or Maclaurin) expansion of the potential energy around the equilibrium point (minimum)
x0 = 0 we get:
∂ 2 U (x) U (x) = U (0) +
· x2 + . . .
(2-195)
2
∂x
x=x0
(The first derivative equals to zero so it does not occur in the expansion.) If we can neglect
higher order terms in the expansion, in this approximation we have obtained the potential
energy of the harmonic oscillator V (x) ≡ U (x) − U (0) = const. · x2 , and the (conservative)
force which moves a particle to the equilibrium is
F =−
∂V (x)
= −kx
∂x
so V (x) = 12 kx2 . A solution of the classical equation of motion (Newton 2nd law) is a
p
coordinate as a harmonic function of time: x(t) = A cos(ωt + ϕ), where ω = k/m and
the amplitude A and the phase ϕ are to be determined from the initial conditions.
2.6.2 Schrödinger equation for harmonic oscillator in one dimension
Quantum mechanical problem of motion of a particle with a mass m in one dimensional potential of a harmonic oscillator with the potential energy V (x) = kx/ 2 is a
solution of Schrödinger equation
∂
1 2
h̄2 ∂ 2
+ kx Ψ(x, t) = ih̄ Ψ(x, t)
(2-196)
−
2
2m ∂x
2
∂t
Str. 2 - 52
2. Schrödinger Eqution I: Introductory Problems
which, after separation of variables
becomes
Ψ(x, t) = f (t)ψ(x)
(2-197)
h̄2 d2
1 2
ih̄ d
1
−
ψ(x) + kx ψ(x) =
f (t).
2
ψ(x)
2m dx
2
f (t) dt
(2-198)
As we had before, the RHS depends only on the spatial coordinate x and the LHS depends
only on time we introduce a (separation) constant of dimension of energy with appropriate
symbol E so the time-dependent part leads to a solution
f (t) = f0 e−iEt/h̄ .
(2-199)
The coordinate x dependent part is
h̄2 d2
1 2
−
+ kx ψ(x) = Eψ(x)
2m dx2
2
2
1
d ψ(x) 2m
+ 2 E − kx2 ψ(x) = 0
2
dx
2
h̄
with k = mω 2 . We introduce a new dimensionless variable
r
mω
2 1/4
2 1/4
2 2
y = (mk/h̄ ) x = (m ω /h̄ ) x =
x
and the symbol
h̄
2E
,
(2Em1/2 /h̄k 1/2 ) = λ =
h̄ω
(2-200)
(2-201)
so (2-200) could be written as
d2 ψ
+ (λ − y 2 )ψ = 0.
dy 2
(2-202)
Eigenfunctions correspond to states with positive energy. The eigenfunctions have to
approach zero for |x| → ∞, because regions with infinite potential have to have zero
probability of finding a particle. Therefore, we have to solve the above equation with
the boundary condition: ψ(y) = 0 for y = ±∞. In that region a dominant term in the
equation is y 2 ψ so we can neglect (for the time being) the term λψ (in comparison to y 2 ψ)
so (2-202) becomes:
′′
ψ∞
= y 2 ψ∞ .
(2-203)
D. Horvat: Physics of Materials: Concepts
Str. 2 - 53
A solution of the equation has a form
1
2
ψ∞ (y) = e± 2 y .
(2-204)
Mathematically are both solutions acceptable, but physics dictates that we accept the one
which provides a square integrable eigenfunction. Therefore, we take the one with the
negative exponent and write a wave function as a solution of the equation (2-202)
ψ(y) = ψ∞ (y) · h(y),
(2-205)
where the function h(y) has ”to cover” (describe) the region |y| < ∞. Plugging this
function back in the equation we get with ψ”∞ − z 2 ψ∞ = −ψ∞ :
e−y
Since e−y
vanish so
2
/2
2
/2
[h′′ (y) − 2yh′ (y) + (λ − 1)h(y)] = 0.
(2-206)
equals to zero only for ±∞, the expression within the parentheses has to
h′′ (y) − 2yh′ (y) + (λ − 1)h(y) = 0.
(2-207)
One of the powerful methods for solving linear differential equations is power series solution.
Here we use this method to solve the differential equation of the second order. So
h(y) =
∞
X
k
ak y ,
′
h (y) =
∞
X
ak ky
k−1
,
h”(y) =
k=2
k=1
k=0
∞
X
ak k(k − 1)y k−2 ,
(2-208)
and substituting these back in the equation (2-207) we get
∞
X
k=2
ak k(k − 1)y
k−2
−2
∞
X
k=1
k
ak ky + (λ − 1)
The above equation we can write, term by term
∞
X
ak y k = 0.
(2-209)
k=0
a2 · 2 · 1 · y 0 − 2 · a1 · 1 · y 1 + (λ − 1) · a0 · y 0 +
+a3 · 3 · 2 · y 1 − 2 · a2 · 2 · y 2 + (λ − 1) · a1 · y 1 +
(2-210)
+a4 · 4 · 3 · y 2 − 2 · a3 · 3 · y 3 + (λ − 1) · a2 · y 2 + · · · .
Since we have linearly independent terms, coefficients by corresponding power have to
vanish
y0
2 · a2 + (λ − 1) · a0 = 0
y1
3 · 2 · a3 − 2 · a1 · 1 + (λ − 1) · a1 = 0
y2
..
.
4 · 3 · a4 − 2 · 2 · a2 + (λ − 1) · a2 = 0
yn
(n + 2)(n + 1) · an+2 − 2 · n · an + (λ − 1) · an = 0
(2-211)
Str. 2 - 54
2. Schrödinger Eqution I: Introductory Problems
In such a way we obtained a recurrence relation for (unknown) coefficients a0 , a1 , a2 , . . . in
the expansion of the function h(y), (2-208). Instead of writing series term by term we can
get a general term by renaming the first sum in (2-209): k−2 → k, k−1 → k+1, k → k+2
(the second term needs not to be renamed) and the term with k = 0 does not contribute
because k occurs in the summed expression so we can shift the beginning of the summation
from k = 1 to k = 0. We get
∞
X
k=0
[ak+2 (k + 2)(k + 1) − 2ak k + (λ − 1)ak ]y k = 0
(2-212)
and we get the relations among the coefficients ai :
2k − (λ − 1)
ak .
(k + 1)(k + 2)
ak+2 =
(2-213)
For a large k the ratio of ak+2 and ak becomes
ak+2
2
≃ .
ak
k
(2-214)
2
If we now look at the power series representation of the function ex we would have
x2
e
=
∞
X
k=even
xk
(k/2)!
(2-215)
we can see that the ratio between two neighboring coefficients is equal 2/k. This means
that the infinite power series expansion of the h(y), (2-208), must be truncated at certain
k = km (the expansion (2-215) is valid for an infinite series). Otherwise this infinite series
2
would represent the function ey , and this would mean that the wave function (2-205),
which is a solution of the equation of (2-202) would have a form
ψ(y) ∼ h(y)e−y
2
/2
2
= ey · e−y
2
/2
−→ ey
2
/2
,
(2-216)
and it does not satisfy the boundary conditions. Because of that the infinite series (2-208)
has to become just a polynomial of n−th order. So let an+2 = 0. The numerator of (2-213)
is zero, and with k → n we get
2k − (λ − 1) = 0,
(2-217)
and we have obtained the relation which quantizes energy
λ → λn = 2n + 1
(2-218)
D. Horvat: Physics of Materials: Concepts
Str. 2 - 55
or
λ=
2mE 1/2
h̄k 1/2
1/2
(λn =
2mEn
)
h̄k 1/2
(2-219)
so we get
En = h̄ω(n + 1/2).
(2-220)
Hence, we have quantized energy and the quantization is here (again) the consequence of
boundary conditions imposed na the wave function. (We should recall that the energy
quantization in the case of infinite square potential well stem from the same condition:
sin kx = 0 for x = a.)
Hermite polynomials
The differential equation (2-207) has a solution in the form of Hermite polynomials Hn
Hn′′ (y) − 2yHn′ (y) + 2nHn (y) = 0
(2-221)
where n is a natural number. Hermite polynomials follow the recurrence relations (i.e.
relations which connect polynomials and their derivatives of diverse order)
Hn′ = 2nHn−1
(2-222)
Hn+1 − 2yHn + 2nHn−1 = 0
1
-2
2
-2
2
-2
2
-2
2
-2
2
-2
2
Figure 2.14 - First six Hermite polynomials, from H0 (x) to H5 (x)
Str. 2 - 56
2. Schrödinger Eqution I: Introductory Problems
4
2
-3
-2
-1
1
2
3
-2
-4
Figure 2.15 - The wave function of harmonic oscillator: e−y
2
/2
· H3 (y).
|ψ(x)| 2
-2
2
Figure 2.16 - Absolute square of a wave function - probability density for harmonic oscillator |ψ(x))|2
First few Hermite polynomials Hn are
H0 (y) = 1,
H1 (y) = 2y
H2 (y) = 4y 2 − 2,
H3 (y) = 8y 3 − 12y
(2-223)
H4 (4) = 16y 4 − 48y 2 + 12, H5 (y) = 32y 5 − 160y 3 + 120y.
They have a property
Z∞
−∞
2
Hm (y)Hn (y)e−y dy =
(
0,
2
√
n
m 6= n
πn!, m = n.
Normalized eigenfunctions ψ(y) are
2
1
ψn (y) = n √
e−y /2 Hn (y),
1/2
(2 πn!)
(2-224)
(2-225)
D. Horvat: Physics of Materials: Concepts
Str. 2 - 57
and taking in account the definition of the variable y and k = mω 2 we can write
ψn (x) =
r
4
2
mω n √
(2 πn!)−1/2 · e−mωx /2h̄ · Hn (
h̄
r
mω
x)
h̄
(2-226)
so that
Z∞
ψn∗ (x)ψ(x) dx = 1.
(2-227)
−∞
If we introduce
b=
r
h̄
mω
we can write down wave functions of the ground state (the lowest energy state) and of two
excited states
1
ψ0 (x) = p √ exp(−x2 /(2b2 )),
b π
2x
1
exp(−x2 /(2b2 )),
ψ1 (x) = p √
b
2b π
1
2x2
ψ2 (x) = p √ ( 2 − 1) exp(−x2 /(2b2 )).
2b π b
(2-228)
Therefore, we have solved the problem of harmonic oscillator in quantum mechanics: we
have found energy which is quantized as a consequence of boundary conditions imposed on
wave functions. It should be noted that for the ground state n = 0, the expression (2-220)
becomes
E0 =
h̄ω
.
2
(2-229)
So even in the ground state there is some energy different from zero. This is a genuine
quantum mechanical result related to before discussed relations which gave products of
uncertainties of coordinate and momentum. This result has to be compared further with
the Planck’s derivation of the spectral density function, where the (unphysical, as we can
see now) assumption was taken that the energy of the ground state is zero.
2.9 Heisenberg uncertainty relation for harmonic oscillator
Str. 2 - 58
2. Schrödinger Eqution I: Introductory Problems
It is now easy to calculate a mean value of coordinate x in the n state of quantum
harmonic oscillator. We have the equation
+∞
Z
2
< x >n ∼
e−y (Hn (y))2 y dy = 0
(2-230)
−∞
since we had to integrate an odd function over the symmetric domain. For the mean value
of coordinate squared we get
hx2 i =
h̄
(2n + 1).
2mω
(2-231)
For the mean value of momentum it is easy to see that it equals to zero since the recurrence
relations for Hermite polynomials (2-222) show that we get an integral over Hn and Hn−1
what is zero. Hence
hpin = 0.
(2-232)
For the square of momentum we would get, using again the recurrence relations the result
hp2 in =
from which
mωh̄
(2n + 1)
2
r
q
mωh̄
2
(2n + 1).
∆pn = hp2 in − (hpin ) =
2
(2-233)
(2-234)
In the similar way we get for the standard deviation of a coordinate
∆xn =
q
2
hx2 in − (hxin ) =
r
h̄
(2n + 1).
2mω
(2-235)
The product of standard deviations or uncertainties is
∆xn ∆pn =
h̄
(2n + 1)
2
(2-236)
and for the ground state (n = 0) this becomes
∆xn ∆pn =
h̄
2
(2-237)
and this is the Heisenberg uncertainty relation for the conjugate variables x and p.
This is already third times that we have product of uncertainties in this form.
D. Horvat: Physics of Materials: Concepts
Str. 2 - 59
These equations represent a mathematical expression of one deep physical principle
which was formulated by W. Heisenberg. This is the principle of uncertainities which says
that it is not possible, in the same time, (or in the same experiment) to measure coordinate
∂
and (its conjugate variable) momentum (x ↔ px = −ih̄ ∂x
) with an infinite precision.
Instead of that, if coordinate has a certain deviation from its mean value - uncertainty ∆x
(or standard deviation), then its conjugate momentum has also a certain uncertainty ∆px
and the product of these two uncertainties equals h̄ or it is larger than h̄. Therefore, we
cannot simultaneously diminish uncertainties of a pair of conjugate variables below a value
which is determined by the boundary given by their product and the boundary equals to
the reduced Planck constant.
Furthermore, for a wave packet we can write ∆x = vg · ∆t and since it is E = p2 /2m
and ∆E = ∆p · (p/m) = vg · ∆p we get (1-65)
∆E · ∆t ≥ h̄
(2-238)
and this is the uncertainty relation for conjugate variables E and t. for the wave packet.
With relations of uncertainty or with principle uncertainty a fundamental boundary
is established for impossibility for simultaneous measurement of two conjugate variables
with arbitrary precision, i.e. the variables which are correlated by the transition ”physical
quantity”→”operator” (Postulate 2). So we will see later that in the case of three dimensional problems we will introduce the angular momentum operator L and its z component
Lz → h̄∂/∂φ where the angle φ is an angle of rotation around the z axis. The uncertainty
relations for the corresponding conjugate variables will be (∆Lz ) · (∆φ) ≥ h̄.
It is important to point out here that the uncertainty relations
have more conceptual character than practical. Still they give an
order of magnitude of conjugate quantities which have to fulfill the
relations, and as such the uncertainty relations are used mostly for
an assessment.
Three dimensional potential well
Consider a particle of a mass m moving in a three dimensional rectangular potential well
Str. 2 - 60
2. Schrödinger Eqution I: Introductory Problems
with impenetrable walls and with sides (ax , ay , az ), and the confining potential is zero
inside the box and infinite outside. Schrödinger equation is
∂
h̄2
∆Ψ(x, t) + V (x, y, z)Ψ(x, t) = ih̄ Ψ(x, t)
−
2m
∂t
(2-239)
and after the separation of time and spatial variables we can write the equation
∆ψ(x) +
2m
[E − V (x, y, z)]ψ(x) = 0
h̄2
(2-240)
where the wave function is written as
i
Ψ(x, t) = e− h̄ Et ψ(x).
(2-241)
It is evident that the problem should be solved in Cartesian coordinates
∂2
∂2
∂2
2mE
ψ(x, y, z) = 0
+
+ 2+
∂x2 ∂y 2
∂z
h̄2
(2-242)
Using again the separation of variables we write the wave function as
ψ(x, y, z) = X(x)Y (y)Z(z)
(2-243)
V (x, y, z) = V (x)V (y)V (z), and V (x) is, for instance
0
for 0 < x < ax
V (x) =
∞
for ax ≥ x ≤ 0
and the same for V (y) and V (z).
(2-244)
since the potential is
After substituting the wave function in Schrödinger equation we get three separated
equations solutions of which we know how to write since they have the same form as the
one dimensional potential well
h̄2
2m
1 d2 X
1 d2 Y
1 d2 Z
+
+
X dx2
Y dy 2
Z dz 2
+E =0
(2-245)
and with E = Ex + Ey + Ez we get
X ′′ +
2m
Ex = 0
h̄2
(2-246)
D. Horvat: Physics of Materials: Concepts
Str. 2 - 61
and the wave function and the energy is
Xn (x) =
r
nx π
2
sin
x
ax
ax
n2x h̄2
.
2ma2x
Ex/n =
(2-247)
The same we get for other components of the wave function Y (y) and Z(z) so the total
wave function is
r s r
nx π
ny π
nz π
2
2
2
ψ(x, y, z) =
sin
x · sin
y · sin
x
(2-248)
ax ay az
ax
ay
az
and the energy
En = Ex/n + Ey/n + Ez/n
n2y
n2z
n2x
+
+
a2x
a2y
a2z
h̄2 π 2
=
2m
!
.
(2-249)
We can set that a = ax = ay = az hence the energy becomes
Enx ,ny ,nz =
h̄2 π 2 2
(n + n2y + n2z ).
2ma2 x
(2-250)
In the above expression the quantum numbers ni are positive and whole numbers. We
can look at one possible combination of these numbers and to calculate the corresponding
energy, for instance
(nx , ny , nz ) = (1, 2, 1) −→ E(1,2,1) =
h̄2 π 2
·6
2ma2
(2-251)
and it is clear that the same energy we can get with and possible permutation of the
numbers (1,2,1). So the same energy we can get using different sets of wave functions. We
say that we have degeneracy when to one energy there correspond different combinations
of wave functions, and number of wave functions giving the same energy is called the order
of degeneracy which is denoted as gd . In the above case we have three different wave
functions giving the same energy so gd = 3. The ground state is obviously non-degenerate
as are for instance states like (2,2,2), (3,3,3) etc.
♠
♠
♠
♠
Was this manual useful for you? yes no
Thank you for your participation!

* Your assessment is very important for improving the work of artificial intelligence, which forms the content of this project

Download PDF

advertisement