# 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 Â 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 Â 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. ♠ ♠ ♠ ♠

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

### Related manuals

Download PDF

advertisement