PY 351 Modern Physics - Lecture notes, 3 Copyright by Claudio Rebbi, Boston University, October 2016. These notes cannot be duplicated and distributed without explicit permission of the author. Time dependence of the wave function The time dependence of the wave function ψ(x, t) is governed by the Schrödinger equation ψ(x, t) ı h t = Ĥψ (1) where Ĥ is the Hamiltonian, or energy operator, for the quantum mechanical system. If we expand ψ(x, t) into a complete set of eigenfunctions ψ n (x) of Ĥ the expansion coefficients will be time dependent ψ(x, t) = c n (t)ψ n (x) (2) n The time dependence will show up in the coefficients c n (t), not in the eigenfunctions ψ n (x) which are the time independent solution to the eigenvalus equation Ĥψ n (x) = E n ψ n (x) (3) (We should note here that in some cases the expansion will be an integral rather than a sum, as in the case of a free particle, but the considerations that follow will still apply.) If we insert the expansion of Eq. 2 into Eq. 1 we get n ı h dc n(t) dt ψ n (x) = n c n (t)ĥψ n(x) = n E n c n (t)ψ n (x) (4) where we have used the fact that the functions ψ n (x) are eigenfunctions of Ĥ (see Eq. 3) and we used the total derivative rather than the partial derivative with respect to time because the expansion coefficients only depend on t. Since the eigenfunctions ψ n (x) are othonormal, the right and left hand sides of the last equation must match term by term and so we obtain the equations ı h dc n(t) = E n c n (t) (5) dt for the time evolution of the coefficients c n (t). These equations are solved by c n (t) = c n (0) e ıent/ h (6) where c n (0) is the initial value of the coefficient c n (its value at time t = 0). Thus the time evolution of the wave function is given by ψ(x, t) = n c n (0) e ıent/ h ψ n (x) (7) 1
In particular, if at t = 0 the system is in the n th eigenstate of energy 1, which means that c n (0) = 1 while all other c n (0) = 0, the system will remain in that eigenstate at all time and its wave function will evolve according to ψ(x, t) = c n (0) e ıent/ h ψ n (x) = e ıent/ h ψ(x, 0) (8) This last equation tells us that the time evolution of an eigenfunction of energy is simply given by a phase factor with the phase rotating with angular velocity E n / h. The initial values of the coefficients c n can be obtained from 2 c n (0) = ψ n (x) ψ(x, 0) (9) (at least in principle, since this requires nowledge of the eigenfunctions ψ n (x) which may not be easy to compute.) What we see from this is that the evolution of ψ(x, t) with time is fully determined by its initial value ψ(x, 0). Free particle The free particle Hamiltonian is Ĥ = 1 2m ˆp2 (10) It is then obvious that the eigenfunctions of momentum ψ(p, x) = 1 2π h e ıpx/ h (11) are also eigenfunctions of Ĥ. (If ˆpψ(p, x) = pψ(p, x) then ˆp 2 ψ(p, x) = ˆp(ˆpψ(p, x)) = ˆp(pψ(p, x)) = pˆpψ(p, x) = p 2 ψ(p.x).) The corresponding eigenvalue is E(p) = p 2 /2m: Ĥe ıpx/ h = p2 2m eıpx/ h (12) It follows that if the initial wave function ψ(x, 0) is the eigenfunction of momentum ψ(p, x) its time evolution will be given by ψ(x, t) = e ıe(p)t/ h ψ(x, 0) = 1 2π h e ıpx/ h ıe(p)t/ h = 1 2π h e ıpx/ h ıp2 t/(2m h) These wave functions may be superimposed to obtain the most general solution of the free particle Schrödinger equation: ψ(x, t) = 1 2π h (13) φ(p) e ıpx/ h ıp2 t/(2m h) dp (14) 1 We say that a system is in an eigenstate of some operator if its wave function is an eigenfunction of that operator. It is just a convenient terminology: e.g. the system is in an eigenstate of energy for the wave function of the system is an eigenfunction of the energy operator. 2 See the second set of lecture notes. 2
where φ(p) are the expansion coefficients of the initial wave function φ(p) = 1 2π h In particular we mae tae for φ(p) a Gaussian centered at p 0 : φ(p) = ψ(t, 0) e ıpx/ h dx (15) 1 π 1/4 b e (p p 0)2 /(2b2 ) This will give origin to a wave function with a Gaussian distribution of momenta with mean p 0 and mean square deviation (16) (p p 0 ) 2 = 1 (p p 0 ) 2 e (p p 0) 2 /b 2 = b2 πb 2 (17) Inserting this expression for φ(p) into Eq. 14 we obtain ψ(x, t) = 1 π 3/4 2 hb e (p p 0) 2 /(2b 2 )+ıpx/ h ıp 2 t/(2m h) dp (18) The Gaussian integral over p can be done analytically and gives the wave function ψ(x, t) of assignment 7. Note: carrying out the integration in Eq. 18 is straightforward but requires great care with the algebra. The exponent is of the form p 2 + ıβp + γ, where the expressions for, β and γ, which may be complex, can be read off Eq. 18. One proceeds by completing the square: p 2 +ıβp+γ = [p ıβ/(2)] 2 β 2 /(4)+γ. The exponential of β 2 /(4)+γ, which does not depend on p, factors out of the integral. The remaining exp [ [p ıβ/(2)] 2] dp can be easily done with the change of variable q = [p ıβ/(2)]. Phase velocity and group velocity: The waves being superimposed in Eq. 14, which are proportional to exp(ıpx/ h ıp 2 t/(2m h) propagate with velocity v ph = p2 t/(2m h) p = p 2m (19) v ph is called the phase velocity of the waves. (We can easily convince ourselves that this is the velocity of propagation looing for the increments x, t which will leave the argument of the exponential (the phase) unchanged. We get the condition ıp x/ h ıp 2 t/(2m h) = 0 from which we get x/ p = p/(2m).) However, as we have seen in assignment 7, the Gaussian wave pacet ψ(x, t) which one obtains from Eq. 18 moves with velocity v gr = p 0 /m, which is also the classical velocity of a particle with momentum p 0 and mass m. v gr is called the group velocity of the wave pacet (because it is the effective velocity of propagation of the group of waves.) Why the difference? We must recall that the waves we superimpose in Eq. 18 (or, more generally, 14) 3
are rapidly oscillating exponentials with imaginary argument, i.e. phase factors of the form exp(ıφ) = cos(φ) + ı sin(φ). The oscillations will lead to cancellations unless the values of x and t are such that the phase φ is stationary with respect to p, i.e. does not vary when p changes, in the neighborhood of the integration values which dominate the integral. This leads to the condition d ( ıpx dp h ıp2 t = 0 2m h) p0 (20) or ıx h ıp 0t m h = 0 (21) from which we deduce a group velocity v gr = p 0 m (22) Heisenberg s uncertainty principle We can establish a relation between the expectation value of x = (x x ) 2 and p = (p p ) 2 by using a mathematical inequality, as explained below. It is convenient to use the notation x 0 and p 0 for x and p, respectively. Consider the integral I() = ψ(x) [ ı(ˆp p 0 ) + x x ][ 0 ı(ˆp p 0 ) + x x 0 ] ψ(x) dx (23) where is a real variable and ψ(x) can be an arbitrary wave function. The crucial observation is that I() is a real, non-negative function of : I() 0 (24) We can prove this using the explicit form of ˆp and an integration by parts: I() = = = ψ(x) ( h d dx + ıp 0 + x x 0 ( h d dx + ıp 0 + x x 0 [( h d dx ıp 0 + x x 0 ) ψ(x) )( h d )ψ(x) ( h d ] [( dx ıp 0 + x x 0 dx ıp 0 + x x 0 h d dx ıp 0 + x x 0 ) ψ(x) dx ) ψ(x) dx ) ] ψ(x) dx (25) The integrand in the last integral is the product of a quantity times its complex conjugate and is therefore non-negative. It follows that the whole integral must be larger than or equal to zero. On the other hand, expanding the product in Eq. 23 we find I() = ψ(x) [ 2 (ˆp p 0 ) 2 + (x x 0) 2 2 ] ı(ˆpx xˆp) ψ(x) dx (26) 4
Let us examine in detail the expression ı(ˆpx xˆp). When this operator acts on a function f(x) it produces ı(ˆpx xˆp)f(x) = h d d xf(x) x h f(x) = hf(x) (27) dx dx Since f(x) is an arbitrary function we conclude that Equation 26 becomes then I() = [ ψ(x) 2 (ˆp p 0 ) 2 + (x x 0) 2 The inequality I() 0 gives us then 2 ı[ˆpx xˆp] = h (28) ] h ψ(x) dx = 2 (ˆp p 0 ) 2 + (ˆx x 0) 2 h (29) 2 2 (ˆp p 0 ) 2 + (ˆx x 0) 2 2 h (30) Let us loo for the minimum with respect to 2 of the expression in the l.h.s. of this inequality. The minimum will occur for the value of 2 which maes its derivative vanish, namely 2 (ˆx x 0 ) = 2 (31) (ˆp p 0 ) 2 Substituting bac into Eq. 30 we find 2 (ˆp p 0 ) 2 (ˆx x 0 ) 2 h (32) or p x h 2 (33) This inequality is the statement of Heisenberg s uncertainty principle : position and momentum cannot be determined with unlimited precision; the product of the standard deviation of these two variables from their average values can never be less than h/2. Confined free particles We will consider two cases in which a particle moves free within a domain of length L. We consider first the case when the particle moves in the domain L/2 x L/2 with periodic boundary conditions, or, equivalently, along a circle of length L, which we already considered when we studied the eigenstates of momentum (lecture 2, pp. 3 to 7). In this case the eigenfunctions of momentum ψ(p n, x) = 1 L e ıpnx h (34) 5
with p n = 2πn h L (n integer, positive, zero or negative) are also eigenfunctions of the energy operator (35) with Ĥ = (ˆp)2 2m Ĥψ(p n, x) = E n ψ(p n, x) = p2 n 2m ψ(p n, x) = 2π2 n 2 h 2 ψ(p ml 2 n, x) (37) The system has an infinite spectrum of energy levels, the lowest one with E 0 = 0 singly degenerate (this means that there is a single state with this energy eigenvalue), all others, with E n = 2π 2 n 2 h 2 /(ml 2 ) and n = 1, 2,... doubly degenerate (this means that there are two states with this energy eigenvalue, corresponding to n and n in Eq. 35.) The other case we will consider is the case of a particle confined to the interval 0 x L by a potential V (x) which is infinite outside of this interval and vanishes inside. The wave function ψ(x) must then vanish for x < 0 and x > 0 and continuity demands (36) ψ(0) = ψ(l) = 0 (38) For 0 x L the eigenfunctions of energy will have to satisfy the eigenvalue equation Ĥψ(x) = h2 d 2 ψ(x) = Eψ(x) (39) 2m dx 2 The solutions to this equation, without regard to the boundary conditions Eq. 38, are of the form f(x) = A sin( 2mE x/ h) + B cos( 2mE x/ h) (40) ψ(0) = 0 rules out the cosine solution, while ψ(l) = 0 imposes the condition sin( 2mE L/ h) = 0 (41) This equation will be satisfied if the argument of the sine function is an integer multiple of π, i.e. if 2mE L = πn (42) h We thus get the eigenvalues of the energy operator Ĥ E n = π2 n 2 h 2 2mL 2 (43) with the corresponding normalized eigenfunctions 2 ψ n (x) = sin(πnx/l) (44) L 6
As in the previous case we find again an infinite spectrum of energy levels with energy proportional to n 2. The states are all singly degenerate and there is no state with zero energy. Moreover the eigenfunctions of energy are no longer also eigenfunctions of momentum, since the boundary conditions 38 are not compatible with momentum eigenstates. The harmonic oscillator The harmonic oscillator Hamiltonian is Ĥ = ˆp2 2m + ˆx2 2 = h2 2m d 2 dx + x2 2 2 In order to obtain the energy eigenvalues and the corresponding wave functions we must solve the equation h2 d 2 ψ(x) + x2 ψ(x) = Eψ(x) (46) 2m dx 2 2 One way to proceed is to start from the eigenfunction (45) ψ(x) = e m x 2 /(2 h) (47) which one may find, for example, by trying a solution of the form exp( x 2 ) and realizing that with a suitable choice of Eq. 46 is satisfied with (lucy guess? inspired guess?) E = E 0 = h 2 m = hω 2 Starting from Eq. 47 one then loos for solutions of the form (48) with f(x) given by a power series expansion, leading to ψ(x) = e m x 2 /(2 h) f(x) (49) ψ(x) = e m x 2 /(2 h) (a 0 + a 1 x + a 2 x 2 + a 3 x 3 + a 4 x 4 +... ) (50) Substituting the above expression into Eq. 46 one obtains a recursive relation among the expansion coefficients by which a 2, a 3, a 4,... can be determined from a 0 and a 1. What happens then is that for a generic value of E the series is infinite and converges to a function which grows at infinity in a manner which maes the whole ψ(x) grow at x = ±, ruling it out as a possible wave function. However, if E taes one of the values ( E n = n + 1 ) hω (51) 2 the series ends at its n th term (a n+1 and all further coefficients turn out to be 0), so that the corresponding solution of the equation is of the form ψ n (x) = e m x 2 /(2 h) H n (x) (52) 7
where H n (x) is a definite polynomial of degree n, called a Hermite polynomial. The function ψ n (x) vanishes at infinity and is normalizable, so it is an acceptable wave function and it is the eigenfunction associated with E n. However, we will obtain the eigenvalues E n and the corresponding eigenfunctions ψ n (x) by following a different route, playing with the properties of some special operators which will lead us to a rather simple, algebraic solution to our problem. We begin from the classical form of the harmonic oscillator Hamiltonian H = 1 2m p2 + 2 x2 (53) where p and x are variables, not operators. If we introduce two complex variables = 1 2m ıp + = 1 2m ıp + 2 x (54) 2 x (55) H can be written as ( H = = 1 )( 1 ) ( 1 ) 2+ ( ıp+ 2m 2 x ıp+ 2m 2 x ) 2 = p 2m 2 x 1 = 2m p2 + 2 x2 (56) We will replicate the same definitions at the operator level introducing two operators â and â given in terms of ˆp and ˆx by expressions similar to Eqs. 54 and 55. It will be useful, however, to have â and â dimensionless ( and carry dimensions of square root of energy), so we introduce hω at denominator in the expressions for â and â, which are thus defined as â = 1 2 hωm ıˆp + â = 1 2 hωm ıˆp + ˆx (57) 2 hω ˆx (58) 2 hω where the symbol in â stands for Hermitian conjugate 3. We may now multiply â by â, as we did with and, but we must be careful since we are dealing with operators. Indeed it will be convenient to consider the two products â â and ââ separately. But first let us rewrite ˆp and ˆx as ı h d/dx and x: â = 1 2 hωm ıˆp + â = 1 2 hωm ıˆp + 2 hω ˆx = h 2ωm 2 hω ˆx = d dx + h 2ωm d dx + 2 hω x (59) 2 hω x (60) 3 Hermitian conjugation plays for operators a role similar to complex conjugation for complex numbers. In our case it is easy to prove with integration by parts that for any ψ(x) ψ (x)â ψ(x) dx is equal to ( ψ (x)âψ(x) dx). This identity may also be taen as the formal definition of the Hermitian conjugate of an operator. 8
With this we find ( h â d )( h â = 2ωm dx + 2 hω x 2ωm h d 2 2ωm dx d 2 2ω 2 m dx x + 2ω 2 m x d dx + = 1 ( h2 d 2 ) hω 2m dx + x2 1 2 2 2 d ) dx + 2 hω x 2 hω x2 (61) d dx x + 1 2 x d dx where we used ω 2 = /m to simplify the last two terms. At this point it would be tempting to say that the last two terms cancel, but it would not be correct. Let us indeed loo at the action of the differential operator represented by the last two terms on any function f(x). We will have ( 1 d 2 dx x + 1 2 x d dx = 1 ( 2 ) f(x) = 1 ( d 2 dx xf(x) + x d ) dx f(x) (62) f(x) x d dx f(x) + x d ) dx f(x) = 1 f(x) (63) 2 and we see that the last two terms in Eq. 62 do not cancel, but, rather, when acting on any function f(x) they give f(x)/2. That is tantamount to saying that 1 2 d dx x + 1 2 x d dx = 1 2 With this result in hand we may recast Eq. 62 in the form â â = 1 ( h2 hω 2m d 2 ) dx + x2 2 2 With a similar calculation one finds ââ = 1 ( h2 d 2 ) hω 2m dx + x2 2 2 1 2 = 1 hω Ĥ 1 2 + 1 2 = 1 hω Ĥ + 1 2 Equations 65 and 66 allow us to easily calculate the whole spectrum, i.e. eigenvalues and eigenfunctions, of Ĥ. To begin with, from Eq. 65 we obtain (64) (65) (66) Ĥ = hωâ â + hω 2 (67) We loo now for a wave function ψ 0 (x) satisfying âψ 0 (x) = 0 (68) or, using the explicit expression, Eq. 59, for â, and after multiplication by 2ωm/ h dψ 0 (x) m + dx h ψ 0(x) = 0 (69) 9
This equation is easily solved and we find the solution of Eq. 46, which we normalize here to ψ(x) 2 = 1, ψ(x) = (m)1/8 m x e 2 /(2 h) (70) (π h) 1/4 With âψ 0 (x) = 0 Eq. 67 gives us Ĥψ 0 (x) = hωâ âψ 0 (x) + hω 2 ψ 0(x) = hω 2 ψ 0(x) (71) and we see that ψ 0 (x) is an eigenfunction of Ĥ with eigenvalue E 0 = hω/2. How about the other eigenfunctions? To find them we tae advantage of the relation ââ â â = 1 (72) which one obtains subtracting Eq. 65 from Eq. 66. Given two operators  and ˆB the expression  ˆB ˆB is called the commutator of  and ˆB and is denoted by [Â, ˆB]. Equation 72 tells us that the commutator of â and â is 1: [â, â ] = 1 (73) and this fact has huge consequences. Indeed, consider now h ψ 1 (x) = â dψ 0 (x) ψ 0 (x) = + 2ωm dx 2 hω x ψ 0(x) (74) (The last expression in this equation could be simplified, but this is not of our concern now.) If we act by Ĥ on ψ 1(x), using the result in Eq. 67, we get Let us focus on â âψ 1 (x). We have Ĥψ 1 (x) = hωâ âψ 1 (x) + hω 2 ψ 1(x) (75) â âψ 1 (x) = â ââ ψ 0 (x) (76) But using the commutator ââ â â = 1, ââ may be written as â â + 1 and thus we get â âψ 1 (x) = â ââ ψ 0 (x) = â (â â + 1)ψ 0 (x) = â â âψ 0 (x) + â ψ 0 (x) = â ψ 0 (x) = ψ 1 (x) (77) since âψ 0 (x) = 0. Inserting our last result into Eq. 75 we get Ĥψ 1 (x) = ( 1 + 1 2) hωψ1 (x) (78) and we see that ψ 1 (x) is an eigenfunction of Ĥ with eigenvalue (1 + 1/2) hω. From ψ 1(x) we could construct another function ψ 2 (x) = â ψ 1 (x) = â â ψ 0 (x) (79) 10
and repeating the argument that led to Eq. 76 we would get The crucial identity in the above derivations is Ĥψ 2 (x) = ( 2 + 1 2) hωψ2 (x) (80) â â(â ) n ψ 0 (x) = â â(â â â... n times )ψ 0 (x) = n(â ) n ψ 0 (x) (81) which can be proved by repeatedly commuting â through â. We illustrate the proof with n = 3: â ââ â â ψ 0 (x) = â (ââ )â â ψ 0 (x) = â (â â + 1)â â ψ 0 (x) = â â (ââ )â ψ 0 (x) + â â â ψ 0 (x) = â â (â â + 1)â ψ 0 (x) + â â â ψ 0 (x) = â â â (ââ )ψ 0 (x) + 2a â â ψ 0 (x) = â â â (â â + 1)ψ 0 (x) + 2a â â ψ 0 (x) since âψ 0 (x) = 0. If we multiply Eq. 81 by hω we get or which shows that â â â â âψ 0 (x) + 3a â â ψ 0 (x) = 3a â â ψ 0 (x) (82) hωâ â(â ) n ψ 0 (x) = (Ĥ hω/2)(â ) n ψ 0 (x) = n hω(â ) n ψ 0 (x) (83) Ĥ(â ) n ψ 0 (x) = (n + 1/2) hω(â ) n ψ 0 (x) (84) is an eigenfunction of Ĥ with eigenvalue (n + 1/2) hω. ψ n (x) = (â ) n ψ 0 (x) (85) From this whole discussion we draw the following very important conclusions: The operators â and â defined in Eqs. 59 and 60 have commutator [â, â ] = 1. The Hamiltonian, or energy operator, of the quantum harmonic oscillator is Ĥ = hω(â â + 1/2). The wave function ψ 0 (x) of the ground state of the oscillator satisfies âψ 0 (x) and is an eigenfunction of Ĥ corresponding to the energy eigenvalue E 0 = hω/2. The excited states of the oscillator are obtained by repeated applications of the operator â. Their (non-normalized) 4 wave functions are given by ψ n (x) = (â ) n ψ 0 (x) and correspond to energy eigenvalues E n = (n + 1/2) hω. Since â increases the excitation level by one quantum of energy hω, â is often referred to as the creation operator (it creates one quantum.) If, for example, the quantum oscillator corresponds to an electromagnetic vibration of frequency ν = ω/(2π) then â would create a photon with energy E = hω = hν. 4 Using the algebra of the â and â operators it is actually rather straightforward to show that the normalized wave functions are (1/ n!)(â ) n ψ 0 (x). 11
If the operator â acts on the wave function ψ n (x) it produces a result proportional to ψ n 1 (x) (we did not show this, but it is easy to prove.) Thus â removes one quantum and is often referred to as annihilation operator. The wave function (â ) n ψ 0 (x) is an eigenstate of the operator ˆN = â â with eigenvalue n (see Eq. 81.) Thus ˆNψ n (x) = nψ n (x) and, in a sense, ˆN counts the number of quanta in the state of the oscillator. ˆN is therefore often referred to as the particle number operator. The algebra of the creation and annihilation operators is very powerful. It provides a full description of the quantum harmonic oscillator and constitutes the foundation for many advanced theoretical developments, such as solid state theory and particle theory. 12