2 x2 where m is the mass of the particle. Consider a particle subjected to a timeindependent

Transcription

1 Warm-up exercises These exercises 1 are a warm-up in the sense that they are supposed to help you revise some material from your previous quantum mechanics studies. It does not mean that the exercises are easy in fact some are pretty tough. Just remember, one learns the most from problems that fight back! Exercise 0.1 (The time-independent Schrödinger equation ) The Schrödinger equation is i 2 Ψ(x, t) = t 2m 2 Ψ(x, t) + V (x, t)ψ(x, t), x2 where m is the mass of the particle. Consider a particle subjected to a timeindependent potential V (x). Assume that a state of the particle is described by a wave function of the form Ψ(x, t) = ψ(x)χ(t). 2 (a) Show that χ(t) = Ae iet/ (A is a constant, E is the energy of the particle), and that ψ(x) must satisfy the time-independent Schrödinger equation 2 2m d 2 ψ(x) + V (x)ψ(x) = Eψ(x). dx2 (b) Prove that these solutions of the Schrödinger equation lead to a timeindependent probability density Ψ(x, t) 2. (Therefore they are called stationary states.) Remember: For complex numbers a 2 = a a. If Ψ(x, t) is the wave function of a particle, then Ψ(x, t) 2 is the probability density of finding that particle at position x, at time t. Therefore we must have + Ψ(x, t) 2 dx = 1, (1) since the probability of the particle being somewhere is 1. Now, this extra condition might seem disturbing - the wave function is supposed to be determined by the Schrödinger equation, right? However, note that if Ψ(x, t) is a solution to the Schrödinger equation (see Exercise 0.1), then so is AΨ(x, t) for any complex 1 (Material based on previous exercises, assembled by L.H. Kristinsdottir, 2014) 2 You might recall from your mathematics studies, that this method to solve a partial differential equation is called separation of variables. You ll find in (a) that χ(t) = Ae iet/ and that ψ(x) satisfies the time-independent Schrödinger equation. Actually, we can tack the integration constant A onto ψ(x) instead of χ(t), since the total solution is anyway ψ(x)χ(t), that is we have χ(t) = e iet/ for convenience later. Imagine you know V (x) (like in Exercise 0.4) and go on and solve the time-independent Schrödinger equation. You ll get some infinite number of solutions ψ n(x) and energies E n. So Ψ n(x, t) = ψ n(x)e ient/ are the separable solutions to the time-dependent Schrödinger equation. The general solution will then be Ψ(x, t) = n cnψn(x, t) = n cnψn(x)e ient/. In fact, if we know that Ψ(x, t 0 ) = n cnψn(x) at some time t 0, then Ψ(x, t) = n cnψn(x)e ien(t t 0)/ for t > t 0. 1

2 constant A. The process of picking A such that eq. (1) holds is called normalizing the wave function. Exercise 0.2 (Normalization ) Let a particle have wave function ψ(x) that is proportional to e mωx2 /2. Normalize ψ(x). Note that Ψ(x, t) = 0 is a solution to the Schrödinger equation and that for some solutions of the Schrödinger equation the integral in eq. (1) becomes infinite. In those cases there is no constant A that we can multiply Ψ with to make eq. (1) hold. Hence, particles cannot be represented with these non-normalizable solutions, they must be rejected. Exercise 0.3 (Constant potential ) Consider a particle of mass m held in one-dimensional potential V (x). Suppose that in some region V (x) is constant, V (x) = V. For this region, find the stationary states of the particle when (a) E > V (hint: define the positive constant k by 2 k 2 /2m = E V ), (b) E < V (hint: define the positive constant κ by 2 κ 2 /2m = V E), (c) E = V, where E is the energy of the particle. Exercise 0.4 (Infinite well ) Consider a particle of mass m confined in an infinite one-dimensional potential well of width a, { 0 0 x a V (x) = otherwise Find the stationary states of the system and the corresponding energies. Harmonic oscillator A one-dimensional harmonic oscillator is characterized by the potential V (x) = 1 2 kx2 where k is a real positive constant. The angular frequency of the oscillator is ω = k m, where m is the mass of the oscillator. The time-independent Schrödinger equation of the system is therefore 2 2m d 2 dx 2 ψ(x) mω2 x 2 ψ(x) = Eψ(x). 2

3 Exercise 0.5 Define K = 2E ω and set ξ = mω x. Show that the eigenvalue equation above becomes d 2 ψ dξ 2 + (K ξ2 )ψ = 0. For very large ξ (i.e., very large x) the constant K is completely dominated by the ξ 2 term, giving the approximation d 2 ψ dξ 2 ξ2 ψ 0. The (approximate) solution is (check it!) ψ(ξ) Ae ξ2 /2 + Be ξ2 /2. Exercise 0.6 Why must we have B = 0? So, for large ξ the wave function ψ goes like e ξ2 /2. Therefore it seems reasonable to write ψ(ξ) = h(ξ)e ξ2 /2. Exercise 0.7 By substituting ψ(ξ) = h(ξ)e ξ2 /2 into the eigenvalue equation in Exercise 0.5, show that h(ξ) satisfies d 2 h dh 2ξ + (K 1)h = 0. dξ2 dξ Exercise 0.8 series Now we assume that the solution is on the form of a power h(ξ) = a k ξ k k=0 (this is called the Frobenius method). Show that the differential equation in the previous exercise becomes [(k + 1)(k + 2)a k+2 (2k + 1 K)a k ] ξ k = 0. k=0 3

4 Since the expansion of a function into power series is unique, the coefficient of every power of ξ in the above series must vanish, (k + 1)(k + 2)a k+2 (2k + 1 K)a k = 0, and therefore we get the recursion formula a k+2 = 2k + 1 K (k + 1)(k + 2) a k. (2) Starting from a 0 one gets all the even-numbered coefficients (a 2, a 4, a 6,...) and starting from a 1 one gets all the odd-numbered ones (a 3, a 5, a 7,...). There is but one problem: Not all the solutions are normalizable! To see this, lets look at what happens at very large ξ. Then the coefficients for very large k dominate, so we approximate the recursion formula for very large k: a k+2 2 k a k, giving a k C (k/2)! where C is some constant. It follows that h(ξ) C 1 (k/2)! ξk C 1 k! ξ2k Ce ξ 2 at very large ξ, which in turn gives ψ(ξ) = h(ξ)e ξ2 /2 Ce ξ2 /2 at very large ξ. But this is precisely the non-normalizable solution we chose to throw away! (See Exercise 0.6.) This means that if we want a normalizable wave function, the power series must terminate: There must exist an integer n such that a n+2 = 0. Looking at the recursion formula (2) we see that this means K = 2n + 1. Note however that with the above condition, only either the even or odd coefficients are terminated. Hence, if n is even a 1 must be zero (making all the odd coefficients zero), and if n is odd a 0 must be zero (making all the even coefficients zero). This way, the whole power series is terminated. Remember that we defined K = 2E ω the allowed energies E n = in Exercise 0.5. We have therefore found ( n + 1 ) ω, where n = 0, 1, 2,... 2 Furthermore, the recursion formula becomes 4 a k+2 = 2(n k) (k + 1)(k + 2) a k.

5 For the first few eigenstates we obtain n = 0 : a 0 0 a 1 = 0 a 2 = 0 a 3 = 0 a 4 = 0 h 0 (ξ) = a 0 ψ 0 (ξ) = a 0 e ξ2 /2 n = 1 : n = 2 : a 0 = 0 a 1 0 a 2 = 0 a 3 = 0 a 5 = 0 a 0 0 a 1 = 0 a 2 = 2a 0 a 3 = 0 a 4 = 0 h 1 (ξ) = a 1 ξ ψ 1 (ξ) = a 1 ξe ξ2 /2 h 2 (ξ) = a 0 (1 2ξ 2 ) ψ 1 (ξ) = a 0 (1 2ξ 2 )e ξ2 /2 Apart from an overall constant, the h(ξ) are the so-called Hermite polynomials, H n (ξ), see Table 1. In general, the normalized wave function reads ψ n (ξ) = ( mω ) 1/4 1 π 2n n! H n(ξ)e ξ2 /2. If we define the characteristic length of the oscillator l 0 = /mω, we can write ξ = x/l 0 and get the wave function in our original variable x ψ n (x) = ( mω ) 1/4 1 π 2n n! H n(x/l 0 )e x2 /2l 2 0. The first four eigenstates of the harmonic oscillator are shown in Figure 1. Examples of the divergent solutions that one finds to the harmonic oscillator when the energy deviates a little from the allowed values are shown in Figure 2. Table 1. The first few Hermite polynomials, H n(ξ). H 0 (ξ) = 1 H 1 (ξ) = 2ξ H 2 (ξ) = 4ξ 2 2 H 3 (ξ) = 8ξ 3 12ξ H 4 (ξ) = 16ξ 4 48ξ H 5 (ξ) = 32ξ 5 160ξ ξ 5

6 Figure 1. The first four eigenstates, ψ n, of the one-dimensional harmonic oscillator. (Figure from D.J. Griffiths, Introduction to Quantum Mechanics, 2nd ed., Pearson Prentice Hall, New Jersey (2005).) Figure 2. The solutions of the harmonic oscillator diverge quickly even if the energy deviates just slightly from its allowed values, here for (a) E = 0.49 ω and (b) E = 0.51 ω. (Figure from D.J. Griffiths, Introduction to Quantum Mechanics, 2nd ed., Pearson Prentice Hall, New Jersey (2005).) 6