Physics 443, Solutions to PS 2

1. Griﬃths 2.12. The raising and lowering operators are 1 a± = √ (∓ipˆ + mωxˆ) 2mωh¯ wherep ˆ andx ˆ are momentum and position operators. Then s h¯ xˆ = (a + a ) 2mω + − s mωh¯ pˆ = i (a − a ) 2 + − The expectation value of the position operator is

Similarly, the expectation value of the momentum operator is

hpi = hψn | pˆ | ψni s * mωh¯ + = ψ | i (a − a ) | ψ n 2 + − n = 0

Expectation values of x2 and p2 are not zero.

1 h¯ 2 = ψ | H | ψ 2mω n hω¯ n h¯ 2 1 = ψ | (n + )¯hω | ψ 2mω n hω¯ 2 n h¯ = (2n + 1) 2mω where we have used the relationship between Hamiltonian operator and a±, namely 1 H =hω ¯ (a a + ) + − 2 1 H =hω ¯ (a a − ) − + 2

Then

2. Griﬃths 2.14. The wave function is initially in the ground state of the oscillator with classical frequency ω. The wave function is

1 mω 4 − mω x2 ψ (x) = e 2¯h 0 πh¯ The spring constant changes instantaneously but the wav function does not. So immediately after the change in spring constant the wave function remains the same. But it is no longer an eigenfunction of the hamiltonian operator. However, any function can be expressed as a 0 linear combination of the new eigenfunctions, ψn and we can write that n=∞ X 0 ψ0(x) = anψn(x) n=0 The probability that we will ﬁnd the oscillator in the nth state, with 0 2 energy En is |an| . After the change, the minimum energy state is 0 1 0 0 E0 = 2 hω¯ =hω ¯ , (since ω = 2ω) so the probablity that a measurement of the energy would still return the valuehω/ ¯ 2 is zero. Since the R 0 0 eigenfunctions are orthonormal ( ψnψmdx = δnm) we can determine the coeﬃcients Z ∞ 0 an = ψn(x)ψ0(x)dx −∞ 1 1 Z ∞ 0 ! 4 0 mω − mω x2 mω 4 − mω x2 a0 = e 2¯h e 2¯h dx −∞ πh¯ πh¯ 1 1 0 ! 4 Z ∞ 0 mω mω 4 −( mω + mω )x2 = e 2¯h 2¯h dx πh¯ πh¯ −∞ √ 1 ! 2 Z ∞ 2mω −( 3mω )x2 = e 2¯h dx πh¯ −∞ √ ! 1 ! 1 2mω 2 2πh¯ 2 = πh¯ 3mω

3 √ 1 2 2! 2 = 3 = 0.971

0 0 The probability of measuring energy E0 =hω ¯ /2 =hω ¯ is the proba- 0 bility that the oscillator is in the state ψ0. The probability that the oscillator is in the ground state is 2 |a0| = 0.943

3. Griffiths 2.26. Using the definition of the Fourier transform, we have 1 Z ∞ Answer = √ dx δ(x) exp(−ikx), 2π −∞ 1 −ikx = √ e |x=0, 2π 1 = √ . 2π We therefore have that 1 Z ∞ 1 δ(x) = √ √ exp(ikx) dk, 2π −∞ 2π 1 Z ∞ = dk eikx. (1) 2π −∞ 4. Griffiths 2.38.

(a) We write the wave function ψ1, which is the ground state wave function for the well of width a as a linear combination of the p eigenfunction, (ψnrime) of the well of width 2a. ∞ X 0 ψ1 = anψn n=1 Solve for the coeﬃcients Z 2a 0 ∗ an = (ψn) ψ1dx 0 s s Z a 2 nπx 2 πx = sin sin dx 0 2a 2a a a

4 We only integrate from 0 to a because ψ1(x) is zero for x < 0 and x > a. Then √ 2 Z π/2 2a an = (sin ny)(sin 2y) dy a 0 π √ 2 2 "sin(2 − n)y sin(2 + n)y #π/2 = − (n 6= 2) π 2(2 − n) 2(2 + n) 0

√ R π/2 2 π 2 If n = 2, the 0 sin 2ydy = 4 and a2 = 2 , and the probability 2 0 ¯h2 π 2 1 to measure E2 = 2m 2a , is |a2| = 2 . Since the sum of prob- abilities for any energy is 1, no other can be more probable than 0 E2.

(b) For all other even n, an = 0. For n odd √ 2 2 "sin(nπ/2) sin(nπ/2)# a = + n 2π (2 − n) (2 + n) √ 4 2 sin(nπ/2) = π (4 − n2)

Then 32 |a |2 = n π2(4 − n2)2 2 0 and |a1| = 0.36 The next most probable result is E1 with probability 0.36. (c)

Z a ∗ hHi = ψ1Hψ1dx 0 h¯2π2 = 2ma2 7. Time dependence. Show that if Qˆ is an operator that does not involve time explicitly, and if ψ is any eigenfunction of Hˆ , that the expectation value of Qˆ in the state of ψ is independent of time.

5 [We start with the Griﬃth’s Equation 3.148

d i dQ hQi = h[H,Q]i + h i, dt h¯ dt Where the last term is zero because Q has no explicit time dependence, d i hQi = hψ|HQ − QH|ψi, dt h¯ = ψhψ|(Q − Q)ψi using ψ is stationary, = 0 (2)

8. Collapse of the wave function. Consider a particle in the inﬁnite square well potential from problem 4.

q 2 nπx (a) Show that the stationary states are ψn(x) = a sin a and the n2π2¯h2 energy spectrum is En = 2ma2 where the width of the box is a. [The time independent Schrodinger’s equation for a particle in an inﬁnite square well is

h¯2 d2ψ − = Eψ 2m dx2

Substitution of the proposed solution ψn(x) gives s s h¯2 d2 2 nπx 2 nπx − sin = E sin 2m dx2 a a a a s s h¯2 nπ 2 2 nπx 2 nπx sin = E sin 2m a a a a a h¯2 nπ 2 → E = ] n 2m a

(b) Suppose we now make a measurement that locates a particle initially in state ψn(x) so that it is now in the position a/2 − /2 ≤ x ≤ a/2 + /2 and described by the state α. In the limit where a, the result of the measurement projects the system onto a superposition of eigenstates of energy. The probability of ﬁnding

6 2 the particle in any eigenstate is P (En) = |hψn | αi| . A reason- √ able estimate of the state | αi is ψα(x) = δ (x − a/2) where () a δ (x−a/2) = 1/ for a/2−/2 ≤ x ≤ a/2+/2 and δ (x− 2 ) = 0 everywhere else. Calculate the probability P (En). [ Any solution to (the time dependent) Schrodinger’s equation can be written as a linear combination of energy eigenstates. (The energy eigenstates form a complete set.) So we can write

∞ X ψα(x) = cnψn n Z ∞ ∗ → cn = ψα(x)ψn(x) dx −∞ Z a/2+/2 ! 1 ∗ = √ ψn(x) dx a/2−/2 s 2 a nπxa/2+/2 = − cos a nπ a a/2−/2 s 2 a nπ nπ nπ nπ = − cos( + ) − cos( − ) a nπ 2 2a 2 2a s 2 a nπ nπ = 2 sin( ) sin( ) a nπ 2 2a ( q (−1)(n−1)/2( 2 ) 2a sin( nπ ), if n is odd, = nπ 2a 0, if n is even.

The probability of ﬁnding the particle in the nth energy eigenstate is ( ( 2 )2 2a sin2( nπ ), if n is odd, P (E ) = |c |2 = nπ 2a (3) n n 0, if n is even. As shrinks, and the particle is more localized, the probability that it will be found in a higher energy state increases. In the limit a, ( 2 , if n is odd, P (E ) → a (4) n 0, if n is even. the particle is equally likely to be found in all n odd states.]