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Physics 610B
HW #3
Fall 2007
Due Thurday, Oct 4, 2007
Problem 3.1 Use WKB to estimate the g.s. (n = 0) energy for the gravitational (linear)
potential, V(y) = mgy. Don’t forget there is a “rigid wall” at y = 0. Also compute the
energy for excited states (n > 0).
Problem 3.2. Shankar 16.2.5; this is actually an application of results from problem 3.2.
Problem 3.3. Return to perturbation theory. Consider a pendulum, which has the
classical Hamiltonian H  12 ml 2 2  mgl cos  . The classical conjugate momentum is
p  ml 2 , hence the quantum Hamiltonian is
1
2 d 2
2
H
p  mgl cos   
 mgl cos  . Expanding the cosine to 2nd order,
2
2
2
2ml
2ml d
2
2

d
1
H 
 mgl  mgl 2 , which is just the harmonic oscillator with a constant
2
2
2
2ml d
and trivial shift.
(a) Write down the g.s. energy for this approximate Hamiltonian.
1
(b) The next term, H   mgl 4 can be thought of as a perturbation. Compute the 1st
24
order correction to the g.s. energy due to this perturbation. You are encouraged to use
creation/annihilation operators, although it is not required.
(c) Finally, compute the second order correction due to this perturbation. You are
encouraged to use creation/annihilation operators, although it is not required.
Problem 3.4 Let’s return to the quartic potential,
2 d 2
1
H 
 x 4 . I discussed this at some length in class. We can introduce an
2
2m dx
24
21 / 3  4 / 3 1 / 3
energy scale  
. With harmonic oscillator ladder operators, the
8 m2/ 3
2
4
1

Hamiltonian becomes Hˆ    aˆ   aˆ   aˆ   aˆ   . With some manipulation, we
6


2
2

2
2
find that Hˆ   Nˆ  3Nˆ  32  23 aˆ  aˆ  13 Nˆ aˆ  Nˆ aˆ 2  aˆ 2 Nˆ  aˆ 2 Nˆ  16 aˆ 4  aˆ 4

 

Confirm for yourself that 0 H 0   , 2 H 2 
3
2
0H 4 
2
3
 
23
2

 , 0 H 2  0 , 4 H 4   , and
59
2
 , as was discussed in class.
Your task: find 2 H 4 and then diagonalize the 3-by-3 matrix to get the ground state
energy. The exact numerical value is 1.4706 ε.