Assignment 5

Problem 1:

Consider the one-dimensional harmonic oscillator with potential V(x) = mω2x2/2.  Use the variational principle to find the ground state energy employing a Gaussian wave function
ψ(x) = (b/π)¼exp(-bx2/2),
which is already normalized to 1.  "b" is a variational parameter that you must optimize.
(a)  Find the kinetic energy <T> in this state.
(b)  Find the potential energy <V> in this state.
(c)  Optimize "b".
(d)  Find the optimal variational energy.  How does the result compare with the exact result for a 1D harmonic oscillator?

Solution:

Problem 2:

In one dimension, the potential energy of a particle of mass m as a function of x is given by
U(x) = (b2|x|)½.  Here b is a positive constant with units energy/length½.
(a)  Use the WKB method to estimate the energy of the particle in the ground state.
(b)  Use the variational method to estimate the energy of the particle in the ground state.
(c)  Which estimate is closer to the true ground-state energy?

0exp(-x2) √x dx = Γ(3/4)/2 = 0.612708

Solution:

Problem 3:

A particle of mass m is constrained to move in an infinitely deep, one-dimensional square well extending from 0 to +a. 
If this particle is under the influence of a delta-function perturbation H' = -αδ(x - a/2), where α is a constant.
(a) (i) Find the first-order correction to the allowed energies.
     (ii) Explain why the energies are not perturbed for even n.
(b)  Find the first three nonzero correction terms in the first-order expansion of the ground state wave function.

Solution:

Problem 4:

A particle of mass m in an infinitely deep square well,
U = 0 for 0 < x < a,  0 < y < a,  U = infinite elsewhere,
is subject to a perturbation H' = -Bxy.
The positive constant B has units energy/length2.
(a)  Calculate the zeroth order energy levels and energy eigenfunctions.
(b)  Determine the effects of the perturbation on the two lowest zeroth order energy levels.
Is there splitting of either of these levels?

Solution:

  12|H'|Φ12> - λ    <Φ12|H'|Φ21>  
   <Φ21|H'|Φ12>  <Φ21|H'|Φ21> - λ   

  = 0.


 

 

12|H'|Φ12> = -(4B/a2)∫0asin2(πx/a) x dx 0asin2(2πy/a) y dy = -Ba2/4 = -C
21|H'|Φ21> = -(4B/a2)∫0asin2(2πx/a) x dx 0asin2(πy/a) y dy  = -Ba2/4 = -C
12|H'|Φ21> = -(4B/a2)∫0asin(πx/a) sin(2πx/a) x dx 0asin(πy/a) sin(2πy/a) y dy

sin(x) sin(2x) = 2sin2(x) cos(x) = 2(1 - cos2(x)) cos(x)
= 2cos(x) - 2 cos3(x) = 2cos(x) - ½cos(3x) - 3/2 cos(x) = ½(cos(x) - cos(3x)).
∫cos(x) x dx = cos(x) + x sin(x).

12|H'|Φ21> = -Ba2(256/(81π4)) = -bC = <Φ21|H'|Φ12>.
C = Ba2/4, b = 1024/(81π4) = 0.13.
det(H' - λI) = 0.  (-C - λ)2 - (bC)2 = 0,  C + λ = ±bC,  λ = -C ± bC.
λ = -C + bC:  Φ = 212 - Φ21],  E2+ = 5π2ħ2/(2ma2) - C + bC.
λ = -C - bC:  Φ = 212 + Φ21],  E2- = 5π2ħ2/(2ma2) - C - bC.
The perturbation removes the degeneracy.  There is splitting.