Infinite Non-Square Well Potential

  • Thread starter dpeagler
  • Start date
  • #1
34
0

Homework Statement



Let ϕn(x) be the complete ortho-normal set of eigenstates of the Hamiltonian H = T +V
and En, n = 0, 1, 2, are the corresponding eigenvalues. The E0 is the ground state energy.
(a) If ϕ(x) is an arbitrary normalized state, show that E0 < = ∫dxϕ(x)Hϕ(x).

(b) Consider one dimensional potential V0(x) = 0, abs|x| ≤ a/2 and = ∞ otherwise. The
second one dimensional potential V1(x) = (4V|x|)/a - V_0, |x|≤ a/2, V > 0 and = ∞otherwise.
If E0 and E1 are the ground state energies of the two potentials respectively. Find which
one (1) E0 < E1 (2) E0 > E1 is correct and prove your judgement.

Homework Equations



Schrodinger's Equation

The Attempt at a Solution



I think I have gotten part A correct. However, I am so confused at a non-square well potential. I know that the infinite potential boundary conditions still apply here, but it throws me off with what is going on inside the well.

Do I just do a normal solution of the T.I.S.E. and let my constant take care of the non-zero potential?

Any help is greatly appreciated.
 

Answers and Replies

  • #2
vela
Staff Emeritus
Science Advisor
Homework Helper
Education Advisor
14,733
1,337
Wouldn't the answer depend on the value of V0? You can shift the value of E1 around by changing V0, but E0 is fixed.
 

Related Threads on Infinite Non-Square Well Potential

  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
11
Views
913
Replies
4
Views
761
Replies
1
Views
3K
Replies
9
Views
6K
Replies
2
Views
5K
Replies
1
Views
2K
Replies
11
Views
5K
Replies
7
Views
3K
  • Last Post
Replies
3
Views
835
Top