Harmonic potential exercise with perturbation theory

  • #1
keyzan
32
11
Homework Statement
Determine the energy spectrum and steady states at order 0 in ##\lambda##
Relevant Equations
##V(x) = \frac{1}{2} k (|x| - \lambda)^2##
Hello there, I'm training with some exercises in view of the July test, so I will post frequently in the hope that someone can help me, since the teacher is often busy and there are no solutions to the exercises.

A particle of mass m in one dimension is subject to the potential:

##V(x) = \frac{1}{2} k (|x| - \lambda)^2##

With ##k, \lambda## positive constants.

1. Determine the energy spectrum and steady states at order 0 in ##\lambda##
Solution:
By eliminating the term in ##\lambda##, the potential becomes:
##V(x) = \frac{1}{2} k x^2##

Consequently the Hamiltonian of the system will be:

##H = \frac{P^2}{2m} + \frac{1}{2} k x^2##

This represents precisely the Hamiltonian of a particle in a harmonic potential and by opting for a whole series of mathematical expedients (which I don't know whether to bring back to the exam) including creation and destruction operators, we can reduce the diagonalization of the Hamiltonian to the problem of diagonalization of ##\hat{N}## which is nothing other than the product of the operators of creation and destruction. In this way we find that the energies will be:

##E_n = \hbar \omega (n + \frac{1}{2}) ##

These energies correspond to stationary states:

##|n\rangle = \frac{(\hat{a}^{+} )^n}{\sqrt{n!}}|0\rangle##

Where 0 is the ground state called "empty" and we take it properly normalized.
The coordinate representation of these stationary states is in general the product of a Hermit polynomial and a quadratic exponential.
Do you consider this first explanation exhaustive?





V(x)=\fract12k(\absx−λ)2
 
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  • #2
2. Determine the shift of the energy of the ground state to the first order of the theory of perturbations in ##\lambda##.

Solution:

At the first order of perturbation theory we have that the energies will be:

##E_n (\lambda) = E_n^{(0)} + \langle \phi_n| \beta W|\phi_n \rangle + O(\lambda^{2})##

Where beta is a constant tending to zero. Let's calculate this average value:

##\langle n | \frac{1}{2} k (-2\lambda x + \lambda^2 ) | n \rangle##

My intuition tells me that since lambda is a very small quantity and since we have it squared in the constant term:

##\frac{1}{2} k \lambda^2##

, this constant term can be considered zero. (Is this correct reasoning? If we had instead had a constant term of the order lambda instead of square lambda, could we have done the same thing and considered it zero?)

Doing this way I have to calculate:
##\langle n | - k\lambda x | n \rangle##
This contribution is zero because we are considering an odd operator between states with equal parity. Therefore there is no first order shift of the energy, much less of the ground state. Correct?

I apologize if my answers are quick but I'm in a bit of a hurry.:nb):smile:
 
  • #3
Is everything ok? Can I continue with the exercise? Although this zero contribution to the first order of perturbation theory seems a bit strange to me
 

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