- #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##

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

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