- #1
indigojoker
- 246
- 0
We know the eigenvalue relation for the Hamiltonian of a SHO (in QM) though relating the raising and lowering operators we get:
[tex]H= \hbar \omega (N+1/2)[/tex]
This is true for [tex] H=\frac{p^2}{2m}+\frac{m \omega^2 x^2}{2}[/tex]
I would like to solve for another case where [tex]V=a\frac{m \omega^2 x^2}{2} [/tex]
where a is some constant
We now have [tex] H=\frac{p^2}{2m}+\frac{ a m \omega^2 x^2}{2}[/tex]
I'm not sure how to go about this. When relating the creation and annihilation operators, we get: [tex]a^{\dagger} a = \frac{m \omega}{2 \hbar} x^2 + \frac{1}{2m \omega \hbar} p^2 -\frac{1}{2} [/tex]
I'm not sure how to incorporate a constant into the potential, any ideas?
[tex]H= \hbar \omega (N+1/2)[/tex]
This is true for [tex] H=\frac{p^2}{2m}+\frac{m \omega^2 x^2}{2}[/tex]
I would like to solve for another case where [tex]V=a\frac{m \omega^2 x^2}{2} [/tex]
where a is some constant
We now have [tex] H=\frac{p^2}{2m}+\frac{ a m \omega^2 x^2}{2}[/tex]
I'm not sure how to go about this. When relating the creation and annihilation operators, we get: [tex]a^{\dagger} a = \frac{m \omega}{2 \hbar} x^2 + \frac{1}{2m \omega \hbar} p^2 -\frac{1}{2} [/tex]
I'm not sure how to incorporate a constant into the potential, any ideas?