Understanding Second Quantization and Its Application in Quantum Mechanics

Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
1 reply · 2K views
QuantumClue
Messages
159
Reaction score
0
I begin with [tex]\int (\bar{\psi}(x) (\mathcal{H} \psi(x)) d^3x[/tex]

This is just

[tex]\int (\bar{\psi}(x) ({\frac{p^2}{2M} + \frac{1}{2}M \omega^2 (x)} \psi(x)) d^3x[/tex]

If one identified that [tex]\bar{\psi}(x)[/tex] and [tex]\psi(x)[/tex] are creation and annihilation operators, I assume that I can simply restate my integral by replacing the appropriate expressions with the following:


[tex]\int (a^{\dagger}a ({\frac{p^2}{2M} + \frac{1}{2}M \omega^2 (x)} aa^{\dagger}) d^3x[/tex]

So that

[tex]\int (\hbar \omega^{-1} \mathcal{H} - \frac{\hbar \omega}{2} ({\frac{p^2}{2M} + \frac{1}{2}M \omega^2 (x)} \hbar \omega^{-1} \mathcal{H} + \frac{\hbar \omega}{2}) d^3x[/tex]

I am just asking if I have assumed to much. Am I allowed to do this, and if not, why not?

Thanks

edit

What am I doing wrong this time, the equations won't show? I love latex, but I hate it sometimes!
 
Last edited:
Physics news on Phys.org
I think I found the invalid move. Sorry about that folks! Latex will show soon I hope :)