- #1
logic smogic
- 56
- 0
Problem
Show that,
[tex]a (a^{ \dagger})^{n} = n (a^{\dagger})^{n-1}+(a^{\dagger})^{n} a[/tex]
Formulae
[tex] a = \sqrt{\frac{m \omega}{2 \hbar}}(x+\frac{\imath p}{m \omega})[/tex]
[tex] a^{\dagger} = \sqrt{\frac{m \omega}{2 \hbar}}(x-\frac{\imath p}{m \omega})[/tex]
[tex] [a,a^{\dagger}]= a a^{\dagger}-a^{\dagger}a=1[/tex]
Attempt
This is just one step in a long derivation from another problem. The author (Goswami, pg 147) uses it without proof, but I would like to modify it – and hence I need to understand where he got it from.
I can see how you might pull [tex] a^{\dagger}[/tex]’s out from the first and third term, and try to use the commutation relation, but the n’s don’t seem to work out right.
Any thoughts?
Show that,
[tex]a (a^{ \dagger})^{n} = n (a^{\dagger})^{n-1}+(a^{\dagger})^{n} a[/tex]
Formulae
[tex] a = \sqrt{\frac{m \omega}{2 \hbar}}(x+\frac{\imath p}{m \omega})[/tex]
[tex] a^{\dagger} = \sqrt{\frac{m \omega}{2 \hbar}}(x-\frac{\imath p}{m \omega})[/tex]
[tex] [a,a^{\dagger}]= a a^{\dagger}-a^{\dagger}a=1[/tex]
Attempt
This is just one step in a long derivation from another problem. The author (Goswami, pg 147) uses it without proof, but I would like to modify it – and hence I need to understand where he got it from.
I can see how you might pull [tex] a^{\dagger}[/tex]’s out from the first and third term, and try to use the commutation relation, but the n’s don’t seem to work out right.
Any thoughts?
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