Commutator relations in simple harmonic oscillator

[brasidas]

Homework Statement

Show that, [tex] [a, \hat H] = \hbar\omega, [a^+, \hat H] = -\hbar\omega [/tex]

Homework Equations

For the SHO Hamiltonian [tex] \hat H = \hbar\omega(a^+a - \frac{\ 1 }{2}) [/tex] with [tex][a^+, a] = 1 [/tex]

[a, b] = -[b, a]

The Attempt at a Solution

I have tried the following:

[tex] [a, \hat H] = a\hat H - \hat Ha = \hbar\omega ( (aa^+a - \frac{\ 1 }{2}a) - (a^+a - \frac{\ 1 }{2})a )
= \hbar\omega (aa^+a -a^+aa) = \hbar\omega [a, a^+] a = - \hbar\omega a [/tex]

And this is nothing like [tex] \hbar\omega [/tex] I am supposed to get. Could anyone point out where I have gone wrong?

 

[CompuChip]

I got that result too:

[tex][a, \hat H] = \hbar \omega [a, a^\dagger a] = \hbar\omega\left( a^\dagger [a, a] + [a, a^\dagger] a \right) = \hbar\omega( 0 - a ) = - \hbar \omega a[/tex]

which agrees perfectly well with the expressions given on Wikipedia.

So having three independent sources with the same result, I suppose that your question is incorrect?