Commutator relations in simple harmonic oscillator

[brasidas]

Homework Statement

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

Homework Equations

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

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

The Attempt at a Solution

I have tried the following:

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

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

 

[CompuChip]

I got that result too:

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

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?