# Proof for commutator [\hat{H},\hat{a}] = - \hbar \omega \hat{a} 1. May 4, 2013 ### 71GA I know how to derive below equations found on wikipedia and have done it myselt: \begin{align} \hat{H} &= \hbar \omega \left(\hat{a}^\dagger\hat{a} + \frac{1}{2}\right)\\ \hat{H} &= \hbar \omega \left(\hat{a}\hat{a}^\dagger - \frac{1}{2}\right)\\ \end{align} where $\hat{a}=\tfrac{1}{\sqrt{2}} \left(\hat{P} - i \hat{X}\right)$ is a annihilation operator and $\hat{a}^\dagger=\tfrac{1}{\sqrt{2}} \left(\hat{P} + i \hat{X}\right)$ a creation operator. Let me write also that: \begin{align} \hat{P}&= \frac{1}{p_0}\hat{p} = -\frac{i\hbar}{\sqrt{\hbar m \omega}} \frac{d}{dx}\\ \hat{X}&=\frac{1}{x_0} \hat{x}=\sqrt{\frac{m\omega}{\hbar}}x \end{align} In order to continue i need a proof that operators $\hat{a}$ and $\hat{a}^\dagger$ give a following commutator with hamiltonian $\hat{H}$: \begin{align} \left[\hat{H},\hat{a} \right] &= -\hbar\omega \, \hat{a}\\ \left[\hat{H},\hat{a}^\dagger \right] &= +\hbar\omega \, \hat{a}^\dagger \end{align} These statements can be found on wikipedia as well as here, but nowhere it is proven that the above relations for commutator really hold. I tried to derive $\left[\hat{H},\hat{a} \right]$ and my result was: $$\left[\hat{H},\hat{a} \right] \psi = -i \sqrt{\frac{\omega \hbar^3}{4m}}\psi$$ You should know that this this is 3rd commutator that i have ever calculated so it probably is wrong, but here is a photo of my attempt on paper. I would appreciate if anyone has any link to a proof of the commutator relations (one will do) or could post a proof here. 2. May 4, 2013 ### strangerep I presume you know the commutation relations between P and X? If so, write them down here. From that, you can work the commutator $[a, a^\dagger]$. From that, you can work out your $[H,a]$, etc. (Hint: apply the Leibniz product rule.) 3. May 4, 2013 ### 71GA \begin{align} \left[\hat{X}, \hat{P}\right] = \left[\tfrac{1}{x_0}\hat{x},\tfrac{1}{p_0}\hat{p}\right]=\tfrac{1}{x_0p_0} \big[\!\!\!\!\!\!\!\!\!\!\!\!\!\ \underbrace{\hat{x},\hat{p}}_{i\hbar \text{... calculated below}} \!\!\!\!\!\!\!\!\!\!\!\! \big]= \frac{1}{x_0 p_0} \hbar i = \sqrt{\frac{m \omega }{\hbar \hbar m \omega}} \hbar i = i \end{align} \begin{align} \left[ \hat{x}, \hat{p} \right] \psi &= \hat{x}\hat{p} \psi - \hat{p} \hat{x} \psi = - x\,i \hbar \frac{d \psi}{d x} - \left(- i\hbar \frac{d}{dx} \left(x \psi \right)\right) = - x\, i \hbar \frac{d \psi}{dx} + i \hbar \underbrace{\frac{d}{dx}\left( x \psi \right)}_{\text{product rule}} =\\ &= - x\, i \hbar \frac{d \psi }{d x} + i\hbar\left( \frac{d x}{d x} \psi + x \frac{d \psi}{d x} \right) = -x\,i\hbar \frac{d \psi}{dx} + i \hbar \psi + x\, i \hbar \frac{d \psi}{dx} = i \hbar \psi \end{align} I don't quite know how to start this. Lets leave this for later and work out $[a, a^\dagger]$ first. 4. May 4, 2013 ### micromass Staff Emeritus Use that the commutator bracket is linear. So $[\alpha A + \beta B, C] = \alpha [A,C] + \beta [B,C]$ and the same thing in the other variable. 5. May 5, 2013 ### jfy4 You are good to go, note $$a=\frac{1}{\sqrt{2}}(\hat{P}-i\hat{X})$$ and $$a^{\dagger} = \frac{1}{\sqrt{2}}(\hat{P}+i\hat{X})$$ now $$[a,a^{\dagger}] = \frac{1}{2}\left[ (\hat{P}-i\hat{X}),(\hat{P}+i\hat{X}) \right]$$ and bust it out using what you just showed us. 6. May 7, 2013 ### 71GA I think i wrote this wrong as it is: $$a=\frac{1}{\sqrt{2}}(\hat{X}-i\hat{P})$$ and $$a^{\dagger} = \frac{1}{\sqrt{2}}(\hat{X}+i\hat{P})$$ 7. May 7, 2013 ### 71GA Thank you all. I just solved this and it feels great. Now that i know following commutators: \begin{align} [\hat{H}, \hat{a}] &= -\hbar \omega \hat{a}\\ [\hat{H}, \hat{a}^\dagger] &= \hbar \omega \hat{a}^\dagger\\ \end{align} On the Wikipedia it is said that these commutators can be used to find eigenstates of Quant. harm. oscilator, but explaination is a bit too fast there. Anyway i strive to be able to derive the equationW_n = \hbar \omega \left(n + \tfrac{1}{2}\right)\$ in full, but first i need to clear why theese two relations hold:

\begin{align}
\hat{H}\hat{a} \psi_n &= (W_n - \hbar \omega) \hat{a} \psi_n\\
\hat{H}\hat{a}^\dagger \psi_n &= (W_n + \hbar \omega) \hat{a}^\dagger \psi_n
\end{align}

I can't see any commutators in above relations, so how do the commutators i just calculated help us to get and solve these two relations? And i am sorry for asking so basic questions but i am a selftaught and a real freshman to commutators algebra.

8. May 7, 2013

### strangerep

Let's stick to the notation used in that Wikipedia page. I.e., let's change your $W_n$ to $E_n$.
As explained on the Wiki page, we have
$$H \psi_n ~=~ E_n \psi_n ~.$$
So to evaluate $H a \psi_n$, use the $[H,a]$ formula to swap the order of $H$ and $a$, and then use the equation above.

BTW, are you self-learning from a textbook? If so, which one? (If not, you probably need to invest in one...) Questions at this level should more properly be posted in the homework forums.

9. May 8, 2013

### physwizard

homework problem. move to other thread.