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I just finished deriving the 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 energy eigenstates of Quant. harm. oscillator, but explanation is a bit too fast there. Anyway i strive to be able to derive the equation $W_n = \hbar \omega \left(n + \tfrac{1}{2}\right)$ in full, but first i need to clarify 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?

I am sorry for asking such a basic questions. I am a self-taught and a real freshman to commutators algebra.

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The commutators in the above expressions are sued to change the order of the Hamiltonian and annihilation or creation operators. I'll show you the first one in some detail, the second one should not give you problems afterwards.

We start from $\hat{H}\hat{a}\psi_n$. Using the commutator $[\hat{H},\hat{a}] = \hat{H}\hat{a}-\hat{a}\hat{H} = -\hbar\omega\hat{a}$, we can write $\hat{H}\hat{a}\psi_n = (\hat{a}\hat{H}-\hbar\omega\hat{a})\psi_n$. Because we have $\hat{H}\psi_n = W_n\psi_n$, we get $(\hat{a}\hat{H}-\hbar\omega\hat{a})\psi_n = (\hat{a}W_n-\hbar\omega\hat{a})\psi_n = (W_n-\hbar\omega)\hat{a}\psi_n$ (note that we can change the order of the annihilation operator and c-numbers $W_n$ and $\hbar\omega$). Therefore, we have $\hat{H}\hat{a}\psi_n = (W_n-\hbar\omega)\hat{a}\psi_n$ and we conclude that $\hat{a}\psi_n$ is an eigenstate of the Hamiltonian with eigenvalue $W_n-\hbar\omega$.

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  • $\begingroup$ I don't quite understand this statement: ˝Because we have $\hat{H}\psi_n = W_n\psi_n$ we get $(W_n-\hbar\omega)\hat{a}\psi_n$˝. $\endgroup$ – 71GA May 7 '13 at 8:44
  • $\begingroup$ I edited the answer to give more detail on this step. Is it understandable now? $\endgroup$ – Ondřej Černotík May 7 '13 at 8:52
  • $\begingroup$ Ok i understand now that you swapped opperator $\hat{H}$ with its eigenvalue $W_n$. All i know from this is that i get an expectation value for energy $\langle W \rangle$ like this: $$\langle W \rangle = \int \limits_{-\infty}^{\infty} \overline{\Psi}\, \left(- \frac{\hbar^2}{2m} \frac{d^2}{d \, x^2} + W_p\right) \Psi \, d x$$ and $$\hat{H} = - \frac{\hbar^2}{2m} \frac{d^2}{d \, x^2} + W_p$$ But i am weak on eigenvalues, eigenvectors and Hilbert space... So how could i connect what i already know to confirm that $\hat{H} = W_n$? $\endgroup$ – 71GA May 7 '13 at 9:26
  • $\begingroup$ Does this mean that eigenvalue in Hilbert space is equivalent of an expectation value??? $\endgroup$ – 71GA May 7 '13 at 9:28
  • $\begingroup$ You can't say $\hat{H} = W_n$ but you can say $\hat{H}\psi_n = W_n\psi_n$ if $\psi_n$ is an eigenstate of $\hat{H}$ with eigenvalue $W_n$. And that is your starting point, together with the commutator, to find what $\hat{H}\hat{a}\psi_n$ is. $\endgroup$ – Ondřej Černotík May 7 '13 at 9:29

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