# Harmonic oscillator's lowering operator acting on state bra

For harmonic oscillator in quantum mechanics we have a lowering operator ($\hat a$) which it's action on state ket is:

$$\hat a\;|n\rangle=\sqrt n \;|n-1\rangle$$

Is following relation true for it's action on state bra?

$$\langle n|\;\hat a=\sqrt n \;\langle n-1|$$

No, that relationship is incorrect. If you start with $$\hat{a}|n\rangle = \sqrt{n}|n-1\rangle$$ and take the conjugate, what you get is $$\langle n|\hat a^\dagger = \sqrt{n}\langle n-1|.$$ To get $\langle n|\hat a$, you need to start instead with an annihilation operator, $$\hat{a}^\dagger|n\rangle = \sqrt{n+1}|n+1\rangle,$$ so that when you take the conjugate you get $$\langle n|\hat a = \sqrt{n+1}\langle n+1|.$$
• here we have $\langle 0|\hat a\neq0$ isn't this a contradiction? – Masoud Mar 13 '17 at 13:26
• No, it's not a contradiction; why would it be? $\hat a$ is not hermitian so the action of its adjoint does not need to coincide with its action. – Emilio Pisanty Mar 13 '17 at 13:44
• Right, $\hat a$ does not annihilate on both its left and right hand side, otherwise $\langle 0|(a a^\dagger - a^\dagger a)|0\rangle$ would have to be simultaneously $0$ (because the first term annihilates the vaccuum on the left, the second term annihilates the vacuum on the right) and $1$ (because $[a, a^\dagger] = 1$). Instead $\langle 0|a = \langle 1|$ and $a^\dagger |0\rangle = |1\rangle$ and we get $1=1,$ much better than $0=1.$ – CR Drost Mar 13 '17 at 16:18
So if we actually write the relation you wrote as a sum of basis matrix elements we get $$a = \sum_k ~\sqrt{k} ~~|k-1\rangle\langle k|~.$$From this you rediscover your relation, for example, by operating on the state $|n\rangle$ to create a Kronecker delta:$$a|n\rangle = \sum_k ~ \sqrt{k} ~~|k-1\rangle\langle k|n\rangle = \sum_k ~ \delta_{kn}~\sqrt{k} |k-1\rangle = \sqrt{n}~|n-1\rangle.$$ Recall that $\delta_{ab}$ is 0 if $a \ne b$ or else 1 when $a = b.$
Practices: show that its conjugate $a^\dagger$ has $|k\rangle\langle k-1|$ in the sum in place of $|k - 1\rangle\langle k|,$ and then use the fact that $\langle \ell - 1|k - 1\rangle = \delta_{(\ell-1)(k-1)} = \delta_{\ell k}$ to prove that $a^\dagger a = \sum_k k~|k\rangle\langle k|,$ the number operator $\hat n$. Then prove similarly that $a a^\dagger = \hat n + 1$ and therefore that $[a, a^\dagger] = 1.$
If you can successfully complete those practice exercises you should also be able to follow the following calculation: $$\langle n|a = \sum_k \sqrt{k} \langle n|k-1\rangle\langle k| = \sum_k \delta_{n(k-1)} \sqrt{k}\langle k|=\sum_k \delta_{(n+1)k} \sqrt{k}\langle k|=\sqrt{n+1}~\langle n+1|.$$