A question about operators in QM and notation

Question

I have a problem understanding a notation written by my professor. He writes:

$$\langle n | a^+ | m \rangle=\langle n | (a^+ | m \rangle) = (\langle n | a) | m \rangle$$

and he wants to demonstrate this equality. To do so, he expands the first and second expressions:

1. $\langle n | (a^+ | m \rangle) = \langle n | \sqrt{m+1} | m+1 \rangle = \sqrt{m+1} \langle n | m+1 \rangle$

2. $(\langle n | a) | m \rangle = \sqrt{n} \langle n-1 | m \rangle$

However, I don't understand why $\langle n | (a^+ | m \rangle)$ is equal to letting $a$ act on the left: $(\langle n | a) | m \rangle$.

To explain my confusion, let me give an example with the time evolution operator:

$\langle n | U | m \rangle = \langle n | (U | m \rangle) = e^{-iE_m t / \hbar} \langle n | m \rangle$

If I let it act on the left, I get:

$(\langle n | U) | m \rangle = e^{-iE_n t / \hbar} \langle n | m \rangle$

and not $(\langle n | U^+ )| m \rangle$, because otherwise I would have:

$(\langle n | U^+ )| m \rangle = e^{+iE_n t / \hbar} \langle n | m \rangle$

What I mean is that $U$ acts on both the right and the left as $U$ and not as $U^+$, otherwise, by acting on the left as $U^+$, I would get $e^{+iE_n t / \hbar} \langle n | m \rangle$ and not the correct $e^{-iE_n t / \hbar} \langle n | m \rangle$.

To clarify this with a slight abuse of notation, I would have:

• Acting on the right: $\langle n | U | m \rangle = \langle n | (U | m \rangle) = \langle n | U m \rangle = \langle n | e^{-iE_m t / \hbar} m \rangle = e^{-iE_m t / \hbar} \langle n | m \rangle$
• Acting on the left: $\langle n | U | m \rangle = (\langle n | U |) m \rangle = \langle n U^+ | m \rangle = \langle n e^{+iE_n t / \hbar} | m \rangle = e^{-iE_n t / \hbar} \langle n | m \rangle$

On the left, it acts as $U$ and not as $U^+$, because the resulting eigenvalue is the negative exponential and not the positive one. If I followed the professor's notation:

• $\langle n | U | m \rangle = \langle n | (U | m \rangle) = (\langle n | U^+ |) m \rangle = \langle n U | m \rangle = \langle n e^{-iE_n t / \hbar} | m \rangle = e^{+iE_n t / \hbar} \langle n | m \rangle$

So...

Returning to the case $\langle n | a^+ | m \rangle$, I would have written that $a^+$ acts on both the right and the left as $a^+$, and thus:

$\langle n | (a^+ | m \rangle) = (\langle n | a^+) | m \rangle$

Could you please explain what I am misunderstanding with the time evolution operator? It seems correct to me, and if I understand this, I would also understand $a$ and $a^+$.

[EDIT] @fulis:

Uh thanks so much, so also in the case of evolution operator where i write:

• Acting on the right: $\langle n | U | m \rangle = \langle n | (U | m \rangle) = \langle n | U m \rangle = \langle n | e^{-iE_m t / \hbar} m \rangle = e^{-iE_m t / \hbar} \langle n | m \rangle$

• Acting on the left: $\langle n | U | m \rangle = (\langle n | U |) m \rangle = \langle n U^+ | m \rangle = \langle n e^{+iE_n t / \hbar} | m \rangle = e^{-iE_n t / \hbar} \langle n | m \rangle$

we properly have (on the left side): $\langle n | U | m \rangle = (U^\dagger | n \rangle )^\dagger | m \rangle = (e^{+iE_n t / \hbar} | n \rangle )^\dagger | m \rangle = e^{-iE_n t / \hbar} (|n \rangle)^\dagger | m \rangle=e^{-iE_n t / \hbar} \langle n | m \rangle$

right? I hope :D

Answer 1

The notation here is a bit loose. We have that

$$ \hat{a}|n\rangle = \sqrt{n}|n-1\rangle $$ and hence

$$ (\hat{a}|n\rangle)^{\dagger} = \langle n| \hat{a}^{\dagger} = \sqrt{n}\langle n-1|. $$

However, when using the raising and lowering operators it's convenient to always think of $\hat{a}$ as lowering and $\hat{a}^{\dagger}$ as raising. Hence, when your prof writes

$$ \bigl(\langle n |\hat{a}\bigr)|m\rangle $$

he means that you treat $\hat{a}$ as if though it is acting on the ket corresponding to the bra $\langle n |$ on right, and is a lowering operator.

The formally correct expression would be

$$ \langle n | \bigl(\hat{a}^{\dagger}|m\rangle\bigr) = (\hat{a}|n\rangle)^{\dagger} |m\rangle $$