0
$\begingroup$

In this paper where the authors derive the formula for coincidence probability in a Hong-Ou Mandel (HOM) interference effect as a function of time delay $\tau$, they arrive at an equation (15) with the following matrix element:

$$\langle\text{vac}|~\hat{c}(\omega)\hat{d}(\omega')~[\hat{c}^{\dagger}(\omega_1)\hat{d}^{\dagger}(\omega_2) - \hat{c}^{\dagger}(\omega_1)\hat{d}^{\dagger}(\omega_2)+\hat{d}^{\dagger}(\omega_1)\hat{c}^{\dagger}(\omega_2)-\hat{d}^{\dagger}(\omega_1)\hat{d}^{\dagger}(\omega_2)]~|\text{vac}\rangle\tag{1}$$

They then use a commutation relation (equation 2) to arrive at a set of delta functions. The commutation relation they exploit is:

$$[\hat{a}(\omega),\hat{a}^{\dagger}(\omega')]=\delta(\omega-\omega')\tag{2}$$

However, if you expand out equation (1), it is clear that all $\hat{c}$'s appear to the left of all $\hat{c}^{\dagger}$'s, so it is completely unclear to me where the commutation relation is coming in!

Can anyone help?

Improve this question
$\endgroup$
4
  • $\begingroup$Hi Cody. What exactly do you mean? Where is the problem--can you elaborate, e.g. via an example calculation? I suppose that in $(2)$ $a$ can be $c$ and $d$, correct? For example, you have $a(\omega)a^\dagger(\omega^\prime) |\mathrm{vac}\rangle\overset{(2)}{=}\delta(\omega-\omega^\prime) |\mathrm{vac}\rangle $. Is that clear? Does that help?$\endgroup$
    – Tobias Fünke
    CommentedAug 2, 2024 at 20:06
  • $\begingroup$@TobiasFünke that does help, thank you. I suppose I just don't see how what you've written counts as a commutation relation, though.$\endgroup$
    – Cody Payne
    CommentedAug 2, 2024 at 20:24
  • $\begingroup$You write $aa^\dagger=\delta + a^\dagger a$, and when applied to the vacuum the second term on the RHS vanishes.$\endgroup$
    – Tobias Fünke
    CommentedAug 2, 2024 at 20:38
  • $\begingroup$Ahh, that makes sense. Thank you a lot!$\endgroup$
    – Cody Payne
    CommentedAug 2, 2024 at 20:39

1 Answer 1

Reset to default
0
$\begingroup$

You need to move the destruction operators to the right, and the creation operators to the left, because $$ \langle \text{vac}\vert \hat a\ne 0\, , \quad \text{but}\quad \hat a\vert\text{vac}\rangle=0\, , $$ and similarly for $\hat a^\dagger$, so $$ \hat a \hat a^\dagger = \hat a \hat a^\dagger -\hat a^\dagger \hat a +\hat a^\dagger \hat a = [\hat a,\hat a^\dagger]+\hat a^\dagger \hat a $$ helps you move one $\hat a$ to the right of an $\hat a^\dagger$ at the cost of adding a $\delta$-function.

Improve this answer
$\endgroup$

    Start asking to get answers

    Find the answer to your question by asking.

    Ask question

    Explore related questions

    See similar questions with these tags.