# How do ladder operators and number states act on multimode states?

The ladder operators for number states, $$\alpha_{\ell}^{\dagger}$$, and $$\alpha_{\ell}$$ have the following properties when working on mode $$\ell$$: $$\begin{array}{l} \hat{\alpha}_{\ell}\left|n_{\ell}\right\rangle=\sqrt{n_{\ell}}\left|n_{\ell}-1\right\rangle \quad \\ \hat{\alpha}_{\ell}\left|0_{\ell}\right\rangle=0 \\ \hat{\alpha}_{\ell}^{\dagger}\left|n_{\ell}\right\rangle=\sqrt{n_{\ell}+1}\left|n_{\ell}+1\right\rangle \end{array}$$

And considering a pure wavefunction $$\psi_{\ell}$$, the expectation value $$\left\langle\alpha_{\ell}^{\dagger} \alpha_{\ell}\right\rangle=\left\langle\psi_{\ell}\left|\alpha_{\ell}^{\dagger} \alpha_{\ell}\right| \psi_{\ell}\right\rangle$$ gives the number of photons in mode $$\ell$$. However, what would be the effect of $$\alpha_{\ell}^{\dagger} \alpha_{\ell}$$ on a wavefunction over some different mode $$\ell'$$? Put differently what would I get if I tried to calculate $$\left\langle\psi_{\ell^{\prime}}\left|\alpha_{\ell}^{\dagger} \alpha_{\ell}\right| \psi_{\ell^{\prime}}\right\rangle$$?

• What is |psi_l> (with or without ')? Mar 30, 2021 at 18:53
• e.g. $|\psi\rangle_{\ell}=\frac{1}{2}\left(\sqrt{3}|n\rangle_{\ell}+|n+1\rangle_{\ell}\right)$ (where I took the $\ell$ outside the bracket with the same meaning)
– Logi
Mar 30, 2021 at 18:59
• But if you have more modes you have to specify what the state is for all modes. And then you know what the operators do. Mar 30, 2021 at 19:02
• I consider mixed states using the density matrix $\hat{\rho}=\sum_{i} p_{i}\left|\psi_{i}\right\rangle\left\langle\psi_{i}\right|$
– Logi
Mar 30, 2021 at 19:03
• Say I want to calculate the expectation value of the operator $\widehat{N}:=\alpha_{\ell}^{\dagger} \alpha_{\ell}+\alpha_{\ell^{\prime}}^{\dagger} \alpha_{\ell^{\prime}}$. The expectation value of a general operator, $\hat{A}$, can be calculated as $\langle\hat{A}\rangle=\sum_{i} p_{i}\left\langle\psi_{i}|\hat{A}| \psi_{i}\right\rangle$. But that means that when calculating $\hat{N}$ I will get "expectation values" of $\alpha_{\ell^{\prime}}^{\dagger} \alpha_{\ell^{\prime}}$ over $\psi_{\ell}$ and vice versa.
– Logi
Mar 30, 2021 at 19:07

Multimode states are product states $$\vert n_1n_2\ldots,n_{\ell'}\ldots,n_\ell\ldots \rangle =\vert n_1\rangle\otimes\ldots \vert n_{\ell'}\rangle \otimes \ldots \otimes \vert n_\ell\rangle\otimes \ldots$$ and the working assumption is probably that if $$\langle n_1\ldots,n_\ell=0,\ldots\vert\psi_i\rangle = 0$$ then $$\hat \alpha_\ell \vert\psi_i\rangle=0$$ so $$\langle \hat \alpha^\dagger_\ell \hat\alpha_\ell\rangle=0$$.

• Now that I think about it, that does make sense, since $\hat{\alpha}_{\ell}\left|0_{\ell}\right\rangle=0$ as written in my question.
– Logi
Mar 30, 2021 at 19:18
• so then it's a matter of figuring out if you multimode state has anything in mode $\ell$. Mar 30, 2021 at 19:20
• And $\alpha_{\ell}$ commutes with $|n\rangle_{\ell^{\prime}}$? So that I can write $\left\langle\left. n\right|_{\ell}\left\langle\left. n\right|_{\ell^{\prime}} \alpha_{\ell}^{\dagger} \alpha_{\ell} \mid n\right\rangle_{\ell^{\prime}} \mid n\right\rangle_{\ell}=\left\langle n\left|\alpha_{\ell}^{\dagger} \alpha_{\ell}\right| n\right\rangle_{\ell}\langle n \mid n\rangle_{\ell^{\prime}}$
– Logi
Mar 30, 2021 at 19:31
• yes. $\hat \alpha_\ell$ and $\hat \alpha^\dagger_\ell$ only "see" the part of the state that refers to mode $\ell$. Mar 30, 2021 at 19:35
• Okay, then I think it adds up. Thank you for the help.
– Logi
Mar 30, 2021 at 19:40