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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 $\left\langle\psi_{\ell^{\prime}}\left|\alpha_{\ell}^{\dagger} \alpha_{\ell}\right| \psi_{\ell^{\prime}}\right\rangle$?

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  • $\begingroup$ What is |psi_l> (with or without ')? $\endgroup$ – Norbert Schuch Mar 30 at 18:53
  • $\begingroup$ 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) $\endgroup$ – Physics101 Mar 30 at 18:59
  • $\begingroup$ 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. $\endgroup$ – Norbert Schuch Mar 30 at 19:02
  • $\begingroup$ I consider mixed states using the density matrix $\hat{\rho}=\sum_{i} p_{i}\left|\psi_{i}\right\rangle\left\langle\psi_{i}\right|$ $\endgroup$ – Physics101 Mar 30 at 19:03
  • $\begingroup$ 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. $\endgroup$ – Physics101 Mar 30 at 19:07
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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 $\langle \hat \alpha^\dagger_\ell \hat\alpha_\ell\rangle=0$.

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  • $\begingroup$ 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. $\endgroup$ – Physics101 Mar 30 at 19:18
  • $\begingroup$ so then it's a matter of figuring out if you multimode state has anything in mode $\ell$. $\endgroup$ – ZeroTheHero Mar 30 at 19:20
  • $\begingroup$ And $\alpha_{\ell}$ commutes with $|n\rangle_{\ell^{\prime}}$? So that I can write $\left\langle\left. n\right|_{\ell}\left\langle\left. n\right|_{\ell^{\prime}} \alpha_{\ell}^{\dagger} \alpha_{\ell} \mid n\right\rangle_{\ell^{\prime}} \mid n\right\rangle_{\ell}=\left\langle n\left|\alpha_{\ell}^{\dagger} \alpha_{\ell}\right| n\right\rangle_{\ell}\langle n \mid n\rangle_{\ell^{\prime}}$ $\endgroup$ – Physics101 Mar 30 at 19:31
  • $\begingroup$ yes. $\hat \alpha_\ell$ and $\hat \alpha^\dagger_\ell$ only "see" the part of the state that refers to mode $\ell$. $\endgroup$ – ZeroTheHero Mar 30 at 19:35
  • $\begingroup$ Okay, then I think it adds up. Thank you for the help. $\endgroup$ – Physics101 Mar 30 at 19:40

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