I'm trying to gain a better understanding of how the electric field operator is used and what it can do. I know that calculating its expectation value tells you that a coherent state is the 'most classical' quantum state of light, and the number states have zero average electric field. The operator appears to both create and destroy every possible mode if acted on a state. But beyond that I'm not really sure what it's useful for, or how you'd go about using it in a given situation (e.g. you want to study a Gaussian wavepacket). If anyone has any general comments on this, please let me know. For now though I have a more specific quesiton. To get a better grasp of the operator I've tried looking at the Hong-Ou-Mandel dip. In the SchrÃ¶dinger picture an input state(adsbygoogle = window.adsbygoogle || []).push({});

[tex]\left|\psi_{\text{in}}\right\rangle=\left|11\right\rangle

=\hat{a}^\dagger_1\hat{a}^\dagger_2\left|00\right\rangle[/tex]

will be transformed by a 50/50 beam splitter operator to an output state of

[tex]\left|\psi_{\text{out}}\right\rangle= \frac{i}{\sqrt{2}}\left(\left|20\right\rangle +\left|02\right\rangle\right).[/tex]

This is easily interpreted as both photons on the same output of the beam splitter, with zero possibility of seeing one photon on each side, (the centre of) the HOM dip.

So next I consider an electric field operator

[tex]\hat{E}=\hat{a}_1\varphi_1 + \hat{a}_2\varphi_2 + \text{ other terms.}[/tex]

FromIntroductory Quantum Opticsby Gerry and Knight, the operator for the beam splitter is

[tex]\hat{U}=\exp\left[i\frac{\pi}{4}\left(\hat{a}_1^\dagger\hat{a}_2 + \hat{a}_1\hat{a}_2^\dagger\right)\right][/tex]

and so the electric field operator transforms to

[tex]\hat{E}'=\hat{U}^\dagger\hat{E}\hat{U} =\frac{1}{\sqrt{2}}\left[ \hat{a}_1\left(\varphi_1+i\varphi_2\right) + ia_2\left(\varphi_1-i\varphi_2\right)\right]+ \text{ other terms.}[/tex]

My question then is, how from this new primed field can we tell that the transformed field cannot have one photon in each of modes 1 and 2 but must have either two in 1 or two in 2? Also, is there anything else interesting / noteworth about the new electric field, from the point of view of gaining conceptual insight to the electric field operator?

**Physics Forums | Science Articles, Homework Help, Discussion**

Dismiss Notice

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Electric field operator (and the HOM dip)

Can you offer guidance or do you also need help?

**Physics Forums | Science Articles, Homework Help, Discussion**