Electric field operator (and the HOM dip)

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SUMMARY

The electric field operator, denoted as &hat;E, plays a crucial role in quantum optics, particularly in understanding coherent states and the Hong-Ou-Mandel (HOM) dip phenomenon. The transformation of the electric field operator using the beam splitter operator &hat;U reveals that the output states from a 50/50 beam splitter will not allow for one photon in each mode but will instead yield either two photons in one mode or none. This behavior is a direct consequence of the superposition principle inherent in quantum mechanics, as illustrated by the transformed electric field operator &hat;E'. The insights gained from this transformation enhance the understanding of photon statistics and quantum state manipulation.

PREREQUISITES
  • Understanding of quantum optics principles
  • Familiarity with electric field operators in quantum mechanics
  • Knowledge of beam splitter operations and their mathematical representation
  • Basic grasp of coherent and number states in quantum theory
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  • Study the mathematical formulation of the Hong-Ou-Mandel effect in detail
  • Explore the implications of electric field operators in quantum state manipulation
  • Learn about the role of beam splitter transformations in quantum optics
  • Investigate the significance of superposition in quantum mechanics
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Quantum physicists, optical engineers, and researchers in quantum optics seeking to deepen their understanding of electric field operators and their applications in photon behavior and quantum state transformations.

mikeu
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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 Schrödinger picture an input state

\left|\psi_{\text{in}}\right\rangle=\left|11\right\rangle<br /> =\hat{a}^\dagger_1\hat{a}^\dagger_2\left|00\right\rangle

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

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

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

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

From Introductory Quantum Optics by Gerry and Knight, the operator for the beam splitter is

\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]

and so the electric field operator transforms to

\hat{E}&#039;=\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.}

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?
 
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The transformation of the electric field operator is simply telling you what happens to the electric field when you apply the beam splitter transformation to it. The two terms in the transformed electric field operator are telling you that the electric field in either output mode is a superposition of the electric fields in the input modes, with one term having a phase shift of $\pi/2$. This is exactly what you would expect from a beam splitter, since it produces a superposition of the incoming fields in the two outputs.From the form of the electric field operator after applying the beam splitter transformation, you can see that there will be no electric field in either output mode if there is an equal number of photons in each input mode. In other words, the HOM dip is a direct consequence of the form of the electric field operator after applying the beam splitter transformation.
 

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