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I'm confused about the form of the quantized electromagnetic field.

Sometimes, one writes the quantized electric field as

$$\hat E = \vec E_0(\hat a + \hat a^†)$$

, which can been seen the website below in the section, "Mathematical formulation 1".

https://en.wikipedia.org/wiki/Jaynes%E2%80%93Cummings_model

But as I know, the quantized electric field as

$$\hat E(r) = i\vec E_0(\hat a(k)e^{ikr} - \hat a^†(k)e^{-ikr})$$

for a given wavevector $k$ and position $r$.

This can be found in the below website.

https://en.wikipedia.org/wiki/Quantization_of_the_electromagnetic_field

I wonder why the two forms are different.

Please help me out.

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  • $\begingroup$ They weren't even in the same picture. The assumption was also very different. $\endgroup$ – ShoutOutAndCalculate Apr 24 at 16:22
  • $\begingroup$ Could you please let me know about that? $\endgroup$ – James Apr 25 at 0:21
  • $\begingroup$ there's wikipedia page about three different pictures. it's standard for grad quantum II. Quantization of EM file alone seemed to be sololy a description of bosonic particles. technically you don't even need QM to do it. Standard EM and statistical mechanics makes more sense. becaue photon itself had number state uncertainty. The best treatment I saw in the standard physics was from Fourier spectrum of wavelets. $\endgroup$ – ShoutOutAndCalculate Apr 25 at 0:32
  • $\begingroup$ Both are time-independent operator in schrodinger picture. I'm not sure if it is a change of picture. $\endgroup$ – James Apr 25 at 1:56
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The first expression, used in the Jaynes-Cummings model, is a description of the electric field at just one single point in space. This is useful when the region of interest (the size of an interacting atom or molecule, for instance) is small compared with the wavelength of the field; $E$ is nearly constant over this region. However, this first expression is really just an approximation of the second expression; the second expression includes the additional spatial dependence that is relevant on longer scales. (The other differences just correspond to phase factor differences in the definitions of the raising and lowering operators, and so have no physical import.)

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  • $\begingroup$ Thank you for your explanation. But I'm confused about the phase terms. Why is it always differently described with different phases? Such as imaginary term, exponential term and the sign difference between creation and annihilation operator? $\endgroup$ – James Apr 25 at 2:46
  • $\begingroup$ And also, the exponential terms in the second equation disappeared in the first equation as the wavelength is large enough so that the exponential term in the second equation vanishes by taylor expansions, right? $\endgroup$ – James Apr 25 at 2:49
  • $\begingroup$ @James 1) The phase of the creation operator is purely a convention, and it is set to what it is for historical reasons. 2) Yes, the long wavelength approximation means that you are only evaluating the field at locations where $r\ll 1\k$, so the exponentials are dominated by their leading term of $1$. $\endgroup$ – Buzz Apr 25 at 16:02
  • $\begingroup$ Thank you very much for your explanation. It was quite helpful. $\endgroup$ – James Apr 26 at 3:13

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