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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. $\endgroup$ Commented Apr 24, 2020 at 16:22
  • $\begingroup$ 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. $\endgroup$ Commented Apr 25, 2020 at 0:32

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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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  • $\begingroup$ @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
    Commented Apr 25, 2020 at 16:02

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