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I'm currently taking a course in Quantum-optics, focusing primarily on the book Introductory to quantum optics by C. Gerry & P. Knight. In the book they first present the quantization of a single mode of the EM field, and the electric field operator is set to be(page 12, equation 2.15) $$\hat{E_1}=\sqrt{\frac{\hslash\omega}{\epsilon_{0}V}}\sin\left(kz\right)\left(\hat{a}+\hat{a}^{\dagger}\right)\vec{e}$$ (Note the index 1 is used to differentiate between the two forms) were $\vec{e}$ is the polarization unit vector.
Later, they present the quantization of the multimode EM field, and the electric field operator of the mode $\vec{k},s$ is set to be (page 24, equation 2.123) $$\hat{E_2}(t)=i\sqrt{\frac{\hslash\omega}{2\epsilon_{0}V}}\left(\hat{a}e^{i\left(\boldsymbol{k}\cdot\boldsymbol{r}-\omega t\right)}-\hat{a}^{\dagger}e^{-i\left(\boldsymbol{k}\cdot\boldsymbol{r}-\omega t\right)}\right)\vec{e_s}$$ where $\boldsymbol{k}$ is the wave vector of the mode and $\vec{e_s}$ is the polarization state unit vector.

I wasn't bothered by the differences until we started talking about the coherent states. First, the book presents the expectation value of $\hat{E_2}$ in the Heisenberg picture, given a coherent state $\left|\alpha\right\rangle$ (page 45, eq 3.13) $$\left\langle \alpha\right|\hat{E_2}(t)\left|\alpha\right\rangle =-\left|\alpha\right|\sqrt{\frac{2\hslash\omega}{\epsilon_{0}V}}\sin\left(\boldsymbol{k}\cdot\boldsymbol{r}-\omega t+\theta\right)$$ which is a regular traveling plane wave.
On the the other hand, when using $\hat{E_1}$, the expectation is found in the Schrödinger picture (page 58, eq 3.35) to be $$\left\langle \alpha e^{-i\omega t}\right|\hat{E_{1}}\left|\alpha e^{-i\omega t}\right\rangle =2|\alpha|\sqrt{\frac{\hslash\omega}{\epsilon_{0}V}}\sin\left(kz\right)\cos\left(\omega t-\theta\right)$$ which is clearly a standing wave.

Can anyone explain the difference between the two operators?
First, I thought it was due to using different modes during the quantization process, but following the quantization once again it appears to me as if the standing waves modes basis was used in both cases.

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2 Answers 2

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Why do you think $\hat E_2$ is a standing wave mode? There is no multiplicative separation of $\vec r$ and $t$ dependence in the operator. On the contrary, they are always grouped in the expression $\vec k\cdot \vec r-\omega t$ so it is rather propagating. Unsurprisingly, this is also seen when taking the expected value of a coherent state.

Meanwhile, the $\hat E_1$ operator does describe a standing wave, made more apparent in the Heisenberg picture where the variables are well separated: $$ \hat E_1 = \sqrt{\frac{\hbar\omega}{\epsilon_0V}}\sin(\vec k \cdot\vec r)(e^{-i\omega t}\hat a+e^{i\omega t}\hat a^\dagger) $$ as is confirmed when taking the expected value as well.

Hope this helps, and tell me if you find some mistakes.

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  • $\begingroup$ Ok, thank you very much! Just to be sure, my initial thoughts were correct and the sole difference between the two operators is in the expansion modes that were used (i.e. plane waves in the case of free space, or standing waves in the case of a cavity)? $\endgroup$
    – Michael
    Apr 25, 2022 at 14:43
  • $\begingroup$ Yes that sums pretty much $\endgroup$
    – LPZ
    Apr 25, 2022 at 14:45
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Expand the sinusoidal term in $E_1$ and we'll directly see the difference in the spatial dependence: $$E_1=\sqrt{\frac{\hbar \omega}{\epsilon_0 V}}\frac{1}{2i}\left(a e^{ikz}-a e^{-ikz}+a^\dagger e^{ikz}-a^\dagger e^{-ikz}\right)\vec{e}.$$ There are two extra terms relative to $E_2$, with opposite signs of $ikz$ in the exponentials. This is because the $E_1$ operator was derived from quantizing an electromagnetic wave inside of a cavity (Fig 2.1, page 11), making it a standing wave, while the $E_2$ operator was derived in free space in terms of plane wave modes (page 18, page 20 2.76).

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