In the quantization of the electromagnetic field (using the "second quantization" method), one says that the transverse vector potential must be

$$A(r,t)=\sum_k (\alpha_k e^{i \vec{k} \vec{r}}+\alpha_{-k}^* e^{-i \vec{k} \vec{r}})$$

Having two terms in order to be real.

But why the second $\alpha$ has that $-k$? Shouldnt be enough to be its complex conjugate? I mean, any complex number plus its conjugate gives a real number.

I dont get it, every book or article I read (for example this one on the end of page 3: https://www.phys.ksu.edu/personal/wysin/notes/quantumEM.pdf) starts to wander how to make the field real, arriving to the obvious conclussion that: $$ \operatorname{something}+\operatorname{something}^*=2 \operatorname{Re}(\operatorname{something}). $$

Im missing something?

  • 3
    $\begingroup$ One does? Not that $\alpha^*_k=\alpha_{-k}$? $\endgroup$ Commented Dec 14, 2017 at 2:26
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    $\begingroup$ You are misreading. Even in the reference which you provide, the second operator (the one with the subscript $-k$)is not conjugated. $\endgroup$
    – J. Murray
    Commented Dec 14, 2017 at 6:50

1 Answer 1


The quirkiness comes from the Fourier transform. The quantization expression is essentially a Fourier transform. It turns out that the Fourier transform of a real function is not a real function anymore. The symmetry also plays a role. If the function is real and symmetric [$f(-x)=f(x)$], then the spectrum is also real and symmetric. However, if the function is real and anti-symmetric [$f(-x)=-f(x)$], then its spectrum is imaginary and anti-symmetric. Combining these two observation, one finds that if $f(x)$ is a general real function, then $$ F^*(a) = F(-a) , $$ where $F(a)$ is the spectrum of $f(x)$.


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