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The point of the nonlinear medium is to create higher harmonics (here: $\omega_2 = 2\omega_1$), i.e. waves with a multiple of the frequency of the initial wave ($k_1,\omega_1$), which do not vanish when the medium ends. This last statement is of the essence here! But I'll come back to that later. Now, the electric field of the intial wave ($k_1,\omega_1$) ...

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A coherent state is a monochromatic sinusoidal field. The electric field pulse is inherently not monochromatic, but instead has a spectrum of frequencies which are superimposed on top of one another so as to all add constructively once every pulse repetition time. Therefore, to represent the field pulse in quantum optics you actually need to bring in more ...

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A coherent state is actually a mathematical idealization of a monochromatic laser. Strictly speaking, any continuous wave laser in the laboratory would be a statistical mixture of phase-randomized coherent states. Furthermore, it would also have a finite linewidth, and different frequency components in that linewidth would have no definite phase relationship ...

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What you are arguing is that you can take a coherent state $\vert\alpha\rangle$, which has energy $|\alpha|^2$ (in units of $\hbar\omega$), remove one photon which has energy $1$, and still be left with $a\vert\alpha\rangle=\vert\alpha\rangle$. Thus, you would have gained $1$ unit of energy (and you could continue doing so). The problem is that applying $a$ ...

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Can someone point me where I am wrong? Consider a state that is a superposition of number states $|n\rangle$ for all $n$. Recalling that $$\hat a |n\rangle = \sqrt{n}|(n-1)\rangle$$ it's clear that \hat a\left(A\sum_{n=0}^\infty \frac{|n\rangle}{\sqrt{n!}}\right) = A\sum_{n=1}^\infty \frac{\sqrt{n}|(n-1)\rangle}{\sqrt{n!}}=A\sum_{n=1}^\infty ...

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