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I have recently been reading about spontaneous parametric down conversion(SPDC). I do clearly understand the process. What has been intriguing lately is the notation. For those of you who are unfamiliar, SPDC is a process which converts one vertically polarized photon to two horizontally polarized photon. (Yeah, it's not 100% accurate, but just to get the idea across.)

Many paper wrote: $ \left|V\right\rangle \rightarrow \left|H\right\rangle\left|H\right\rangle$

I was just wondering about the meaning of $\left|H\right\rangle\left|H\right\rangle$ since it's neither the inner product nor the outer product. Additionally, if $\left|H\right\rangle$ could be represented by the vector $(1,0)$, what could be the implications?

Thanks!

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It is a tensor product. (At least it always was when I encountered such notations, I can't speak with authority about SPDC specifically)

Let $\mathcal{H}_1$ be the Hilbert space of polarization states for a single photon. Then the space of states for a two photon system is $\mathcal{H}_2 = \mathcal{H}_1 \otimes \mathcal{H}_1$, and the state you consider in your OP should be really written as $|H\rangle \otimes |H\rangle$. In applications, people often leave out the tensor product sign and any zero states. If $|H\rangle = (1,0)$, then $|H\rangle \otimes |H\rangle = (1,0,0,0)$.

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Isn't the tensor product the same as outer product? Also, I thought that the outer product is represented by |H><H|. –  krismath Jul 9 at 16:06
    
@krismath: You are correct that the outer product is also a tensor product. The outer product lies in the tensor product space $\mathcal{H}^* \otimes \mathcal{H}$, as is indicated by the presence of a bra and a ket in its notation. Outer products are operators, they do not lie in a space of states. –  ACuriousMind Jul 9 at 16:41

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