Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?


share|cite|improve this question
up vote 1 down vote accepted

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 $|H\rangle \otimes |H\rangle = (1,0,0,0)$.

share|cite|improve this answer
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 '14 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 '14 at 16:41

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.