0
$\begingroup$

I got a bit confused about the transition of notation between the first and second quantization. When a state is written as: $\rho =a\vert H \rangle \langle H \vert + b\vert V \rangle \langle V \vert+ c\vert H \rangle \langle V \vert + d\vert V \rangle \langle H \vert$, does it imply a single particle in this state, i.e. in the second quantized form it is: $\rho =a\vert 1_H 0_V \rangle \langle 1_H 0_V \vert + b\vert 0_H 1_V\rangle \langle 0_H 1_V \vert+ c\vert 1_H 0_V\rangle \langle 0_H 1_V \vert + d\vert 0_H 1_V \rangle \langle 1_H 0_V \vert$?

And is it equivalent to write

$\rho =\left(a\vert H \rangle \langle H \vert + b\vert V \rangle \langle V \vert+ c\vert H \rangle \langle V \vert + d\vert V \rangle \langle H \vert \right)\otimes|\alpha \rangle \langle \alpha\vert$

and

$\rho =a\vert \alpha_H 0_V \rangle \langle \alpha_H 0_V \vert + b\vert 0_H \alpha_V\rangle \langle 0_H \alpha_V \vert+ c\vert \alpha_H 0_V\rangle \langle 0_H \alpha_V \vert + d\vert 0_H \alpha_V \rangle \langle \alpha_H 0_V \vert$, where $|\alpha\rangle$ is some state in a Fock space (i.e. a coherent state)?

$\endgroup$
  • $\begingroup$ What are the $\left|0_H\right>$, $\left|0_V\right>$, etc.? $\endgroup$ – coconut Dec 19 '16 at 17:24
  • $\begingroup$ The vacuum states. $\endgroup$ – Ilya Dec 19 '16 at 22:35
1
$\begingroup$

Yes for the first equivalence, it is exactly what it means.

Not at alls for the second (the one with $|α\rangle$): The first writing supposes three modes (One horizontally polarized, one vertically polarized, and a third one (a different spatial mode ?), the latter being in the state $|α\rangle$, while the second one only deals with two modes. Another way to see there is a difference, is to look at the probability to have 2 photons in the mode $H$ : It is null for the first state, and equal to $|\langle 2_H | α_H\rangle |^2 ≠ 0$ for the second writing.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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