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

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


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

  • $\begingroup$ What are the $\left|0_H\right>$, $\left|0_V\right>$, etc.? $\endgroup$
    – coconut
    Dec 19, 2016 at 17:24
  • $\begingroup$ The vacuum states. $\endgroup$
    – Ilya
    Dec 19, 2016 at 22:35

1 Answer 1


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.


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