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In a text book for quantum communication, I learned that one generates optical pulses (wavepackets), each of which contains only one photon. For instance, the state of two wavepackets are described by $|H\rangle_1|V\rangle_2$, where $|H\rangle_1$ ($|V\rangle_2$) means that a single-photon is in horizontal (vertical) polarization and temporal mode 1 (2). I guess, here $|H\rangle_1|V\rangle_2$ is highly temporally-localized states, i.e., it is in a superposition of many single-photon states with different single-frequency plane waves such that the temporal overlap is negligible.

On the other hand, I know that photons are boson, and any bosonic state has to be invariant under the particle exchange. But, the above state $|H\rangle_1|V\rangle_2$ is not symmetric nor anti-symmetric under the exchange because its exchanged state $|V\rangle_1|H\rangle_2$ is clearly different from $|H\rangle_1|V\rangle_2$.

My question is that why we are allowed to consider $|H\rangle_1|V\rangle_2$ despite the fact that any bosonic state is invariant under the particle exchange? Why the state is not a symmetric state, i.e., $(|H\rangle_1|V\rangle_2+|V\rangle_1|H\rangle_2)/\sqrt{2}$?

Ketty
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2 Answers2

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When you write a state like $|H\rangle_a \otimes |V\rangle_b = \hat{a}^{\dagger}_H\hat{b}^{\dagger}_V|\mathrm{vac}\rangle$ you're using the formalism of second quantization. In this formalism the symmetry of the multi-particle wavefunction is guaranteed by the commutation relations of the creation and annihilation operators. The photonic creation and annihilation operators obey the bosonic commutation relations:

$$ [\hat{a}_i,\hat{a}_j^{\dagger}] = \delta_{ij}, \qquad [\hat{a}_i,\hat{a}_j] = [\hat{a}_i^{\dagger},\hat{a}_j^{\dagger}] = 0, $$

and these ensure that the wavefunction is symmetric under particle exchange. The state $|H\rangle_a \otimes |V\rangle_b$ is not the explicit form that wavefunction, it is a more efficient representation that doesn't enumerate all the different ways to place identical particles in the various states.

When you do the exchange $|H\rangle_a \otimes |V\rangle_b \to |V\rangle_a \otimes |H\rangle_b$ you are not exchanging the particles, you are exchanging the states. The symmetrized two-photon wavefunction in the first-quantization picture would be:

$$ \Psi = \frac{\psi_{a,H}(\mathbf{r}_1)\psi_{b,V}(\mathbf{r}_2) + \psi_{b,V}(\mathbf{r}_1)\psi_{a,H}(\mathbf{r}_2)}{\sqrt{2}}. $$

Here we are writing out the different ways in which the different identical particles can be in the different states. Particle exchange corresponds to swapping $\mathbf{r}_1 \leftrightarrow \mathbf{r}_2$, and it's easy to see that this doesn't change the state.

In second quantization we don't explicitly say which identical particle is in which state, because this isn't a meaningful question (they are identical). Therefore, we also can't do the particle exchange in this picture.

fulis
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The question is whether the photons are distinguishable in another degree of freedom. If so, then the state must not be complete symmetric.

In this case, the photons are distinguished by their temporal mode. Therefore, the state need not be symmetric under exchange of the two photons.

Quantum Mechanic
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