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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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  • One a general note: a photon is not a wave packet. It's only a quantum of energy. A classical wave packet is an approximation of special kind of uncorrelated many-photon state. The fact that there are no correlations lets us approximate that many-photon state as an ensemble of single photon states. None of these descriptions are either physically or mathematically equivalent, even though we like to sloppily glance over the differences. – FlatterMann May 25 '23 at 13:41
  • Thank you so much for your note. My understanding is that if we superpose single-photon states, each of which has different energy (or equivalently frequency), then we can construct a wave packet containing only one photon. Please kindly correct me if I am wrong. – Ketty May 25 '23 at 14:19
  • One photon has an angular momentum of one Planck unit. A multi-photon state requires multiple energy exchanges with one Planck unit each, so one can't build a wave packet that behaves like a single photon. – FlatterMann May 25 '23 at 20:41

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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  • Thank you so much for your time to answer. I think $|H\rangle_a\otimes |V\rangle_b$ has nothing to do with the second quantization, but this is just a tensor product state of two photons (for distinguishable photons according to my new understanding thanks to one of the answers here). The second quantization is a process to find out a basis for a symmetric subspace of the space spanned by ${|H\rangle_a\otimes |H\rangle_b, |H\rangle_a\otimes |V\rangle_b, |V\rangle_a\otimes |H\rangle_b, |V\rangle_a\otimes |V\rangle_b}$. Please kindly correct me if I am wrong. – Ketty May 25 '23 at 14:24
  • Identical particles can occupy distinguishable states, and in these cases the symmetrization of the wavefunction can be ignored in practice (it does not affect the observables of the system) since the particles are effectively distinguishable. However, formally the wavefunction is still symmetric or anti-symmetric. See answers to this and this question and related links. – fulis May 25 '23 at 16:07
  • The notation $|H\rangle_a\otimes |V\rangle_b$ it is ambiguous, but in the context of quantum optics it almost certainly refers to a state expressed in secnd quantization. If you choose to interpret it as an unsymmetrized state, then the answer to your question is that it is formally not a physical state, but in practice it behaves as one. You can also write down other unphysical states like $(|H\rangle_a\otimes |V\rangle_a-|V\rangle_a\otimes |H\rangle_a)/\sqrt{2}$, whereas in s.q. this state is zero for bosons since it has the wrong symmetry. The formalism only lets you define physical states. – fulis May 25 '23 at 16:23
  • Thank you so much, and I am convinced that $|H\rangle_1|V\rangle_2$ should be represented by $|H, r_1\rangle_1|V, r_2\rangle_2+|V, r_2\rangle_1|H, r_1\rangle_2$, where "1" and "2" represent labels for two single-photons, and $r_1$ and $r_2$ indicate temporal modes. If single-photon states satisfy $\langle r_2|r_1\rangle=0$, i.e., the temporal modes are perfectly localized, then the state can be regarded as $|H\rangle_{r_1}|V\rangle_{r_2}$, but when we have some overlap, this is not the case. I should have been careful about the difference between a label for a photon and its mode. Am I right? – Ketty May 27 '23 at 13:53
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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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  • Thank you so much for your swift answer and pointing out what I am missing. In my example, I think you are right that two photon can be distinguished by their temporal mode, i.e., they do not overlap each other, and therefore, the state need not be symmetric under exchange.

    The temporal mode is one of the examples of "another degree of freedom" to distinguish the two photons, but is there other examples? Can any mode, such as frequency and so on, be such an example? To be honest, I didn't fully understand what it means by identical particles.

     – Ketty May 25 '23 at 14:14
  • @Ketty exactly! Frequency, spatial mode, temporal mode, polarization... – Quantum Mechanic May 25 '23 at 15:35