This question regards the reconciliation of QIT with what I have learned separately in lectures about quantum physics. My understanding is that indistinguishable bosons are always described by symmetric states. An example of such a 2-boson state is given by $|{𝜓}\rangle=\frac{|{0}\rangle|{1}\rangle+|{1}\rangle|{0}\rangle}{\sqrt{2}}$ for arbitrary basis vectors $|{0}\rangle$ and $|{1}\rangle$. Despite this, in QIT we often consider the antisymmetric bell state $|{𝜓'}\rangle=\frac{|{0}\rangle|{1}\rangle-|{1}\rangle|{0}\rangle}{\sqrt{2}}$. How is it permissible to consider such a state if you are using photons for qubits? Is it as simple as the fact that an implicit labelling of the respective subspaces of each qubit allows us to 'distinguish' the bosons and hence allows them to occupy a symmetric state? How can this be permissible if bosons are indistinguishable by definition? Am I mistakenly conflating two unrelated concepts?

  • $\begingroup$ The '\ket' commands were being interpreted as intended in the preview and now appear as red text in the post - if anybody knows how to fix this please let me know. $\endgroup$ Commented Mar 14, 2023 at 13:13
  • $\begingroup$ Use \rangle instead of \ket (Just another way to represent it) $\endgroup$ Commented Mar 14, 2023 at 13:53
  • $\begingroup$ ^ I've made these edits - they just need approval. Instead of using '\ket{x}' you can use '|x\rangle'. $\endgroup$
    – jth
    Commented Mar 14, 2023 at 14:43

2 Answers 2


The confusion here arises because of not being careful about what the labels inside the kets mean.

When we talk about the Bell states (or about the phenomena of entanglement), we consider two different degrees of freedom of the particles: usually (1) their location, and (2) their spin or polarization. Let location be either $A$ or $B$, and spin be either 0 or 1.

Then, in plain language, one of Bell states is one in which either (a) the particle at $A$ has spin 0 and particle at $B$ has spin 1, or (b) the particle at $A$ has spin 1 and the particle at $B$ has spin 0.

Often, this is written in some short form as follows (ignoring normalization) $$ |\psi_{b} \rangle = |0 \rangle_A |1 \rangle_B - |1 \rangle_A |0 \rangle_B. $$ This is confusing way of labeling states, because it does not clearly incorporate the formalism of bosons and fermions. To get back on the right track, note that the Hilbert space of one particle is $H^{\text{spin}} \otimes H^{\text{position}}$. Meaning, the Hilbert space of two particles is $H^{\text{spin 1}} \otimes H^{\text{position 1}} \otimes H^{\text{spin 2}} \otimes H^{\text{position 2}}$.

The Bell states are non-trivial states within this space. One way of writing a Bell state is

$$ |\psi_{b} \rangle = |0, A, 1, B \rangle - |1, A, 0, B\rangle. $$

Before, proceeding, I want to emphasize that I am just labeling states. There is no physics in the labeling itself, but in the manipulation of the labels.

The interesting part comes about, when we ask ourselves what is $|0, A, 1, B \rangle$ for a pair of bosons. Two bosons, must have a symmetric state. Which means, we can write it as $$ |0, A, 1, B \rangle = |0\rangle |A \rangle |1\rangle |B \rangle + |1\rangle |B \rangle |0\rangle |A \rangle, $$ where the kets are identifying states inside each of the Hilbert spaces inside $H^{\text{spin 1}} \otimes H^{\text{position 1}} \otimes H^{\text{spin 2}} \otimes H^{\text{position 2}}$.

The physics here comes in if we swap the two bosons. Physically, this means swapping both their position and spin, and obtaining the same state. Mathematically, we swap with the $P_{12}$ operator as $$ P_{12}|0, A, 1, B \rangle = |1, B, 0, A \rangle = |1\rangle |B \rangle|0\rangle |A \rangle + |0\rangle |A \rangle|1\rangle |B \rangle. $$ It should be clear from the last expression above that $|0, A, 1, B \rangle = |1, B, 0, A \rangle$, which is our required symmetry condition for bosons.

In the same way, the second term in $|\psi_b\rangle$ is $$ |1, A, 0, B\rangle = |1\rangle |A \rangle |0\rangle |B \rangle + |0\rangle |B \rangle |1\rangle |A \rangle, $$ which is also symmetric.

So, in the Bell state is $|\psi_{b} \rangle = |0, A, 1, B \rangle - |1, A, 0, B\rangle$ each individual term is symmetric, so the superposition of the two terms is also symmetric, making this is valid state for bosons.

A different way

A different way of seeing why the anti-symmetric Bell state actually symmetric for bosons is to substitute the individual terms as follows, $$ |\psi_b \rangle = |0\rangle |A \rangle |1\rangle |B \rangle + |1\rangle |B \rangle |0\rangle |A \rangle - |1\rangle |A \rangle |0\rangle |B \rangle - |0\rangle |B \rangle |1\rangle |A \rangle. $$ If we reoder the Hilbert spaces as $H^{\text{spin 1}} \otimes H^{\text{spin 2}} \otimes H^{\text{position 1}} \otimes H^{\text{position 2}}$, this becomes $$ |\psi_b \rangle = (|0\rangle |1\rangle - |1\rangle |0\rangle ) (|A\rangle |B\rangle - |B\rangle |A\rangle) $$ In simple words, we have the tensor product of two anti-symmetric states, one for the spin part of the bosons and one for their position. Bosonic states are symmetric, when we exchange all degrees of freedom. Doing such a swap means that each component state will pick up a minus sign, but overall $|\psi_b \rangle$ will retain a positive sign.

So, really, when we write something like $|0\rangle |1\rangle - |1\rangle |0\rangle$, we are considering the spin part of the Bosonic state, and ignoring the space part. Linearity ensures that our calculations continue to work.


Finally, let's connect this whole picture back to experimental situation. In photonic Bell state experiments, typically (but not always) we create polarization-entangled states. i.e. $$ |\psi_{b} \rangle = |H, P_1, V, P_2 \rangle - |V, P_1, H, P_1\rangle, $$ where $H,V$ are the polarization states. And $P_1, P_2$ are paths along which the photons can move.


You're exactly right that indistinguishable bosons are represented by symmetric states. However, when we think about a photons in QIS, we often think about them interacting with beamsplitters. When a single photon interacts with a 50-50 beamsplitter, it ends up in an equal superposition (that differ by an $i$ phase, but that's more than I want to discuss) of being transmitted vs reflected. In other words, that photon is now in a superposition of traveling down both paths, but by putting a detector along both paths, we can measure that photon and determine which path it went down. In other words, that spatial separation means we need not constrain ourselves to symmetric states. However, it's worth mentioning that when we bring the two paths back together like in a Mach-Zehnder interferometer, it's the identicality of the photons at the second beamsplitter (since they're now colocated again) that causes them to interfere.


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