I am curious whether there is a sense in which each of the EPR particles is in an eigenstate of some observable.

Consider a pair of EPR particles 1 and 2, of which combined state is given by

$|\Psi\rangle_{12} = \frac{1}{\sqrt{2}}(|A\rangle_{1}|A\rangle_{2}+|B\rangle_{1}|B\rangle_{2})$.

From this one gets a reduced density matrix $\frac{1}{2}(|A\rangle\langle A| + |B\rangle\langle B|)$, which supposedly represents the state of only one of the EPR particles.

But if this reduced density matrix really represents the 'state' of a single particle, despite it not being a vector, what is the observable of which this state is an eigenstate (or something analogous to an eigenstate)?

Since $(|A\rangle\langle A| + |B\rangle\langle B|) \frac{1}{2}(|A\rangle\langle A| + |B\rangle \langle B|) = \lambda(\frac{1}{2}(|A\rangle\langle A| + |B\rangle \langle B|))$ with $\lambda = 1$, does it make sense to take $(|A\rangle\langle A| + |B\rangle\langle B|) $ as an observable for which $\frac{1}{2}(|A\rangle\langle A| + |B\rangle \langle B|)$ serves as something analogous to an eigenstate? So that we can say that each EPR particle is still in an eigenstate of some observable?

Or should we say that each constituent of an EPR pair does not have any eigenstate in any extended sense?

  • 1
    $\begingroup$ Neither a wave function nor a density matrix describe the state of a single system. They are describing quantum mechanical and quantum mechanical/statistical ensembles of the system. Since the individual system does not have a state, then neither of its sub-systems can have one, either. This is not just so for EPR but for any quantum mechanical system. EPR is merely a good way to confuse yourself about this. $\endgroup$ Commented Apr 13, 2023 at 11:34

2 Answers 2


I think we can make the situation more concrete.

Let $\mathcal{H} \cong \mathcal{H}_1 \otimes \mathcal{H}_2$ be a composite system of two spin-1/2 particles. Then, the maximally entangled Bell state is given by $$\lvert \Psi \rangle=\frac{1}{\sqrt{2}}(\lvert00\rangle+\lvert11\rangle).$$

This is similar to the state you write in your post. We can then compute the density operator $\lvert \Psi \rangle \langle \Psi \lvert$ and trace out of subsystem 2. This will result in a maximally mixed state $$\frac{1}{2}(\lvert 0 \rangle_1 \langle 0 \lvert_1 + \lvert 1 \rangle_1 \langle 1 \lvert_1) \tag{1}$$ because $\lvert \Psi \rangle$ is maximally entangled.

Importantly, (1) is a mixed state, so it is a statistical mixture of pure states of particle 1. We can see in this more concrete example that each pure state composing this mixed state is an eigenstate of $\hat{S}_z$. However, I would think this mixed state is not an eigenstate of an observable (Spin-z, etc.) of system 1 since all of these observables have pure states as their eigenstates.

What you are doing in your post is essentially projecting a density operator on itself, analogous to projecting a state ket on itself. Using bra-ket notation, it is clear that such projections will give you the original state, whether density matrix or ket, back.


There is no non-trivial observable for which the reduced density matrix of one half of an EPR system is an eigenstate. Your candidate observable $|A\rangle\langle A| + |B\rangle\langle B|$ is just the identity operator and all of its possible values are 1, so there is no need to do any measurement to know its value.


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