If we consider a bipartie system that is entangled,

$$|\psi^{AB}\rangle=\frac{|{\uparrow_z\downarrow_z}\rangle-|{\downarrow_z\uparrow_z}\rangle}{\sqrt{2}} $$

And we want to know the probability distribution of an experiment that involves only system A, we can use the reduced density matrix:

$$\rho^A=\frac12 |{\uparrow_z}\rangle\langle{\uparrow_z}|+\frac12|{\downarrow_z}\rangle\langle{\downarrow_z}|$$

Which results from "ignoring" the rest of the bipartite system.

I'm struggling to make sense of this state. The density matrix describes a mixed ensemble, which may contain an even mixture of ups and down in any axis with equal probability. But here we have a single electron for example which can't be a statistical mixture.

Normally if I want to describe a quantum state, I think of an ensemble of states that are all prepared exactly the same and then I measure different operators on the states.But when I want to use this way of thinking for the state considered here, someone might say: well, if you entangle a lot of particles and you collect always the second particle and use these as your ensemble, you got a statistical mixture of pure states.

-> But this can't be true since none of the electrons is in a pure state?

Also, if someone else measures the spin of the other particles and therefore makes the entangled state collapse, now my ensemble "really is a mixture" of pure states, but I can't know that until I recieve the information from him.

So my two questions are:

  • Is there a difference between my ensemble before the other person has measured his particles and after the measurement?

  • How can the state of system A of the entanglement be a mixed state if it is only one particle? Is the description by the reduced density matrix incomplete?

  • $\begingroup$ Can the downvoter please explain? I think this is a perfectly reasonable and natural question.It is stated clearly and I have explained what I already thought about it and how it leads to a problem. $\endgroup$
    – curio
    Commented Jan 10, 2018 at 13:26

1 Answer 1


The reduced density matrix of one electron in an entangled pair includes all of the information you can get by measuring that electron without measuring the other electron and comparing the results. The entangled electron also contains other information that can't be accessed by measuring that electron in isolation. This information can be described by the electron's Heisenberg picture observables, see:


  • $\begingroup$ This is very interesting and exactly what I was looking for.The paper is hard for me to understand though. Could you elaborate a little more? $\endgroup$
    – curio
    Commented Jan 10, 2018 at 19:17
  • $\begingroup$ What did you find hard to understand? $\endgroup$
    – alanf
    Commented Jan 11, 2018 at 8:57
  • $\begingroup$ The heisenberg picture formalism.Ive seen it before and I have calculated the harmonic oscillator with it but the application on quantum gates is new for me. $\endgroup$
    – curio
    Commented Jan 11, 2018 at 9:38
  • $\begingroup$ Are you familiar with the Pauli matrices? $\endgroup$
    – alanf
    Commented Jan 11, 2018 at 11:47
  • $\begingroup$ yes we can use them to make unitary transformations of 2 state systems and also as hermitian observables. $\endgroup$
    – curio
    Commented Jan 11, 2018 at 15:21

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