Start with an entangled pair of spin one-half particles:

$|\psi\rangle = \frac{1}{\sqrt 2}(|\uparrow\rangle_z |\downarrow\rangle_z + |\downarrow\rangle_z |\uparrow\rangle_z)$

This cannot be factored as a product of separate states for the two particles, so therefore is entangled. If we looked at the first particle in isolation, we'd have a mixed state. In particular, if we put it through a Stern-Gerlach analyzer, we'd get 50% up and 50% down. But we don't put it through it yet.

Instead, take the second particle and put it through the following Stern-Gerlach analyzer loop:

enter image description here

where, on the $\uparrow$ branch, the particle is unchanged, but on the $\downarrow$ branch, we have a uniform magnetic field that flips it to $\uparrow$. So, whether we put $\uparrow$ or $\downarrow$ in, we get up $\uparrow$ out.

Now our state is:

$|\psi\rangle = \frac{1}{\sqrt 2}(|\uparrow\rangle_z |\uparrow\rangle_z + |\downarrow\rangle_z |\uparrow\rangle_z)$

$ = \frac{1}{\sqrt 2}(|\uparrow\rangle_z + |\downarrow\rangle_z) \otimes |\uparrow\rangle_z$

$ = |\uparrow\rangle_x |\uparrow\rangle_z$

Which is no longer entangled, and is now a product state. In particular, if we put the first particle through an x-oriented SG analyzer, it looks like it would always come out the $\uparrow$ side.

I think this is wrong, and the first particle would stay in a mixed state. But my question is: what are the rules around when we take the complex sum of states, possibly getting interference, before taking a measurement? This is similar to having "which-way" information but then erasing it without measuring it. Information is lost, we destroy the information in the spin state of the second quantum particle. So now, when the first particle goes through an analyzer, we can't tell which branch it takes, so they should interfere, acting as a pure state rather than a mixture. So I'm confused.

  • 2
    $\begingroup$ Going through the magnetic field is unitary transformation. It can not take both $\uparrow$ and $\downarrow$ to $\uparrow$. What you described is like a projective measurement (with post selection), which can indeed disentangle the pair. An example of unitary would be $|\uparrow\rangle\rightarrow |\downarrow\rangle, |\downarrow\rangle\rightarrow |\uparrow\rangle$ on one of the spin and indeed it does not change the entanglement. More generally, for a bipartite system, any unitary acting on one subsystem does not change the entanglement between the two subsystems. $\endgroup$
    – Meng Cheng
    Commented Sep 6, 2022 at 20:09
  • $\begingroup$ @MengCheng thanks. Indeed, as you say, the magnetic field flips whatever arrow it sees, up to down and down to up. However, it only ever sees down. There's a Stern-Gerlach analyzer in front of the magnetic field, that splits the incoming beam, directing all $\uparrow$ particles away from the field, and only $\downarrow$ particles to the field. $\endgroup$ Commented Sep 6, 2022 at 20:15
  • $\begingroup$ @MengCheng I agree my entire diagram taken as a whole is not unitary, since it's not surjective (onto). However, I don't see what would go wrong if we actually tried to construct it. $\endgroup$ Commented Sep 6, 2022 at 20:35
  • $\begingroup$ Your final displayed equation is missing a term. $\endgroup$
    – WillO
    Commented Sep 6, 2022 at 20:51

1 Answer 1


Your analyzer effectively observes the particle, which changes the state of your system from $(UD+DU)$ to either $UD$ or $DU$. In the former case, you follow with a unitary transformation that converts $UD$ to $UU$. Your system overall is now in one of the two states $UU$ and $DU$, with equal probabilities.

  • $\begingroup$ "Your analyzer effectively observes the particle." How is that? There's no way to tell which path the particle takes through my analyzer. I agree my analyzer isn't unitary, but I don't see what would prevent someone from physically building it, or where the which-way information would be if we did. $\endgroup$ Commented Sep 6, 2022 at 20:55
  • $\begingroup$ @MartinC.Martin You are choosing to flip the spins only of spin up particles. To do that, you have to measure the spin first to decide whether you are going to flip the spin or not. (When I say "you" here, I mean your apparatus.) $\endgroup$
    – d_b
    Commented Sep 6, 2022 at 22:39
  • 2
    $\begingroup$ The issue that seems to be bothering you has nothing to do with entanglement --- you can see the same issue with a single (unentangled) qubit that enters your apparatus, is measured as either up or down, and is deflected through a magnetic field if down, changing it to up. All qubits that enter this apparatus emerge in the up state, therefore the apparatus is not unitary. $\endgroup$
    – WillO
    Commented Sep 7, 2022 at 3:54

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