Suppose I have a source that produces pairs of spin-1/2 spins in the entangled state

$$ \frac{1}{\sqrt{2}} \left( \mid\uparrow_z\uparrow_z\rangle + \mid\downarrow_z\downarrow_z\rangle \right). $$

The particles are sent opposite ways into respective Stern-Gerlach (SG) detectors where the spins are measured. One of the particles is measured first (doesn't matter which one).

If the Stern-Gerlach representing the first measurement is oriented vertically, to align with the z-axis, then the measurement (meaning the detection of the first particle after it passes through) causes the particle to assume one of the two states $\mid\uparrow_z\rangle$ or $\mid\downarrow_z\rangle$. This in turn causes the other particle to assume the same state, and therefore both particles come out of the same channel of their respective Stern-Gerlach setups. I'm on board with this.

Now suppose we turn both Stern-Gerlachs to a new angle. The same thing will happen, meaning the two particles will always come out in the same channel. I have an image in my head of an SG measurement effectively disturbing the particle and destroying the old information we (might have) had while forcing the spin to pick a state in the new measurement basis defined by the SG's orientation. So how can the correlations survive a measurement in a basis other than z? What is the intuition for this? How can I understand intuitively why being entangled in the z-direction means that measuring one particle at some angle $\theta$ causes both particles to collapse to $\mid\uparrow_\theta\rangle$ (for example)?

Happy to edit or answer clarifying questions.

  • $\begingroup$ I'm confused by "how can the entanglement survive a measurement in a basis other than z?" Measuring one spin destroys the entanglement, no matter what basis the measurement is performed in. Do you mean how do the correlations in measurement outcomes survive, or something like that? $\endgroup$
    – user34722
    Mar 23, 2022 at 6:00
  • $\begingroup$ @user34722 Yeah, you're right, I do mean something like "how do the correlations survive." Thanks for catching this. $\endgroup$ Mar 23, 2022 at 10:45

1 Answer 1


When you write down a quantum state, you necessarily pick a basis to write it in. Sometimes the full symmetry of the state isn't easy to see in a particular basis. In this case, it looks like the $z$ direction is special, so it's surprising to find perfect correlation in other bases as well. But changing measurement basis is equivalent to rotating the state, and look what happens when we rotate both spins about the $y$ axis:

$$e^{i(\sigma_{y,1} + \sigma_{y,2})\theta} \frac{|\uparrow_z \uparrow_z\rangle + |\downarrow_z \downarrow_z\rangle}{\sqrt{2}} = \frac{|\uparrow_z \uparrow_z\rangle + |\downarrow_z \downarrow_z\rangle}{\sqrt{2}}$$

because the state is an eigenvector of $\sigma_{y,1}+\sigma_{y,2}$. Note that if you instead had a classical mixture of $|\uparrow_z \uparrow_z\rangle$ and $|\downarrow_z \downarrow_z\rangle$, then rotation would change the state. So this is a really nice illustration of how quantum correlations i.e. entanglement differ from classical correlations.

An even more striking example is the singlet state $|\uparrow_z \downarrow_z\rangle - |\downarrow_z \uparrow_z\rangle/\sqrt{2}$, which exhibits perfect anti-correlation in all basis. This comes from the fact that it has total angular momentum of zero, so it is invariant under (global) spatial rotations. In contrast, a classical mixture of $|\uparrow_z \downarrow_z\rangle$ and $|\downarrow_z \uparrow_z\rangle$ is only perfectly correlated in the z basis.

  • $\begingroup$ So suppose I was producing the entangled state in the question and then sending the particles into respective Stern-Gerlachs in the x direction. You're saying that state would always give me particles that come out in the same port, but $\mid\uparrow_z\uparrow_z\rangle$ would NOT? I have more trouble with that, because I'm thinking about both the particles still "having the same state." What happens when we measure the first particle that I'm not getting intuitively? $\endgroup$ Mar 23, 2022 at 10:50
  • $\begingroup$ Correct! Measuring along $x$ is the same as rotating about $y$ and then measuring $z$. If we do a $\pi/2$ rotation on $|\uparrow_z\uparrow_z\rangle$, then each qubit winds up in $(|\uparrow_x\rangle + |\downarrow_x\rangle)/\sqrt{2}$, so measuring one particle does nothing to the other. But when add a $|\downarrow_z\downarrow\rangle = |\uparrow_x\uparrow_x\rangle - |\uparrow_x\downarrow_x\rangle - |\downarrow_x\uparrow_x\rangle + |\downarrow_x\downarrow_x\rangle$, the $|\uparrow_x\downarrow_x\rangle$ and $|\downarrow_x\uparrow_x\rangle$ components cancel i.e. destructively interfere. $\endgroup$
    – user34722
    Mar 23, 2022 at 23:32
  • $\begingroup$ This is a great example to build intuition for quantum (as opposed to classical) correlations. Definitely worth playing around with and writing down a few examples. Try the singlet state! $\endgroup$
    – user34722
    Mar 23, 2022 at 23:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.