I am studying for my quantum mechanics exam and I came across the following question, I hope I reason correctly.

Consider two spin-$\frac{1}{2}$ systems one of which is measured by Alice and one measured by Bob, their states are called $|\chi\rangle_A$ and $|\phi\rangle_B$ respectively.

Suppose the system is in the state $|s\rangle=\frac{1}{\sqrt{2}}\left(|{\uparrow}\rangle_A|{\downarrow}\rangle_B - |{\downarrow}\rangle_A |{\uparrow}\rangle_B\right)$ (both in $S^z$ basis).

Now, I want to know what happens when Alice and Bob start to measure $S^x$ and $S^z$.

  1. If Alice measures $S^z$ whilst Bob doesn't measure, she measures $\pm\frac{\hbar}{2}$ with probability $\frac{1}{2}$ each. If Bob measures and Alice does not, he finds the same. If Bob measures after Alice has measured, Bob's outcome depends on Alice's measurement. If Alice measures $\frac{\hbar}{2}$, the system has collapsed into $|{\uparrow\downarrow}\rangle$ so Bob measures $-\frac {\hbar}{2}$.

  2. I want to do the same for $S^x$. If Alice is the only one measuring, she will measure $-\frac {\hbar}{2}$ with probability 1 since $\frac{1}{\sqrt{2}}(|{\uparrow}\rangle-|{\downarrow}\rangle)$ is an eigenstate of $S^x$ with eigenvalue $-\frac{\hbar}{2}$. If Bob measures only, he measures also $-\frac{\hbar}{2}$ since Bob also measures an eigenstate. Now, I let Bob measure after Alice. Since Alice was already in an eigenstate, nothing will change in this setup.

Is this correct? I am quite certain about $S^z$, but I am wondering whether I reasoned the case $S^x$ correctly.


2 Answers 2


Nothing that Alice does has any impact on experiments that happen locally within Bob's system. The only effect that you get if Alice measures $S_z$ and Bob simultaneously measures $S_z$, is that they both get completely randomized lists of $↑$ and $↓$ results, but if after the measurements are done Alice sends her lists of results to Bob, Bob will see that it's completely anticorrelated with his results.

As for what happens if they both measure $S_x$, consider the following:

  • if you define $|{+}⟩=\tfrac{1}{\sqrt{2}} \big(|{↑}⟩+|{↓}⟩\big)$ and $|{-}⟩=\tfrac{1}{\sqrt{2}} \big(|{↑}⟩-|{↓}⟩\big)$, then you also have $|{↑}⟩=\tfrac{1}{\sqrt{2}} \big(|{+}⟩+|{-}⟩\big)$ and $|{↓}⟩=\tfrac{1}{\sqrt{2}} \big(|{+}⟩-|{-}⟩\big)$
  • and if you then put in those identities into your expression for the singlet state, $$|s\rangle=\frac{1}{\sqrt{2}}(|{\uparrow}\rangle_A|{\downarrow}\rangle_B - |{\downarrow}\rangle_A |{\uparrow}\rangle_B),$$ to get what it looks like in the $S_x$ basis, you get... $$|s\rangle=\frac{1}{\sqrt{2}}(|{+}\rangle_A|{-}\rangle_B - |{-}\rangle_A |{+}\rangle_B),$$ i.e. exactly the same but in the $S_x$ basis.

You can work from there to infer whether there will be differences in the measurement results when they both measure $S_x$ vs. when they both measure $S_z$.


Your answer is wrong. To calculate what happens when you measure $S_x$ on system $A$, you have to work out the state written in terms of the eigenstates of $S_x$ on system $A$. The $S_x$ eigenstates $|+\rangle,|-\rangle$ can be written in terms of the $S_z$ eigenstates as $$|+\rangle=\tfrac{1}{\sqrt{2}}(|\uparrow\rangle+|\downarrow\rangle)$$ and $$|-\rangle=\tfrac{1}{\sqrt{2}}(|\uparrow\rangle-|\downarrow\rangle).$$ You can use this to calculate the the $S_z$ eigenstates in terms of the $S_x$ eigenstates and substitute the result into the state $|s\rangle$. You will then be able to work out the probabilities of the different results and the probabilities of finding a correlation between the results when you compare them. The real answers don't match the answer you gave.

  • $\begingroup$ Thank you :). Is $S^z$ correct? $\endgroup$
    – James
    Commented Jan 28, 2019 at 9:09
  • $\begingroup$ For an exam I think you should stick to giving the probabilities of each result and the probabilities of correlations between results. What is actually happening in reality to bring those results about is controversial and making statements beyond the probabilities of results and correlations may result in you losing marks. $\endgroup$
    – alanf
    Commented Jan 28, 2019 at 9:18
  • $\begingroup$ What I think is actually happening is explained here physics.stackexchange.com/questions/190670/… If you say anything even vaguely resembling this answer in an exam you will lose marks. $\endgroup$
    – alanf
    Commented Jan 28, 2019 at 9:24

Your Answer

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

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