I'm having some trouble understanding something concerning sequential measurements.

Suppose we have a wave function consisting of a superposition of spin up and spin down in the z direction (spin=1/2). We measure the spin in the z direction, then in x and then in z again and ask what is the probability that the spin will be up.

I know the answer is half, as once I measure in the x direction, there's a 1/2 chance to have spin up x and 1/2 for spin down x, regardless of the original wave function, and the same when going back to measuring spin z again.

However, when I perform $S_zS_xS_z$ on the original wave function, then project it on spin up (z) and calculate the absolute value squared (the probability for spin up), I get the same probability as with the original wave function for the first time I measured spin in the z direction. So suppose I originally had 1/3 probability to have spin up in the first measurement, then I will have the same the second time even after I've measured spin in the x direction inbetween.

Obviously just multiplying the original wave function by the operators $S_zS_xS_z$ doesn't work, but why? Why is this fundamentally wrong?

Thanks in advance

(I'd write the matrices and calculate them here, but my computer is broken and doing so on the phone isn't the easiest)


$\let\a=\alpha \let\b=\beta \def\ket#1{|#1\rangle} \def\braket#1#2{\langle#1|#2\rangle} \def\half{{\textstyle{1 \over 2}}}$

Obviously just multiplying the original wave function by the operators SzSxSz doesn't work, but why? Why is this fundamentally wrong?

Simply because measuring an observable on a given state is not to apply the operator to the state vector. Reminding you the basic QM postulate concerning measurement is in order:

The measurement of observable $A$ on state $\ket s$ gives as a result an eigenvalue $a$ of $A$ and leaves the system in the corresponding eigenstate $\ket a$ of $A$. Every eigenvalue may be obtained, each with probability $|\braket as|^2$. (I have given the simplest form of the postulate, holding for non-degenerate eigenvalues.)

Let's apply it to our case. The initial ket is $$\a\,\ket{z+} + \b\,\ket{z-} \qquad |\a|^2 + |\b|^2 = 1$$ (I assume notation is self-explanatory).

Measurement of $s_z$ on this state will leave the system in one of two states:

  • $\ket{z+}$ with probability $|\a|^2$
  • $\ket{z-}$ with probability $|\b|^2.$

Now for measurement of $s_x$ on $\ket{z+}$. Results are

  • $\ket{x+}$ with probability $\half$
  • $\ket{x-}$ with probability $\half$

and the same happens for $\ket{z+}$. Therefore after measuring $s_x$ we may have

  • $\ket{x+}$ with probability $\half\,(|\a|^2 + |\b|^2) = \half$
  • $\ket{x-}$ with probability $\half\,(|\a|^2 + |\b|^2) = \half.$

I leave you finishing the exercise.

  • $\begingroup$ So as far as I understand, what I did is: find the probability of measuring z+ with a totally different operator and in general it's to some extent a coincidence that z+ even is an eigenvalue for the operator? $\endgroup$ – Mageer Oct 9 '18 at 10:47
  • $\begingroup$ @Mageer I'm not certain to understand your question. Furthermore, I'm afraid my notation was not so self-explanatory, after all. When I write $|z+\rangle$ I mean the eigenvector of $s_z$ to eigenvalue +1/2. Could you please rephrase your question? $\endgroup$ – Elio Fabri Oct 9 '18 at 14:19
  • $\begingroup$ your notation was self-explanatory and you've helped me solve the problem, thanks :) $\endgroup$ – Mageer Oct 9 '18 at 15:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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