# Sequential spin measurement misunderstanding

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?

(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.

• 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? – Mageer Oct 9 '18 at 10:47
• @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? – Elio Fabri Oct 9 '18 at 14:19
• your notation was self-explanatory and you've helped me solve the problem, thanks :) – Mageer Oct 9 '18 at 15:57