Sunday, December 31, 2017

quantum mechanics - What is the spin state of a spin-1/2 particle when it comes out of a Stern-Gerlach apparatus?


Having a particle entering the apparatus with spin state $|+\rangle$, for which $\hat S_x|+\rangle=+\frac\hbar 2|+\rangle$, I have a question about how to express the spin state when it comes out. I mean: will the spin state be an eigenstate of the spin $z$ component? or would it be a superposition of the $z$ component's eigenstate ?



Answer



A Stern-Gerlach apparatus oriented along the $z$ axis acts as a measurement on the basis of the $z$ component of the particle's spin. What that means is that the particle will always come out in an eigenstate of $\hat S_z$. You don't know which one, of course, as that is decided probabilistically.


Even if you don't actually observe the outcome of the measurement (in which case the system would be in the corresponding eigenstate), it is not accurate to say that the particle comes out in a superposition of $z$ eigenstates. While what comes out is a probabilistic mixture, it is a mixed state in which the coherence between the two contributions has been lost.


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