Consider the context of the Stern-Gerlach experiment. As is stated on numerous sources (e.g. Feynman Lectures, MIT Lecture), a silver atom in the state of $\vert+z\rangle$ that is put through a Stern-Gerlach apparatus in the $x$ direction followed another apparatus oriented in the opposite direction will recombine the two beams and restore the state as $\vert+z\rangle$.

My question is why shouldn't the new state be the mixed state given by the following density operator? $$\rho = \frac{1}{2}|+x\rangle\langle+x|+\frac{1}{2}|-x\rangle\langle-x|$$

Now, consider the scenario where I have a friend standing in between the two $x$-oriented Stern-Gerlach apparatus to observe that individual atom's $x$-spin, but does not tell me the observations. Clearly a measurement has been done, so why shouldn't I use a mixed state from that point on? Now what if I don't even know if anyone is standing between the two apparatus making an observation?

In general, the density operator of the full ensemble after measurement, when not knowing the measurement result, is given by $$\rho'=\Sigma_i P_i\rho P_i$$ (source: Wikipedia) where $P_i$ is the projection oeprator onto the $i^{th}$ eigenspace. Why is it that this new density operator is not applicable to the case of the Stern-Gerlach scenario above where I throw the two beams back into a single full ensemble? When is the new density operator applicable?

  • 1
    $\begingroup$ Related: physics.stackexchange.com/q/282410/109928 $\endgroup$ Commented Dec 1, 2016 at 12:02
  • $\begingroup$ @StéphaneRollandin Nice! $\endgroup$
    – OON
    Commented Dec 1, 2016 at 12:16
  • $\begingroup$ Mixed with respect to which basis ? $\endgroup$
    – Our
    Commented Jan 19, 2019 at 12:49

1 Answer 1


In the first case you and no one performs the measurement, and more generally, you can neglect decoherence of the atom state.

The reason decoherence happens is that atom becomes entangled with some degrees of freedom of the environment that we don't track. If you treat all those degrees of freedom as part of some big closed quantum system (which at some level would require you to talk about quantum state of the universe which may make sense or may not), in the ideal limit $\langle \mathrm{env},-|\mathrm{env},+\rangle=0$ you can describe its state as, \begin{equation} |\mathrm{env},0\rangle\otimes\frac{1}{\sqrt{2}}\Big(|+x\rangle+|-x\rangle\Big)\mapsto \frac{1}{\sqrt{2}}|\mathrm{env},+\rangle\otimes|+x\rangle+\frac{1}{\sqrt{2}}|\mathrm{env},-\rangle\otimes|-x\rangle \end{equation} So in some sense each $|\pm x\rangle$ becomes "marked" and don't make coherent superposition anymore. Indeed the state of the atom is now described by mixed density matrix you wrote.

Some degree of decoherence always happens. However the situation described above corresponds to the idealized limit of total decoherence, that is assumed to take place if you perform the ideal projective measurement - some degree of atom should become perfectly correlated with degrees of freedom of the measurement device. However sometimes $|\mathrm{env},\pm\rangle$ are so far from being orthogonal, then you can forget environment entirely and use ordinary quantum mechanics of closed systems to describe objects of interest.

Exactly this kind of situation happening in the Stern-Gerlach experiment - although atom interacting with device and other stuff around somewhat changes their states and therefore becomes entangled, this effect is so tiny that atom remains in practically perfect coherent superposition until it hits the screen. But in this case we don't measure the spin of the atom in any sense until the very end. If you add some sort of the active measurement device in a way (even without friend) it will ruin this approximation and you will have to describe atom state using mixed density matrix $\rho'$.


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