Distinguishing between prepared and unprepared states in the Stern-Gerlach experiment

Question

$ \newcommand{\bra}[1]{\left\langle #1 \right|} \newcommand{\ket}[1]{\left| #1 \right\rangle} \newcommand{\braket}[2]{\left\langle #1 \middle| #2 \right\rangle}$I have a problem and am confused as to what the correct answer is. The question given is:

$\mathsf{A}$ is a beam of spin-$\frac{1}{2}$ atoms prepared to be spin-aligned along the +$x$-axis. $\mathsf{B}$ is a beam of similar unpolarized atoms. $\mathsf{A}$ and $\mathsf{B}$ are separately passed through a Stern-Gerlach experiment aligned along $z$. In each case you get two emerging beams coming out of the Stern-Gerlach apparatus. Is there any difference between the 2 cases? If so, how could you detect that experimentally?

Now, atoms in the $\mathsf{A}$ beam have pure quantum state:

$$\ket{\psi} = \ket{\uparrow_{x}} = \frac{1}{\sqrt{2}}\left(\ket{\uparrow_{z}} + \ket{\downarrow_{z}}\right)$$

And therefore:

$$P(\ket{\psi'}=\ket{\uparrow_{z}})=|\braket{\uparrow_{z}}{\psi}|^{2} = \frac{1}{2}$$

However, for beam $\mathsf{B}$ we have an unpolarized beam and thus we have that the density matrix is given by:

$$\mathbf{\rho}=\frac{1}{2}\begin{pmatrix}1 & 0 \\ 0 & 1\end{pmatrix}$$

And therefore the probability of measuring $\ket{\uparrow_{z}}$ after passing it through the Stern-Gerlach experiment is the same. Therefore, I do not see how there can be a possibility of distinguishing between the two states after passing them through the Stern-Gerlach apparatus. Yet the phrasing of the question has made me think I am misunderstanding something.

Am I missing something here?

Answer 1

The usual answer--and I think what the question is looking for--is that there is a difference, but you can't see it with a $z$-axis Stern-Gerlach apparatus. Imagine you took the output from the $z$-axis SGA, merged the two beams back together, and sent it through an $x$-axis SGA. Then one of them is still in the $| + x \rangle$ eigenstate, so will always yield a $+\hbar/2$ measurement, but the other is still an unpolarized beam in this axis, just as it was before. The $z$-axis measurement will only disturb the $x$ state if you actually stop and measure and that point; if you discard the information by putting them all back together it's no harm no foul.

Answer 2

If you have a pure state, then its most general form is $$ \vert \psi\rangle = \cos(\textstyle\frac{1}{2}\theta)\vert +\rangle + \sin(\frac{1}{2}\theta)e^{i\varphi}\vert -\rangle $$ where $\vert \pm\rangle$ refers to some reference orientation (usually the $\hat z$) axis. Such a state is the eigenstate of \begin{align} \hat n\cdot\vec \sigma&=n_x\sigma_x+n_y\sigma_y+n_z\sigma_z\, ,\\ &= \sin\theta\cos\varphi \sigma_x +\sin\theta\sin\varphi\sigma_y + \cos\theta\sigma_z \end{align} so that, by aligning your magnet in the $\hat n$ direction, you will get 100% of the beam deflecting in the "up" direction. In other words, you can find a rotation of your reference axis $\hat z$ so that your state $\vert\psi\rangle$ is an eigenstate of $\hat n\cdot\vec\sigma$, the Pauli matrix "in this direction".

On the other hand, you cannot find such as rotation if you have any mixed state, i.e. you cannot find a direction of the Stern-Gerlach apparatus that will result in 100% of your beam deflected in a single direction. In your specific example, your $\rho$ is the so-called garbage state so whatever orientation you give to you SG magnet you will always get 50% of the beam going up and another 50% going down.

This extends to finite dimensional spaces. A pure state will be the eigenstate of some operator of the form $U^\dagger D U$, where $D$ is diagonal and $U$ is unitary. It might not be trivial to actually implement this operator but in principle it's possible as per

Park, J.L. and Band, W., 1971. A general theory of empirical state determination in quantum physics: Part I. Foundations of Physics, 1(3), pp.211-226.