$ \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?


2 Answers 2


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.


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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