# I Stern-Gerlach "interferometry"

#### WWCY

Summary
State passing through Stern-Gerlach magnet w/o projective measurement. How to treat it?
Suppose we have a state $|\psi\rangle \sim |\uparrow z\rangle + |\downarrow z\rangle$ passing through a Stern-Gerlach magnetic field oriented in the $\hat{z}$ direction such that $|\uparrow z\rangle$'s are pulled up and $|\downarrow z\rangle$'s are pulled down. Then suppose we place some operation $M_\uparrow$ on the top path and $M_\downarrow$ on the bottom path and both paths are later recombined.

If we treated the whole set-up as some kind of black-box where we don't obtain read-outs for $M_{\uparrow/\downarrow}$, what state would we observe post-recombination?

Would it be $|\psi\rangle \sim M_\uparrow|\uparrow z\rangle + M_\downarrow|\downarrow z\rangle$? Or even $|\psi\rangle \sim M_\uparrow|\uparrow z\rangle|\text{u}\rangle + M_\downarrow|\downarrow z\rangle|\text{d}\rangle$, where $u,d$ represent "path degrees-of-freedom".

My guess is that it's the former since $|\uparrow z\rangle$ "intrinsically" specifies the upwards path in such a setup. But I don't really have anything concrete to check this guess against. Assistance would be greatly appreciated!

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#### vanhees71

Gold Member
For your operations $M_{\uparrow}$ and $M_{\downarrow}$ you can take some Stern-Gerlach experiment with the field oriented in some other direction.

What happens by the magnet in the SGE is that the spin component in direction of the field gets entangled with the component of the particle's position. Dynamically this happens, because the magnetic field leads to the rapid precession of the spin components perpendicular to this field, so that their dynamical effect during the particle motion is (almost) irrelevant. That's why in the two partial beams the spin component in direction of the field is (almost) perfectly determined, while the two perpendicular components are maximally indetermined. This is independent of the initial state the particle enters the field.

Now the same holds true for a 2nd SGE performed on one of the beams with determined $s_z$-component. Say, you direct its field in $x$-direction. Again the magnet will separate the beam in two partial beams with determined spin components, but this time of course $s_x$ is determined. All you can say from the preparation in a given $s_z=\pm \hbar/2$ state is that the probability with which a particular particle will end up with $s_x=\hbar/2$ is 1/2 and to end up with $s_z=-\hbar/2$ is also 1/2, but now the $s_x$ component is entangled with the $x$ component of position and now the $s_z$ component is completely indetermined.

#### WWCY

Thanks for the response.

When you say
What happens by the magnet in the SGE is that the spin component in direction of the field gets entangled with the component of the particle's position
Do you mean that any initial state expanded in the $\hat{e}$ basis: $|\psi\rangle = \alpha_1 |\uparrow e\rangle + \alpha_2 |\downarrow e\rangle$ is mapped to the state* $|\psi '\rangle =\alpha_1 |\uparrow e\rangle|+\hat{e}\rangle + \alpha_2 |\downarrow e\rangle|-\hat{e}\rangle$ after it passes through a SG magnet with a field along the $\hat{e}$ axis? Would your suggestion of applying a successive SG magnet oriented along axis $\hat{x}$ (say on the $+\hat{e}$ path) then be tantamount to applying this map again but in terms of the particle's spin components in the $\hat{x}$ direction?

*I am using $\pm \hat{e}$ to refer to position components.

Cheers.

#### PeroK

Homework Helper
Gold Member
2018 Award
I get the feeling from both of your posts that you are fundamentally misunderstanding something about QM here.

For example, a state is a vector. In your OP you say:

Suppose we have a state $|\psi\rangle \sim |\uparrow z\rangle + |\downarrow z\rangle$
That's like saying:

Suppose we have a vector $\vec{v} \sim \hat{x} + \hat{y}$

Which doesn't make a lot of sense to me.

And, when you say:

*I am using $\pm \hat{e}$ to refer to position components.
What are "position components" in the context of quantum spin?

#### WWCY

Thanks for the response. Perhaps I should have worded my question better:

Suppose we have a vector $\vec{v} \sim \hat{x} + \hat{y}$
I should have wrote it as $|\psi\rangle = a_1|\uparrow \rangle + a_2|\downarrow \rangle$

What are "position components" in the context of quantum spin?
By this I meant the path the particle takes after it passes through the SG magnet, not a position-representation of spin. So $\{ |+\hat{e}\rangle, |-\hat{e}\rangle \}$ are states representing the path the particle takes after the SG magnet. If $|+\hat{e}\rangle$, the particle is deflected upwards along $\hat{e}$, if $|-\hat{e}\rangle$ it is deflected downwards. My question in my preceding post would then be whether or not the spin components are entangled with the path "states" in the manner I wrote.

Cheers.

#### PeroK

Homework Helper
Gold Member
2018 Award
My question in my preceding post would then be whether or not the spin components are entangled with the path "states" in the manner I wrote.

Cheers.
That's not what entanglement means. The full wave-function of an electron is a combination of its position-space wave-function and its spin state.

In the case of SG the electron is either in state $\psi_1 \chi_+$ or state $\psi_2 \chi_-$. Where $\psi$ is a position-space wave-function and $\chi$ is a spin state.

#### WWCY

That's not what entanglement means. The full wave-function of an electron is a combination of its position-space wave-function and its spin state.

In the case of SG the electron is either in state $\psi_1 \chi_+$ or state $\psi_2 \chi_-$. Where $\psi$ is a position-space wave-function and $\chi$ is a spin state.

Do $\psi_{1/2}$ refer to distinct positions?

#### PeroK

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2018 Award
Do $\psi_{1/2}$ refer to distinct positions?
They are the two position-space wave-functions associated with the two trajectories out of the SG apparatus.

#### Nugatory

Mentor
If we treated the whole set-up as some kind of black-box where we don't obtain read-outs for $M_{\uparrow/\downarrow}$....
That bit about not obtaining a readout suggests that you are thinking of the $M$ devices as macroscopic devices that could in principle tell us which path the particle takes. If so.....

As @vanhees71 says above, the interaction with the S-G device entangles the particle spin and the particle position. Thus, the the particle emerges in a state $|\psi_{\uparrow} \chi_+\rangle+|\psi_\downarrow \chi_-\rangle$ where the two $\chi$ functions are "positive spin" and "negative spin" and the two $\psi$ functions are "deflected up" and "deflected down". (There are some pitfalls in treating a continuous observable such as position in this manner, but we can get away with ignoring them here).

You have placed the two $M$ devices at different positions, so the interaction with either one will be a position measurement. This will collapse the wavefunction to one of "spin up on the upper path and affected by $M_{\uparrow}$" and "spin down on the lower path and and affected by $M_\downarrow$". As with all quantum mechanical observations, whether we actually get a readout from the $M$ is irrelevant - the interaction itself collapses the wave function.

Now when you recombine the beams you'll have prepared a mixed state (a bunch of particles half in one state and half in the other) instead of a superposition. This mixed state cannot be written as a state vector at all - you'll need a density matrix.

This thought experiment is somewhat analogous to a double-slit experiment with detectors at the slits; the superposition of paths is eliminated by the interaction with the detectors.

#### WWCY

Thanks a lot for the comments, I'm starting to see the bigger picture now.

I still have a question regarding the position-spin wavefunction as I've never really learnt it properly.

If I start with the initial electron wavefunction $\phi = \psi_1 \chi_+$, how would I write the Hamiltonian that governs the time-evolution of the electron as it passes through the SG magnet? i.e. How do I show that $\phi$ propagates in the $+z$ direction (and $-z$ for a state with opposite spin)?

Also,

Now when you recombine the beams you'll have prepared a mixed state (a bunch of particles half in one state and half in the other) instead of a superposition. This mixed state cannot be written as a state vector at all - you'll need a density matrix.
Does this mean that entire setup can be represented by a completely positive (CP), trace-preserving map that is a sum of two CP maps?

Cheers.

#### vanhees71

Gold Member
It's
$$\hat{H}=\frac{1}{2M} \hat{\vec{p}}^2 + \frac{g \mu_{\text{B}}}{\hbar} \hat{\vec{s}} \cdot \vec{B}(\hat{\vec{x}}),$$
where
$$\mu_{\text{B}}=\frac{e}{2m} \hbar>0$$
is the Bohr magneton (the additional - sign in front of the potential energy is due to the fact that in the silver atom the magnetic moment comes from an electron with a negative charge $q_{\text{e}}=-e$). The gyro-factor of the electron $g \simeq 2$ (with small deviations due to radiative QED corrections. The precise value of the gyro-factor of the electron is in fact among the most accurate predictions of the Standard Model).