Problem involving a sequential Stern-Gerlach experiment

[AndersF]

TL;DR Summary

Determining the spin state of an electron beam after going through two sequential Stern-Gerlach devices

An electron beam with the spin state $|\psi\rangle = \frac{1}{\sqrt{3}}|+\rangle+\sqrt{\frac{2}{3}}|-\rangle$, where $\{|+\rangle,|-\rangle\}$ is the eigenstates of $\hat S_z$, passes through a Stern-Gerlach device with the magnetic field oriented in the $Z$ axis. Afterwards, it goes through another Stern-Gerlach device, this time with the magnetic field oriented in the $X$ axis.

My book states that the number of emerging electrons after the second S_G with spin-states $S_{x+}$ and $S_{x-}$ will be the same. I have done the math and, indeed, I have found this to be true, but I don't get this: won't the coefficients of $|+\rangle$ and $|-\rangle$ in $|\psi\rangle$ affect at all the result?

 

[PeroK]

Summary:: Determining the spin state of an electron beam after going through two sequential Stern-Gerlach devices

An electron beam with the spin state $|\psi\rangle = \frac{1}{\sqrt{3}}|+\rangle+\sqrt{\frac{2}{3}}|-\rangle$, where $\{|+\rangle,|-\rangle\}$ is the eigenstates of $\hat S_z$, passes through a Stern-Gerlach device with the magnetic field oriented in the $Z$ axis. Afterwards, it goes through another Stern-Gerlach device, this time with the magnetic field oriented in the $X$ axis.

My book states that the number of emerging electrons after the second S_G with spin-states $S_{x+}$ and $S_{x-}$ will be the same. I have done the math and, indeed, I have found this to be true, but I don't get this: won't the coefficients of $|+\rangle$ and $|-\rangle$ in $|\psi\rangle$ affect at all the result?

I presume only one of the z-beams is passed through the x-magnet? In which case, both $|+ \rangle$ and $|-\rangle$ are equal superpositions of $S_{x+}$ and $S_{x-}$.

 

[PeroK]

PS the original beam is, of course, not an equal superposition of $S_{x+}$ and $S_{x-}$.

 

[AndersF]

Yes, the first beam splits in two after passing through the first S-G, and then both beams pass through the second S-G.

 

[PeroK]

Yes, the first beam splits in two after passing through the first S-G, and then both beams pass through the second S-G.

Loosely, by splitting the beam according to z-spin, you prevent the z-spin-amplitudes from interfering. With the full beam you effectively get interference between the z+ and z- states. This, in many shapes and forms, is what QM is all about and where it differs from classical mechanics.

 

[PeterDonis]

the first beam splits in two after passing through the first S-G, and then both beams pass through the second S-G.

How do you pass both beams through the second S-G magnet? They come out of the first one going in different directions.

 

[Demystifier]

How do you pass both beams through the second S-G magnet? They come out of the first one going in different directions.

I think there are two second SG magnets.

 

[AndersF]

The statement of the problem doesn't specify it. I guess that Demystifier is right and that there are two second SG magnets.

 

[Demystifier]

My book states that the number of emerging electrons after the second S_G with spin-states $S_{x+}$ and $S_{x-}$ will be the same. I have done the math and, indeed, I have found this to be true, but I don't get this: won't the coefficients of $|+\rangle$ and $|-\rangle$ in $|\psi\rangle$ affect at all the result?

Why do you expect that the result should depend on coefficients? Do you expect that it should depend on their absolute values, or on their phases?

If you expect a dependence on phases, then you expect interference and @PeroK in #5 explained why there is no interference. If you expect a dependence on absolute values, then have in mind that absolute values of $\langle+_x|+\rangle$, $\langle-_x|+\rangle$, $\langle+_x|-\rangle$ and $\langle-_x|-\rangle$ are all the same.

 

[romsofia]

So, to clarify, this is the set up:
You have an oven which heats up the silver atoms, and we run it through the inhomogenous magnetic field and find the probabilities $S_{z+} = \frac{1}{3}$ and $S_{z-}=\frac{2}{3}$. We then take both of these outputs and put them into an analyzer on the x-axis, and youre asking why don’t the probabilities from the z analyzer matter when analyzing on the x?

If so, calculate $[S_z,S_x]$ and see why! If you get a non-zero answer, then knowing $S_z$ changes what you knew about $S_x$ (ala a “collapse” of information) even if in this case it is nothing, since it a fresh system, but the point remains. The observables don’t commute (that is for two operators $[S_z,S_x]= S_zS_x-S_xS_z \neq 0$) so it really doesn't matter what my probabilities are on the z-axis, according the quantum theory! But maybe the following will help more?

An even more fun “paradox”, IMHO, is to justify the following: Take one output from the Z analyzer (ie spin up) throw it into an analyzer on the X axis, then take one output from that (ie spin down), and take only that beam and throw it back to an analyzer on the z axis, youll see that you get back a probablistic outcome of both up/down (due to the argument above, the operators don't commute). Now, repeat the experiment, but now take BOTH outputs from the x-analyzer and throw them into the z-analyzer, you will no longer get a probabilistic outcome! Youll get whatever output you chose on the original z analyzer, almost as if no measurement took place. How can that be?

I think we can justify it by commuting operators, which I wrote, but then your question actually seems to counter it, it is similar to the “paradox” I posed, that is, if I take both beams of information, did I measure anything? If you initially take both outputs from your initial measurement, throw them on the x-axis, did $\psi$ actually change? If so, then commuting variables justify your question. If not, then the “paradox” answer it, and it is as if you never measure on the z-axis to begin with!

Now, a more important question to look into, IMO, is how exactly do you set up “sequential” Stern-Gerlach experiments in the lab!

Hope this long winded post helps, these questions have always perplexed me as well, so youre in good company!

 

[vanhees71]

Are you referring to the "humpty-dumpty setup"? Then have a look at the marvelous papers by Schwinger et al:

https://link.springer.com/article/10.1007/BF01909939
https://link.springer.com/article/10.1007/BF01384847
https://journals.aps.org/pra/abstract/10.1103/PhysRevA.40.1775

 

[Demystifier]

Are you referring to the "humpty-dumpty setup"? Then have a look at the marvelous papers by Schwinger et al:

https://link.springer.com/article/10.1007/BF01909939
https://link.springer.com/article/10.1007/BF01384847
https://journals.aps.org/pra/abstract/10.1103/PhysRevA.40.1775

I didn't know that Schwinger worked on such stuff.

 

[PeterDonis]

The statement of the problem doesn't specify it.

What statement of the problem? You're the one that originally stated the problem in the OP of this thread. You gave no reference. Where did you get it from?

 

[Demystifier]

What statement of the problem? You're the one that originally stated the problem in the OP of this thread. You gave no reference. Where did you get it from?

He said "my book".

 

[vanhees71]

It's important to clearly state both the preparation and the measurement, i.e., the entire measurement. That's what QT is all about!

Under "squential Stern-Gerlach experiments" one usually understands that one uses one SG magnet to prepare a beam with a given spin component in the direction given by the direction of the magnetic field and then using only the corresponding beam for another Stern-Gerlach experiment, i.e., one uses a state preparation according to what's called a von Neumann filter measurement (which I'd rather take as a preparation procedure).

Of course you can also think about other scenarios, which are much more challenging like the question of reversing the first SG experiment, i.e., to take the two splitted beams and let them run through a exactly reversed other SG magnet to recombine the beams and restoring the original state. That's possible in principle but very challenging in practice. For (theoretical) details see the above cited "Humpty Dumpty papers" by Englert, Schwinger, and Scully.

 

[PeterDonis]

He said "my book".

Which, as your smiley clearly indicates, does not convey the necessary information.

 

[AndersF]

Which, as your smiley clearly indicates, does not convey the necessary information.

Rather than a book, I should have said notes, it's a text written by the professors for the Quantum Mechanics course I am taking (and not a very good one, by the way). I'm sorry, I don't have more information...

 

[PeroK]

Rather than a book, I should have said notes, it's a text written by the professors for the Quantum Mechanics course I am taking (and not a very good one, by the way). I'm sorry, I don't have more information...

The important question is whether you understand the key point about superposition and interference?

 

[AndersF]

The important question is whether you understand the key point about superposition and interference?

Well, with the answers I have read, I think I have managed to clarify myself with this topic (also, looking at vanhees71's articles, it seems that the topic is extensive). Thank you very much :)

 

[PeroK]

Well, with the answers I have read, I think I have managed to clarify myself with this topic (also, looking at vanhees71's articles, it seems that the topic is extensive). Thank you very much :)

We are all used to classical probabilities. Suppose we have an equal mixture of A and B, each of which has an equal chance of having the property X. It's an easy matter to calculate the probability of X in the mixture is:
$$P(X) = P(X|A)P(A) + P(X|B)P(B) = \frac 1 2 \cdot \frac 1 2 + \frac 1 2 \cdot \frac 1 2 = \frac 1 2$$
And we see that the probabilities combine independently, with no "interference".

In QM we use complex probabilitiy amplitudes, with the probability being the square of the modulus of the amplitude. This means that the terms in the superposition may not combine independently and do not behave like classical probabilities above. If we take $x$ and $z$ spin, we have:
$$|z+ \rangle = \frac 1 {\sqrt 2}(|x+ \rangle + |x- \rangle)$$ $$|z- \rangle = \frac 1 {\sqrt 2}(|x+ \rangle - |x- \rangle)$$
We see that the states of $|z+ \rangle$ and $|z- \rangle$ both have a equal probability of giving $|x+ \rangle$ and $|x- \rangle$ upon measurement of $x$ spin. But, a simple superposition of these states may cancel out either the $|x+ \rangle$ or $|x- \rangle$ state. And we see that it is not the case that every supeposition of $|z+ \rangle$ and $|z- \rangle$ has an equal probability of giving $|x+ \rangle$ and $|x- \rangle$ upon measurement of $x$ spin.

To be specific, we have: $$|x+ \rangle = \frac 1 {\sqrt 2}(|z+ \rangle + |z- \rangle)$$ $$|x- \rangle = \frac 1 {\sqrt 2}(|z+ \rangle - |z- \rangle)$$ In other words, the $|x+ \rangle$ state is an equal superposition of two states, each of which individually would yield $|x+ \rangle$ with only 50% probability.

This, in a nutshell, is where QM probability amplitudes behave very differently from classical probabilities.

 

[PeroK]

PS the two states $z+ \rangle$ and $|z- \rangle$ form a basis for the space of spin states. Any spin state, therefore, can be expressed as a linear combination (superposition) of these two states. The same is true for $x+ \rangle$ and $|x- \rangle$, and the $y$ spin states and, indeed, the spin states about any axis.

I would encourage you to study the spin 1/2 system until you really understand it, as this simple system contains the heart of QM.

 

[vanhees71]

A very nice book using the "spin-1/2-first" approach is

J.J. Sakurai, S.F. Tuan, Modern Quantum Mechanics, Revised Edition

The newer 2nd and 3rd editions coauthored by Napolitano are also fine, but it contains the socalled "relativistic quantum mechanics", which should rather be substituted by a treatment in terms of relativistic QFT ;-).

 

[PeterDonis]

Rather than a book, I should have said notes, it's a text written by the professors for the Quantum Mechanics course I am taking (and not a very good one, by the way).

If we can't see it, it can't serve as a basis for a PF discussion. We need a reference that we can all see.

If your text is not very good, that's even more of a reason not to use it as a reference for discussion. You might try looking at some other QM textbooks.

 

[CelHolo]

A very nice book using the "spin-1/2-first" approach is

Schwinger's own book "The Symbolism of Atomic Measurement" is also quite nice in this regard. Especially the first few chapters.

 

[vanhees71]

Particularly the very first chapter is a gem!