The experiment is shown below. How do I calculate the probability of observing a count in detector A, B, or C? Sakurai's text for example starts out describing how to calculate the outcome of simpler chains of S-G devices but I can't find any reference that explains how to handle only slightly-more complicated situations like this.



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For the proper method you only invoke the born rule at the very end just in case the SG machines are super sensitive and preserve a well defined phase between the outcomes. But in your case it won't affect the answers to invoke it at every device. So you could imagine starting with 2,4,8, 16, 32, or 64 particles, whichever makes your calculations easier. Send them through your machines each time, have the correct fraction come out each side.

There is an easy way where you only use the Born rule to find out how much is in C, then how much is in A, and then note that the rest end up in B. However, that trick won't work for more complicated setups, so we can sue that as a check again the full solution (however if you haven't yet learned the full method, the following might seem confusing).

Start by writing $|X_+\rangle$ as a linear combination of $|Z_+\rangle$ and $|Z_-\rangle$, then do the same for $|X_-\rangle$. Then repeat again to write $|Z_+\rangle$ as a linear combination of $|X_+\rangle$ and $|X_-\rangle$, then do the same for $|Z_-\rangle$.

Now you are ready. Whenever something enters a SG machine write it as a linear combination of the states associated with the machine, then send each part of the sum to the different branches. So in general the scalars in front of each would get smaller as you branched more and more, but in one case you fed two results into the same machine. The result then depends on the exact speed and length of the paths, but in this case you are sending X states into an X SG so it just sends them through.

How to compute with the Born rule. For numbers $|a+b|^2\neq|a|^2+|b|^2$ but for vectors, they can ... if they are orthogonal. After a strong measurement, there are a bunch of distinct orthogonal outcomes. You can find the squared length of each of them, which is the probability of getting each distinct outcome, and it is all about whether they are distinct (i.e. orthogonal). The $|X_+\rangle$ and $|X_-\rangle$ are orthogonal, and distinct.

Each time you send a particle in you started with a state $|+000\rangle$ where the + says that you have an $|X_+\rangle$ and the three zeros say that each detector is in the ready state. After the machines you end up with a combination like $a|+100\rangle$+$b|+010\rangle$+$c|-010\rangle$+$d|-001\rangle$ where you get the a,b,c, and d from the steps I hinted at in the third and fourth paragraph. The two middle ones both describe the middle detector firing, but they are orthogonal states (since $|X_+\rangle$ and $|X_-\rangle$ are orthogonal), so the sum of the squares is the square of the sum.

  • $\begingroup$ What I'm interested in is the "proper" method you first mention, but then only later give a vague account of. Do you mind explaining in more detail? What is the observable being measured here really (that gives A B or C), and why can't I just write the wave function in a basis of its eigenvalues and use the amplitudes to find the probabilities? Naively the probabilities should be 1/4, 1/2, 1/4, but on the other hand if the amplitudes add when feeding the 2 results into the same machine, then you'd think by the Born rule the probs would actually be 1/, 4/6, 1/6 because the amplitude is squared. $\endgroup$ – user1247 Mar 14 '15 at 1:14
  • $\begingroup$ @user1247 i edited my answer, the naive probabilities are correct, I just want to make sure your technique works for harder problems. If it helps, think about squaring the whole sate vector. $\endgroup$ – Timaeus Mar 14 '15 at 1:40
  • $\begingroup$ Thanks! Your edit helps a lot (although I think the c|+010> should be c|-010>, right?). To make sure I understand, the probabilities would be 1/6, 1/4, 1/6 if I turned one of the S-G upside down so that instead of feeding |X+> and |X-> to the last S-G I fed |X+> and |X+> to it, right? Then the naive intuition would be wrong. $\endgroup$ – user1247 Mar 14 '15 at 4:35
  • $\begingroup$ @user1247 I was trying to set it up so that the analogous computation with half silvered mirrors and nonabsorbing polarizers would work right. But maybe SG devices can't recombine those beams the way a halfsilvered mirror would. If the SG device changes in response to how it was used, then they are still orthogonal even when both got $|X+\rangle$. $\endgroup$ – Timaeus Mar 21 '15 at 21:52
  • $\begingroup$ That makes sense, but can you do something analogous instead with halfsilvered mirrors? The context of my question was the one here. Basically I'm trying to find a realistic example where you can split and recombine/interfere equal-sized parts of the wave function so that the Born rule generates probabilities that suggest a non-uniform branch-counting measure. I just wanted to check this criticism of the MWI for myself, and intriguingly can't find a physical example. $\endgroup$ – user1247 Mar 24 '15 at 6:20

The probability of detecting a particle at A and C follow from the simple cases given in Sakurai or elsewhere. The only sticking point is detector B. Since it measure the combination of two orthogonal states, you can add the probability of detecting either an $|X_+\rangle$ or an $|X_-\rangle$ from the step before as the two states do not interfere with each other. So, the result is that you have a 25% chance of triggering A or C and a 50% chance of triggering B.

  • $\begingroup$ So the general procedure is piecemeal, there is no way to just write the final state down and use the amplitudes to calculate the probabilities for each outcome at once? What if I sent |X+> and |X+> into the last S-G instead of |X+> and |X->, then would the final probs be 1/6, 4/6, 1/6, since the the amplitude at C is now double and with Born gets squared ? $\endgroup$ – user1247 Mar 14 '15 at 1:20
  • $\begingroup$ You can write the final state down and get the final answer that way, but you still have to consider the effects of each Stern-Gerlach device on the state separately. For your second question, when you combine states that are not orthogonal, you have to consider interference due to differences in phase. For example, if instead of combining the +|X+> with the +|X-> you combined the +|X-> and the -|X-> beams, you would have 0 probability of activating detector B. $\endgroup$ – ajctt Mar 14 '15 at 3:12
  • $\begingroup$ Thanks, and if I combined +|x+> and +|x+> (by just turning one of the S-G upside-down in my above picture) then the probabilities would be 1/6, 4/6, 1/6, right? $\endgroup$ – user1247 Mar 14 '15 at 21:41

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