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I am wondering about the following hypothetical example regarding quantum measurements made by probes:

An open, two-plate capacitor, in a LC circuit, has charge, q, that we want to measure. Each of the capacitor's plates is in the x-z plane and they separated by a distance along the y axis. There is no enclosure between the two plates:

--------- (top cap. plate, in the x-z plane)

       (space between plates that a probe can enter from the left and exit to the right)

---------(bottom cap. plate, in the x-z plane)

As you know, each element of the diagonal of the capacitor's density matrix (in Hilbert Space) represents a charge state, and the the value of each of those diagonal elements is the probability (or probability density) for the charge to be measured in that state. Let's suppose that the initial density matrix of the capacitor's charge has a diagonal whose elements are all zero (representing zero probability for those states), except for the charge states between state q1 and state q6 (i.e., state q0 and state q7, on the diagonal, are both outside of that range and, hence, have zero probability). Of course, there could be an infinite number of charge states within the continuous range between q1 and q6.

Suppose electron 1 (a probe) flies through the capacitor, parallel to the x axis, from left to right. When the electron exits the capacitor, the capacitor's charge, q, will be entangled with electron 1's y-momentum (this example, so far, is from Braginsky's book on quantum measurement, so please forgive the improbable idea of an electron flying through an open two-plate capacitor).

Now, suppose electron 2 also flies through the capacitor, entering immediately upon the exit of electron 1 from the capacitor, so back action effects of electron 1 on q can mostly be ignored. Upon exit, electron 2's y-momentum is now also entangled with the charge of the capacitor.

Suppose electron 2 is then detected (its momentum measured) before electron 1 is detected. That would be a non-orthogonal measurement; the result is not a single eigenstate of capacitor charge, but a small continuous range of charge values with non-zero probabilities. That is, as a result of the measurement, let us assume that the density matrix of the capacitor's charge now has a diagonal whose elements all have the value of zero, except for the charge states between state q3 and state q6 (so the non-zero range along the diagonal of the capacitor's density matrix is smaller than before).

Now, let's assume that electron 1 is measured, immediately after electron 2 was measured, so that we can ignore the back action of electron 2 on q after it is measured.

When electron 1 is measured (immediately after electron 2 is measured), is electron 1 still entangled with the capacitor's charge and does its measurement therefore act to further refine the measurement made by electron 2 (further narrow the non-zero range along the diagonal of the capacitor's density matrix)2? That is, is q now non-zero along the diagonal in a range that is smaller than the range between q3 and q6? For example, after electron 1 is measured, the range of non-zero states along the diagonal might be between q4 and q5. If so, that would mean that electron1's measurement of charge refined the measurement of charge from a different probe (electron 2) which became entangled with the charge after electron 1 did.

David
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  • A way to say "parallel to x axis and perpendicular to y axis" is "in the x-z plane". – DanielSank Nov 17 '15 at 06:16
  • The cutaway diagram is actually in the x-y plane, where each of the two dashed lines is parallel to x, and the perpendicular between the dashed lines is parallel to y. – David Nov 17 '15 at 06:21
  • The second measurement is refinement because you said that the first measurement has less than infinite precision. The key phrase to Google is "Quantum Bayes inference" or some such. It's just like making multiple imprecise classical measurements, but with some quantum wrinkles. – DanielSank Nov 17 '15 at 06:22
  • Right, so the pates are in the x-z plane. – DanielSank Nov 17 '15 at 06:23
  • anyway, just wondering if the measurement of electron 2 (whose y-momentum was entangled with the capacitor's charge) breaks the entanglement of electron 1's y-momentum with the capacitor's charge. – David Nov 17 '15 at 06:23
  • yes, plates are in x-z plane. sorry for the confusion – David Nov 17 '15 at 06:23
  • if, after electron 2 is measured, electron 1's y-momentum is still entangled with the capacitor charge, then subsequent measurement of electron 1 would be refining the measurement made by electron 2, whose interaction with the capacitor was after the interaction of electron 1 with the capacitor. Another example of "quantum weirdness". – David Nov 17 '15 at 06:27
  • I really dislike the phrase "quantum weirdness" as it promotes a feeling of mysticism rather than understanding. In my opinion, it's not that weird once we understand that 1) quantum mechanics is a theory about information, and 2) measurement requires physical interaction. Obviously, I'm not explaining anything here. If I can cobble together a truly good answer (which this question deserves) I'll write one. For now let me promise that this isn't all that complicated if you approach it with a density matrix description and keep your wits about you. – DanielSank Nov 17 '15 at 06:30
  • I put "quantum weirdness" in quotes to indicate that it is a questionable expression. However, it also is a common one that conveys the point of non-intuitiveness. – David Nov 17 '15 at 06:34
  • Oh, gotcha. Yeah, it's common, but I don't like it :\ – DanielSank Nov 17 '15 at 06:38
  • 1 of 2: Anyway, I agree with you that it would be a refinement for the following reason: Electron 1 and 2 must be entangled, as each is entangled with the capacitor (after each has exited it and before either is measured) and the capacitor cannot be included in two entangle wave functions, each with a total probability of 100%. Therefore, there must be a single composite entangled wave function that includes the capacitor and both electrons. – David Nov 17 '15 at 06:38
  • 2 of 2: Then, once electron 2 is measured, that composite wave function is altered to contain a lot more elements whose value is zero and therefore limits the values that electron 1 can subsequently measure out to. Then, once electron 1 is measured, the composite entangled wave function has even fewer non-zero elements, thereby further refining the measurement made by electron 2. – David Nov 17 '15 at 06:40
  • Woah, wait a minute. I said that the seconds measurement is a refinement specifically because you said that the measurement leaves the capacitor charge in a range. I figured this was meant to indicate that you can't measure the electron momentum perfectly, and so the inferred charge of the capacitor is not known to arbitrary precision. If you did measure the electron momenta exactly then the first measurement would kill the remaining entanglement. – DanielSank Nov 17 '15 at 06:40
  • I typed my previous comment before I saw "2 of 2". I think you get it. – DanielSank Nov 17 '15 at 06:41
  • Cool, I needed someone to do a sanity check. Thank you! – David Nov 17 '15 at 06:42
  • There's actually a well defined way to analyze cases like this. Unfortunately, it's not in many books. – DanielSank Nov 17 '15 at 06:54
  • Do you have a link to it? – David Nov 17 '15 at 07:04
  • Unfortunately not. I believe if you look up papers by Alexander Korotkov you will find relevant things. Also, I posted a new question trying to get at the heart of the matter. – DanielSank Nov 17 '15 at 07:19

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