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I should begin by saying that I am a total newbie when it comes to Quantum Mechanics. Therefore my question might sound metaphysical to people who know their stuff. So please forgive.

What I am trying to understand is the basic system of an electron in a box. Suppose I place a detector at a fixed point inside and try to detect the presence of an electron. There are two possibilities:

(1) The detector detects an electron. The Copenhagen interpretation explains this by saying that the act of measurement forced the wave function to collapse at that point.

(2) The detector did not detect anything. Would this "not detecting an electron" qualify as a true measurement in the Copenhagen sense? I mean, if this qualified as a measurement would it change the wave function?

What led me to this confusion is the explanation in Griffiths' Quantum Mechanics. He states that the same experiment when done soon after possibility (1) gives us the same result i.e, observe the electron (which does not confuse me.) But what happens when you do the experiment soon after possibility (2)? Would I still not observe the electron?

What if I extend this to the two-slit experiment and place the detector at only one slit. If I didn't observe the electron, it means that it passed through the other hole, which doesn't lead to the pattern. So again, does this make "not detecting the electron" a true observation which collapses the wave function?

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Although a lot of these quantum interpretation questions are metaphysical, I think this one is sufficiently precise that it's fine. –  David Z Jul 7 '11 at 18:53
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This question is much less metaphysical than I expected from the title. –  C.R. Jul 8 '11 at 1:32
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3 Answers 3

I am going to give the straight Copenhagen answer, as did the other posters, but in the dual, «wave picture», which (in the Copenhagen view) is equivalent to the particle picture.

The electron is measured if it interacts with the detector or if it interacts with a, say, photographic emulsion located some distance away from the slits. There are two possibilities: if the detector at one slit measures it there, the wave function collapses (actually the electron is probably absorbed, as has been pointed out, but to make life interesting, suppose it has not been absorbed or otherwise destroyed). If not, then this negative result was still an interaction between the electron's matter wave and that detector, it simply was an interaction that did not result in a detection. But that is still a result, still an interaction, and so now the electron wave collapses to this degree: the components of the matter wave that would correspond to it's having been at the detector's slit are subtracted from the matter wave function. It diffracts through the other slit, however, and begins spreading. Now if it interacts with the photographic plate behind the two slits, it is measured again and the spot on the plate that develops localises the wave again so the electron is localised at that spot (and probably absorbed ...). The statistical distribution of those spots will not show interference because in all of those matter waves, the component associated to «passed through the other slit» were suppressed. And in all the electrons that were detected and passed through the other slit, their matter waves now only possess that component, which is orthogonal to all the rest, and that is why there is no interference.

Orthogonal things do not interfere, that is why I do not interfere with you: I am localised over here and so all those components are orthogonal to your wave function, which only has components associated with your being over there.

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The quantum formalism actually has a very nice feature for exactly the kind of question you're asking. They're called projection operators.

Usually, of course, you talk about measuring some Hermitian operator. But you can also take one of the eigenstates of the operator (or any state really) and turn it into a projection operator, which corresponds to asking the system "Are you in state $S$"? Given a state $|S \rangle$ the corresponding projection operator is simply $|S\rangle \langle S|$. You can check that its an observable with eigenvalues 1 (yes) and 0 (no). If you get "yes", then you know the state is $S$, but if you get "no" then you've projected the system onto the entire subspace of possibilities that are orthogonal to $S$.

You can also ask the system very specific kinds of questions. Say you have an electron in a 1-D well, and you know it is somewhere between x=0 and x=2. Suppose you want to ask "Are you somewhere between 0 and 1?" You can formalize this question as the sum (integral) of the projection operators $|x\rangle \langle x|$ between x=0 and x=1. If you get yes (1), then you know the particle is between 0 and 1, and if you get no (0) then you know the particle is somewhere between 1 and 2. So there is a great deal of freedom to ask the system very detailed questions, neglecting of course the practical difficulties which may be associated with these particular measurements. These are in fact the most general type of quantum mechanical measurement and they're called positive operator-valued measures (POVM).

Note: Sorry, I had to use parentheses for my bras and kets because the site didn't like the proper ones with arrows. They seemed to break everything.

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Yes, not detecting an electron is also a fully-qualified measurement, provided you make sure the measurement would have detected the electron, had it been there. That condition is usually not given in real quantum mechanics experiments. Keep in mind that "measuring an electron" actually means you let something interact with it; it's not possible to register an electron going through a slit in the way a hidden camera can register a car passing a bridge. Pretty much the only way to make sure an electron will be detected if it's there is to use a bulky solid-state detector, which will completely absorb the electron. If you have a box with an electron and such a detector and have not measured the electron yet, you can be sure that it is "still outside of the detector" and you will not measure it if you do another measurement very quickly after the first one, either (you need to be really quick though: the state "outside of the detector" is not an energy eigenstate and therefore unstable in time). If you have measured the electron, you can be sure it's not out in the box anymore, simply because it's stuck in the detector!

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You said that one has to use a bulky detector and do the experiment real quick. But what if I use a ZnS screen which always detects the presence of the electron if its there? (thereby repeating the experiment endlessly) –  Bernhard Heijstek Jul 8 '11 at 15:47
    
You can not repeat the experiment endlessly with a $\mathrm{Zn\,S}$ screen which always detects the presence of an electron. Such a screen is one of those "bulky solid state detectors", it will absorb the electron. –  leftaroundabout Jul 8 '11 at 18:50
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