Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Double-slit experiment, thought experiment

  1. Jun 20, 2013 #1
    I'm trying to understand the implications of the double-slit experiment, and I have been considering what QM theory would predict as the outcome of the following experiment:

    Consider an electron gun firing electrons towards a detection screen two metres away (gun perpendicular to screen), and there is a blocking partition halfway between the gun and the detection screen. We shall consider the screen and the partition to be infinite planes, for simplicity.

    Now, we open up a slit in the partition, directly in front of the gun, whereupon we start to see electrons being detected on the screen. Now I know that if a second slit is opened up, the familiar diffraction pattern is seen, consistent with wave-like behaviour, and we can't really state which slit the electrons have travelled through. But what if the second slit is opened up, let's say, 500m along the screen from the first? I understand that any diffraction pattern would probably be almost unnoticeable, but let's say the instruments are extremely precise. My thinking is this: if the electrons "passing through" the second slit were having to travel sqrt(500^2 + 1) metres before getting to the screen, then at an average speed of, say 10^6 m s^-1, wouldn't we already notice some difference in arrival time? If the individual blips detected on the screen differed greatly in travel time, wouldn't this give us a clear indication of which slit the electron had travelled through?

    Thanks in advance, this subject is a little freaky to me!
  2. jcsd
  3. Jun 20, 2013 #2
    I would say that you won't obtain interferance pattern (when the slit is at 500 m). But If you ajust the gun in a way that it is in middle of the two silts (eg slits are at 250 m upo and down) then you will obtain an interferance pattern on the screen, but it will be very light.

    This might help you make the subject less freaky (I hope) http://ravimohan.me/2013/06/11/linear-vectors-and-quantum-mechanics-ii/
    Last edited: Jun 20, 2013
  4. Jun 23, 2013 #3
    Hmm...interesting. So if the double slit experiment is performed again, but this time instead of having a 500m gap between slits, you have a normal size gap (let's say 1m), will you only get the interference pattern if the electron gun is positioned half-way between the slits?
    I would posit that no, you will get the interference pattern if the gun is positioned in front of one of the slits. (Otherwise, how do you know you are exactly halfway between the slits? To what degree of accuracy?) So if you repeated the experiment and kept increasing the gap distance (with electron gun always in front of the first slit) there must be a point at which the interference pattern stops. What makes the interference pattern disappear above a certain limit? Let's ignore the technical difficulties and accuracy of measurement for the purposes of this argument.

    EDIT: By the way, I like the blog...
  5. Jun 23, 2013 #4
    No, I never said this :). I just answered the thought experiment designed by you
    The answer you gave was right
    But accuracy of instruments has nothing to do with this observation.

    Let me provide some mathematical basis for my answer. I assume you are comfortable with equation
    [itex]\langle y|s \rangle = \langle y|1 \rangle\langle 1|s \rangle + \langle y|2 \rangle\langle 2|s \rangle[/itex]
    [itex]\langle y|s \rangle = \varphi_1(y)\langle 1|s \rangle + \varphi_2(y)\langle 2|s \rangle[/itex]

    Now Feynman says the if a free particle in moving with momentum [itex]p[/itex] then the amplitude of the particle to go from [itex]\vec{r}_1[/itex] to [itex]\vec{r}_2[/itex] is given by
    [itex]\langle \vec{r}_2|\vec{r}_1 \rangle = \exp{\frac{\iota \vec{p}.\vec{r}_{12}}{\hbar}}/|\vec{r}_{12}|[/itex]

    Now in order to have appericiable inteference pattern, [itex]\langle[/itex][itex] 1[/itex][itex]|s \rangle[/itex] and [itex]\langle[/itex][itex] 2[/itex][itex]|s \rangle[/itex] must of comparable order which, further, depend inversely on the distance.
  6. Jun 23, 2013 #5
    Thanks for your reply - I can see that the diffraction pattern will not be "appreciable" due to the inverse ##r## relationship, but (correct me if I'm wrong) there will be some interference due to quantum effects, no matter how slight. They may not be significant enough to be noticed as the pattern of detected particles turns up on the screen (which is why I mentioned the accuracy of the equipment in showing the pattern), but mathematically the interference term is still there.

    Which brings me to the real point of my thought experiment: If the electron gun fires just one electron, then quantum theory (if I've understood correctly) claims that there will be interference despite only one electron being fired. (Which would build up into the interference pattern if we fired enough electrons). As far as I'm aware, it also claims that we cannot determine which slit the electron passed through. Now given the distance between the two slits, can we not use the speed of the electron leaving the gun and the distance between the electron gun and the screen to determine which slit it passed through, using a simple distance-speed-time relationship? Or would the uncertainty in the speed of the electron leaving the gun be large enough to make this unknown?

    Do you see what I'm getting at?
  7. Jun 23, 2013 #6


    User Avatar
    Science Advisor

    Only if the slit distance is smaller than the coherence length of whatever is fired at the double slit (photon, electron, buckyball, caterpillar, whatever). The coherence length/volume is roughly a measure of the distance/volume inside which all particles of the same kind are indistinguishable or equivalently the volume inside which a single particle may "hide" (over which it is delocalized in a coherent manner) at a certain time. This is why Young's original double slit experiment did not work just with sunlight. A single pinhole was needed to filter the light, giving the filtered light large spatial coherence which then in turn shows a decent interference pattern at reasonable slit distances.

    If you make the slit distance large enough, quantum mechanics predicts that there will be no interference pattern at all. This is why the sources are usually placed in the middle of the double slit. It makes the description way easier. You do not have to deal with phase shifts on the way to the slits, unequal intensities and so on.
  8. Jun 23, 2013 #7
    Without putting some measurement apparatus at slits, true.
    I dont think you can do calculations like this. You cannot assume an electron taking a well defined path and apply usual distance-speed-time relationship. That is classical concept. What you can evaluate is the odds (probability) of electron being detected at certain point at ceratin time when it was fired from certain point with certain momentum at certain time and for that you will have to use the number

    [itex]\langle \vec{r}_2(t = t_2)|\vec{r}_1(t = t_1) \rangle [/itex]
    which can be found from schrodingers equation :). I believe you trust this equation. If you do then tell me that [itex]\vec{r}_{12}[/itex] is fixed and still (in general) you are getting different probabilities (at different times) for for same initial and final points. Does the usual distance-speed-time relationship hold?

    Do you see what I am getting at :)
  9. Jun 23, 2013 #8
    You are right, but I believe that is not the case in totally isolated systems (there wont be any decoherence without external influences no matther how long the volume of the apparatus be. Of course slit wall-electron gun-electron interaction is not considered.)
    Last edited: Jun 23, 2013
  10. Jun 23, 2013 #9


    User Avatar
    Science Advisor

    Well, there is always decoherence with respect to the particle source. The particle emission time cannot be determined better than the coherence time of the particle is, which is why tomwilliam2's trick of backtracking via electron speed and slit position does not work: the exact emission time is not well defined. Unless you consider a completely delocalized electron in an almost permanent superposition of already being emitted and still being in the gun, you always have finite coherence properties, especially for electrons. You may consider rather delocalized photons, but even here, this is more of academic interest.

    For electrons, the assumption of no decoherence is pretty unphysical as they behave very particle-like (meaning that the coherence time and length scales are pretty short), which means that you can assume things like a certain emission time.
  11. Jun 24, 2013 #10
    Thanks very much both of you. I think the point about coherence length is a very good explanation of why the quantum effects wouldn't be seen with such a massive distance between slits.

    @Ravi: You said "I dont think you can do calculations like this. You cannot assume an electron taking a well defined path and apply usual distance-speed-time relationship."
    Yes, I know that treating the electron in a classical way is erroneous...but like anyone experiencing QM for the first time, I'm trying to see what the result that the classical approach gives, rather than just taking the tenets of QM on trust. I quite believe that I could predict probabilities accurately using the established theory of QM and Schrödinger's equation, but that is not what I'm trying to do. I'm trying to work out exactly where (and how) the classical theory breaks down. My textbook tells me that you can't determine which slit the electron went through unless you try to detect it there, so my first instinct is to think of ways of doing that and see why they fail. One idea was to give the two possible paths a huge difference in trajectory distance, so that you could tell which slit by how long the electron took to be detected. If we still saw quantum interference in this experiment it would surely then be possible to determine the trajectory! But as Cthugha says (this is actually what I suspected to be the case) you wouldn't be able to apply the speed-distance-time approach because the length scales involved will result in emission time being less well defined.

    So thanks for all your time - it is a very interesting subject, which I look forward to reading more about.
  12. Jun 24, 2013 #11
    Just curious, how do we exactly find the parameters like coherence length of the electron?
  13. Jun 24, 2013 #12


    User Avatar
    Science Advisor

    Good question. I am an optics guy. In optics that is simple: Coherence length is what is measured by a double slit. So you can place the source somewhere, vary the slit distance and watch the interference pattern disappear.

    However, coherence length is not an inherent property of the light field, but also of the setup in question. It is basically related to the inverse angular width of the source as seen at the double slit. A narrower range of emission angles arriving at the double slit means larger coherence length, so you can increase it by just placing it further away from the slit or filtering it.

    I assume the same procedure works for electrons as well. Vary the double slit distance and watch the interference pattern disappear and you will get your coherence length for the given geometry.
  14. Jun 24, 2013 #13
    Ok, thanks for explaining it. I am also eager to know how one can calculate coherence length theoretically.
  15. Jun 24, 2013 #14


    User Avatar
    Science Advisor

    Ok, there are longitudinal and transverse coherence length. For light, the longitudinal coherence length is just the coherence time times the speed of light. This is just related to the width of the Fourier transform of the power spectral density of the light field. This is not what you measure in a double slit, but in a Michelson interferometer.

    For transverse spatial coherence, it is easier expressed in terms of coherence area, which is defined as the squared mean wavelength of the light field divided by the solid angle of the source.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook