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Inside the Wavefunction

  1. Jun 7, 2007 #1
    "Inside the Wavefunction"


    I was thinking about the wavefunction when I intercepted an interesting thought experiment that I couldn't quite formulate the answer to:

    Say that we imagine a small particle that is small enough so that it is about the size of an electron (the particle having no charge). Let's say that we want to "explore" the inside of the wavefunction with this particle by inserting it into the probability cloud that surrounds the nucleus of the hydrogen atom. By signaling off another particle when struck by another electron, this particle should reveal whether or not it has come in contact with an electron. If we insert the particle in a tiny bit of the probability sphere, and if for instance, we receive no signal, then we can identify that the electron was not in the same place that our particle entered.

    Given that the absence of a signal is information about the electron, the wavefunction should effectively collapse in two perspectives: one belonging to us, the observers, and one belonging to the electron itself. To us, we can identify that the electron was not in the same place as our detector particle, but because we have only tested a very small amount of probabilities in the cloud available for us to try, the electron should still be in a state of uncertainty relative to us. Whereas our knowledge of the electron improves very little, the electron itself has its own collapse, forming the other perspective. In this perspective, the electron has been collapsed to a path that was not testable in the observer's apparatus.

    My question is this: after our first "measurement", does the probability still exist in the observer's perspective and in the electon's perspective? So, even though we have collected data about one are of the cloud, the other areas are still unknown, and that portion of the probability cloud should be in a state of uncertainty. But for the electron, it has collapsed itself, and has settled into an eigenstate, and has no more interference. Given these two scenarios, is there an effective paradox, what can we observers say about the probability cloud now? Thanks for any comments,

  2. jcsd
  3. Jun 8, 2007 #2


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    A lot of what you say isn't meaningful in QM terms. You haven't defined what you mean by 'signalling', or 'come into contact with an electron' or 'insert the particle in a tiny bit of the probabilty sphere'.

    For example, if you want to localise a particle in a region 'tiny' compared to the size of the 'probability cloud' ( which actually doesn't have a size as in the 'size' of a grapefruit)
    then the particle must necessarily have far higher energy that the system it's 'exploring' and will probably just blow it apart.

    I hope I'm not being being too disparaging, it's good to think imaginatively about these things. But the notions you're used to in classical mechanics, like 'position' and 'momentum' no longer apply. Keep on studying QM and you'll get the quantum picture.
    Last edited: Jun 8, 2007
  4. Jun 8, 2007 #3


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    In addition to what Mentz said, I also have a feeling that you don't know what a QM "wavefunction" is.

  5. Jun 9, 2007 #4
    Hi regent,

    That's a good question. In general, if you really do only test a small region of position space, you will *not* collapse the wavefunction into a position eigenstate. In other words, your view of the situation from "the observer's perspective" is the correct one.

    In detail, what happens here is that you are performing a measurement, but not the measurement corresponding to the position operator x. Instead, the operator is more like
    [tex] x P(A) = \sum_{x \in A} x \ |x><x| [/tex]
    where A is the region you're looking in, and P(A) is the projection operator onto that region. The right-hand is just the normal expansion of the position operator in the position eigenbasis, except we restrict the sum to the locations in the subset of space we're looking in.

    As you can see, the eigenstates of this operator includ the usual states |x> with values x within the region A, but all position eigenstates |x> outside A are degenerate eigenstates with eigenvalue 0.

    When you do such a measurement, what happens as usual is that the state of the system gets projected onto the eigenstate(s) corresponding to the measured eigenvalue. If you get a nonzero result from your measurement (ie we see the particle in the region we measured) then the system is the corresponding position eigenstate |x>. If we don't see the particle, then the particle gets projected onto the eigenstate(s) for the zero eigenvalue, [tex]\sum_{x \notin A} |x>[/tex].

    More simply put, if we don't see the particle, its wavefunction gets "zeroed out" in A. ie, it goes to:
    [tex]\psi(x) = \left\{ \begin{array}{cc} 0,&\mbox{ if } x \in A \\ \psi_0(x), & \mbox{ otherwise }\end{array}\right. [/tex]

    If you think about it, this is exactly analagous to what happens when we, say, measure the position of a particle without measuring its spin; or we measure just the first particle in an entangled two-particle system. We just project onto the eigenstates of that observable. To visualize it consider measuring just the z-component of a particle's position. The resultant wavefunction will be just a single horization "slice" of the original wavefunction at the measured z-value.... Everywhere else it will be zero. Similarly, in your example, not finding the particle in the region A "zeroes out" the wavefunction in that area.
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