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Wave vector collapse

  1. Feb 23, 2015 #1
    Hi everyone

    When a momentum operator followed by a position operator acts on a wave vector what does it give? (or the other wave around, changing the order)
    Is this the collapse of a wave function? And if so, can we solve this to predict the answer or not?

    I tried but got stuck in the math,
    It is difficult to show the math here.. but the procedure that i used is:
    assume a state that is a superposition of both position eigen state and momentum eigen state. (I hope that can form a state, i mean it should)
    Now when this state is acted upon by the position operator: well it should pick out the position state (collapse)
    but actually the momentum state is fourier transform of X vector, so X° can operate on that too...
    and then P° can act on it...

    Does this seem fine? I was expecting X° will pick out X state and then P° will have to act on that.... like collapse..
    I am guessing my idea of a collapse of wave function is flawed.
    (I studied QM using the Schrodinger's wave function approach and these kets n operators confuse me..)
    I will be happy if someone could help me with these ideas and point out exactly what is wrong.

  2. jcsd
  3. Feb 23, 2015 #2


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    Any position eigenstate is a superposition of momentum eigenstates, and vice versa. So this state exists, but it can be written as a superposition of just position eigenstates or just momentum eigenstates by expanding the position eigenstate in terms of momentum eigenstates or vice versa. One way it's easy to calculate what will happen when you apply the position operator to the state, the other way it's easy to calculate what will happen when you apply the momentum operator to the state, but it's the same state either way so you will choose whichever representation is easy for what you're going to do.

    No, because
  4. Feb 23, 2015 #3


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    Really you need to see an axiomatic treatment of QM.

    Get a hold of Ballentine - Quantum Mechanics - A Modern Development:

    Read the first three chapters and all will be clear.

    But basically its a simple understanding of the two axioms of QM that I will state:

    Axiom 1
    Associated with each measurement we can find a Hermitian operator O, called the observations observable, such that the possible outcomes of the observation are its eigenvalues

    Axiom 2 - called the Born Rule
    Associated with any system is a positive operator of unit trace, P, called the state of the system, such that expected value of of the outcomes of the observation is Trace (PO).

    In fact they can be reduced even further to one axiom using a very beautiful theorem called Gleason's Theorem:

    See post 137.

    But that is just by the by out of interest.

  5. Feb 24, 2015 #4
    Sure. Take a plane wave, split the beam and pass one part through a converging lens. Recombine with the original beam. Where the light comes to a focus, the wavefunction is a summation of positional and momentum states as required.

    I am hesitant to call it a superposition as some people prefer to reserve that term for a summation of eigenstates of the same operator. In fact I think that is where your problem lies - you specifically added eigenstates of two different operators. Such eigenstates are not orthogonal so your observation cannot be expected to "pick out" one component over the others. Hence Nugatory's statement that you need to decompose the momentum state into a superposition of positional states. A real-world observation does this for you: the screen shows a background patch of even illumination (the plane wave or momentum state) plus a bright spot in the middle (the focussed wave or positional state).
  6. Feb 24, 2015 #5


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    Never seen any textbook do that, and such would be false.

    The principle of superposition applies to any pure states and conversely any pure state can be decomposed in an infinite number of ways into the superposition of other states. It simply reflects pure states vector space structure.

  7. Feb 24, 2015 #6
    okk... So to be precise I shouldn't call such a state : a superposition of state.
    Leaving that aside, I want to write an equation that says when a measurement (of either position or momentum) is made the wave vector collapses.. how is that written and solved?
    I was assuming that by the action of an operator on a general state we should be left with the eigenstate of that operator. But if i can't have a general state (which has not been measured before) I am no longer sure about how to write such an equation.
    I was reading about that preparation of a state vector is position type and the basis of the vector for measurement is momentum type for double slit experiment. Does this answer my question why I can't have a general state (because only one will be prepared at a time.)?

    I will go through the Book mentioned, but until then I will appreciate if someone could explain it.
  8. Feb 24, 2015 #7


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    I can write a given state as a sum of position eigenstates or as a sum of momentum eigenstates but it's the same state either way, just as ##(\vec{N}+\vec{E})## is the same vector as ##\sqrt{2}(\vec{NE})##.

    If I apply the position operator to a given state, I will get collapse to a position eigenstate and the result of my position measurement will be the corresponding position eigenvalue; if I apply the momentum operator to that state I will get collapse to a momentum eigenstate and the result of my momentum measurement will be the corresponding momentum eigenvalue. Thus, if I want to calculate the probability of getting a particular result from a position measurement, the first thing I will do is do whatever math is needed to write the state as a sum of position eigenstates, and likewise for momentum.
    (Warning - I have run roughshod glossed over some important mathematical subtleties of the continuous-spectrum position and momentum operators here).

    State preparation works the same way. If I prepare a bunch of electrons by passing them through a vertically-oriented Stern-Gerlach device and keep only the ones that were deflected upwards, it's easy to write the resulting prepared state as ##|up\rangle##. However, if I'm then going to measure their spin along some other axis, I'm going to do some math to rewrite that state in a basis more appropriate for that calculation. For example, if I want to know how they'll behave if passed through another SG device that is oriented horizontally, I'll take advantage of the fact that ##|up\rangle=(|left\rangle+|right\rangle)/\sqrt{2}##. Either way, it's the same state and it's telling me that particles prepared in this way have a 100% chance of being spin-up and a 50/50 chance of being spin-left or spin-right.
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