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What is momentum in QM?

  1. Mar 11, 2005 #1


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    I know quantum mechanics is strange, but I'd like to think the strangeness is well defined. So can someone explain exactly what we mean when we talk about the momentum of a quantum particle? The reason I'm confused is that when the particle is in a position eigenstate, it has a definite position. If you take the measurement now, get a value, and then take it again later without disturbing the system in between, you'll get the same value. By heisenbergs principle, the uncertainty in momentum is infinite. But how can it be that there is any uncertainty in momentum when the particle is sitting perfectly still at the same spot? It's not moving, so what is this momentum it might have?

    Also, the usual qualitative way of describing uncertainty, (that you need to shine a very energetic photon to get an accurate position measurement, but that disturbs momentum, etc.) doesn't seem to make sense, because how can you disturb momentum but still maintain a fixed position?

    I know I must be thinking about momentum the wrong way, ie, too classically. Because mathematically, position and momentum are completely symmetrical in QM, but intuitively, momentum tells how the position is changing, but position is irrelevant to the momentum.
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  3. Mar 12, 2005 #2
    Sorry I have never hear of a "position eigenstate."
    What system is this true in? I don’t think it will be sitting perfectly still after measurement. How could you take a measurement and not impart some momentum to it? In what situation do you find a particle sitting perfectly still?

    It will not maintain a fixed position and never had a fixed position.

    If you have solved the SE of a particular system in position space you can transfer it to momentum space. Qualitatively it is like performing a Fourier transform.

    Some if you have a Gaussian represent the probability function of some particle you find that a narrow position will result in a wide momentum probability function and vs versa.

    For a particle holding still in one exact spot in one dimension. This can be represented by delta function. The Fourier transform of a delta function is a flat function with equal probability everywhere. So for a particle in one exact spot the momentum has an equal likelihood of being anywhere from minus infinity to positive infinity. This of course is not possible and neither is a perfectly still particle.

    Does this help?
  4. Mar 12, 2005 #3


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    As for momentum,it is the quantum observable associated to the classical one.If you can write the classical Hamiltonian of the system,then u need to apply the second postulate (the postulate of quantization) to it.

  5. Mar 12, 2005 #4
    Yes. I don't think you can understand what momentum in quantum mechanics is unless you understand the whole framework. If you do, you will understand Daniel's answer. If you don't understand the whole framework, then any answer I can think of won't really explain it that much, just give you more words for a concept that you won't understand.
  6. Mar 12, 2005 #5


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    The issue here isn't the uncertainty in a SINGLE measurement (this is limited to the resolution of your instrument and is NOT the uncertainty in QM). The issue here is your ability to predict the outcome of the next measurement, and the next, and the next, and the next....

    When you have made a measurement of the exact position, you destroy your ability (to some degree) to predict what you will get when you measure its momentum. You CAN measure it, and measure it with very high accuracy, but if you prepare an identical system, measure the SAME position, you will see that the momentum value you obtain will vary, and can vary widely! If you do this repeatedly, you will see a very large SPREAD in the value of the momentum. Based on this, you cannot say that you know very well what the NEXT value of momentum is going to be - thus, you have a large uncertainty in predicting the dynamics of the system.

    Again, I hate being tacky, but I have tackled this misconception about the HUP in one of my Journal entry, where I illustrated this using the single-slit measurement. You may want to read this rather than have me repeat the whole thing here.

  7. Mar 16, 2005 #6
    Your confusion is soundly motivated. Einstein had similar reservations.

    He proposed this: You measure the time at which a particle passes
    two different detectors, A and B. When it passes B, you now know
    what the particle's position AND momentum were all along the way
    between A and B.

    The problem really IS that you cannot determine momentum and position in
    the same place and at the same time.

    In QM, momentum is literally determined by measuring the distance
    between crests and troughs of the wave function, i.e. the wavelength.

    If you have a wave packet which is crammed into one tiny spot AND
    the wavelength is about the same size as the spot, you can't count
    a lot of crests/troughs, so the momentum cannot be determined accurately.

    At the other exteme, knwoing the momentum exactly means having an
    infinte number of crests/trhoughs to count. But that means the wave
    function is an infinitely long wave in space, so you loose all position

    This uncertainty principle is not magical. It happens in a radar when
    you try to determine the freqeucy of a very short pulse. If you can only
    count a few cycles, you can only determine frequecy with limited accuracy.
    Last edited: Mar 16, 2005
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