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Momentum Measurement Confusion

  1. Sep 26, 2011 #1
    Hi everyone,

    I was thinking about what it means to talk about the "momentum of a particle" and I tried to come up with a little scheme to help me, but as of now it has only made more confusion... This is how my thoughts went.

    I know from quantum mechanics can one can relate classical quantities, like momentum, to expectation values. I was also under the impression that a particle never really has a value for momentum until it is measured, just a probability distribution; that the value of a particle's momentum isn't already existent, it comes about after measurement. With these in mind I was thinking about what
    \eta_{\alpha\beta}p^\alpha p^\beta =(mc)^2
    means quantum mechanically. Does it mean
    \langle p^0\rangle^2-\langle p^1\rangle^2-\langle p^2\rangle^2-\langle p^3\rangle^2=(mc)^2
    It seems to me this might be true, but that this isn't the whole story, and can't be the last word. This only says the average has to be equal to [itex](mc)^2[/itex], not the individual measurements, but I feel like if I took the square sum of the four measured values of an arbitrary particle that they would equal [itex](mc)^2[/itex] every time. So then I thought that perhaps the above formula applies to the individual measurements of a particle's momentum. Then [itex]p^\alpha =\langle p^0,\vec{p}\rangle[/itex] are measured values. But I also know something else about a particle's momentum: measurements of different components of a particle's momentum leave the other components alone, I believe [itex][\hat{p}_\alpha , \hat{p}_\beta ]=0[/itex] tells us this.

    Now here I take a step and assume that the equation [itex]\eta_{\alpha\beta}p^\alpha p^\beta =(mc)^2[/itex] holds true for individual measurements of a particles momentum. I also assume that a measurement of a particles momentum in the [itex]\alpha[/itex] direction does not affect a measurement of the particle's momentum in the [itex]\beta[/itex] direction. Then lets say I make measurements of a particles momentum for [itex]p_0, p_1, p_2[/itex], then using [itex]p_\alpha p^\alpha =(mc)^2[/itex] I can find [itex]p_3[/itex] without having to measure it. To me this seems to be saying one of two things: either

    1) After I measured the three components of the particle's momentum, the [itex]p_3[/itex] component was decided, in which case the previous measurements of the various components of the particle's momentum did affect the measurements of other components.


    2) the particle had [itex]p_3[/itex] to begin with.

    Neither of these seem correct to me. Where along this did my thinking go askew.

  2. jcsd
  3. Sep 30, 2011 #2


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    Science Advisor

    Of the four components of pμ only three are independent. You can specify a state by specifying the three-momentum, in which case p0 = E, is determined, since it is a function of the other three. The relation ημνpμpν is not just an expectation value, it's an operator relation, holding whether the pμ have been measured or not.
  4. Sep 30, 2011 #3
    Thanks Bill,

    That is just what Dr. Polyzou told me also. There seem to be some cool consequences for this then. Consider that if we measure two different spatial components, [itex]p_1, p_2 [/itex], then the relativistic equation reads
    p_{3}^2 - p_{0}^2 = p_{5}^2
    where [itex]p_{5}^2=-(mc)^2-p_{2}^2 - p_{1}^2 [/itex] and is a negative number. The this is the equation for a hyperbola, with [itex]p_{0}[/itex] on the x-axis . The cool part is that the values for [itex]-p_{5}< p_0 < p_5[/itex] are no longer accessible which means that the probability for these values in the probability distribution should be zero. So after each measurement, the probability distributions, lose in a sense, some of the values. For me this was a learning experience about the power of relativity even over probability.
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