Momentum Measurement Confusion

[jfy4]

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 $(mc)^2$, 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 $(mc)^2$ every time. So then I thought that perhaps the above formula applies to the individual measurements of a particle's momentum. Then $p^\alpha =\langle p^0,\vec{p}\rangle$ 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 $[\hat{p}_\alpha , \hat{p}_\beta ]=0$ tells us this.

Now here I take a step and assume that the equation $\eta_{\alpha\beta}p^\alpha p^\beta =(mc)^2$ holds true for individual measurements of a particles momentum. I also assume that a measurement of a particles momentum in the $\alpha$ direction does not affect a measurement of the particle's momentum in the $\beta$ direction. Then let's say I make measurements of a particles momentum for $p_0, p_1, p_2$, then using $p_\alpha p^\alpha =(mc)^2$ I can find $p_3$ 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 $p_3$ 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.

or

2) the particle had $p_3$ to begin with.

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

Thanks,

 

[Bill_K]

Of the four components of p_μ 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 η^μνp_μp_ν is not just an expectation value, it's an operator relation, holding whether the p_μ have been measured or not.

 

[jfy4]

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, $p_1, p_2$, then the relativistic equation reads
$$p_{3}^2 - p_{0}^2 = p_{5}^2$$
where $p_{5}^2=-(mc)^2-p_{2}^2 - p_{1}^2$ and is a negative number. The this is the equation for a hyperbola, with $p_{0}$ on the x-axis . The cool part is that the values for $-p_{5}< p_0 < p_5$ 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.