- #1
jfy4
- 649
- 3
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
[tex]
\eta_{\alpha\beta}p^\alpha p^\beta =(mc)^2
[/tex]
means quantum mechanically. Does it mean
[tex]
\langle p^0\rangle^2-\langle p^1\rangle^2-\langle p^2\rangle^2-\langle p^3\rangle^2=(mc)^2
[/tex]
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 let's 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.
or
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.
Thanks,
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
[tex]
\eta_{\alpha\beta}p^\alpha p^\beta =(mc)^2
[/tex]
means quantum mechanically. Does it mean
[tex]
\langle p^0\rangle^2-\langle p^1\rangle^2-\langle p^2\rangle^2-\langle p^3\rangle^2=(mc)^2
[/tex]
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 let's 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.
or
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.
Thanks,