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- Summary
- This is my first semester of quantum mechanics, reading Griffith's intro book, and I'm confused. I seek to clarify some misconceptions/lack of conceptions regarding <x> and <p>. I can't bring myself to move forward if I can't get a clear physical description of these two. How am I going to understand energy if I can't understand momentum. Losing sleep. A lot to unpack here, so I really appreciate the help anyone can spare.

Here's my understanding. Please let me know what I'm missing or what I've misunderstood. <x> represents the average of position measurements on some N, identically prepared particles. If those measurements are thought of as a distribution, then that distribution's form should match the wave function's probability density; the bigger N is, the better the match. If this is true, then any laboratory measurement of <x> (as described above) should be consistent with the <x> that is mathematically calculated. To me, <x> is a sort of composite image of a particle's position at some instant in time. I don't understand why <x> is an important quantity? It's astonishing that It can be measured and i'm sure it forms a foundation of more interesting theory I'll learn later. But, is that the limit of <x>'s function? Or is there something more insightful about <x> that I can understand now? What does <x> enable us to do/understand? Also, how is a particle's position measured in experiment, and is it precise?

<x> can change with respect to time, so the derivative of <x> can yield some nonzero value. By definition d<x>/dt which I'll call <v> represents the instantaneous rate of change of <x>. It's a measure of how fast the average of x changes, and seems unrelated to classical velocity. Griffiths makes it sound that particle's don't really have velocities anyway. That they can't be measured. It seems that <v> is not like <x> in that <x> can be verified by experiment.

What really, really confuses me is <p> = d<x>/dt * m. If my definition of <v> is correct, then it seems reasonable to want to measure it. But <p> seems like a weird thing to calculate. Perhaps it's because I'm thinking classically, but it's difficult not to when the operators look so similar to their classical counterparts. Is <p> what we would expect if we measured the momentum in N identically prepared experiments and took their average (how is this possible if <v> can't be measured in that way)? Is it a measure of the rate of change of <x> scaled by mass? Or are those two things one and the same?

Thanks a bunch.

<x> can change with respect to time, so the derivative of <x> can yield some nonzero value. By definition d<x>/dt which I'll call <v> represents the instantaneous rate of change of <x>. It's a measure of how fast the average of x changes, and seems unrelated to classical velocity. Griffiths makes it sound that particle's don't really have velocities anyway. That they can't be measured. It seems that <v> is not like <x> in that <x> can be verified by experiment.

What really, really confuses me is <p> = d<x>/dt * m. If my definition of <v> is correct, then it seems reasonable to want to measure it. But <p> seems like a weird thing to calculate. Perhaps it's because I'm thinking classically, but it's difficult not to when the operators look so similar to their classical counterparts. Is <p> what we would expect if we measured the momentum in N identically prepared experiments and took their average (how is this possible if <v> can't be measured in that way)? Is it a measure of the rate of change of <x> scaled by mass? Or are those two things one and the same?

Thanks a bunch.