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Momentum Operator Derivation Questions

  1. Sep 3, 2015 #1
    Hello,

    This is probably a very easy questions about the one-dimension momentum operator derivation. So you take the d<x>/dt to find the "velocity" of the expectation value. At one point in the derivation early on, you bring in the d/dt into the integral of the expectation value. The book I'm going off of just basically takes the x out of the derivative because I'm guessing it isn't a function of time.

    My question is, how do I know it's not a function of time? When I was trying to do the derivation alone before I looked at the book I used the chain rule here, and obviously made a mess. How do you know that x isn't really x(t)? I dunno I used to make this mistake in classical mechanics, and always assume x or theta wasn't a function of time, so I wouldn't treat them accordingly when doing derivatives. How do I know this x is different?

    If anyone would like to help me out here, but wants to see more lines of the derivation I could post them. I just didn't take the time now because I again forget all the latex I learned from the last time I used it, probably a year ago. Just let me know if it would help you out.

    Thanks anyone for your time.
     
  2. jcsd
  3. Sep 3, 2015 #2

    Geofleur

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    This is the difference between a theory of point particles and a field theory. In classical mechanics, we have a point particle with trajectory ## \textbf{r}(t) ##, while in quantum mechanics we have a wave function ## \Psi(\textbf{r},t) ##. The ## \textbf{r} ## is not a function of time here - the wave function is a field-like quantity and it depends on ## \textbf{r} ## as well as ## t ##. To get the probability of finding a particle in a volume ## d^3\textbf{r} ##, you use ## \Psi ## to get a probability field ## | \Psi |^2 d^3\textbf{r}##. ## \textbf{r} ## just marks the position at which you are considering such probabilities, and it can be anywhere in space.
     
  4. Sep 3, 2015 #3
    Ok wow, after reading your response a few times it makes A LOT of sense. Thanks! I think it answers a lot of other little questions I had about earlier stuff too. I just never had the berries to ask when I probably should have.
     
  5. Sep 4, 2015 #4
    If x doesn't depend on t, how can its average value <x> depend on it?
    (It's to understand better what x means in this context).

    Edit. I ask this because don't know if the OP has clear that "x" here means two different things: "operator" in on case and "variable" in another.

    --
    lightarrow
     
    Last edited: Sep 4, 2015
  6. Sep 4, 2015 #5
    I see the difference after Geoflur's post. I hadn't noticed it was all in the notation of the wave function pretty much.
     
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