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Heisenberg momentum uncertainty

  1. Oct 8, 2014 #1
    1. The problem statement, all variables and given/known data
    An electron is trapped in a one-dimensional well of width 0.132 nm. The electron is in the n = 10 state.
    a) What is the energy of the electron?
    b) What is the uncertainty in the momentum?
    c) What is the uncertainty in the position?

    2. Relevant equations
    En = h2n2/8mL2
    Delta(x)Delta(p) > hbar/2

    3. The attempt at a solution
    The answers to this problem are in the back of our book, I got 2,160 eV for A, which was correct.
    The book gave us a hint for part B: use Delta(x)Delta(p) > hbar/2, but here I have one equation with two unknowns. (I tried using 0.132 nm as delta(x) but that didn't give the right answer. Also, why would they ask for delta(x) in part C if it was given... so

    I tried finding the momentum using

    p = (1/c)sqrt(E2-(mc2)2). And my answer for momentum is the same as the book's answer for the uncertainty in momentum. I guess I can't see how to relate delta(p) to p. Also, why doesn't the length that the electron is trapped in count for delta(x)? Thanks for reading,

    Lee
     
  2. jcsd
  3. Oct 8, 2014 #2

    td21

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    Hi!
    Uncertainity (S.D.) in momentum and position can be found from the square root of the expectation value of the (p - pbar)^2 and (x-xbar)^2, respectively. Note that pbar is zero and xbar is a/2 in 1 dimensional infinite square well.
    Yes, it happens that the S.D. of momentum is equal to momentum in this case. Can you guess the reason behind?
     
  4. Oct 8, 2014 #3
    I'm not 100% clear on what you mean.. I've never actually taken a statistics class yet. The formula my book uses is

    delta p = sqrt((p2)avg - (pav)2). If the average momentum is zero, and the formula reduces to:

    sqrt(p2)avg, would I just use the momentum I found with relativistic dynamics...and that's why the standard deviation is the same as the momentum value? What does that mean physically? Thanks for the reply.

    Lee
     
  5. Oct 9, 2014 #4
    Just noticed I got moved, I didn't think homework problems from a modern physics class where considered introductory physics, but I guess I will post all of my questions here from now on. Sorry about that.
     
  6. Oct 9, 2014 #5

    BvU

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    Never mind the moving to introductory. Perhaps they'll move it back once this is indeed about relativistic QM (it's not, I should think)..

    We're more used to writing ##p=i\hbar {\partial \over \partial x}## in simple QM, where E=p2/(2m) if V=0. Same difference. But <p> = 0 for sure.

    ##\Delta p\Delta x \ge \hbar/2## is not an equation but something else (an inequality). Why they provide it as a hint is a mystery to me (at n=10 it's a lot more than ##\hbar/2##).

    If your potential well is from 0 to L, I expect <x> to come out L/2 too.... Look up the wave function and check. Then do <x2>. You will find that ##\Delta x \propto L##, so L does count!
     
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