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Measuring x and ##p_{y}## precisely

  1. Mar 26, 2015 #1

    Maylis

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    1. The problem statement, all variables and given/known data
    upload_2015-3-26_23-32-13.png

    2. Relevant equations


    3. The attempt at a solution

    1.
    $$ [\hat {x}, \hat {p}_{y}] = \hat {x}\hat {p}_{y} - \hat {p}_{y} \hat {x}$$
    $$= \hat {x}(-i \hbar \frac {\partial f}{\partial y}) - (- i \hbar \frac {\partial (fx)}{\partial y}) $$
    Since x is not a function of y, it can be taken out of the derivative, so the commutator is zero.

    However, I have no clue how to answer the next two questions. I am very confused regarding measurements in quantum mechanics. Does it have something to do with Heisenberg Uncertainty principle?
     
  2. jcsd
  3. Mar 26, 2015 #2
    In this context it indeed does. It can be shown that the product of the imprecisions in the measurements of two observables A and B follows the Heisenberg uncertainty relation:

    $$\Delta A \Delta B \geq \frac{1}{2}\left | \left\langle \left[ \hat{A}, \hat{B} \right] \right\rangle \right |$$
    With this you can answer the second question.

    For the third one, do the exact same thing as you did for the first question: operate with the commutator on a test function f and see if you can rearrange the order of the derivatives so that the result is zero.
     
  4. Mar 30, 2015 #3

    Maylis

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    Since the commutator is zero, does that mean that I can measure it precisely? Because ##\sigma_{x} \sigma_{p_{y}} \ge 0##?
     
  5. Mar 30, 2015 #4
    That's right, their measurements are not constrained by the uncertainty relation.
     
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