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Expected value for p

  1. Sep 7, 2015 #1
    I have done a lot of counts but I'm sure that there will be a quicker way.. Can you suggest me?

    I have followed the way ##\bar{p}=-i \hbar\int \psi(x)* \cdot \frac{\partial}{\partial x} \psi(x) \ dx##...

    Can I follow a quicker way?
     
  2. jcsd
  3. Sep 7, 2015 #2

    blue_leaf77

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    Is that integral too lengthy to compute?
     
  4. Sep 7, 2015 #3
    It is, but what it asks me to compute after that is even longer.. ##\bar {p}^2##!!

    Can't I express the state at t=0 as: ##| \psi(x,0)>=\frac{1}{\sqrt2} (|1>+e^{i \gamma} |2>##?
     
  5. Sep 7, 2015 #4

    blue_leaf77

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    How is it lengthy, it's just integral of product sinusoids, you can make use of their (anti)symmetry property to eliminate some terms without actually calculating it.
     
  6. Sep 7, 2015 #5
    mmh.. for example?
     
  7. Sep 7, 2015 #6

    Avodyne

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    ##p^2## is much easier than ##p##. I just computed its expectation value in my head.

    This is because ##p^2## is a special operator for this problem. It's specialness is related to the definition of ##\phi_n##.
     
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