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Expectation values r and x for electron in H2 ground state

  1. May 13, 2015 #1

    Maylis

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    1. The problem statement, all variables and given/known data
    upload_2015-5-13_23-4-52.png

    2. Relevant equations
    $$ \psi_{100} = \frac {1}{\sqrt{\pi a^{3}}} e^{-r/a} $$

    3. The attempt at a solution
    a)
    $$\langle r \rangle = \frac {1}{\pi a^{3}} \int_0^{2 \pi} d \phi \int_{0}^\pi d \theta \int_0^{\infty} r^{3} e^{-2r/a} dr$$
    This comes out to be ##\frac {3}{2}a##

    $$\langle r^{2} \rangle = \frac {1}{\pi a^{3}} \int_0^{2 \pi} d \phi \int_{0}^\pi d \theta \int_0^{\infty} r^{4} e^{-2r/a} dr$$
    Which comes out as ##3a^{2}##

    b)
    I know ##r^{2} = 3x^{2}##, so the answer for the expectation value of ##x^{2}## is one third the expectation value of ##r^{2}##, therefore ##\langle x^{2} \rangle = a^{2}##

    However, I am confused how to find ##\langle x \rangle##. Do I just say ##x = r sin \theta##, therefore ##dx = sin \theta dr + r cos \theta d \theta##?
     
    Last edited: May 13, 2015
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  3. May 13, 2015 #2

    mfb

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    <x> is easy to find by symmetry, but you can calculate the integral if you want.
     
  4. May 13, 2015 #3

    Maylis

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    Yes, I am confused how to exploit the symmetry
     
  5. May 13, 2015 #4

    mfb

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    Do you expect the electron to be on average more on the left or the right side? Does that question even make sense as you don't know where left and right are?
     
  6. May 13, 2015 #5

    Maylis

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    Okay, I suppose that makes sense, but then why would the square x coordinate not be zero then? By the same reasoning
     
  7. May 14, 2015 #6

    PeroK

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    Because a measurement of ##x^2## is positive so the expected value can't be 0.
     
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