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Simple expectation value calculation

  1. Jan 15, 2013 #1
    1. The problem statement, all variables and given/known data

    This problem comes from the second edition of Griffiths's, Introduction to Quantum Mechanics.

    Given the Gaussian Distribution: p(x) = Aec(x-a)2

    find <x>, that is, the expectation (or mean) value of x.

    Clearly, to do this you evaluate the following integral: ∫xp(x)dx on (-∞,∞). I've tried performing this integration analytically, and it seems to be more bothersome than a rectal exam. The solution should be x = a since, by definition, p is centered at x = a (this is evident graphically as well). Is a qualitative solution sufficient here? Another method (which obviously does not work in general) is to set p'(x) = 0 and solve for x since we know the Gaussian distrubtion contains only one extremum. Any advice on the best means to proceed?


    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Jan 15, 2013 #2

    haruspex

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    ∫Axec(x-a)2.dx
    subst x+a for x:
    ∫A(x+a)ecx2.dx = ∫Axecx2.dx + a∫Aecx2.dx
    = (1/2)∫Aecx2.dx2 + a
     
  4. Jan 15, 2013 #3
    I tried this already. Wouldn't the differential be required to change also? That is, dx becomes d(x+a).
     
    Last edited: Jan 15, 2013
  5. Jan 15, 2013 #4
    EDIT* sorry, sorry, the differential is dx'/dx = d/dx[x+a], so dx' = dx.
     
    Last edited: Jan 15, 2013
  6. Jan 15, 2013 #5
    Now let's talk about <x2>, I seem to encounter an indeterminate form upon evaluating the integral at bounds +/- ∞ (see the integral of x2e-cx2)
     
  7. Jan 15, 2013 #6

    haruspex

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    Same substitution gets rid the -a. ∫x2e-cx2.dx (with a suitable multiplier to get total prob of 1) is just var(X), a standard result. To work it out for yourself, use integration by parts.
     
  8. Jan 15, 2013 #7
    I obtain the integral in the attached photo. Obviously there is a problem evaluating it at +/- ∞ because of the second (subtracted) term, no?
     

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  9. Jan 15, 2013 #8

    haruspex

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    The attachment is illegible for me. Much too small. Did you try integration by parts?
    ##\int_{-\infty}^{+\infty} x^2e^{-cx^2}.dx = [x \int^x ye^{-cy^2}.dy]_{-\infty}^{+\infty} - \int_{-\infty}^{+\infty} \left(\int^x ye^{-cy^2}.dy\right) dx ##
     
  10. Jan 15, 2013 #9
    I think you're misunderstanding me. Look here for the integral of interest (the one you're attempting to solve by parts): http://integral-table.com/ and you'll see the "problem" I'm encountering with evaluating the result at +/- infinity.
     
  11. Jan 15, 2013 #10

    haruspex

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    Number (70)? erf(-∞) = -1, erf(+∞) = +1. The other term vanishes at both extremes, leaving √(π/a)/(2a)
     
  12. Jan 15, 2013 #11
    Why does the other term vanish at both extremes? Isn't 0*infinity an indeterminate form?
     
  13. Jan 16, 2013 #12

    haruspex

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    e-x2 tends to zero much faster than x tends to infinity.
     
    Last edited: Jan 16, 2013
  14. Jan 16, 2013 #13
    There's an x multiplied by that too however, i.e. xecx2, so at each extreme it becomes +/- infinity * e^-infinity
     
    Last edited: Jan 16, 2013
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