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Expectation Value <x^2>

  1. Oct 15, 2011 #1
    1. The problem statement, all variables and given/known data
    1. What is <x[itex]^{2}[/itex]>, in terms of position and expectation values.
    2. How can I use the correspondence principal to explain the quantum vs classical results (below).

    My textbook (Serway, Modern Physics) uses <x> as the expectation value, meaning the average position of a particle. It also uses <x[itex]^{2}[/itex]> in one example, and wants me to calculate it. I have calculated it by using the formula for <x>, but have no idea what it represents. The book does not define it, and I can't find it elsewhere.

    The problem is to compare my calculated classical <x> and <x[itex]^{2}[/itex]> values vs the books example calculating them for a quantum situation (particle in infinite square well of length L).



    2. Relevant equations
    <x> = [itex]\int[/itex]x (1/L)dx from 0 to L

    <x[itex]^{2}[/itex]> = = [itex]\int[/itex]x[itex]^{}[/itex] (1/L)dx from 0 to L




    3. The attempt at a solution
    The probability density (classical) is given as 1/L.
    My classical <x> agrees with that of the book's quantum = L/2

    My classical <x[itex]^{2}[/itex]> is L[itex]^{2}[/itex]/3 while the book's quantum value is L[itex]^{2}[/itex]/3 -L[itex]^{3}[/itex]/(2 pi[itex]^{2}[/itex]).

    I am also to use the correspondence principal to discuss the findings, but I cannot see how the quantum value would be altered (no n factor to increase) as the quantum world approaches the macro.

    Thank you for your time.
     
  2. jcsd
  3. Oct 15, 2011 #2

    vela

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    The expectation value of a function g(X) of a continuous random variable X is given by
    [tex]\langle g(x)\rangle = \int g(x) f(x)\,dx[/tex]where f(x) is the probability density function for X.

    For the classical case, you have f(x)=1/L, so
    \begin{align*}
    \langle x \rangle &= \int_0^L x \frac{1}{L}\,dx \\
    \langle x^2 \rangle &= \int_0^L x^2 \frac{1}{L}\,dx
    \end{align*}
    The only difference for the quantum case is that you now have f(x)=|ψ(x)|2.
     
  4. Oct 16, 2011 #3
    O.k. thank you. I used the equation with x[itex]^{2}[/itex] to arrive at my answer for the <x[itex]^{2}[/itex]> part of the question. So if <x> is the average position, <x[itex]^{2}[/itex]> is just the average of the position squared? I don't seem to understand the reason for the squaring. I assume <x[itex]^{2}[/itex]> means the same thing for the classical and quantum, like <x> = average position, in both.
     
  5. Oct 16, 2011 #4

    vela

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    Yes, you got it. As for the reason why you'd calculate <x2>, it's simply because the math calls for it. For example, the characteristic width of the wave function is [itex]\Delta x = \sqrt{\langle (x-\langle x \rangle)^2 \rangle} = \sqrt{\langle x^2 \rangle - \langle x \rangle^2}[/itex]. In plain old probability and statistics, this is how you'd calculate the standard deviation of a random distribution.
     
  6. Oct 16, 2011 #5
    Great, thanks for the help.

    I am still puzzled as to how I can use these results to "discuss" Bohr's correspondence principal. I see the classical and quantum results agree for the <x> value, which makes sense, because in in infinite well they are both forever confined within. However, I cannot explain the discrepancy in the <x[itex]^{2}[/itex]>value. There is no "n" quantum number for me to increase ad thus see what happens as the quantum scale approaches the classical.
     
  7. Oct 16, 2011 #6

    vela

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    Did your book calculate the expectation value for a particular value of n? If not, it's a typo. There should be an n in the answer.
     
  8. Oct 16, 2011 #7
    You are correct, the books quantum expression was for the ground state (n=1).
    There was an n in every answer, just hidden as n=1, in the denominator. So as n[itex]\rightarrow[/itex]infinity, both quantum terms go to zero. Thus predicting a zero average position and average position squared. Does this mean the particle is now localized at the origin?
     
  9. Oct 16, 2011 #8

    vela

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    That's not correct. In the limit as n goes to infinity, the quantum mechanical result should go to the classical result.
     
  10. Oct 17, 2011 #9
    I made a mistake in following the n through the books calculations. You are correct, the 2nd term drops out (in the <x^2>, so that it approaches macroscopic reality as n gets large.
     
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