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Expected Values in a Harmonic Oscillator

  1. Jul 1, 2008 #1
    Problem
    Show that in the [tex]n[/tex]th state of the harmonic oscillator


    [tex]\langle x^2 \rangle = (\Delta x)^2[/tex]
    [tex]\langle p^2 \rangle = (\Delta p)^2[/tex]​

    Solution
    This seems too simple... I'm not sure if it's correct...

    It is obvious that [tex]\langle x \rangle = 0[/tex]... this is true because the parity of the square of the eigenfunction is [tex]1[/tex] (in other words, the probabiliity density is an even function). Now, we know that [tex](\Delta x)^2 = \langle x^2 \rangle - \langle x \rangle ^2 [/tex], but [tex]\langle x \rangle = 0[/tex], so by substitution, the desired result follows. A similar argument can be made for the momentum. [tex]\blacksquare[/tex]
     
  2. jcsd
  3. Jul 1, 2008 #2

    G01

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    Homework Helper
    Gold Member

    You can always explicitly show that <x> = 0 in the nth state if you feel you need to show more work.

    To get started try representing x in terms of the raising and lowering operators in the following line, letting the operators act on any kets to their right, and simplifying:

    <n|x|n> = ... |n> is the eigenket for the nth eigenstate.


    You should end up with delta functions that have to be zero.
     
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