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Harmonic oscillator kinetic energy

  1. Jun 6, 2009 #1
    1. The problem statement, all variables and given/known data

    Calculate the expected value of the kinetic energy being

    [tex]\varphi(x,0)=\frac{1}{\sqrt{3}}\Phi_0+\frac{1}{\sqrt{3}}\Phi_2-\frac{1}{\sqrt{3}}\Phi_3[/tex]

    2. Relevant equations

    [tex]K=\frac{P^2}{2m}[/tex]

    3. The attempt at a solution

    I tried to solve it using two diffrent methods and they don't give the same result, so something is wrong with one or both of them.
    The first method is using matrix representations. I use the matrix of P and generate P^2.
    Then I compute

    [tex]\varphi(x,0)^*,P^2,\varphi(x,0)[/tex]

    and end up with an expresion like

    [tex]-3+2/3*\sqrt{2}cos (2wt)[/tex]

    The second method is using the creation and annhilitation operators a+ and a.

    [tex]P^2=a^{+}a^{+}+aa-a^{+}a-aa^{+}[/tex]

    [tex]\phi(x)_na^{+}a^{+}\phi(x)_m=\sqrt{(n+1)(n+2)}\phi(x)_{n,m+2}[/tex]
    [tex]\phi(x)_naa\phi(x)_m=\sqrt{(n-1)(n)}\phi(x)_{n,m-2}[/tex]
    [tex]\phi(x)_na^{+}a\phi(x)_m=n\phi(x)_{n,m}[/tex]
    [tex]\phi(x)_naa^{+}\phi(x)_m=(n+1)\phi(x)_{n,m}[/tex]

    But here I only get values for

    [tex]\phi(x)_0,\phi(x)_0;\phi(x)_2,\phi(x)_2;\phi(x)_3,\phi(x)_3[/tex]
    those are giving a number and
    [tex]\phi(x)_2,\phi(x)_0[/tex]
    which is giving an expression like a*exp(2jwt)

    The question is: where is the other exp(-2jwt) to form the cos(2wt) of the other solution and why I don't get the same result ... I guess I have to normalise somewhere.
     
  2. jcsd
  3. Jun 7, 2009 #2

    Cyosis

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    Homework Helper

    How do you get a time dependence in your answer? You want to calculate the kinetic energy of a harmonic oscillator so using the creation and annihilation operator is a smart thing to do.
    Ignoring all constants, you want to calculate [itex]\langle \varphi |p^2|\varphi \rangle[/itex]. You know that, assuming [itex]\phi_n[/itex] is normalized, that [itex]\langle \phi_n |\phi_m\rangle=\delta_{nm}[/itex]. Therefore every term will just be a number.

    How did you get this?
     
  4. Jun 7, 2009 #3
    You are right, the energy should be constant in time.

    Then

    [tex]
    \phi(x)_0a^{+}a^{+}\phi(x)_2=0
    [/tex]

    because the functions of the basis are orthonormal?
     
  5. Jun 7, 2009 #4

    Cyosis

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    Homework Helper

    Yes that is correct. On top of that [itex]a^{+}a^{+}\phi(x)_2[/itex] would yield [itex]const*\phi(x)_4[/itex], which isn't even an allowed state in this problem. Same for [itex]a \phi_0[/itex] etc.
     
  6. Jun 7, 2009 #5
    yes, i wanted to put

    [tex]

    \phi(x)_2a^{+}a^{+}\phi(x)_0=0

    [/tex]

    this gives an allowed state right?
     
  7. Jun 7, 2009 #6

    Cyosis

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    Homework Helper

    Yes, because [itex]a^{+}a^{+}\phi(x)_0=const*\phi(x)_2[/itex] and [itex]\phi(x)_2[/itex] is a part of [itex]\varphi(x,0)[/itex]. The entire thing should yield [itex]\sqrt{2}[/itex].
     
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