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I Expectation for the Harmonic Oscillator ( using dirac)

  1. Jun 12, 2018 #1
    I've been trying to form a proof using , using majorly dirac notation.There has been claims that its much better to use in QM.

    The question i wanted to generally show that the expected value is Zero for all odd energy levels.I believe i have solved the question but im a bit Iffy about a step i took:

    <x>n = <Ψn|x|Ψn> = L
    for a given Ψn = (A+)n(n!)-2
    Energy eigen functions have definite parity, assume for all odd n's if one is zero the rest should follow.
    Take n = 1
    => L = <(A+)(n!)-2|x|(A+)(n!)-2>
    = (n!)-1 <(A+)|x|(A+)>
    B = <(A+)|x|(A+)>
    Define A+ = Lx + iC : B,C are Real
    => <Lx+iC|x|Lx+iC>
    (Bit iffy after these steps)
    = <Lx|x|Lx|> + <iC|x|iC>
    = <L|x3|L>+<C|x|C>
    as ∫x2n+1dx for limits [-∞,∞] and n =0,1,2,3....
    => 0
    we find B=0
    <x>n = 0
    For odd ns.
    Last edited: Jun 12, 2018
  2. jcsd
  3. Jun 12, 2018 #2


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    What about even values of ##n##? You are going about it backwards. Instead of replacing ##\psi_n##, write operator ##x## in terms of ##a^{\dagger}## and ##a## and use your knowledge of what ##a^{\dagger}|\psi_n>## and ##a|\psi_n>## are equal to.
  4. Jun 14, 2018 #3
    Surely you wouldnt get an actual value for even values.That would be extremely counter intuitive, that said I will try your advice.
  5. Jun 14, 2018 #4


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    You wrote a couple of things that don't make sense.
    This should be
    |\psi_n\rangle = \frac{(A^\dagger)^n}{\sqrt{n!}} | 0 \rangle

    The notation here doesn't work. You can't have an operator in a bra or a ket. You should end up with something like
    \langle 0 | A x A^\dagger |0 \rangle
    and so on.
  6. Jun 14, 2018 #5


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    Why is it so counter intuitive? In your proof, which needs fixing as @DrClaude suggested, you have the integral ∫x2n+1dx where n is odd. What would happen to this integral if n were even? Say n = 2k?
  7. Jun 14, 2018 #6


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    What do you mean by "energy", i.e. what is the Hamiltonian?
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