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Expectation values and commutation relations

  1. Mar 25, 2017 #1
    1. The problem statement, all variables and given/known data
    I am trying to calculate the expectation value of ##\hat{P}^3## for the harmonic oscillator in energy eigenstate ##|n\rangle##

    2. Relevant equations


    3. The attempt at a solution

    ##\hat{P}^3 = (i \sqrt{\frac{\hbar \omega m}{2}} (\hat{a}^\dagger - \hat{a}))^3 = -i(\frac{\hbar \omega m}{2})^{\frac{3}{2}}(\hat{a}^\dagger \hat{a}^\dagger \hat{a}^\dagger + \hat{a} \hat{a}\hat{a}^\dagger - \hat{a} \hat{a}^\dagger \hat{a}^\dagger - \hat{a}^\dagger \hat{a} \hat{a}^\dagger - \hat{a}^\dagger \hat{a}^\dagger \hat{a} - \hat{a} \hat{a} \hat{a} + \hat{a} \hat{a}^\dagger \hat{a} + \hat{a}^\dagger \hat{a} \hat{a})##

    I have used the trick of adding a zero such that ##\hat{a} \hat{a} \hat{a}^\dagger = \hat{a} \hat{a} \hat{a}^\dagger - \hat{a} \hat{a}^\dagger \hat{a} + \hat{a} \hat{a}^\dagger \hat{a} = \hat{a} [\hat{a}, \hat{a}^\dagger] + \hat{a} \hat{a}^\dagger \hat{a} = \hat{a} + \hat{a} \hat{a}^\dagger \hat{a}## on the terms ##\hat a \hat a \hat{a}^\dagger, \hat a \hat{a}^\dagger \hat{a}^\dagger, \hat{a}^\dagger \hat{a}^\dagger \hat a, \hat{a}^\dagger \hat a \hat a## and ended up with the expression (dropping the prefactors)

    ##\hat P^3 \propto (\hat{a}^\dagger \hat{a}^\dagger \hat{a}^\dagger + 3 \hat a \hat{a}^\dagger \hat a - 3 \hat{a}^\dagger \hat a \hat{a}^\dagger + \hat a \hat a \hat a)##

    which I don't think can be simplified further using the trick above. Are there any tricks I can use to tackle the central and outside terms?

    Although it looks like the expectation value of all that will be zero due to the orthonormality of states...
     
  2. jcsd
  3. Mar 25, 2017 #2

    kuruman

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    Don't look for tricks. Start with the expanded 8-term expression you have and throw out the operators that you know give zero expectation values. Remember that these are also known as ladder operators. What is the result when they operate on | n >?
     
    Last edited: Mar 25, 2017
  4. Mar 25, 2017 #3
    ##\hat a |n\rangle = \sqrt{n} |n-1\rangle##

    ##\hat a \hat a |n\rangle = \sqrt{n} \hat a |n-1\rangle = \sqrt{n} \sqrt{n-1} |n-2\rangle##

    ##\hat a \hat a \hat a |n\rangle = \sqrt{n} \sqrt{n-1} \sqrt{n-2} |n-3\rangle##

    ##\langle n| \hat a \hat a \hat a |n\rangle = \sqrt{n} \sqrt{n-1} \sqrt{n-2} \langle n|n-3\rangle = 0## due to the orthonormality of states.

    I think all of the others go to zero as well, since all the combinations involve uneven numbers of the creation and annihilation operators, which leads to the scalar product of a state with a different state, all of which are orthonormal.
     
  5. Mar 25, 2017 #4

    kuruman

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    You got it. :smile:
     
  6. Mar 25, 2017 #5
    awesome, thanks for your help.

    I'm still getting to grips with Dirac notation, but it's benefits are immediately obvious.
     
  7. Mar 25, 2017 #6

    kuruman

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    Yesss. It allows you to integrate without actually doing any integrals (in some cases).
     
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