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Matrix elements in the quantum oscillator

  1. Oct 5, 2014 #1
    1. The problem statement, all variables and given/known data

    For a quantum oscillator find all non-zero matrix elements of the operators ##\hat{x}^3## and ##\hat{x}^4##

    2. Relevant equations

    ##\hat{x} = \sqrt{\frac{\hbar}{2m\omega}}\left(a+a^\dagger\right)##

    ##a^\dagger |n\rangle = \sqrt{n+1}|n+1\rangle##
    ##a |n\rangle = \sqrt{n}|n-1\rangle##

    3. The attempt at a solution
    Okay so first it is necessary to compute ## \langle n |\hat{x}^3|n'\rangle## and then from that outcome determine the non-zero elements.

    $$x^3 = \left(\frac{\hbar}{2m\omega}\right)^{3/2}(a + a^\dagger)^3 =\left(\frac{\hbar}{2m\omega}\right)^{3/2}(a + a^\dagger)(a _ + a^\dagger)(a + a^\dagger) = $$

    $$=\left(\frac{\hbar}{2m\omega}\right)^{3/2}(a a a + a^\dagger a a + a a^\dagger a + a^\dagger a^\dagger a + a a a^\dagger + a^\dagger a a^\dagger + a a^\dagger a^\dagger + a^\dagger a^\dagger a^\dagger) $$

    So then we compute

    $$\langle n|\hat{x}^3|n' \rangle = $$
    $$\left(\frac{\hbar}{2m\omega}\right)^{3/2} \langle n|(a a a + a^\dagger a a + a a^\dagger a + a^\dagger a^\dagger a + a a a^\dagger + a^\dagger a a^\dagger + a a^\dagger a^\dagger + a^\dagger a^\dagger a^\dagger)|n'\rangle$$

    And then from here I am having trouble evaluating this, however, I am pretty sure that the majority of the terms above go to zero.

    Any help would be greatly appreciated.
     
  2. jcsd
  3. Oct 6, 2014 #2
    You just have to evaluate the action of the operators on the states sequentially. Lets take the term [itex]a a^{\dagger} a[/itex] for example:
    [tex]\left\langle n\right| a a^{\dagger} a \left|n'\right \rangle = \left\langle n\right| a a^{\dagger} \sqrt{n'} \left|n' - 1\right \rangle
    = \left\langle n\right| a \sqrt{n'} \sqrt{n'} \left|n'\right \rangle
    = \left\langle n\right| \sqrt{n'} \sqrt{n'} \sqrt{n'} \left|n'-1\right \rangle
    = \left(n'\right)^{3/2} \delta_{n,n'-1}
    [/tex]
     
  4. Oct 6, 2014 #3
    Thanks for your response. Why wouldn't the above be: $$\langle n|aa^\dagger a|n'\rangle = \langle n|aa^\dagger \sqrt{n'}|n'-1\rangle = \langle n|a\sqrt{n'+1}\sqrt{n'}|n'\rangle = \langle n|\sqrt{n'+1}n'|n'-1\rangle = n'\sqrt{n'+1}\delta_{n,n'-1}$$

    Becuase you would apply the lowering operator and then the raising and then the lowering, which would give you that factor ##\sqrt{n'+1}## instead of 3 factors of ##\sqrt{n'}##. Or am I misguided in my thinking?
     
  5. Oct 6, 2014 #4

    BvU

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    I seem to remember that ##a^\dagger a = N## and ##a a^\dagger = N+1## ...
     
  6. Oct 6, 2014 #5
    Okay so just to make sure that I am going about this right here is what I have for the first four terms:
    $$\langle n|a a a|n'\rangle = \langle n|a a \sqrt{n'}|n'-1\rangle = \cdots = (n')^{3/2}\delta_{n,n'-3}$$
    $$\langle n|a^\dagger a a|n'\rangle = \langle n|a^\dagger a \sqrt{n'}|n'-1\rangle = \langle n|(n')^{3/2}|n'-2\rangle = (n')^{3/2}\delta_{n,n'-2}$$
    $$\langle n|a a^\dagger a|n'\rangle = \langle n|a n'|n'\rangle = (n')^{3/2}\delta_{n,n'-1}$$
    $$\langle n|a^\dagger a^\dagger a|n'\rangle = \langle n |a^\dagger n'|n'\rangle = n'\sqrt{n'+1}\delta_{n,n'+1}$$
     
  7. Oct 6, 2014 #6

    BvU

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    As I said (you can check with your post #1) ##a^\dagger a = N ##, so ##a^\dagger a \, a |n> = a^\dagger a \, \sqrt N\, |n-1> = \sqrt N a^\dagger a \, |n-1> = (N-1)\sqrt N |n-1>##, meaning I get something else for your second term...
    And for the coefficient of the third ...
    And for the fourth I get the same as you :-)
     
  8. Oct 6, 2014 #7
    Because you are applying the raising operator on the state [itex]\left|n'-1\right\rangle[/itex] instead of [itex]\left|n'\right\rangle[/itex], and so you obtain a factor [itex]\sqrt{(n'-1) +1} = \sqrt{n'}[/itex].
    This also sort of explains why you got the wrong answer for the second term: you need to take note of the state that you are operating on.
     
  9. Oct 6, 2014 #8
    Okay after about five minutes of staring at it I think it makes sense now. What you are both saying is that $$\langle n|a^\dagger a a|n'\rangle =\underbrace{ \langle n |\sqrt{n'}a^\dagger a |n'-1\rangle}_\textrm{Now we are acting on n-1 state and thus the outcome of the operator will be not just n' but n'-1}$$

    It clicked now, the notation was slightly throwing me off, but also I clearly wasn't getting it. Thank you for your help.
     
  10. Oct 6, 2014 #9

    BvU

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    Nice exercise. In my book (Merzbacher, QM, 1970 :-) ) there is a matrix for x that still looks decent. But for higher powers it will look a bit less nice; still it's a diagonal band matrix.
     
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