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I am currently trying to understand the impact of lowering and raising operators on the expectation values of certain operators in a Harmonic Oscillator when the energy eigenstate has been stated as $|n\rangle$ e.g. $\hat{p}^3$

If I have a combination of lowering and raising operators I know the aim to compare their expectations value to the following

$\langle{n}|m\rangle=0$ when $m\neq{n}$

for example I have done the following

$\langle{n}|\hat{a}^3|n\rangle=\sqrt{(n)(n-1)(n-2)}\langle{n}|n-3\rangle=0$ because clearly $n\neq{n-3}$

however my difficulties are when I have combinations of lowering and raising operators, like so

$\langle{n}|\hat{a}{\hat{a}^\dagger}^2|n\rangle$

another example is

$\langle{n}|\hat{a}^2\hat{a}^\dagger|n\rangle$

I need an explanation of what $\hat{a}$, $\hat{a}^2$, $\hat{a}^\dagger$ and ${\hat{a}^\dagger}^2$ become when combined with each other in an expectation value like the ones given above.

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In general given $\hat A \hat B \vert \psi\rangle$, the product of operators acting on a state is defined as:

  1. compute first $\hat B\vert \psi \rangle$ and call the result $\vert\psi'\rangle$,
  2. compute $\hat A \vert \psi'\rangle=\hat A\left[\hat B\vert\psi\rangle\right]$

For one of your examples this yields:

\begin{align} \langle n\vert \hat a\,(\hat a^\dagger)^2\vert n\rangle & = \langle n\vert \hat a\,\hat a^\dagger\left[\hat a^\dagger \vert n\rangle \right]\, ,\\ &=\sqrt{n+1}\, \langle n\vert \hat a\,\hat a^\dagger \vert n+1\rangle\, ,\\ & = \sqrt{n+1}\sqrt{n+2}\, \langle n\vert \hat a\,\vert n+2\rangle\, , \\ &=\sqrt{n+1}(n+2)\,\langle n\vert n+1\rangle =0 \end{align}

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  • $\begingroup$ Can you explain the reasons for this happening please its not the answer I am in need of but an applicable method $\endgroup$ – Sam Mar 25 '17 at 23:29
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    $\begingroup$ Not sure I follow. Apply the operators in sequence one after the others as above. $\endgroup$ – ZeroTheHero Mar 25 '17 at 23:30
  • $\begingroup$ So what happens to the $\hat{a}$ between the 3rd and 4th line and why ? and what changes when $\hat{a}$ is squared ? $\endgroup$ – Sam Mar 25 '17 at 23:31
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    $\begingroup$ $\sqrt{(n+1)(n+2)}\,\hat a\vert n+2\rangle= \sqrt{(n+1)(n+2)}\sqrt{n+2}\vert n+1\rangle=\sqrt{n+1}(n+2)\vert n+1\rangle$. If you have $\hat a^2\vert n\rangle$ then this is just $\hat a \left[\hat a\vert n\rangle\right]=\hat a\vert n-1\rangle\sqrt{n}$. For any operator $\hat A$ then $(\hat A)^2$ means you apply it twice. $\endgroup$ – ZeroTheHero Mar 25 '17 at 23:37
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From quantum mechanics operators, we know taht \begin{align} \hat a \vert n\rangle = \sqrt{n} \vert n-1\rangle \end{align} \begin{align} \hat{a}^\dagger \vert n\rangle = \sqrt{n+1} \vert n+1\rangle \end{align} and any operator $\hat{A}$ and $\hat{B}$ are satisfies \begin{align} \hat{A} \hat{B}\vert \psi \rangle = \hat{A} (\hat{B}\vert \psi \rangle) \end{align}

To calculate $\langle{n}|\hat{a}^2\hat{a}^\dagger|n\rangle$ \begin{equation} \langle{n}|\hat{a}^2\hat{a}^\dagger|n\rangle=\\ \sqrt{n+1}\langle{n}|\hat{a}^2|n+1\rangle= \\ (n+1)\langle{n}|\hat{a}|n\rangle=\\ (n+1)\sqrt{n}\langle{n}|n-1\rangle=(n+1)\sqrt{n}\delta_{n,n-1}=0 \end{equation}

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