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This question refers to a harmonic oscillator with length parameter a. In terms of raising and lowering operators, the operator $\hat{p}_x^{4}$ for such an oscillator can be expressed as

$\hat{p}_x^4 = \frac{\hbar^4}{4a^4}(\hat{A}-\hat{A^†})^4$

(a) Explain why any term (such as $\hat{A}\hat{A^†}\hat{A^†}\hat{A}$) with a lowering operator on the extreme right has zero expectation value in the ground state of a harmonic oscillator. (5 marks)

(b) Explain why any term (such as $\hat{A}\hat{A^†}\hat{A^†}\hat{A^†}$) with unequal numbers of raising and lowering operators has zero expectation value in the ground state of a harmonic oscillator. (8 marks)

My answer (a): In a harmonic oscillator, the lowest energy of the eigenfunction is called the zero-point energy of the oscillator. A term with $\hat{A}$ on the extreme right such as $\hat{A}\hat{A^†}\hat{A^†}\hat{A}$ will always have $\hat{A}$ operate on the eigenfunction first. However, $\hat{A}\psi_0(x) = 0$ and the operators that follow will operate on $0$ giving $0$ again. As a result, any term with $\hat{A}$ on the extreme right will result in an expectation value of $0$.

and (b): For a term with unequal numbers of raising and lowering operators, evaluating this will result in an eigenfunction with a quantum number different from the one it started with, having not been raised and lowered in equal amount. In the expectation value integral, we will be left with the multiplication of two eigenfunctions with different quantum numbers. For example,

$\int_{-\infty}^{\infty}\psi_0^*(x)\psi_1^*(x) dx$

This leaves us with orthonormal eigenfunctions in the integral and hence the expectation value will be $0$. This is the case when applied in the context of the ground state of a harmonic oscillator.

I would appreciate critique of my answer and/or additional information to flesh out my answer because I don't feel like my current one justifies the marks. I must be missing some crucial information I'm guessing.

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    Your answer looks perfectly correct. I don't get what you're saying about the marks (sorry, american). Did you not get full credit for this and are wondering why? – 2017-01-24
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    It just doesn't feel like it's enough to just a total of 13 marks between them, but at least it is correct! – 2017-01-24
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    $\hat{A}\left\vert 0\right\rangle = 0$. – 2017-01-25

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$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ $\ds{\large\mathbf{\pars{a}}}$ Because $\ds{\hat{A}\left\vert 0\right\rangle = 0\left\vert\mbox{null vector}\right\rangle}$

$\ds{\large\mathbf{\pars{b}}}$ Define $\ds{\hat{A}\pars{\phi} \equiv \exp\pars{\ic\hat{A}^{\dagger}\hat{A}\phi}A \exp\pars{-\ic\hat{A}^{\dagger}\hat{A}\phi}}$ where $\ds{\phi \in \mathbb{R}}$. Then, \begin{align} \ic\,\totald{\hat{A}\pars{\phi}}{\phi} & = \bracks{\hat{A}\pars{\phi},\hat{A}^{\dagger}\pars{\phi}\hat{A}\pars{\phi}} = \hat{A}\pars{\phi}\implies\hat{A}\pars{\phi} = \hat{A}\expo{-\ic\phi} \end{align} Moreover, \begin{align} \hat{A}\hat{A}^{\dagger}\hat{A}^{\dagger}\hat{A}^{\dagger} &= \exp\pars{-\ic\hat{A}^{\dagger}\hat{A}\phi}\hat{A}\pars{\phi} \hat{A}^{\dagger}\pars{\phi}\hat{A}^{\dagger}\pars{\phi} \hat{A}^{\dagger}\pars{\phi}\exp\pars{\ic\hat{A}^{\dagger}\hat{A}\phi} \\[5mm] &= \bracks{\exp\pars{-\ic\hat{A}^{\dagger}\hat{A}\phi}\hat{A} \hat{A}^{\dagger}\hat{A}^{\dagger} \hat{A}^{\dagger}\exp\pars{\ic\hat{A}^{\dagger}\hat{A}\phi}}\expo{2\ic\phi} \end{align} The ground state average $\ds{\left\langle\hat{A}\hat{A}^{\dagger}\hat{A}^{\dagger}\hat{A}^{\dagger} \right\rangle}$ satisfies $$ \left\langle\hat{A}\hat{A}^{\dagger}\hat{A}^{\dagger}\hat{A}^{\dagger} \right\rangle = \left\langle\hat{A}\hat{A}^{\dagger}\hat{A}^{\dagger}\hat{A}^{\dagger} \right\rangle\expo{2\ic\phi}\,\quad\color{#f00}{\forall\ \phi \in \mathbb{R}} \implies \bbx{\ds{\left\langle\hat{A}\hat{A}^{\dagger}\hat{A}^{\dagger}\hat{A}^{\dagger} \right\rangle = 0}} $$