10.4: Simple Example
- Page ID
- 14812
\( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)
\( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)
\( \newcommand{\dsum}{\displaystyle\sum\limits} \)
\( \newcommand{\dint}{\displaystyle\int\limits} \)
\( \newcommand{\dlim}{\displaystyle\lim\limits} \)
\( \newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\)
( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\)
\( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)
\( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\)
\( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)
\( \newcommand{\Span}{\mathrm{span}}\)
\( \newcommand{\id}{\mathrm{id}}\)
\( \newcommand{\Span}{\mathrm{span}}\)
\( \newcommand{\kernel}{\mathrm{null}\,}\)
\( \newcommand{\range}{\mathrm{range}\,}\)
\( \newcommand{\RealPart}{\mathrm{Re}}\)
\( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)
\( \newcommand{\Argument}{\mathrm{Arg}}\)
\( \newcommand{\norm}[1]{\| #1 \|}\)
\( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)
\( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\AA}{\unicode[.8,0]{x212B}}\)
\( \newcommand{\vectorA}[1]{\vec{#1}} % arrow\)
\( \newcommand{\vectorAt}[1]{\vec{\text{#1}}} % arrow\)
\( \newcommand{\vectorB}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)
\( \newcommand{\vectorC}[1]{\textbf{#1}} \)
\( \newcommand{\vectorD}[1]{\overrightarrow{#1}} \)
\( \newcommand{\vectorDt}[1]{\overrightarrow{\text{#1}}} \)
\( \newcommand{\vectE}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{\mathbf {#1}}}} \)
\( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)
\(\newcommand{\longvect}{\overrightarrow}\)
\( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)
\(\newcommand{\ket}[1]{\left| #1 \right>}\)
\(\newcommand{\bra}[1]{\left< #1 \right|}\)
\(\newcommand{\braket}[2]{\left< #1 \vphantom{#2} \right| \left. #2 \vphantom{#1} \right>}\)
\(\newcommand{\braopket}[3]{\left< #1 \vphantom{#2}\vphantom{#3} \right| #2 \vphantom{#1}\vphantom{#3} \left| #3 \vphantom{#1}\vphantom{#2} \right>}\)
\(\newcommand{\qmvec}[1]{\mathbf{\vec{#1}}}\)
\(\newcommand{\op}[1]{\hat{\mathbf{#1}}}\)
\(\newcommand{\expect}[1]{\langle #1 \rangle}\)
\(\newcommand{\dfn}[1]{\emph{\textbf{#1}}}\)
The best way to clarify this abstract discussion is to consider the quantum mechanics of the Harmonic oscillator of mass \(m\) and frequency \(\omega\),
\[\hat{H}=-\frac{\hbar^2}{2 m} \frac{d^2}{d x^2}+\frac{1}{2} m \omega^2 x^2\]
If we assume that the wave function at time \(t=0\) is a linear superposition of the first two eigenfunctions,
\[\begin{aligned}
\psi(x, t=0) & =\sqrt{\frac{1}{2}} \phi_0(x)-\sqrt{\frac{1}{2}} \phi_1(x) \\[8pt]
\phi_0(x) & =\left(\frac{m \omega}{\pi \hbar}\right)^{1 / 4} \exp \left(-\frac{m \omega}{2 \hbar} x^2\right) \\[8pt]
\phi_1(x) & =\left(\frac{m \omega}{\pi \hbar}\right)^{1 / 4} \exp \left(-\frac{m \omega}{2 \hbar} x^2\right) \sqrt{\frac{m \omega}{\hbar}} x
\end{aligned}\]
(The functions \(\phi_0\) and \(\phi_1\) are the normalised first and second states of the harmonic oscillator, with energies \(E_0=\frac{1}{2} \hbar \omega\) and \(E_1=\frac{3}{2} \hbar \omega\).) Thus we now kow the wave function for all time:
\[\psi(x, t)=\sqrt{\frac{1}{2}} \phi_0(x) e^{-\frac{1}{2} i \omega t}-\sqrt{\frac{1}{2}} \phi_1(x) e^{-\frac{3}{2} i \omega t} .\]
In figure \(\PageIndex{1}\) we plot this quantity for a few times.
The best way to visualize what is happening is to look at the probability density,
\[\begin{aligned}
\mathcal{P}(x, t) & =\psi(x, t)^* \psi(x, t) \\
& =\frac{1}{2}\left(\phi_0(x)^2+\phi_1(x)^2-2 \phi_0(x) \phi_1(x) \cos \omega t\right)
\end{aligned}\]
This clearly oscillates with frequency \(\omega\).
Question: Show that \(\int_{-\infty}^{\infty} \mathcal{P}(x, t) d x=1\).
Another way to look at that is to calculate the expectation value of \(\hat{x}\) :
\[\begin{aligned}
\langle\hat{x}\rangle & =\int_{-\infty}^{\infty} \mathcal{P}(x, t) d x \\[8pt]
& =\frac{1}{2} \underbrace{\int_{-\infty}^{\infty} \phi_0(x)^2 x d x}_{=0}+\frac{1}{2} \underbrace{\int_{-\infty}^{\infty} \phi_1(x)^2 x d x}_{=0}-\cos \omega t \int_{-\infty}^{\infty} \phi_0(x) \phi_1(x) x d x \\[8pt]
& =-\cos \omega t \sqrt{\frac{\hbar}{m \omega}} \frac{1}{\sqrt{\pi}} \int_{-\infty}^{\infty} y^2 e^{-y^2} \\[8pt]
& =-\frac{1}{2} \sqrt{\frac{\hbar}{m \omega}} \cos \omega t .
\end{aligned} \]
This once again exhibits oscillatory behaviour!


