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8.2 Expectation value of $ \hat{x}^{2}_{}$ and $ \hat{p}^{2}_{}$ for the harmonic oscillator

As an example of all we have discussed let us look at the harmonic oscillator. Suppose we measure the average deviation from equilibrium for a harmonic oscillator in its ground state. This corresponds to measuring $ \hat{x}$. Using

$\displaystyle \phi_{0}^{}$(x) = $\displaystyle \left(\vphantom{ \frac{m \omega}{\hbar \pi} }\right.$$\displaystyle {\frac{m \omega}{\hbar \pi}}$ $\displaystyle \left.\vphantom{ \frac{m \omega}{\hbar \pi} }\right)^{1/4}$exp$\displaystyle \left(\vphantom{ - \frac{m \omega}{2 \hbar} x^2 }\right.$ - $\displaystyle {\frac{m \omega}{2 \hbar}}$x2$\displaystyle \left.\vphantom{ - \frac{m \omega}{2 \hbar} x^2 }\right)$ (8.17)

we find that

$\displaystyle \langle$x$\displaystyle \rangle$ = $\displaystyle \int_{-\infty}^{\infty}$$\displaystyle \left(\vphantom{ \frac{m \omega}{\hbar \pi} }\right.$$\displaystyle {\frac{m \omega}{\hbar \pi}}$ $\displaystyle \left.\vphantom{ \frac{m \omega}{\hbar \pi} }\right)^{1/2}$xexp$\displaystyle \left(\vphantom{ - \frac{m \omega}{ \hbar} x^2 }\right.$ - $\displaystyle {\frac{m\omega}{\hbar}}$x2$\displaystyle \left.\vphantom{ - \frac{m \omega}{ \hbar} x^2 }\right)$dx = 0. (8.18)

Qn Why is it 0? Sinilarly, using $ \hat{p}$ = - i$ \hbar$$ {\frac{d}{dx}}$ and

$\displaystyle \hat{p}$$\displaystyle \left(\vphantom{ \frac{m \omega}{\hbar \pi} }\right.$$\displaystyle {\frac{m \omega}{\hbar \pi}}$ $\displaystyle \left.\vphantom{ \frac{m \omega}{\hbar \pi} }\right)^{1/4}$exp$\displaystyle \left(\vphantom{ - \frac{m \omega}{2 \hbar} x^2 }\right.$ - $\displaystyle {\frac{m \omega}{2 \hbar}}$x2$\displaystyle \left.\vphantom{ - \frac{m \omega}{2 \hbar} x^2 }\right)$ = im$\displaystyle \omega$x$\displaystyle \left(\vphantom{ \frac{m \omega}{\hbar \pi} }\right.$$\displaystyle {\frac{m \omega}{\hbar \pi}}$ $\displaystyle \left.\vphantom{ \frac{m \omega}{\hbar \pi} }\right)^{1/4}$exp$\displaystyle \left(\vphantom{ - \frac{m \omega}{2 \hbar} x^2 }\right.$ - $\displaystyle {\frac{m \omega}{2 \hbar}}$x2$\displaystyle \left.\vphantom{ - \frac{m \omega}{2 \hbar} x^2 }\right)$, (8.19)

we find

$\displaystyle \langle$p$\displaystyle \rangle$ = 0. (8.20)

More challenging are the expectation values of x2 and p2. Let me look at the first one first:
$\displaystyle \langle$x2$\displaystyle \rangle$ = $\displaystyle \int_{-\infty}^{\infty}$$\displaystyle \left(\vphantom{ \frac{m \omega}{\hbar \pi} }\right.$$\displaystyle {\frac{m \omega}{\hbar \pi}}$ $\displaystyle \left.\vphantom{ \frac{m \omega}{\hbar \pi} }\right)^{1/2}$x2exp$\displaystyle \left(\vphantom{ - \frac{m \omega}{ \hbar} x^2 }\right.$ - $\displaystyle {\frac{m\omega}{\hbar}}$x2$\displaystyle \left.\vphantom{ - \frac{m \omega}{ \hbar} x^2 }\right)$dx  
  = $\displaystyle \left(\vphantom{ \frac{\hbar}{m \omega} }\right.$$\displaystyle {\frac{\hbar}{m\omega}}$ $\displaystyle \left.\vphantom{ \frac{\hbar}{m \omega} }\right)$$\displaystyle \pi^{-1/2}_{}$$\displaystyle \int_{-\infty}^{\infty}$y2exp$\displaystyle \left(\vphantom{ - y^2 }\right.$ - y2$\displaystyle \left.\vphantom{ - y^2 }\right)$dy  
  = $\displaystyle \left(\vphantom{ \frac{\hbar}{m \omega} }\right.$$\displaystyle {\frac{\hbar}{m\omega}}$ $\displaystyle \left.\vphantom{ \frac{\hbar}{m \omega} }\right)$$ {\frac{1}{2}}$. (8.21)

Now for $ \hat{p}^{2}_{}$,

$\displaystyle \hat{p}^{2}_{}$exp$\displaystyle \left(\vphantom{ - \frac{m \omega}{2 \hbar} x^2 }\right.$ - $\displaystyle {\frac{m \omega}{2 \hbar}}$x2$\displaystyle \left.\vphantom{ - \frac{m \omega}{2 \hbar} x^2 }\right)$ = [- (m$\displaystyle \omega$x)2 + $\displaystyle \hbar$m$\displaystyle \omega$]exp$\displaystyle \left(\vphantom{ - \frac{m \omega}{2 \hbar} x^2 }\right.$ - $\displaystyle {\frac{m \omega}{2 \hbar}}$x2$\displaystyle \left.\vphantom{ - \frac{m \omega}{2 \hbar} x^2 }\right)$. (8.22)

Thus,
$\displaystyle \langle$$\displaystyle \hat{p}^{2}_{}$$\displaystyle \rangle$ = $\displaystyle \int_{-\infty}^{\infty}$$\displaystyle \left(\vphantom{ \frac{m \omega}{\hbar \pi} }\right.$$\displaystyle {\frac{m \omega}{\hbar \pi}}$ $\displaystyle \left.\vphantom{ \frac{m \omega}{\hbar \pi} }\right)^{1/2}$[- (m$\displaystyle \omega$x)2 + $\displaystyle \hbar$m$\displaystyle \omega$]exp$\displaystyle \left(\vphantom{ - \frac{m \omega}{ \hbar} x^2 }\right.$ - $\displaystyle {\frac{m\omega}{\hbar}}$x2$\displaystyle \left.\vphantom{ - \frac{m \omega}{ \hbar} x^2 }\right)$dx  
  = $\displaystyle \hbar$m$\displaystyle \omega$$\displaystyle \pi^{-1/2}_{}$$\displaystyle \int_{-\infty}^{\infty}$[1 - y2]exp$\displaystyle \left(\vphantom{ - y^2 }\right.$ - y2$\displaystyle \left.\vphantom{ - y^2 }\right)$dy  
  = $\displaystyle \hbar$m$\displaystyle \omega$$ {\frac{1}{2}}$. (8.23)

This is actually a form of the uncertainity relation, and shows that

$\displaystyle \Delta$x$\displaystyle \Delta$p $\displaystyle \geq$ $ {\frac{1}{2}}$$\displaystyle \hbar$! (8.24)


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© 1998 Niels Walet, UMIST
Email Niels.Walet@umist.ac.uk