- #1
- 8,446
- 5,592
I'm trying to derive something which shouldn't be too complicated, but I get different results when doing things symbolically and with actual operators and wave functions. Some help would be appreciated.
For the hydrogenic atom, I need to calculate ##\langle \hat{H}\hat{V} \rangle## and ##\langle \hat{V}\hat{H} \rangle##, where ##\hat{H} = p^2/2m + \hat{V}## and ##\hat{V} = -Z\hbar^2/(m a_0 r)## for a state that is an eigenstate of ##\hat{H}##, ##\hat{H}|n\rangle = E_n |n\rangle##. Since both ##\hat{H}## and ##\hat{V}## are hermitian, it follows that
$$
\begin{align*}
\langle n | \hat{V}\hat{H} | n \rangle &= \langle n | \hat{V} E_n | n \rangle \\
&= E_n \langle n | \hat{V} | n \rangle = - E_n Ze^2 \langle n | \frac{1}{r} | n \rangle \\
\langle n | \hat{H}\hat{V} | n \rangle &= \left( \langle n | \hat{H} \right) \hat{V} | n \rangle \\
&= E_n \langle n | \hat{V} | n \rangle = - E_n \frac{Z\hbar^2}{m a_0} \langle n | \frac{1}{r} | n \rangle
\end{align*}
$$
from which I conclude that ##\langle \hat{H}\hat{V} \rangle = \langle \hat{V}\hat{H} \rangle = E_n \frac{Z\hbar^2}{m a_0} \langle 1/r \rangle##.
But when I try to calculate the expectations values explicitely on the 1s wave function,
$$
\psi_{1\mathrm{s}} = \frac{1}{\sqrt{\pi}} \left( \frac{Z}{a_0} \right)^{3/2} e^{-Z r /a_0}
$$
I find that ##\langle \hat{H}\hat{V} \rangle = \frac{5}{2} \frac{\hbar^4 Z^4}{m^2 a_0^4}##, while ##\langle \hat{V}\hat{H} \rangle = \frac{1}{2} \frac{\hbar^4 Z^4}{m^2 a_0^4}##. Since for hydrogenic atoms we have
$$
\left\langle \frac{1}{r} \right\rangle = \frac{1}{n^2} \frac{Z}{a_0}
$$
and
$$
E_n = - \frac{\hbar^2 Z^2}{2 m a_0^2} \frac{1}{n^2}
$$
I would expect ##E_n \frac{Z\hbar^2}{m a_0} \langle 1/r \rangle = \hbar^4 Z^4 / (2 m^2 a_0^2)##, which is the value I get for ##\langle \hat{V}\hat{H} \rangle##, but not ##\langle \hat{H}\hat{V} \rangle##.
By the way, I did the integrations with the actual wave function both by hand and with Mathematica, so I know there is no mistake there.
Can somewhat figure out what I did wrong?
For the hydrogenic atom, I need to calculate ##\langle \hat{H}\hat{V} \rangle## and ##\langle \hat{V}\hat{H} \rangle##, where ##\hat{H} = p^2/2m + \hat{V}## and ##\hat{V} = -Z\hbar^2/(m a_0 r)## for a state that is an eigenstate of ##\hat{H}##, ##\hat{H}|n\rangle = E_n |n\rangle##. Since both ##\hat{H}## and ##\hat{V}## are hermitian, it follows that
$$
\begin{align*}
\langle n | \hat{V}\hat{H} | n \rangle &= \langle n | \hat{V} E_n | n \rangle \\
&= E_n \langle n | \hat{V} | n \rangle = - E_n Ze^2 \langle n | \frac{1}{r} | n \rangle \\
\langle n | \hat{H}\hat{V} | n \rangle &= \left( \langle n | \hat{H} \right) \hat{V} | n \rangle \\
&= E_n \langle n | \hat{V} | n \rangle = - E_n \frac{Z\hbar^2}{m a_0} \langle n | \frac{1}{r} | n \rangle
\end{align*}
$$
from which I conclude that ##\langle \hat{H}\hat{V} \rangle = \langle \hat{V}\hat{H} \rangle = E_n \frac{Z\hbar^2}{m a_0} \langle 1/r \rangle##.
But when I try to calculate the expectations values explicitely on the 1s wave function,
$$
\psi_{1\mathrm{s}} = \frac{1}{\sqrt{\pi}} \left( \frac{Z}{a_0} \right)^{3/2} e^{-Z r /a_0}
$$
I find that ##\langle \hat{H}\hat{V} \rangle = \frac{5}{2} \frac{\hbar^4 Z^4}{m^2 a_0^4}##, while ##\langle \hat{V}\hat{H} \rangle = \frac{1}{2} \frac{\hbar^4 Z^4}{m^2 a_0^4}##. Since for hydrogenic atoms we have
$$
\left\langle \frac{1}{r} \right\rangle = \frac{1}{n^2} \frac{Z}{a_0}
$$
and
$$
E_n = - \frac{\hbar^2 Z^2}{2 m a_0^2} \frac{1}{n^2}
$$
I would expect ##E_n \frac{Z\hbar^2}{m a_0} \langle 1/r \rangle = \hbar^4 Z^4 / (2 m^2 a_0^2)##, which is the value I get for ##\langle \hat{V}\hat{H} \rangle##, but not ##\langle \hat{H}\hat{V} \rangle##.
By the way, I did the integrations with the actual wave function both by hand and with Mathematica, so I know there is no mistake there.
Can somewhat figure out what I did wrong?