Homework Help: Calculating potential energy of hydrogen atom

1. Mar 14, 2017

Leechie

1. The problem statement, all variables and given/known data
Calculate $\left< \frac 1 r \right>$ and $\left< \frac 1 {r^2} \right>$ and the expectation value and uncertainty of the potential energy of the electron and proton for a hydrogen atom in the given state.

The given state is:
$$\psi_{2,1,-1} \left( r,\theta,\phi \right) = \left( \frac 1 {64 \pi a^3_0} \right)^{1/2} \frac r {a_0} e^{-r/2a_0} \sin(\theta) e^{-i\phi}$$
2. Relevant equations

3. The attempt at a solution
I've calculated the expectation values for $\left< \frac 1 r \right>$ and $\left< \frac 1 {r^2} \right>$ which I think are right. The values I've got are:
$$\left< \frac 1 r \right> = \frac 1 {4a_0}$$ $$\left< \frac 1 {r^2} \right> = \frac 1 {12a^2_0}$$
The expectation for the potential energy I've got is:
$$\left< E \right> = - \frac {e^2} {4 \pi \varepsilon_0} \left< \frac 1 r \right> = -\frac {e^2} {16 \pi \varepsilon_0 a_0 }$$
This is where I'm getting a bit confused. I'm trying to use $\Delta E = \sqrt {\left< E^2 \right> - \left< E \right>^2 }$ to find the uncertainty. I thought it should equal zero since the question relates to a stationary state but I'm getting:

For $\left< E^2 \right>$ I've got:
$$\left< E^2 \right> = - \frac {e^4} {256 \pi^2 \varepsilon^2_0 a^2_0 } \left< \frac 1 {r^2} \right> = -\frac {e^4} {3072 \pi^2 \varepsilon^2_0 a^2_0 }$$
So for the uncertainty I've got:
\begin{align} \Delta E & = \sqrt {\left< E^2 \right> - \left< E \right>^2 } \nonumber \\ & = \sqrt {-\frac {e^4} {3072 \pi^2 \varepsilon^2_0 a^2_0 } - \left( -\frac {e^2} {16 \pi \varepsilon_0 a_0 } \right)^2 } \nonumber \\ & = \sqrt {-\frac {e^4} {3072 \pi^2 \varepsilon^2_0 a^2_0 } + \frac {e^4} {256 \pi^2 \varepsilon^2_0 a^2_0 } } \nonumber \\ & = \sqrt {-\frac 1 {3072} + \frac 1 {256} } \nonumber \\ & = \sqrt {\frac {11} {3072} } \end{align}
Can anyone help with this please.

2. Mar 14, 2017

TSny

You're working with the potential energy V, rather than the total energy E. So, it might be less confusing to use the symbol V rather than E.
The wavefunction is not an eigenstate of the potential energy operator $V$. So, do you expect $\Delta V$ to be zero?
The factor of 256 in this expression does not look correct.
[Edit: Also, the overall dimensions of this expression are incorrect for the square of an energy.]

Last edited: Mar 14, 2017
3. Mar 15, 2017

Leechie

You're right about changing the the symbol $E$ to $V$. I think using $E$ instead of $V$ is what was confusing me when thinking about the uncertainty.

I think I seen where I've gone wrong with my value for $\left< V^2 \right>$. I've corrected it to:
$$\left< V^2 \right> = \frac {e^4} {16 \pi^2 \varepsilon^2_0 } \left< \frac {1} {r^2} \right> = \frac {e^4} {192 \pi^2 \varepsilon^2_0 a^2_0 }$$
So my uncertainty equation now looks like:
\begin{align} \Delta V & = \sqrt {\left< V^2 \right> - \left< V \right>^2} \nonumber \\ & = \sqrt { \frac {e^4} {192 \pi^2 \varepsilon^2_0 a^2_0 } - \left( - \frac {e^2} {16 \pi \varepsilon_0 a_0 } \right)^2 } \nonumber \\ & = \sqrt { \frac {e^4} {192 \pi^2 \varepsilon^2_0 a^2_0 } - \frac {e^4} {256 \pi^2 \varepsilon^2_0 a^2_0 } } \nonumber \\ & = \frac {e^2} {\sqrt {768} \pi \varepsilon_0 a_0 } \nonumber \end{align}
Which I think looks better, would you agree?

4. Mar 15, 2017

TSny

Yes, that looks good to me.

5. Mar 15, 2017