# 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