- #1
Leechie
- 19
- 2
Homework Statement
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:
$$ \psi_{2,1,-1} \left( r,\theta,\phi \right) = \left( \frac 1 {64 \pi a^3_0} \right)^{1/2} \frac r {a_0} e^{-r/2a_0} \sin(\theta) e^{-i\phi}$$
Homework Equations
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.