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Calculating potential energy of hydrogen atom

  1. Mar 14, 2017 #1
    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. jcsd
  3. Mar 14, 2017 #2

    TSny

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    Homework Helper
    Gold Member

    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
  4. Mar 15, 2017 #3
    Thanks for your reply TSny.

    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?
     
  5. Mar 15, 2017 #4

    TSny

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    Yes, that looks good to me.
     
  6. Mar 15, 2017 #5
    Great. Thanks for your help!
     
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