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Energy of Hydrogen 1s using simplified Schrodinger equation

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Homework Statement


[/B]
The Hamiltonian and wavefunction for the ground state of the hydrogen atom H(1s1) are given,
in atomic units, as ## \hat {H} = - \frac{1}{2} \nabla^2 - \frac {1}{r} ## and ## \phi(1s) = \sqrt {\frac {1}{\pi }} e^{-r} ## . Using the radial portion of the Laplacian in the simplified form ## \nabla^2 = \frac{\mathrm{d^2} }{\mathrm{d} r^2} \frac{\mathrm{d} }{\mathrm{d} r} ## (Basically ignoring the Legendrian)

Verify that the total energy equals -0.5 (Hartree) for H (1s1)

I have a couple of issues here:

The first issue comes from the way question shows the simplified form of the Laplacian (polar coordinates). Looking up the simplified Laplacian (just the radial component) online shows me ## \nabla^2 = \frac {1}{r^2} \frac{\mathrm{d} }{\mathrm{d} r} r^2 \frac{\mathrm{d} }{\mathrm{d} r} ## I don't see how they are the same thing?

Additionally, from my reading, I understand that there should be a centrifugal term in the Hamiltonian here. Usually of the form ## - \frac {l(l+1)}{r^2} ## . I assume its omitted because a s-electron has l=0 ?

My notes/online resources show how to work this problem with centrifugal term included and the common simplified Laplacian, ## \nabla^2 = \frac {1}{r^2} \frac{\mathrm{d} }{\mathrm{d} r} r^2 \frac{\mathrm{d} }{\mathrm{d} r} ##. I know the fist step (once you set up the SE) is to make use of a decaying exponential solution but I'm just unsure of how to get there in the way the question is set up. Any help would be much appreciated.

Homework Equations




The Attempt at a Solution



## \psi(r,\theta,\phi) = R(r)Y(\theta,\Phi) ##

## -\frac{1}{2} \left \{\frac{\mathrm{d^2} }{\mathrm{d} r^2} + \frac {2}{r}\frac{\mathrm{d} }{\mathrm{d} r} - \frac {1}{r} \right \}R(r) = ER(r) ##

## \alpha = -2E ##

## \frac{\mathrm{d^2} }{\mathrm{d} r^2}R(r) + \frac {2}{r}\frac{\mathrm{d} }{\mathrm{d} r}R(r) - \frac {1}{r} R(r) = \alpha R(r) ##
 

Answers and Replies

  • #2
Orodruin
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Using the radial portion of the Laplacian in the simplified form ## \nabla^2 = \frac{\mathrm{d^2} }{\mathrm{d} r^2} \frac{\mathrm{d} }{\mathrm{d} r} ## (Basically ignoring the Legendrian)
Is this really what the problem states? If so it is clearly a typo, the laplacian is a second order differential operator, not a third order one.
I assume its omitted because a s-electron has l=0 ?
Yes.

Finding the energy should just be a matter of insertion into the SE and solving for E.
 
  • #3
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My apologies, its not third order. The question gives the laplacian as :

$$ \nabla^2 = \frac{\mathrm{d^2} }{\mathrm{d} r^2} + \frac {2}{r} \frac{\mathrm{d} }{\mathrm{d} r} $$
 
  • #4
Orodruin
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My apologies, its not third order. The question gives the laplacian as :

$$ \nabla^2 = \frac{\mathrm{d^2} }{\mathrm{d} r^2} + \frac {2}{r} \frac{\mathrm{d} }{\mathrm{d} r} $$
That is correct and indeed the same thing as
$$
\frac{1}{r^2}\frac{d}{dr} r^2 \frac{d}{dr}.
$$
In order to see that they are indeed the same differential operator, try applying the latter to any function ##f(r)## and see what comes out in terms of ##r## and derivatives of ##f(r)##.
 
  • #5
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I have worked through it and managed to convert between the two. I guess I just needed confirmation that they are indeed the same. Thank you so much!
 

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