# Energy of Hydrogen 1s using simplified Schrodinger equation

## Homework Statement

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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.

## The Attempt at a Solution

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

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

$\alpha = -2E$

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

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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.

• Xilus1
My apologies, its not third order. The question gives the laplacian as :

$$\nabla^2 = \frac{\mathrm{d^2} }{\mathrm{d} r^2} + \frac {2}{r} \frac{\mathrm{d} }{\mathrm{d} r}$$

Orodruin
Staff Emeritus
Homework Helper
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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 $f(r)$.

• Xilus1
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!