STO-1G for H-atom

[scarecrow]

I have a Gaussian trial wavefunction for the ground 1s state of H atom:

\psi (r)= A Exp[-c r^2],

where A is the normalization constant and c is the variational parameter.

I'm trying to calculate the variational integral W(c) = < \psi (r) | H | \psi (r)>, where H is the Hamiltonian for the H-atom.

My question is wouldn't the angular momentum term operator, L^2, in the Hamiltonian disappear since the trial wavefunction only depends on r?

(And by the way, how do I use all those mathematical typesettings on this board?)

[jtbell]

(And by the way, how do I use all those mathematical typesettings on this board?)

http://www.physicsforums.com/showthread.php?t=8997

[Meir Achuz]

I have a Gaussian trial wavefunction for the ground 1s state of H atom:

F(r) = A Exp[-c r^2],

where A is the normalization constant and c is the variational parameter.

I'm trying to calculate the variational integral W(c) = < F(r) | H | F(r) >, where H is the Hamiltonian for the H-atom.

My question is wouldn't the angular momentum term operator, L^2, in the Hamiltonian disappear since the trial wavefunction only depends on r?

Yes, L^2=0 for spherically symmetric state.

[scarecrow]

I'm confused about what my prof. said: he said make sure you are doing the proper 3D integration for the variational integral, but how is this a 3D integration when H only depends on r and so does \psi (r)?

Thanks jtbell for the link.

[Gza]

I'm confused about what my prof. said: he said make sure you are doing the proper 3D integration for the variational integral, but how is this a 3D integration when H only depends on r and so does \psi (r)?

Sorry for the typesetting errors...

think about it in spherical coordinates; yes it only depends on r, but you're still running the integration w.r.t a spherical differential volume:
dv=r^2*sin(theta)*dtheta*dphi. when the integrations carried over, the angular coordinates integrate easy enough, leaving nothing more than an extra r^2 factor in your integral.

[dextercioby]

The Lebesgue measure in L^{2}\left(\mathbb{R}^{3}, d^{3}x\right) is of course "d^{3}x" and, as you can see, it coincides with the Riemann measure in \mathbb{R}^{3} .


Daniel.

[scarecrow]

Okay, can someone please tell me what I'm doing wrong:

\psi (r) = A Exp[- \alpha r^2]

H = \frac{h^2}{2\mu} (\frac{d^2}{dr^2} + \frac{2 d}{r dr} ) - \frac{e^2}{r^2}

W(\alpha) = < \psi (r) | H | \psi (r) >

W = \int_{0}^{\infty} \psi (r) (H \psi (r) ) r^2 dr \int_{0}^{\pi} \sin \theta d \theta \int_{0}^{2 \pi} d \phi

[reilly]

So far so good. Carry out the angular integrals, and you are ready for bear, or whatever else you might want to be ready for.

Regards,
Reilly Atkinson

[scarecrow]

I carried out the integration and I get something disturbing...:yuck:

I get a W that doesn't even depend on \alpha. This doesn't make any sense because the next step is to minimize W with respect to \alpha to find the \alpha _{0} that corresponds to the minimum of W.

:confused:

Is this because the Hamiltonian (H-atom) is written incorrectly?

[t!m]

In the Hamiltonian, the coulomb potential goes as 1/r, not 1/r^2 ... and don't forget the 1/(4*pi*epsilon_0).

[scarecrow]

Thanks t!m...I didn't catch that. I ran the integrations on Mathematica. However, now W is inversely proportional to the square root of alpha.

And when you take the derivative of W w.r.t to alpha and set it equal to, there's obviously a problem since alpha is in the denominator and I'm trying to find the value of alpha where W is a minimum.

I'm sorry but nothing seems to be working. Once this integration works, then I'll know how to do the rest of the problem.

[dextercioby]

For the minimum you need to set the partial derivative wrt alpha to 0, find the alpha from the resulting eqn and then make sure the second partial derivative is positive on the set of solutions.

Daniel.

[CarlB]

1) Remember that the constant "A" depends on alpha.

2) There is a possibility that mathematica is screwing up the integral.

Back in my day we had to do these integrals by hand, in our heads, while walking to school in 1 meter deep snow. AND we had to work on the farm in the summer, 2 meters deep in bull doo-doo.

[scarecrow]

I already solved for A and already had taken the partial derivative wrt to alpha and set it to 0. I also checked the second derivative and that still gives me an alpha in the denominator.

This problem seemed so easy...

Can someone try to do the integral and tell me if you're getting the same results?

\psi (r) = (\frac{2 \alpha}{\pi})^\frac{1}{4} Exp[-\alpha r^2]

[dextercioby]

The matrix element of the Coulomb potential is

I=-4\pi e^2 \int_{0}^{\infty} r{}e^{-2\alpha r^2} {}dr

Can you perform such an integration?

Daniel.

[scarecrow]

Yes I do, I did the exact integral. Like I said, the answer makes no sense because alpha is in the denominator, and if you take the derivative W wrt to alpha and set it to 0, you get an impossible scenario. Alpha must be greater than 0.

Do you see what I'm trying to explain?

[dextercioby]

I'm not gonna do the calculations for you, no way, but the question still remains: Are absolutely sure that once you solve the equation for alpha you get alpha =0 ?

Daniel.

[scarecrow]

W =-4\pi e^2 \int_{0}^{\infty} r{}e^{-2\alpha r^2} {}dr = \frac{-e^2}{\alpha}

W(\alpha) = \frac{-e^2}{\alpha}

\frac{dW}{d\alpha} = \frac{e^2}{\alpha^2} = 0

Now \alpha > 0. Otherwise, it will blow up.

[cristo]

Yes I do, I did the exact integral. Like I said, the answer makes no sense because alpha is in the denominator, and if you take the derivative W wrt to alpha and set it to 0, you get an impossible scenario. Alpha must be greater than 0.

Do you see what I'm trying to explain?

Why don't you post your solution to the integral. People are more likely to look through your work and point out errors than sit down and calculate those integrals for you!!

[scarecrow]

Why don't you post your solution to the integral. People are more likely to look through your work and point out errors than sit down and calculate those integrals for you!!
:smile: I posted my answer right when you made your post.

[dextercioby]

If the H-atom hamiltonian you've written is correct, then

\alpha=\frac{e^{4}}{\pi^{3}\left(\frac{\sqrt{2}}{1 6}\frac{\hbar^{2}}{\mu}+\frac{\sqrt{2}}{2}\right)^ {2}}

Daniel.