How to Derive the Hydrogen Atom Hamiltonian in Spherical Coordinates?

FloridaGators
Messages
11
Reaction score
0
The Hamiltonian for a Hydrogen atom in Cartesian Coordinates (is this right?):
\hat{H} = - \frac{\bar{h}^2}{2m_p}\nabla ^2_p - \frac{\bar{h}^2}{2m_e}\nabla ^2_e - \frac{e^2}{4\pi\epsilon _0r}
In Spherical Coordinates do I just use:
x=r sin θ cos φ, y = r sin θ sin φ, and z = r cos θ?
 
Last edited:
Physics news on Phys.org
It is \hbar instead of h. But that is essentially correct. You might want to convert it to the center of mass reference frame before you do any work on it though. There are tons of sites out there that solve it as well and show all the work.
 
FloridaGators said:
The Hamiltonian for a Hydrogen atom in Cartesian Coordinates (is this right?):
\hat{H} = - \frac{h^2}{2m_p}\nabla ^2_p - \frac{h^2}{2m_e}\nabla ^2_e - \frac{e^2}{4\pi\epsilon _0r}

First, this form has no explicit reference to Cartesian coordinates.

Second, this is only correct if you define r to be the separation between the proton and the electron; not the distance from the origin.

In Spherical Coordinates do I just use:
x=r sin θ cos φ, y = r sin θ sin φ, and z = r cos θ?

There are many sites and texts that derive expressions for \nabla^2 in Spherical coordinates.
 
Thank you for helping. Do you mind my asking what your search inquiry in google was to find that?
 
Last edited:
##|\Psi|^2=\frac{1}{\sqrt{\pi b^2}}\exp(\frac{-(x-x_0)^2}{b^2}).## ##\braket{x}=\frac{1}{\sqrt{\pi b^2}}\int_{-\infty}^{\infty}dx\,x\exp(-\frac{(x-x_0)^2}{b^2}).## ##y=x-x_0 \quad x=y+x_0 \quad dy=dx.## The boundaries remain infinite, I believe. ##\frac{1}{\sqrt{\pi b^2}}\int_{-\infty}^{\infty}dy(y+x_0)\exp(\frac{-y^2}{b^2}).## ##\frac{2}{\sqrt{\pi b^2}}\int_0^{\infty}dy\,y\exp(\frac{-y^2}{b^2})+\frac{2x_0}{\sqrt{\pi b^2}}\int_0^{\infty}dy\,\exp(-\frac{y^2}{b^2}).## I then resolved the two...
Hello everyone, I’m considering a point charge q that oscillates harmonically about the origin along the z-axis, e.g. $$z_{q}(t)= A\sin(wt)$$ In a strongly simplified / quasi-instantaneous approximation I ignore retardation and take the electric field at the position ##r=(x,y,z)## simply to be the “Coulomb field at the charge’s instantaneous position”: $$E(r,t)=\frac{q}{4\pi\varepsilon_{0}}\frac{r-r_{q}(t)}{||r-r_{q}(t)||^{3}}$$ with $$r_{q}(t)=(0,0,z_{q}(t))$$ (I’m aware this isn’t...
Hi, I had an exam and I completely messed up a problem. Especially one part which was necessary for the rest of the problem. Basically, I have a wormhole metric: $$(ds)^2 = -(dt)^2 + (dr)^2 + (r^2 + b^2)( (d\theta)^2 + sin^2 \theta (d\phi)^2 )$$ Where ##b=1## with an orbit only in the equatorial plane. We also know from the question that the orbit must satisfy this relationship: $$\varepsilon = \frac{1}{2} (\frac{dr}{d\tau})^2 + V_{eff}(r)$$ Ultimately, I was tasked to find the initial...
Back
Top