# Kinetic energy of a hydrogen atom in its ground state

1. Jan 31, 2014

### Nikitin

1. The problem statement, all variables and given/known data
While playing around with basic QM, I tried using the hamilton operator to find the kinetic energy of a hydrogen atom in its ground state. I assume the wave function $\psi_1$ is known. However, I of course ran into problems...

1) in my solution attempt below, I end up with a kinetic energy 3 times the correct one.
2) Since I am dividing the momentum by the electron mass, am I finding only the kinetic energy of the electron and not the whole hydrogen atom? I mean, doesn't the proton have spin energy?

So while discounting the spin-energy of the proton may sound idiotic, I tried it anyway and ended up with an answer which isn't ridiculously wrong, so I at least did something right. The thing is though I feel I am calculating blindly as I don't know how the wave-function I use was derived.
3. The attempt at a solution

My plan:
1) find the kinetic energy as a function of r using the hamilton operator on the ground-state wavefunction for Hydrogen: $\psi_1 = \frac{1}{\sqrt{\pi a_0^3}} \cdot e^{-r/a_0}$.
2) integrate to find the expected kinetic energy.

Look at the uploaded picture for details:

https://fbcdn-sphotos-g-a.akamaihd.net/hphotos-ak-ash3/t1/q80/s720x720/1011747_10202355134216366_233891779_n.jpg

Can somebody point out my mistakes?

Last edited: Jan 31, 2014
2. Jan 31, 2014

I think people are more likely to respond if you took some time and laid out your work neatly and legibly. For example what does "oppgave2forts" mean?

3. Jan 31, 2014

### Nikitin

It means assignment 2 continued. Just some irrelevant Norwegian.

And yes, you are right, my writing is somewhat illegible (for instance, one of my 2s in the second line look like a curly d) as I wrote with a bruised thumb. I'll upload a new picture in a minute

EDIT: OK I updated the OP with more clear writing. I hope it's easier to understand now.

Last edited: Jan 31, 2014
4. Jan 31, 2014

### vela

Staff Emeritus
If you're off by a constant factor, it probably means you made a sign mistake (2+1 = 3 instead of 2-1 = 1) or you made an arithmetic mistake.

5. Jan 31, 2014

### Nikitin

Yes I thought so too, but did you read 2) in the OP? Is my plan a good one in the first place? I wanna know if there is any point in looking for a calculation mistake, because it might be my logic that is mistaken.

PS: I uploaded a new picture in the OP

6. Jan 31, 2014

Hello Nikitin. Have you looked at the LaTex system for typing equations? I guess you are a student and it will probably be to your future advantage to familiarise yourself with the system. I don't know how to use it but it's on my list of things to look at.

7. Jan 31, 2014

### Nikitin

I am familiar with latex, but you still can't read what I wrote in the uploaded picture?

8. Jan 31, 2014

It's good and I can read it. I just didn't know if you were familiar with LaTex.
I'm sorry but I can't help with your specific question. I used to know some of that sort of stuff years ago but I have forgotten most of it because I haven't used it.

9. Jan 31, 2014

### vela

Staff Emeritus
I know that posting pictures is easier for you, but the helpers generally prefer that you type in your work rather than force them to open another window to see your work, which they can't quote and is often hard to read. I tend to skip over threads where the poster uses images.

10. Jan 31, 2014

### Staff: Mentor

You have a sign problem when taking the derivative. First, the minus in $-\hbar^2 / 2m_e$ disappears. Second, you should get different signs for $1/a_0$ and $2/r$. (This is where quoting can be useful for everybody, as vela pointed out )

There is no such thing as "spin energy". In the absence of external fields, there is no difference in energy between the two spin orientations of H. (Spin of the nucleus has an effect on the energy levels of the electron through the hyperfine interaction, but that's probably not what you're thinking about here.)

The original Hamiltonian this comes from is that for an electron interacting with a fixed (infinitely heavy) nucleus. This is why only the mass of the electron appears. A much better result is obtained by replacing it by the reduced mass for the proton plus the electron (this also changes the constant $a_0$).

Also be careful that you are not considering the Hamiltonian of the system, but only the kinetic energy operator. Your first equation should not start with $\hat{H} \psi$, but rather $\hat{T} \psi$ (or $\hat{K} \psi$, or whatever convention you want to use to denote the kinetic energy operator).

11. Feb 1, 2014

### Nikitin

OK thanks for the help guys. And yes, I'll keep the latex thing in mind for next time.