# Yukawa Potential

nickjer
Got a bit confused now, this is what I have noted down in addition to the previous post (I think this is more correct):

$$\left<\psi V_{\gamma}(r)\psi\right> = \int^{\infty}_{0}\left[-\left(\frac{1}{\pi r a^{3}}\right)e^{-r\left(\frac{2}{a}-\gamma\right)}\right]dr$$

so then get this:

$$=\left[\frac{1}{\pi a^{3}}\left(\frac{2}{a}-\gamma\right)e^{-r\left(\frac{2}{a}-\gamma\right)}\right]^{\infty}_{0}$$

and after input limits:

$$=\frac{1}{\pi a^{3}}\left(\frac{2}{a}-\gamma\right)$$

.. any good?

You are still leaving out constants, like charge. Also, you completely changed the solution. Not sure what happened but now things are worse.

Go back and carefully step through everything on paper that way you can keep track of everything.

Hart
These are my exact full calculations:

$$V_{\gamma}(r) = -\left(\frac{q^{2}}{4\pi \epsilon_{0}r}\right)e^{-\gamma r}$$

$$\left<\psi(r)|V_{\gamma}(r)|\psi(r)\right> = \left<\psi(r)|-\left(\frac{q^{2}}{4\pi \epsilon_{0}r}\right)e^{-\gamma r}|\psi(r)\right> = 4\pi \int^{\infty}_{0}\psi(r)V_{\gamma}(r)\psi(r)r^{2}dr$$

$$=\left(-\frac{4\pi q^{2}}{4\pi \epsilon_{0}}\right)\int^{\infty}_{0}r e^{-\gamma r}\psi(r)\psi(r) dr$$

$$=\left(-\frac{q^{2}}{\epsilon_{0}\pi a^{3}}\right)\int^{\infty}_{0}r e^{-\left(\frac{2}{a}+\gamma\right)r} dr$$

$$=\left(-\frac{q^{2}}{\epsilon_{0}\pi a^{3}}\right)\left(-\frac{4 \pi}{\frac{2}{a}+\gamma}\right)$$

$$=\frac{4 q^{2}}{\epsilon_{0}a^{3}\left(\frac{2}{a}+\gamma\right)}$$

.. I'm hoping this is along the right lines, though I can't see that I've missed anything else now, but I'm unsure what I need to do further now.

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nickjer
The solution to the integral in the 2nd to last line is incorrect. Scroll back a page and see how we solved it.

Hart
so..

$$\int_{0}^{\infty} r e^{-\left(\frac{2}{a}+\gamma\right)r}dr = \left(\left(-\frac{r}{\alpha}-\frac{1}{\alpha^{2}}\right)e^{-\left(\frac{2}{a}+\gamma\right)r}\right)\right|^{\infty}_{0}$$

??

And then inputting the limits:

$$\implies = \left(0-\frac{1}{\alpha^{2}}\right)$$

??

Therefore:

$$\implies = \left(-\frac{4 \pi }{\alpha^{2}}\right)$$

.. better?

Yes, now put back in all the constants. And it should look very similar to the ground state energy of a hydrogen atom using the coulomb potential.

.. that's what I've used, obviously with the value of $\alpha$ substituted.

.. OH wait, it's $\alpha^{2}$ not $\alpha$, so then the result should be:

$$=\frac{4 q^{2}}{\epsilon_{0}a^{3}\left(\frac{2}{a}+\gamma\right)^{2}}$$

correct now?

nickjer
Then where did the 4*pi come from since you used it in line 2 already. And why isn't the denominator squared?

nickjer
Also, once you finished this, you will need to do the same for the coulomb potential. Since you need to do

$$<\psi_0|V_\lambda-V|\psi_0>=<\psi_0|V_\lambda|\psi_0>-<\psi_0|V|\psi_0>$$

But that will be much easier to solve, since we solved it basically already. You would replace $$\alpha$$ from your previous integral with just $$2/a_0$$

Hart
so it should be:

$$\left<\psi(r)|V_{\gamma}(r)|\psi(r)\right> = \left<\psi(r)|-\left(\frac{q^{2}}{4\pi \epsilon_{0}r}\right)e^{-\gamma r}|\psi(r)\right> =\left(-\frac{q^{2}}{\epsilon_{0}\pi a^{3}}\right)\left(-\frac{1}{\left(\frac{2}{a}+\gamma}\right)^{2}\right)$$

better?

nickjer
.. of course! what a silly mistake there! :grumpy:

so..

$$\int_{0}^{\infty} r e^{-\left(\frac{2}{a}+\gamma\right)r}dr = \left(\left(-\frac{r}{\alpha}-\frac{1}{\alpha^{2}}\right)e^{-\left(\frac{2}{a}+\gamma\right)r}\right)\right|^{\infty}_{0}$$

??

And then inputting the limits:

$$\implies = \left(0-\frac{1}{\alpha^{2}}\right)$$

??

Therefore:

$$\implies = \left(-\frac{4 \pi }{\alpha^{2}}\right)$$

.. better?

Forgot to mention, the solution is positive. You subtract the last limit of r=0, making it positive.

Hart
erm.. yep. so should be:

$$\int_{0}^{\infty} r e^{-\left(\frac{2}{a}+\gamma\right)r}dr = \left(\left(\frac{r}{\left(\frac{2}{a}+\gamma\right)}-\frac{1}{\left(\frac{2}{a}+\gamma\right)^{2}}\right)e^{-\left(\frac{2}{a}+\gamma\right)r}\right)\right|^{\infty}_{0} = \frac{4\pi}{\left(\frac{2}{a}+\gamma\right)^{2}}$$

right?

but that doesn't help simplify this:

$$\left<\psi(r)|V_{\gamma}(r)|\psi(r)\right> = \left<\psi(r)|-\left(\frac{q^{2}}{4\pi \epsilon_{0}r}\right)e^{-\gamma r}|\psi(r)\right> =\left(-\frac{q^{2}}{\epsilon_{0}\pi a^{3}}\right)\left(\frac{1}{\left(\frac{2}{a}+\gamma}\right)^{2}\right)$$

does it?

nickjer
There is no 4*Pi in the first line, since you already pulled out the angular integrals. The last line is now correct. To simplify it, I would recommend pulling out the 2/a from the denominator so it more resembles the ground state of the coulomb potential.

Now go back and solve this again for just the coulomb potential. You will notice a similarity.

Hart
um, don't get how I can rearrange that how you said

.. also, so to get the final answer I need also caclulate this:

$$\int_{0}^{\infty} r e^{-\left(\frac{2}{a_{0}}\right)r}dr = \left(\left(-\frac{a_{0}r}{2}-\frac{a_{0}^{2}}{4}\right)e^{-\left(\frac{a_{0}}{2}\right)r}\right)\right|^{\infty}_{0} = \left(\frac{a_{0}r}{2}+\frac{a_{0}^{2}}{4}\right)$$

Then take this away from what we're currently working out, so that the final answer is:

$$<\psi_0|V_\lambda-V|\psi_0>=<\psi_0|V_\lambda|\psi_0>-<\psi_0|V|\psi_0>$$

correct??

nickjer
That is correct (in theory). But you solved the integral wrong. It is the exact same integral as before but without the gamma.

Hart
..huh?

nickjer
You have:

$$\int r e^{-\alpha r}dr$$

Use,
$$r = u$$
$$e^{-\alpha r} dr = dv$$

That gives,
$$dr = du$$
$$\frac{-1}{\alpha} e^{-\alpha r} = v$$

Use,
$$\int u dv = uv - \int v du$$
$$\int r e^{-\alpha r}dr = \frac{-r}{\alpha} e^{-\alpha r}+\int \frac{1}{\alpha} e^{-\alpha r}dr = \frac{-r}{\alpha} e^{-\alpha r}-\frac{1}{\alpha^2}e^{-\alpha r}$$

I left out the bounds for simplicity, but you should include them. I also used alpha to make my work easier.

The same as this but now alpha = 2/a and not 2/a+gamma.

Hart
.. so you mean for the second part it will be this:

$$\int_{0}^{\infty} r e^{-\left(\frac{2}{a}\right)r}dr = \left(\frac{r}{\left(\frac{2}{a}\right)}-\frac{1}{\left(\frac{2}{a})^{2}}\right)\right|^{\infty}_{0} = \frac{4\pi}{\left(\frac{2}{a}\right)^{2}} = \pi a^{2}$$

??

Going back to the first bit, how do I rearrange that result better then? I didn't get what you said about pulling the $\frac{2}{a}$ term out?!