## Yukawa Potential

 Quote by Hart .. still can't figure out how to do part 3 of this question This is as far as I've got: $$E_{n}^{1} = E_{0}' - E_{0} \approx \left< \psi_{n}^{0}|\omega|\psi_{n}^{0}\right>$$ so.. $$E_{n}^{1} = E_{0}' - E_{0} \approx \left< \psi_{n}^{0}|V_{\gamma}-V_{0}|\psi_{n}^{0}\right>$$ .. and then I need to do some sort of Taylor expansion thing with the perturbation?! Really don't get this

Pertubation expansion is done by so:

$E_n=\sum^{\infty}_{j=0} \lambda^{j}E^{j}_{n}$

Its wave function is expanded in terms given by:

$|\psi>=\sum^{\infty}_{j=0} \lambda^{j}|\psi_{n}>$

This all works when dealing with Eigenvalues dealing with $H=H_{o}+ \lambda H$ where $\lambda$is the expansion parameter.

 Quote by Hart 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.

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

 Quote by 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?
 Quote by nickjer 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?

 Then where did the 4*pi come from since you used it in line 2 already. And why isn't the denominator squared?
 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 $$2/a_0$$
 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?

 Quote by Hart .. of course! what a silly mistake there! 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.

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

 Quote by 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.

 .. 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?!