# Yukawa Potential

by Hart
Tags: potential, yukawa
 P: 169 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.
 P: 675 The solution to the integral in the 2nd to last line is incorrect. Scroll back a page and see how we solved it.
P: 169
 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?
 P: 675 Then where did the 4*pi come from since you used it in line 2 already. And why isn't the denominator squared?
 P: 675 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 $$2/a_0$$
 P: 169 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?
P: 675
 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.
 P: 169 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?
 P: 675 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.
 P: 169 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??
 P: 675 That is correct (in theory). But you solved the integral wrong. It is the exact same integral as before but without the gamma.
 P: 169 ..huh?
P: 675
 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.
 P: 169 .. 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?!

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