# Yukawa Potential

by Hart
Tags: potential, yukawa
 P: 676 You made a mistake in the 4th line. You might want to double check your 2nd term after the equal sign in that integration. Also, not sure what you mean by integrate twice more. And where are the limits on your integral. The integral should be taken from 0 to infinity.
 P: 169 Ah, so it should be.. $$\int_{0}^{\infty} r e^{-\left(\frac{2}{a}+\gamma\right)r}dr = \left[-\left(\frac{r}{\frac{2}{a}+\gamma}\right)e^{-\left(\frac{2}{a}+\gamma\right)r}\left] + \left[e^{-\left(\frac{2}{a}+\gamma\right)r}\right] = \left[\left(1-\left(\frac{r}{\frac{2}{a}+\gamma}\right)\right)e^{-\left(\frac{2}{a}+\gamma\right)r}\right]^{\infty}_{0} = -1$$ ..since $e^{-\infty} = 0$ ?? and I meant, do I need to integrate this again as your expression had $\int\int\int$
 P: 676 Your 2nd term after the first equal sign is still incorrect. You might want to double check your math. Yes, it was a volume integral, but since there were no angular components in the potential or wavefunction. So I already integrated over theta and phi, and I got 4*Pi.
 P: 169 I don't see how? I was using product rule, so.. $$\int_{0}^{\infty} r e^{-\left(\frac{2}{a}+\gamma\right)r}dr = U.V$$ Therefore: $$U.dV = \left[-\left(\frac{r}{\frac{2}{a}+\gamma}\right)e^{-\left(\frac{2}{a}+\gamma\right)r}\left]$$ And: $$V.dU = (1).e^{-\left(\frac{2}{a}+\gamma\right)r} = e^{-\left(\frac{2}{a}+\gamma\right)r}$$ .. I can't see where the mistake is (sorry!). Right, thanks for clarifying that, so whatever this integral equals then multiplied by $4\pi$ is the answer?
 P: 676 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.
 P: 169 Oh right OK I see, so it should be this then.. $$\int_{0}^{\infty} r e^{-\left(\frac{2}{a}+\gamma\right)r}dr = \left(\left(-\frac{r}{\alpha}-\frac{1}{\alpha^{2}}\right)e^{-\left(\frac{2}{a}+\gamma\right)r}\right)\right|^{\infty}_{0}$$ ?? And then inputting the limits: $$\implies = \left(\frac{r}{a}-\frac{1}{\alpha^{2}}\right)$$ ?? Therefore: $$\implies = 4 \pi \left(\frac{r}{a}-\frac{1}{\alpha^{2}}\right)$$ ??
 P: 676 You seem to still have an 'r' leftover. You need to set the limits for all the 'r's in the equation.
 P: 169 .. 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?
 P: 676 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.
 P: 169 Um, right so this: $$\frac{q^{2}}{4\pi \epsilon_{0}}$$ Hence: $$<\psi|V_{\gamma}(r)|\psi> = -\left(\frac{q^{2}}{\alpha^{2}\epsilon_{0}}\right)$$ ??
 P: 676 There are more constants out front than just that. You should write out the whole equation from the beginning. And keep track of all the constants. Also, you need to substitute back in what alpha is. You can't have alpha there, since I made it up to make the work simpler.
 P: 169 I've got that: $$\alpha = \frac{2}{a}+\gamma$$ Therefore: $$-\left(\frac{4 \pi }{\alpha^{2}}\right) = -\left(\frac{4 \pi }{\left(\frac{2}{a}+\gamma\right)^{2}}\right)$$ .. but then don't see where to go from that.
 P: 676 Like I said, stop doing it in pieces. You need to write out the whole thing with every constant.
 P: 169 $$\int \psi^{\dagger}(r) V(r) \psi(r) r^2 dr = \int_{0}^{\infty}\left(\frac{-1}{\pi r a^{3}}\right)e^{-\left(\frac{2}{a}+\gamma\right)r^{2}}dr = \left[\left(\frac{2r^{2}}{\pi a^{3}}\left(\frac{2}{a}-\gamma\right)\right)e^{-\left(\frac{2}{a}+\gamma\right)r^{2}}\right]^{\infty}_{0}$$
 P: 169 $$\int \psi^{\dagger}(r) V(r) \psi(r) r^2 dr = \int_{0}^{\infty}\left(\frac{-1}{\pi r a^{3}}\right)e^{-\left(\frac{2}{a}+\gamma\right)r^{2}}dr = \left[\left(\frac{-1}{\pi r a^{3}}\right)\left(-\frac{2}{a}+\gamma\right)\left(2r\right)e^{-\left(\frac{2}{a}+\gamma\right)r^{2}}\right]^{\infty}_{0} =\left[\left(\frac{2r^{2}}{\pi a^{3}}\left(\frac{2}{a}-\gamma\right)\right)e^{-\left(\frac{2}{a}-\gamma\right)r^{2}}\right]^{\infty}_{0}$$ .. but then I can't see where I've gone wrong, because obviously inputting limits will give 0 as the result.
 P: 169 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?
P: 136
 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.
P: 676
 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.

 Related Discussions Quantum Physics 14 Quantum Physics 2 Advanced Physics Homework 3 Advanced Physics Homework 6 Advanced Physics Homework 7