Yukawa Potential


by Hart
Tags: potential, yukawa
Hart
Hart is offline
#37
Mar7-10, 07:44 PM
P: 169
These are my exact full calculations:

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

[tex]\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[/tex]

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

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

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

[tex]=\frac{4 q^{2}}{\epsilon_{0}a^{3}\left(\frac{2}{a}+\gamma\right)}[/tex]

.. 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.
nickjer
nickjer is offline
#38
Mar7-10, 08:22 PM
P: 676
The solution to the integral in the 2nd to last line is incorrect. Scroll back a page and see how we solved it.
Hart
Hart is offline
#39
Mar7-10, 08:27 PM
P: 169
Quote Quote by Hart View Post

so..

[tex]\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}[/tex]

??

And then inputting the limits:

[tex]\implies = \left(0-\frac{1}{\alpha^{2}}\right)[/tex]

??

Therefore:

[tex]\implies = \left(-\frac{4 \pi }{\alpha^{2}}\right)[/tex]

.. better?
Quote Quote by nickjer View Post
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 [itex]\alpha[/itex] substituted.

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

[tex]=\frac{4 q^{2}}{\epsilon_{0}a^{3}\left(\frac{2}{a}+\gamma\right)^{2}}[/tex]

correct now?
nickjer
nickjer is offline
#40
Mar7-10, 08:28 PM
P: 676
Then where did the 4*pi come from since you used it in line 2 already. And why isn't the denominator squared?
nickjer
nickjer is offline
#41
Mar7-10, 08:31 PM
P: 676
Also, once you finished this, you will need to do the same for the coulomb potential. Since you need to do

[tex]<\psi_0|V_\lambda-V|\psi_0>=<\psi_0|V_\lambda|\psi_0>-<\psi_0|V|\psi_0>[/tex]

But that will be much easier to solve, since we solved it basically already. You would replace [tex]\alpha[/tex] from your previous integral with just [tex]2/a_0[/tex]
Hart
Hart is offline
#42
Mar7-10, 08:36 PM
P: 169
so it should be:

[tex]\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)[/tex]

better?
nickjer
nickjer is offline
#43
Mar7-10, 08:41 PM
P: 676
Quote Quote by Hart View Post
.. of course! what a silly mistake there!

so..

[tex]\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}[/tex]

??

And then inputting the limits:

[tex]\implies = \left(0-\frac{1}{\alpha^{2}}\right)[/tex]

??

Therefore:

[tex]\implies = \left(-\frac{4 \pi }{\alpha^{2}}\right)[/tex]

.. better?
Forgot to mention, the solution is positive. You subtract the last limit of r=0, making it positive.
Hart
Hart is offline
#44
Mar7-10, 08:47 PM
P: 169
erm.. yep. so should be:

[tex]\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}}[/tex]

right?

but that doesn't help simplify this:

[tex]\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)[/tex]

does it?
nickjer
nickjer is offline
#45
Mar7-10, 08:54 PM
P: 676
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
Hart is offline
#46
Mar7-10, 08:59 PM
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:

[tex]\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)[/tex]

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

[tex]<\psi_0|V_\lambda-V|\psi_0>=<\psi_0|V_\lambda|\psi_0>-<\psi_0|V|\psi_0>[/tex]

correct??
nickjer
nickjer is offline
#47
Mar7-10, 09:08 PM
P: 676
That is correct (in theory). But you solved the integral wrong. It is the exact same integral as before but without the gamma.
Hart
Hart is offline
#48
Mar7-10, 09:12 PM
P: 169
..huh?
nickjer
nickjer is offline
#49
Mar7-10, 09:13 PM
P: 676
Quote Quote by nickjer View Post
You have:

[tex]\int r e^{-\alpha r}dr[/tex]

Use,
[tex]r = u[/tex]
[tex]e^{-\alpha r} dr = dv[/tex]

That gives,
[tex]dr = du[/tex]
[tex]\frac{-1}{\alpha} e^{-\alpha r} = v[/tex]

Use,
[tex]\int u dv = uv - \int v du[/tex]
[tex]\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}[/tex]

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
Hart is offline
#50
Mar7-10, 09:13 PM
P: 169
.. so you mean for the second part it will be this:

[tex]\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}[/tex]

??

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


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