
#19
Mar610, 07:39 AM

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. 



#20
Mar610, 08:42 AM

P: 169

Ah, so it should be..
[tex]\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 [/tex] ..since [itex]e^{\infty} = 0[/itex] ?? and I meant, do I need to integrate this again as your expression had [itex]\int\int\int[/itex] 



#21
Mar610, 07:07 PM

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. 



#22
Mar610, 07:23 PM

P: 169

I don't see how?
I was using product rule, so.. [tex]\int_{0}^{\infty} r e^{\left(\frac{2}{a}+\gamma\right)r}dr = U.V[/tex] Therefore: [tex] U.dV = \left[\left(\frac{r}{\frac{2}{a}+\gamma}\right)e^{\left(\frac{2}{a}+\gamma\right)r}\left] [/tex] And: [tex]V.dU = (1).e^{\left(\frac{2}{a}+\gamma\right)r} = e^{\left(\frac{2}{a}+\gamma\right)r}[/tex] .. I can't see where the mistake is (sorry!). Right, thanks for clarifying that, so whatever this integral equals then multiplied by [itex]4\pi[/itex] is the answer? 



#23
Mar610, 07:31 PM

P: 676

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. 



#24
Mar610, 07:49 PM

P: 169

Oh right OK I see, so it should be this then..
[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(\frac{r}{a}\frac{1}{\alpha^{2}}\right)[/tex] ?? Therefore: [tex]\implies = 4 \pi \left(\frac{r}{a}\frac{1}{\alpha^{2}}\right)[/tex] ?? 



#25
Mar610, 07:51 PM

P: 676

You seem to still have an 'r' leftover. You need to set the limits for all the 'r's in the equation.




#26
Mar610, 08:06 PM

P: 169

.. 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? 



#27
Mar610, 08:15 PM

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.




#28
Mar610, 08:26 PM

P: 169

Um, right so this:
[tex] \frac{q^{2}}{4\pi \epsilon_{0}}[/tex] Hence: [tex]<\psiV_{\gamma}(r)\psi> = \left(\frac{q^{2}}{\alpha^{2}\epsilon_{0}}\right)[/tex] ?? 



#29
Mar610, 09:55 PM

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. 



#30
Mar710, 04:08 AM

P: 169

I've got that:
[tex]\alpha = \frac{2}{a}+\gamma[/tex] Therefore: [tex]\left(\frac{4 \pi }{\alpha^{2}}\right) = \left(\frac{4 \pi }{\left(\frac{2}{a}+\gamma\right)^{2}}\right)[/tex] .. but then don't see where to go from that. 



#31
Mar710, 08:10 AM

P: 676

Like I said, stop doing it in pieces. You need to write out the whole thing with every constant.




#32
Mar710, 02:35 PM

P: 169

[tex]\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}[/tex]




#33
Mar710, 02:39 PM

P: 169

[tex]\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}[/tex] .. but then I can't see where I've gone wrong, because obviously inputting limits will give 0 as the result. 



#34
Mar710, 03:05 PM

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):
[tex]\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[/tex] so then get this: [tex]=\left[\frac{1}{\pi a^{3}}\left(\frac{2}{a}\gamma\right)e^{r\left(\frac{2}{a}\gamma\right)}\right]^{\infty}_{0}[/tex] and after input limits: [tex]=\frac{1}{\pi a^{3}}\left(\frac{2}{a}\gamma\right)[/tex] .. any good? 



#35
Mar710, 04:29 PM

P: 136

Pertubation expansion is done by so: [latex]E_n=\sum^{\infty}_{j=0} \lambda^{j}E^{j}_{n}[/latex] Its wave function is expanded in terms given by: [latex]\psi>=\sum^{\infty}_{j=0} \lambda^{j}\psi_{n}>[/latex] This all works when dealing with Eigenvalues dealing with [latex]H=H_{o}+ \lambda H[/latex] where [latex]\lambda[/latex]is the expansion parameter. 



#36
Mar710, 04:48 PM

P: 676

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


Register to reply 
Related Discussions  
Yukawa Potential  Quantum Physics  14  
1)yukawa Potential  Quantum Physics  2  
Yukawa Potential  Advanced Physics Homework  3  
Yukawa potential  Advanced Physics Homework  6  
yukawa potential ,confused  Advanced Physics Homework  7 