Yukawa Potential Homework: Sketch, Schrodinger Eq & Perturbation Theory

[Hart]

Homework Statement

The Yukawa potential is given by:

V_{\gamma}(r) = -\frac{q^{2}}{4\pi \epsilon_{0}r}e^{-\gamma r}

Where \gamma is a constant. This describes a screened Coulomb potential.

I. Sketch the radial dependence of this potential.

II. State the radial Schodinger equation for this potential.

III. Assume that \gamma is a small parameter. Use first order perturbation theory to estimate the energy shift of the ground state of the hydrogen atom when the Coulomb potential is replaced by the Yukawa potential. Hint: The perturbation is:

\omega = V_{\gamma}-V_{0}.

Homework Equations

Within the question statement / answers.

The Attempt at a Solution

I don’t seem to be able to find much useful information about this potential which I can understand and/or use to try and answer this question.

I. Obviously I know that it’s hard to show what it looks like over these forums, but any description of what it looks like would be great.. I don’t know.

II. Help and advice needed!

III. Again, help and advice needed!

.. sorry I don't have really any input at the moment, but I really don't get what's going on here and havn't been able to find much info elsewhere.

 

[nickjer]

I) Just plot it in a graphing calculator or some software. It will look like:

\frac{-1}{r}e^{-r}

2) Just write out the Schrodinger equation in spherical coords using this potential for V(r)

3) Expand that perturbation to first order in lambda using a Taylor expansion. Then operate on this perturbation with the ground state:

E_0^{(1)} = <0|V_{\lambda}-V_0|0>

where |0> is the ground state of the hydrogen atom.

 

[Hart]

1. I don't have a graphical calculator and I havn't really managed to find any good free graphing software from online searches.. any useful links or advice?

2. So the Schrodinger equation in spherical polar coordinates:

E \psi \left(r, \theta, \phi \right) = \left[-\frac{\hbar^{2}}{2mr^{2}}\frac{\partial}{\partial r}\left(r^{2}\frac{\partial}{\partial r}\right) - \frac{\hbar^{2}}{2mr^{2}}\left(\frac{1}{sin \theta}\frac{\partial}{\partial \theta}\left(sin \theta \frac{\partial}{\partial \theta}\right) + \frac{1}{sin^{2}\theta}\frac{\partial^{2}}{\partial \phi^{2}}\right) + V(r)\right] \psi \left( r, \theta, \phi \right)

Hence with input of the potential defined:

E \psi \left(r, \theta, \phi \right) = \left[-\frac{\hbar^{2}}{2mr^{2}}\frac{\partial}{\partial r}\left(r^{2}\frac{\partial}{\partial r}\right) - \frac{\hbar^{2}}{2mr^{2}}\left(\frac{1}{sin \theta}\frac{\partial}{\partial \theta}\left(sin \theta \frac{\partial}{\partial \theta}\right) + \frac{1}{sin^{2}\theta}\frac{\partial^{2}}{\partial \phi^{2}}\right) + <br /> -\frac{q^{2}}{4\pi \epsilon_{0}r}e^{-\gamma r}\right] \psi \left( r, \theta, \phi \right)

3. I'm not sure how to do the Taylor expansion / perturbation thing :|

 

[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}&#039; - E_{0} \approx \left&lt; \psi_{n}^{0}|\omega|\psi_{n}^{0}\right&gt;

so..

E_{n}^{1} = E_{0}&#039; - E_{0} \approx \left&lt; \psi_{n}^{0}|V_{\gamma}-V_{0}|\psi_{n}^{0}\right&gt;

.. and then I need to do some sort of Taylor expansion thing with the perturbation?!

Really don't get this

 

[nickjer]

What is the Taylor Expansion for an exponential?

 

[nickjer]

Also when they ask for the radial Schrodinger Eqn, I think they only want it in terms of 'r' and not theta and phi. You can separate out theta and phi using spherical harmonics.

 

[gabbagabbahey]

1. I don't have a graphical calculator and I haven't really managed to find any good free graphing software from online searches.. any useful links or advice?

At large r, e^{-r} falls off much faster than any polynomial of r, so you would expect it to look a lot like an inverted (because of the negative sign in front) decaying exponential there.

For small r, e^{-r}\approx 1-r so you would expect it to (again inverted) decay like 1/r from -\infty there.

In between the two extremes, you would expect it to decay slightly faster than an exponential, from a large negative value towards zero.

As for graphing software, gnuplot should be more than enough. I suggest you do three plots and superimpose them. Plot -\frac{1}{r}, -e^{-r} and -\frac{1}{r}e^{-r} together so you can compare their behavior at large and small r. It is also worth noting that \frac{1}{r}e^{-r}=1 at r\approx 0.567, so you may want to adjust the range of at least one of your graphs to examine the behavior near this point.

 

[Hart]

Taylor Expansion for an Exponential:

Isn't it this..

e^{x} = \sum_{n=0}^{\infty}\frac{x^{n}}{n!} = 1 + x + \frac{x^{2}}{2!} + \frac{x^{3}}{3!}+ ...

??!

If so, presumably then have to put the function:

\frac{-1}{r}e^{-r}

into it? where x = r?

Radial Schodinger Equation:

.. oh, so you think it should be this instead:

ER = -\frac{\hbar^{2}}{2m_{e}}\frac{1}{r^{2}}\frac{d}{dr}\left(r^{2}\frac{dR}{dr}\right) + \left[V(r) + \frac{l(l+1)\hbar}^{2}{2m_{e}r^{2}}\right]R

then input given potential:

V_{\gamma}(r) = -\frac{q^{2}}{4\pi \epsilon_{0}r}e^{-\gamma r}

hence:

ER = -\frac{\hbar^{2}}{2m_{e}}\frac{1}{r^{2}}\frac{d}{dr}\left(r^{2}\frac{dR}{dr}\right) + \left[-\frac{q^{2}}{4\pi \epsilon_{0}r}e^{-\gamma r} + \frac{l(l+1)\hbar^{2}}{2m_{e}r^{2}}\right]R

?!


Yukawa Plot:

I plotted the three functions as suggested, and using the software suggested, this is a plot of them:

(nb1: apologies, can't figure out how to make the images smaller.)
(nb2: had to use x as variable not r, it wouldn't let me plot it otherwise)

So the Yukawa Potential should look like this:

.. yes?

 

[nickjer]

I) You don't need to plot negative 'r' since the radius is from 0 to infinity.

III) Actually you can solve it analytically without the need of an expansion to first order in \lambda

Try solving:
&lt;\psi_0|-\frac{q^{2}}{4\pi \epsilon_{0}r}e^{-\gamma r}|\psi_0&gt;

Using the hyrdrogen atom's ground state wavefunction, which should just be an exponential. Just be sure to integrate it in spherical coords.

That energy should be very similar to the ground state energy but with an extra \lambda term.

To check your answer. Let \lambda go to 0, and see if it matches the ground state energy when using the Coulomb potential.

 

[Hart]

1. OK so on that second graph, which will be the one I actually use and sketch to represent the Yukawa potential, there only needs to be the right hand side of it yes? i.e. only need the line going from y = -\infty to x = \infty?

2. Could you give a bit more guidance on how to do that? Just to get me started.

 

[nickjer]

&lt;\psi|V(r)|\psi&gt; = \int \int \int \psi^{\dagger}(r,\theta,\phi) V(r) \psi(r,\theta,\phi) r^2 sin(\theta) dr d\theta d\phi

Hint: The ground state wave function for your problem has no angular dependence. So the angular integrals will factor out and you get 4\pi from them:

&lt;\psi|V(r)|\psi&gt; = 4\pi \int \int \int \psi^{\dagger}(r) V(r) \psi(r) r^2 dr

 

[Hart]

.. so in that expression:

V(r) = \frac{-1}{r}e^{-r}

and

\psi(r)^{+} = \psi(r) = \frac{1}{\sqrt{\pi a^{3}}} e^{\left(\frac{-r}{a}\right)}

??

Then just need to input these - hence multiply them together? and then obviously integrate.

 

[nickjer]

Your V(r) has more terms in it than just that.

 

[Hart]

.. how?/why? .. the potential is given as:

V_{\gamma}(r) = -\frac{q^{2}}{4\pi \epsilon_{0}r}e^{-\gamma r}

but I thought that could take the factor out just to leave the exponential to put within the integrals?

 

[nickjer]

You need to solve the integral with the \gamma still in the potential. You can pull the other terms out front, but they will still be included in the final energy.

 

[Hart]

.. ok, so it should be this then:

V(r) = \frac{-1}{r}e^{-\gamma r}

as can take this bit outside the integral calculation:

\frac{q^{2}}{4\pi \epsilon_{0}}

do I just treat \gamma as a constant?

and also

\psi(r)^{+} = \psi(r) = \frac{1}{\sqrt{\pi a^{3}}} e^{\left(\frac{-r}{a}\right)}

??

 

[nickjer]

Yes, \gamma is a constant.

Also, that looks like the ground state of the hydrogen atom. So you are good there.

 

[Hart]

Right so using these:

V(r) = \frac{-1}{r}e^{-\gamma r} and \psi(r)^{+} = \psi(r) = \frac{1}{\sqrt{\pi a^{3}}} e^{\left(\frac{-r}{a}\right)}

I've got to that:

\psi^{\dagger}(r) V(r) \psi(r) = -\left(\frac{1}{\pi r a^{3}}\right)e^{-\left(\frac{2}{a}+\gamma\right)r}

Therefore:

\int \psi^{\dagger}(r) V(r) \psi(r) r^2 dr = \int -\left(\frac{r}{\pi a^{3}}\right)e^{-\left(\frac{2}{a}+\gamma\right)r}dr = -\left(\frac{1}{\pi a^{3}}\right)\int r e^{-\left(\frac{2}{a}+\gamma\right)r}dr

So:

\int r e^{-\left(\frac{2}{a}+\gamma\right)r}dr = \left[-\left(\frac{r}{\frac{2}{a}+\gamma}\right)e^{-\left(\frac{2}{a}+r\right)r}\left] + \left[e^{-\left(\frac{2}{a}+r\right)}\right]

Hence:

\int \psi^{\dagger}(r) V(r) \psi(r) r^2 dr = -\left(\frac{1}{\pi a^{3}}\right)\int r e^{-\left(\frac{2}{a}+\gamma\right)r}dr = -\left(\frac{1}{\pi a^{3}}\right)<br /> <br /> <br /> \left(1- \left(\frac{r}{\frac{2}{a}+\gamma}\right)\right)e^{-\left(\frac{2}{a}+r\right)r}

.. hopefully this is going along the right lines?! :shy:


.. if so, I then have to integrate twice more?

 

[nickjer]

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.

 

[Hart]

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

 

[nickjer]

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.

 

[Hart]

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?

 

[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<br /> = \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.

 

[Hart]

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)

??

 

[nickjer]

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

 

[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?

 

[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.

 

[Hart]

Um, right so this:

\frac{q^{2}}{4\pi \epsilon_{0}}

Hence:

&lt;\psi|V_{\gamma}(r)|\psi&gt; = -\left(\frac{q^{2}}{\alpha^{2}\epsilon_{0}}\right)

??

 

[nickjer]

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.

 

[Hart]

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.

 

[nickjer]

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

 

[Hart]

\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}

 

[Hart]

\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 = <br /> <br /> \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}<br /> <br /> =\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.

 

[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&lt;\psi V_{\gamma}(r)\psi\right&gt; = \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?

 

[ManyNames]

.. 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}&#039; - E_{0} \approx \left&lt; \psi_{n}^{0}|\omega|\psi_{n}^{0}\right&gt;

so..

E_{n}^{1} = E_{0}&#039; - E_{0} \approx \left&lt; \psi_{n}^{0}|V_{\gamma}-V_{0}|\psi_{n}^{0}\right&gt;

.. 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&gt;=\sum^{\infty}_{j=0} \lambda^{j}|\psi_{n}&gt;

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

 

[nickjer]

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&lt;\psi V_{\gamma}(r)\psi\right&gt; = \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.

 

[Hart]

These are my exact full calculations:

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

\left&lt;\psi(r)|V_{\gamma}(r)|\psi(r)\right&gt; = \left&lt;\psi(r)|-\left(\frac{q^{2}}{4\pi \epsilon_{0}r}\right)e^{-\gamma r}|\psi(r)\right&gt; = 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.

 

[nickjer]

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

 

[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?

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?

 

[nickjer]

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

 

[nickjer]

Also, once you finished this, you will need to do the same for the coulomb potential. Since you need to do

&lt;\psi_0|V_\lambda-V|\psi_0&gt;=&lt;\psi_0|V_\lambda|\psi_0&gt;-&lt;\psi_0|V|\psi_0&gt;

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

 

[Hart]

so it should be:

\left&lt;\psi(r)|V_{\gamma}(r)|\psi(r)\right&gt; = \left&lt;\psi(r)|-\left(\frac{q^{2}}{4\pi \epsilon_{0}r}\right)e^{-\gamma r}|\psi(r)\right&gt; =\left(-\frac{q^{2}}{\epsilon_{0}\pi a^{3}}\right)\left(-\frac{1}{\left(\frac{2}{a}+\gamma}\right)^{2}\right)

better?

 

[nickjer]

.. 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.

 

[Hart]

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&lt;\psi(r)|V_{\gamma}(r)|\psi(r)\right&gt; = \left&lt;\psi(r)|-\left(\frac{q^{2}}{4\pi \epsilon_{0}r}\right)e^{-\gamma r}|\psi(r)\right&gt; =\left(-\frac{q^{2}}{\epsilon_{0}\pi a^{3}}\right)\left(\frac{1}{\left(\frac{2}{a}+\gamma}\right)^{2}\right)

does it?

 

[nickjer]

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]

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:

&lt;\psi_0|V_\lambda-V|\psi_0&gt;=&lt;\psi_0|V_\lambda|\psi_0&gt;-&lt;\psi_0|V|\psi_0&gt;

correct??

 

[nickjer]

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

 

[Hart]

..huh?

 

[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<br /> = \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.

 

[Hart]

.. 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?!