# Zero-point energy from Schroedinger equation using Ritz variaton principle

1. Aug 27, 2011

### Juqon

1. The problem statement, all variables and given/known data
Calculate the zero-point energy of the hydrogen atom using the Ritz variaton principle and the approach $\psi_{\alpha}$.
Hint: The stationary Schroedinger equation in spherical coordinates is ...

2. Relevant equations
$$\left\{ -\frac{\hbar^{2}}{2\mu}\left[\frac{\partial^{2}}{\partial r^{2}}+\frac{2}{r}\frac{\partial}{\partial r}\right]+\frac{\hat{\vec{L}}^{2}}{2\mu r^{2}}+V(r)\right\} \psi=E\psi$$$$\psi_{\alpha}(\vec{r})=C\exp\left\{ -\alpha r\right\} ;C>0;\alpha>0; C,\alpha=const.$$ $$\int_{0}^{\infty}dr\cdot r\cdot\exp\left\{ -2\alpha r\right\} =\frac{1}{4}\alpha^{2}$$$$\int_{0}^{\infty}dr\cdot r^{2}\cdot\exp\left\{ -2\alpha r\right\} =\frac{1}{4}\alpha^{3}$$

3. The attempt at a solution

$$\left\{ -\frac{\hbar^{2}}{2\mu}\left[\frac{\partial^{2}}{\partial r^{2}}+\frac{2}{r}\frac{\partial}{\partial r}\right]+\frac{\hat{\vec{L}}^{2}}{2\mu r^{2}}+V(r)\right\} \psi=E\psi$$$$\Leftrightarrow \left\{ -\frac{ \hbar^{2} }{2\mu} \left [\frac{\partial^{2}} {\partial r^{2}}+ \frac{2}{r} \frac {\partial} {\partial r} \right]+ \frac {\left(\hbar^{2} l(l+1) \right)^{2}} {2\mu r^{2}}+\frac{1} {4\pi\epsilon_{0} } \frac{qQ}{r} \cdot C \cdot \exp -\alpha r \right\} \cdot C\cdot\exp\left\{ -\alpha r\right\} = E \cdot C\cdot\exp\left\{ -\alpha r\right\}$$$$\Leftrightarrow \left\{ -\frac{\hbar^{2}}{2\mu}\left [\alpha^{2}-\frac{2}{r}\alpha\right] +\frac{\hbar^{4}l^{2} (l^{2}+2l+1)}{2\mu r^{2}} +\frac{1}{4\pi\epsilon_{0}}\frac{qQ} {r}\right\} \cdot C\cdot\exp\left\{ -\alpha r\right\} = E \cdot C\cdot\ exp\left\{ -\alpha r\right\}$$$$\Leftrightarrow E=\left\{ -\frac{\hbar^{2}}{2\mu}\left[\alpha^{2}-\frac{2}{r}\alpha \right]+\frac{\hbar^{4}l^{2}(l^{2}+2l+1)}{2\mu r^{2}}+\frac{1}{4\pi\epsilon_{0}} \frac{qQ}{r}\right\}$$

Where do I need to integrate at all?

Last edited: Aug 27, 2011
2. Aug 27, 2011

### vela

Staff Emeritus
Your Schrodinger equation seems to be missing a plus or minus sign.

All you've done is apply the Hamiltonian to the trial function. Your estimate for E is supposed to be a constant. You still have r's and l's in it as well as the parameter alpha. You need to review what you're supposed to do to solve for the estimate of the ground state energy. This example might help you see what the general procedure is:

http://quantummechanics.ucsd.edu/ph130a/130_notes/node378.html#example:hopoly

3. Aug 27, 2011

### Juqon

Ok, I will correct the sign, sorry.
The example was very helpful, thanks. Did I use the integral measure correctly? And is the result really that complicated?

$$E_{0}=\frac{<\psi|\hat{H}|\psi>}{<\psi|\psi>}= \frac{\int_{-\infty}^{\infty} dr\cdot r^{2}\cdot\psi^{*}\left\{ -\frac{\hbar^{2}}{2\mu}\left[\alpha^{2}-\frac{2}{r}\alpha\right]\psi+\frac{\hbar^{4}l^{2}(l^{2}+2l+1)}{2\mu r^{2}}\psi+\frac{1}{4\pi \epsilon_{0} } \frac{qQ}{r}\psi\right\} }{\int_{-\infty}^{\infty}dr\psi^{*}\psi}$$

$$=\frac{2\int_{0}^{\infty}dr\cdot r^{2}\cdot\exp\left\{ -2\alpha r\right\} \left\{ \frac{ \hbar^{2}}{2\mu}\frac{2}{r}\alpha+\frac{ \hbar^{4}l^{2}(l^{2}+2l+1)}{2\mu r^{2}}-\frac{ \hbar^{2}}{2\mu}\alpha^{2}+\frac{1}{4\pi \epsilon_{0} } \frac{qQ}{r}\right\} }{2\int_{0}^{\infty}dr\cdot r^{2}\cdot\exp\left\{ -2\alpha r\right\} }$$

$$=\frac{2\int_{0}^{\infty}dr\cdot\exp\left\{ -2\alpha r\right\} \left\{ \frac{\hbar^{2}}{2\mu}2r\alpha+r^{2}\frac{\hbar^{4}l^{2}(l^{2}+2l+1)}{2\mu}-r^{2}\frac{\hbar^{2}}{2\mu}\alpha^{2}+\frac{r}{4 \pi \epsilon_{0} } \frac{qQ}{1}\right\} }{2\int_{0}^{\infty}dr\cdot r^{2}\cdot\exp\left\{ -2\alpha r\right\} }$$

$$=\frac{2\int dr\cdot\exp\left\{ -2\alpha r\right\} \cdot r\left\{ \frac{\hbar^{2}}{2\mu}2\alpha+\frac{1}{4\pi \epsilon_{0} } \frac{qQ}{1}\right\} +2\int dr\cdot\exp\left\{ -2\alpha r\right\} \cdot r^{2}\cdot\left(\frac{\hbar^{4}l^{2}(l^{2}+2l+1)}{2\mu}-\frac{\hbar^{2}}{2\mu}\alpha^{2}\right)}{2\int_{0}^{\infty}dr\cdot r^{2}\cdot\exp\left\{ -2\alpha r\right\} }$$

$$=\frac{2\cdot\frac{1}{4}\alpha^{2}\left\{ \frac{\hbar^{2}}{2\mu}2\alpha+\frac{1}{4\pi \epsilon_{0} } \frac{qQ}{1}\right\} +2\frac{1}{4}\alpha^{3}\cdot\left(\frac{\hbar^{4}l^{2}(l^{2}+2l+1)}{2\mu}-\frac{\hbar^{2}}{2\mu}\alpha^{2}\right)}{2\frac{1}{4}\alpha^{3}}$$

$$=\frac{\frac{1}{2}\alpha^{2}\left\{ \frac{\hbar^{2}}{2\mu}2\alpha+\frac{1}{4\pi \epsilon_{0} } \frac{qQ}{1}\right\} +\frac{1}{2}\alpha^{3}\cdot\left(\frac{\hbar^{4}l^{2}(l^{2}+2l+1)}{2\mu}-\frac{\hbar^{2}}{2\mu}\alpha^{2}\right)}{\frac{1}{2}\alpha^{3}}$$

$$E_{0}=\frac{1}{4\pi \epsilon_{0} } \frac{qQ}{\alpha}+\frac{ \hbar^{2}}{2\mu}2+\frac{\hbar^{4}l^{2}(l^{2}+2l+1)-\hbar^{2}\alpha^{2}}{2\mu}$$

$$\frac{\partial E_{0}}{\partial\alpha}=- \frac{1}{4\pi \epsilon_{0} } \frac{qQ}{\alpha^{2}}+\frac{\hbar^{4}l^{2}(l^{2}+2l+1)-\hbar^{2}\alpha}{\mu}=0\Leftrightarrow\frac{\hbar^{4}l^{2}(l^{2}+2l+1)-\hbar^{2}\alpha}{\mu}=\frac{1}{4\pi \epsilon_{0} } \frac{qQ}{\alpha^{2}}$$

$$\Leftrightarrow\alpha^{3}= \frac{qQ}{4\pi \epsilon_{0} } \frac{\mu}{\hbar^{4}l^{2}(l^{2}+2l+1)- \hbar^{2}}\Leftrightarrow\alpha= \sqrt[3]{\frac{qQ}{4\pi \epsilon_{0} } \frac{\mu}{\hbar^{2}\left(l^{2}(l^{2}+2l+1)-1\right)}}$$

$$E_{0}=\frac{1}{4\pi \epsilon_{0} } \frac{qQ}{\sqrt[3]{\frac{qQ}{4\pi \epsilon_{0} } \frac{\mu}{\hbar^{2}\left(l^{2}(l^{2}+2l+1)-1\right)}}}+ \frac{\hbar^{2}}{2\mu}2+ \frac{\hbar^{4}l^{2}(l^{2}+2l+1) - \hbar^{2}\left(\frac{qQ}{4\pi \epsilon_{0} } \frac{\mu}{\hbar^{2}\left(l^{2}(l^{2}+2l+1)-1\right)}\right)^{\frac{2}{3}}}{2\mu}$$

4. Aug 27, 2011

### vela

Staff Emeritus
It's a bit simpler than what you have. First, find out what the L2 operator looks like in spherical coordinates. That'll simplify things quite a bit.

You used $\hat{L}^2\psi = \hbar^2[l(l+1)]^2\psi$, which is incorrect. If you have an eigenfunction $\psi_{nlm}$, then $\hat{L^2}\psi_{nlm} = \hbar^2 l(l+1) \psi_{nlm}$. The factor l(l+1) isn't squared.

Where did those factors of 2 in front of each integral come from?

5. Aug 28, 2011

### Juqon

Wikipedia says $L^2 = -\hbar^2 \left(\frac{1}{\sin\theta}\frac{\partial}{\partial \theta}\left( \sin\theta \frac{\partial}{\partial \theta}\right) + \frac{1}{\sin^2\theta}\frac{\partial^2}{\partial \phi^2}\right)$. But that would be more complicated, because I don't see it go away by a derivation. $\psi$ is still dependent on the angles. Besides, there is an r² too much in front of $\frac{\hbar^{4}l^{2}(l^{2}+2l+1)}{2\mu r^{2}}$ in my solution.
An integral from -infinity to +infinity is two times the same integral from 0 to infinity.

6. Aug 28, 2011

### vela

Staff Emeritus
You seem to have some misconceptions related to spherical coordinates. You need to get those straightened out first, so you should go back and review the basics of spherical coordinates. Note that the vector $\vec{r}$ and the coordinate $r$ are not the same thing, so in the equations you find in your textbook, you need to understand which one is being referred to.

Say you have a generic wave function $\psi$. In Cartesian coordinates, the quantity $\langle \psi \vert \psi \rangle$ would be given by$$\langle \psi \vert \psi \rangle = \int_{-\infty}^\infty \int_{-\infty}^\infty \int_{-\infty}^\infty \psi^*(x,y,z)\psi(x,y,z)\,dx\,dy\,dz$$How would you express this quantity using spherical coordinates?

7. Aug 28, 2011

### Juqon

I'd say that is in spherical coordinates:
$\langle \psi \vert \psi \rangle = \int_{-\infty}^\infty \int_{-\infty}^\infty \int_{-\infty}^\infty r²\cdot sin² (\theta) \psi^*(r, \theta,\phi)\psi(r,\theta,\phi)\ ,dr\,d \theta\,d\phi$.

I understand of course that $\vec{r}=(r cos \theta cos \phi, r cos \theta sin ..., cos)$ or something like that. Still don't see the connection.

8. Aug 28, 2011

### vela

Staff Emeritus
Close, but you need to understand what values the coordinates can take on in each system. In Cartesian coordinate, you have to let x, y, and z each run from -∞ to +∞ to cover all of space. What about in spherical coordinates? Remember you don't want to cover a point multiple times. There should be only one combination of r, θ, and φ for each point in space.

9. Aug 28, 2011

### Juqon

$$\langle \psi \vert \psi \rangle = \int_{0}^\infty \int_{0}^\pi \int_{0}^{2\pi} r²\cdot sin² (\theta) \psi^*(r, \theta,\phi) \psi(r,\theta,\phi)\ ,dr\,d \theta\, d\phi$$

10. Aug 28, 2011

### vela

Staff Emeritus
That's correct except sin θ shouldn't be squared.

So now back to the problem. You have $\psi = C e^{-\alpha r}$, so$$\langle \psi \vert \psi \rangle = \int_0^\infty \int_0^\pi \int_0^{2\pi} e^{-2\alpha r}r^2\sin \theta \,dr\,d\theta\,d\phi$$What do you get when you evaluate this integral?

11. Aug 28, 2011

### Juqon

$\langle \psi \vert \psi \rangle = \int_0^\infty \int_0^\pi \int_0^{2\pi} e^{-2\alpha r}r^2\sin \theta \,dr\,d\theta\,d\phi=\frac{1}{4}\alpha^{3} \cdot 2\pi²$
And with $$L^2 = -\hbar^2 \left(\frac{1}{\sin\theta}\frac{\partial}{\partial \theta}\left( \sin\theta \frac{\partial}{\partial \theta}\right) + \frac{1}{\sin^2\theta}\frac{\partial^2}{\partial \phi^2}\right)$$ the whole L term goes away?

12. Aug 28, 2011

### vela

Staff Emeritus
That's not correct. Try again.
Yes. Can you explain why?

13. Aug 28, 2011

### Juqon

$$\langle \psi \vert \psi \rangle = \int_0^\infty \int_0^\pi \int_0^{2\pi} e^{-2\alpha r}r^2\sin \theta \,dr\,d\theta\,d\phi=\frac{1}{4}\alpha^{3} \cdot 2\pi \cdot\left[-cos\right]^\pi _0 = \frac{1}{4}\alpha^{3} \cdot 2\pi \cdot\left[-(-1) - (-(1))\right]=\frac{1}{4}\alpha^{3} \cdot 4\pi =\alpha^{3}\pi$$
It goes away because it is differentiated with respect to $\theta$ and $\varphi$, but these are not in the result any more, is a constant with respect to the angles and thus 0.

14. Aug 28, 2011

### vela

Staff Emeritus
Excellent! Now onto the numerator...

15. Aug 28, 2011

### Juqon

Thanks, now here is my newest problem.
[EDIT: I think I found it.]

$$E_{0}=\frac{<\psi|\hat{H}|\psi>}{<\psi|\psi>}= \frac{\int_{-\infty}^{\infty}dr\cdot r^{2}\cdot\psi^{*}\left\{ -\frac{\hbar^{2}}{2\mu}\left[\alpha^{2}-\frac{2}{r}\alpha\right]\psi+\frac{1}{4\pi\epsilon_{0}} \frac{qQ}{r}\psi\right\} }{\int_{-\infty}^{\infty}dr\psi^{*}\psi}$$

$$=\frac{2\int_{0}^{\infty}dr\cdot r^{2}\cdot\exp\left\{ -2\alpha r\right\} \left\{ \frac{\hbar^{2}}{2\mu}\frac{2}{r}\alpha-\frac{\hbar^{2}}{2\mu}\alpha^{2}+\frac{1}{4 \pi \epsilon_{0}}\frac{qQ}{r}\right\} }{2\int_{0}^{\infty}dr\cdot r^{2}\cdot\exp\left\{ -2\alpha r\right\} }$$

$$=\frac{2\int_{0}^{\infty}dr\cdot\exp\left\{ -2\alpha r\right\} \left\{ \frac{\hbar^{2}}{2\mu}2r\alpha-r^{2}\frac{\hbar^{2}}{2\mu}\alpha^{2}+\frac{r}{4 \pi \epsilon_{0}}\frac{qQ}{1}\right\} }{2\int_{0}^{\infty}dr\cdot r^{2}\cdot\exp\left\{ -2\alpha r\right\} }$$

$$=\frac{2\int dr\cdot\exp\left\{ -2\alpha r\right\} \cdot r\left\{ \frac{\hbar^{2}}{2\mu}2\alpha+\frac{1}{4 \pi \epsilon_{0}}\frac{qQ}{1}\right\} +2\int dr\cdot\exp\left\{ -2\alpha r\right\} \cdot r^{2}\cdot\left(-\frac{\hbar^{2}}{2\mu}\alpha^{2}\right)}{2\int_{0}^{\infty}dr\cdot r^{2}\cdot\exp\left\{ -2\alpha r\right\} }$$

$$=\frac{2\cdot\frac{1}{4}\alpha^{2}\left\{ \frac{\hbar^{2}}{2\mu}2\alpha+\frac{1}{4 \pi \epsilon_{0}}\frac{qQ}{1}\right\} +2\frac{1}{4}\alpha^{3}\cdot\left(-\frac{\hbar^{2}}{2\mu}\alpha^{2}\right)}{2\frac{1}{4}\alpha^{3}}$$

$$=\frac{\frac{1}{2}\alpha^{2}\left\{ \frac{\hbar^{2}}{2\mu}2\alpha+\frac{1}{4 \pi \epsilon_{0}}\frac{qQ}{1}\right\} +\frac{1}{2}\alpha^{3}\cdot\left(-\frac{\hbar^{2}}{2\mu}\alpha^{2}\right)}{\frac{1}{2}\alpha^{3}}$$

$$E_{0}=\frac{1}{4 \pi \epsilon_{0}}\frac{qQ}{\alpha}+\frac{\hbar^{2}}{ \mu}$$

$$\frac{\partial E_{0}}{\partial\alpha}=-\frac{1}{4 \pi \epsilon_{0}}\frac{qQ}{\alpha^{2}}=0$$

Last edited: Aug 28, 2011
16. Aug 28, 2011

### vela

Staff Emeritus
You don't see a disconnect between what you wrote in post 13 and what you wrote in post 15?

17. Aug 28, 2011

### Juqon

This is my newest version. I think I showed in #13 what the result of the integral ist, but now that differentiations make it 0 I deleted the whole L term.

$$E_{0}= \frac{<\psi|\hat{H}|\psi>}{<\psi|\psi>}=\frac{\int_{-\infty}^{\infty}dr\cdot r^{2}\cdot\psi^{*}\left\{ -\frac{\hbar^{2}}{2\mu}\left[\alpha^{2}-\frac{2}{r}\alpha\right]\psi-\frac{e^{2}}{r}\psi\right\} }{\int_{-\infty}^{\infty}dr\psi^{*}\psi}$$

$$=\frac{2\int_{0}^{\infty}dr\cdot r^{2}\cdot\exp\left\{ -2\alpha r\right\} \left\{ \frac{\hbar^{2}}{2\mu}\frac{2}{r}\alpha-\frac{\hbar^{2}}{2\mu}\alpha^{2}-\frac{e^{2}}{r}\right\} }{2\int_{0}^{\infty}dr\cdot r^{2}\cdot\exp\left\{ -2\alpha r\right\} }$$

$$=\frac{2\int_{0}^{\infty}dr\cdot\exp\left\{ -2\alpha r\right\} \left\{ \frac{\hbar^{2}}{2\mu}2r\alpha-r^{2}\frac{\hbar^{2}}{2\mu}\alpha^{2}-e^{2}\right\} }{2\int_{0}^{\infty}dr\cdot r^{2}\cdot\exp\left\{ -2\alpha r\right\} }$$

$$=\frac{2\int dr\cdot\exp\left\{ -2\alpha r\right\} \cdot r\left\{ \frac{\hbar^{2}}{2\mu}2\alpha-e^{2}\right\} +2\int dr\cdot\exp\left\{ -2\alpha r\right\} \cdot r^{2}\cdot\left(-\frac{\hbar^{2}}{2\mu}\alpha^{2}\right)}{2\int_{0}^{\infty}dr\cdot r^{2}\cdot\exp\left\{ -2\alpha r\right\} }$$

$$=\frac{2\cdot\frac{1}{4}\alpha^{2}\left\{ \frac{\hbar^{2}}{2\mu}2\alpha-e^{2}\right\} +2\frac{1}{4}\alpha^{3}\cdot\left(-\frac{\hbar^{2}}{2\mu}\alpha^{2}\right)}{2\frac{1}{4}\alpha^{3}}$$

$$=\frac{\frac{1}{2}\alpha^{2}\left\{ \frac{\hbar^{2}}{2\mu}2\alpha-e^{2}\right\} +\frac{1}{2}\alpha^{3}\cdot\left(-\frac{\hbar^{2}}{2\mu}\alpha^{2}\right)}{\frac{1}{2}\alpha^{3}}$$

$$E_{0}=-\frac{e^{2}}{\alpha}+\frac{\hbar^{2}}{\mu}-\frac{\hbar^{2}}{2\mu}\alpha^{2}$$

$$\frac{\partial E_{0}}{\partial\alpha}=\frac{e^{2}}{\alpha^{2}}-\frac{\hbar^{2}}{\mu}\alpha=0\Leftrightarrow\frac{\hbar^{2}}{\mu}\alpha=\frac{e^{2}}{\alpha^{2}}$$

$$\Leftrightarrow\alpha^{3}=e^{2}\cdot\frac{\mu}{ \hbar^{2}}\Leftrightarrow\alpha=\sqrt[3]{e^{2}\cdot\frac{\mu}{\hbar^{2}}}$$

$$E_{0}=- \frac{e^{2}}{\sqrt[3]{e^{2}\cdot\frac{\mu}{\hbar^{2}}}} + \frac{\hbar^{2}}{\mu}-\frac{\hbar^{2}}{2\mu} \sqrt[3]{e^{2}\cdot\frac{\mu}{\hbar^{2}}}^{2}$$

18. Aug 28, 2011

### vela

Staff Emeritus
Look at the denominators in the very first line. How is that consistent with what you have in post #13?

19. Aug 29, 2011

### Juqon

Mh, I didn't change the other integrals.

$$E_{0}= \frac{<\psi|\hat{H}|\psi>}{<\psi|\psi>}=\frac{\int_{-\infty}^{\infty}d\vec{r}\psi^{*}\left\{ -\frac{\hbar^{2}}{2\mu}\left[\alpha^{2}-\frac{2}{r}\alpha\right]\psi-\frac{e^{2}}{r}\psi\right\} }{\int_{-\infty}^{\infty}d\vec{r}\psi^{*}\psi}$$

$$=\frac{\int_{-\infty}^{\infty}d\vec{r}\exp\left\{ -2\alpha r\right\} \left\{ \frac{\hbar^{2}}{2\mu}\frac{2}{r}\alpha-\frac{\hbar^{2}}{2\mu}\alpha^{2}-\frac{e^{2}}{r}\right\} }{\int_{-\infty}^{\infty}d\vec{r}\exp\left\{ -2\alpha r\right\} }$$

$$=\frac{\int_{0}^{\infty}dr \int_{0}^{\pi}d\theta\int_{0}^{2\pi}d\phi\cdot\sin\theta\cdot\exp\left\{ -2\alpha r\right\} \cdot r\left\{ \frac{\hbar^{2}}{2\mu}2\alpha-e^{2}\right\} +\int_{0}^{\infty}dr\int_{0}^{\pi}d\theta\int_{0}^{2\pi}d\phi\cdot\sin\theta\cdot\exp\left\{ -2\alpha r\right\} \cdot r^{2}\cdot\left(-\frac{\hbar^{2}}{2\mu}\alpha^{2}\right)}{\int_{0}^{\infty}dr\int_{0}^{\pi}d\theta\int_{0}^{2\pi}d \phi\cdot r^{2}\sin\theta\cdot\exp\left\{ -2\alpha r\right\} }$$

$$=\frac{\pi\alpha^{2}\left\{ \frac{\hbar^{2}}{\mu}\alpha-e^{2}\right\} +\pi\alpha^{3}\cdot\left(-\frac{\hbar^{2}}{2\mu}\alpha^{2}\right)}{\pi\alpha^{3}}$$

$$E_{0}=-\frac{e^{2}}{\alpha}+\frac{\hbar^{2}}{\mu}\left(1-\frac{\alpha^{2}}{2}\right)$$

 $$\frac{ \partial E_{0}}{\partial\alpha}=\frac{e^{2}}{\alpha^{2}}-\frac{\hbar^{2}}{\mu}2\alpha=0 \Leftrightarrow\frac{\hbar^{2}}{\mu}2\alpha=\frac{e^{2}}{\alpha^{2}}$$

 $$\Leftrightarrow \alpha^{3}=e^{2}\cdot\frac{\mu}{2\hbar^{2}} \Leftrightarrow\alpha=\sqrt[3]{e^{2}\cdot\frac{\mu}{2 \hbar^{2}}}$$

$$E_{0}=-\frac{e^{2}}{\sqrt[3]{e^{2}\cdot\frac{\mu}{2\hbar^{2}}}}+\frac{\hbar^{2}}{\mu2}\left(2-\sqrt[3]{e^{2}\cdot\frac{\mu}{2\hbar^{2}}}^{2}\right)$$

Last edited: Aug 29, 2011
20. Aug 29, 2011

### vela

Staff Emeritus
The integrals should be
\begin{align*}
\int_0^\infty r e^{-2\alpha r}\,dr = \frac{1}{4\alpha^2} &\\
\int_0^\infty r^2 e^{-2\alpha r}\,dr = \frac{1}{4\alpha^3} &
\end{align*}
With those corrections, it should work out.