# Angular momentum of wavefunction in Hydrogen atom

1. Jan 15, 2014

### skrat

1. The problem statement, all variables and given/known data
Electron in Hydrogen atom can be described with wavefunction $\psi =\frac{1}{2}(\psi _1+\psi _2+\psi _3+\psi _4)$ where $\psi _1=\psi _{200}$, $\psi _2=\frac{1}{\sqrt{2}}(\psi _{211}+\psi _{21-1})$, $\psi _3=\frac{i}{\sqrt{2}}(\psi _{211}-\psi _{21-1})$ and $\psi _4=\psi _{210}$.

Wavefunctions $\psi _{nlm}$ are all normalized. Calculate the $<l^2>$ and $<l_z^2$.

2. Relevant equations
$\psi _{200}=\frac{1}{4\sqrt{2\pi }r_B^{3/2}}(2-\frac{r}{r_B})e^{-\frac{r}{2r_B}}$

$\psi _{210}=\frac{1}{4\sqrt{2\pi }r_B^{5/2}}re^{-\frac{r}{2r_B}}cos\vartheta$

$\psi _{21\pm 1}=\frac{1}{8\sqrt{2\pi }r_B^{5/2}}re^{-\frac{r}{2r_B}}sin\vartheta e^{\pm i\varphi }$

operator $\hat{l}^2=-\hbar ^2(\frac{\partial^2 }{\partial \vartheta ^2}+ctg\vartheta \frac{\partial }{\partial \vartheta }+\frac{1}{sin^2\vartheta }\frac{\partial^2 }{\partial \varphi ^2})$

and

$\hat{l}_z^2=-\hbar ^2\frac{\partial^2 }{\partial \varphi ^2}$

3. The attempt at a solution

I don't know... I am really blind or is there no other way than actually doing all the integrals? :/

If possible, please answer this question first, before I start writing all the equations on forum.

I tried calculating everything and got $<\hat{l}^2>=\frac{5}{4}\hbar ^2$ which I guess is the right result... Or maybe it is wrong by factor 1/2, or maybe even worse..

Last edited: Jan 15, 2014
2. Jan 15, 2014

### Staff: Mentor

[STRIKE]The first problem is that $\psi$ is not normalized.[/STRIKE] Forget that. $\psi$ is normalized.

What does $nlm$ stand for in $\psi _{nlm}$? Can you use that to quickly calculate how $\hat{l}^2$ and $\hat{l}_z^2$ operate on the $\psi _{nlm}$'s?

3. Jan 15, 2014

### skrat

4. Jan 15, 2014

### Staff: Mentor

They are not only numbers, they are quantum numbers. They are not chosen arbitrarily!

You can calculate $\hat{l}^2 \psi_{nlm}$ without even knowing what $\hat{l}^2$ or $\psi_{nlm}$ look like in spherical coordinates, in about 2 seconds...

5. Jan 15, 2014

### skrat

Well yes, of course they are not arbitrary.

n= principal quantum number
l= azimuthal quantum number and
m= should be something like projection of l to z axis.

How do I that? Sorry for not guessing. I don't want to just come up with some number and not know how I go it.

6. Jan 15, 2014

### Staff: Mentor

I don't know how you learned about the eigenstates of hydrogen, but the link between the quantum numbers $l$ and $m$ and the operators $\hat{l}$ and $\hat{l}_z$ should've been obvious. You may need to revised the derivation of the eigenstates of hydrogen, or find a better textbook!

Since $\psi_{nlm}$ corresponds to a state with quantum numbers $l$ and $m$, then it is an eigenstate of $\hat{l}$ and $\hat{l}_z$ :
$$\hat{l} \psi_{nlm} = \sqrt{l(l+1)} \hbar \psi_{nlm}$$
$$\hat{l}_z \psi_{nlm} = m \hbar \psi_{nlm}$$

7. Jan 15, 2014

### skrat

Ok, I've tried that BUT I don't what happens to $\frac{1}{\sqrt{2}}$ in front of $\psi _{nlm}$ ?

like for $\psi _2=\frac{1}{\sqrt{2}}(\psi _{211}+\psi _{21-1})$

than $\hat{l} \psi_{2} = \frac{1}{\sqrt{2}}( \hat{l}\psi _{211}+ \hat{l}\psi _{21-1})$

$\hat{l} =\frac{1}{\sqrt{2}} (2 \sqrt{2} \hbar)$ and

$\hat{l}^2=2\hbar ^2$

Is that the idea here?

8. Jan 15, 2014

### Staff: Mentor

Almost. Remember that
$$\langle \hat{l}^2 \rangle = \int \psi^* \hat{l}^2 \psi \,\mathrm{d}\tau$$

9. Jan 15, 2014

### skrat

I guess solution is something like:

$\hat{l}\psi =\frac{1}{2}(\hat{l}\psi _{200}+\frac{1}{\sqrt{2}}(\hat{l}\psi _{211}+\hat{l}\psi _{21-1})+\frac{i}{\sqrt{2}}(\hat{l}\psi _{211}-\hat{l}\psi _{21-1})+\hat{l}\psi _{210})$

$\hat{l}=\frac{1}{2}(0+\frac{1}{\sqrt{2}}2\sqrt{2}\hbar+0+\sqrt{2} \hbar )$ and finally

$\hat{l}^2 =\frac{1}{4}(4\hbar^2+2\hbar^2)=\frac{3}{2}\hbar^2$

Than $\langle \hat{l}^2 \rangle = \int \psi^* \hat{l}^2 \psi \,\mathrm{d}\tau=\int \psi^* \hat{l}^2 \psi \,\mathrm{d}\tau =\frac{3}{2}\hbar^2 \int \left | \psi \right |^2d\tau=\frac{3}{2}\hbar^2$

That's should be it for $\langle \hat{l}^2 \rangle$

10. Jan 15, 2014

### Staff: Mentor

is incorrect: going from $\hat{l}$ to $\hat{l}^2$, you squared the terms individually, which is not what you would get if you took the square of the equation for $\hat{l}$. To make sure that you don't make a mistake, calculate directly
$$\hat{l}^2 \psi_{nlm} = l(l+1)\hbar^2 \psi_{nlm},$$
such that
$$\hat{l}^2\psi =\frac{1}{2}(\hat{l}^2\psi _{200}+\frac{1}{\sqrt{2}}(\hat{l}^2\psi _{211}+\hat{l}^2\psi _{21-1})+\frac{i}{\sqrt{2}}(\hat{l}^2\psi _{211}-\hat{l}^2\psi _{21-1})+\hat{l}^2\psi _{210})$$
and so on.

Also, you mixed together $\hat{l}^2\psi$ and $\langle \hat{l}^2 \rangle$ (the $\psi$'s disappeared in the middle, to reappear later in the integral). This results in an error, which actually cancelled out an error earlier in the calculation, which gave you the right final answer.

I suggest that you start by rewriting $\psi$ as a sum of the four $\psi_{nlm}$, with a single coefficient in front of each. Then use the orthogonality of the $\psi_{nlm}$'s to get a short equation for $\langle \hat{l}^2 \rangle$. I may not be clear, so I'll give you an example:

Take
$$\psi = a_1 \psi_1 + a_2 \psi_2 = a_1 (c_1 \psi_{210} + c_2 \psi_{320} ) + a_2 (c_3 \psi_{210} + c_4 \psi_{320} )$$
Then
$$\psi = (a_1c_1 + a_2c_3)\psi_{210} + (a_1c_2 + a_2c_4) \psi_{320}$$
therefore
$$\langle \hat{l}^2 \rangle = \left|a_1c_1 + a_2c_3\right|^2 \langle \hat{l}^2 \rangle_{210} + \left|a_1c_2 + a_2c_4\right|^2 \langle \hat{l}^2 \rangle_{320}$$
where I used orthogonality of the $\psi_{nlm}$'s and defined
$$\langle \hat{l}^2 \rangle_{nlm} = \int \psi_{nlm}^* \hat{l}^2 \psi_{nlm} \, d\tau = l(l+1) \hbar^2$$
Finally,
$$\langle \hat{l}^2 \rangle = \left|a_1c_1 + a_2c_3\right|^2 2 \hbar^2 + \left|a_1c_2 + a_2c_4\right|^2 6 \hbar^2$$

11. Jan 15, 2014

### skrat

Ok, yep. I get it now. This makes more sense.

Than you very much for all the help!

12. Jan 15, 2014

### Staff: Mentor

I should point out one more thing: $\psi$ is not an eigenfunction of $\hat{l}^{2}$ (because it is made up of a superposition of eigenstates of $\hat{l}$ with different quantum numbers $l$). Therefore, it is incorrect to write
$$\int \psi^* \hat{l}^2 \psi \,\mathrm{d}\tau =\frac{3}{2}\hbar^2 \int \left | \psi \right |^2d\tau$$
as you can't write $\hat{l}^2 \psi$ as $c \psi$ (where c is a number).

13. Jan 15, 2014

### skrat

Hmmm, one last question:

$\langle \hat{l}^2 \rangle = \int \psi^* \hat{l}^2 \psi \,\mathrm{d}\tau=\int \frac{1}{2}(\psi _1+\psi _2+\psi _3+\psi _4)^{*} \hat{l}^2 \frac{1}{2}(\psi _1+\psi _2+\psi _3+\psi _4) \mathrm{d}\tau$
$\int \frac{1}{4}(\psi _1+\psi _2+\psi _3+\psi _4)^{*}2\hbar^2(\psi _2+\psi _4)\mathrm{d}\tau=\frac{\hbar^2}{2}\int (\left | \psi _2 \right |^2+\left | \psi _4 \right |^2)\mathrm{d}\tau=\frac{\hbar^2}{2}(\frac{1}{2}(1+1)+1)=\hbar^2$

Where did I get it wrong, except that the result should be $\frac{3}{2}\hbar^2$?

14. Jan 15, 2014

### Staff: Mentor

You forgot $\psi_3$ (more probably, incorrectly took $\hat{l}^2 \psi_3 = 0$, which is not true).

15. Jan 15, 2014

### skrat

This is killing me.

$\int \frac{1}{4}(\psi _1+\psi _2+\psi _3+\psi _4)^{*}2\hbar^2(\psi _2+\psi _3+\psi _4)\mathrm{d}\tau=\frac{\hbar^2}{2}\int (\left | \psi _2 \right |^2+\left | \psi _3 \right |^2+\left | \psi _4 \right |^2)\mathrm{d}\tau=\frac{\hbar^2}{2}(\frac{1}{2}(1+1)-\frac{1}{2}(1+1)+1)=\frac{\hbar^2}{2}$

????

$\hat{l}^2\psi _n=2\hbar^2\psi _n$ That is true for all the wavefunctions $\psi _2$, $\psi _3$ and $\psi _4$ i am delaing with.
$\hat{l}^2\psi _1=0$ therefore
$\hat{l}^2\psi =\hat{l}^2(\psi _1+\psi _2+\psi _3+\psi _4)=2\hbar^2(+\psi _2+\psi _3+\psi _4)$

Than $\left | \psi _2 \right |^2=\frac{1}{2}(1+1)=1$ and
$\left | \psi _3 \right |^2=\frac{-1}{2}(1+1)=-1$ and finally
$\left | \psi _4 \right |^2=1$

What stupidity did i do this time?

16. Jan 15, 2014

### Staff: Mentor

An absolute value can't be negative

17. Jan 15, 2014

### skrat

Please hit me in the face. -.-

DrClaude, again, thank you for your help and patience!

cheers