# Mean values for hydrogen atom

1. Jul 22, 2011

### dingo_d

1. The problem statement, all variables and given/known data

So I've been racking my brain around the hydrogen mean values.

$$\left\langle \frac{1}{r}\right\rangle=\frac{1}{a_0n^2}$$, that I can solve with the recurrence relation in Schaum:

$$\left\langle r^k\right\rangle=\int_0^\infty r^{k+2}|R_{nl}(r)|^2dr$$

by simply putting in the radial part of hydrogen wave function. But when I do the exact same thing for

$$\left\langle r\right\rangle$$

I get integral:

$$\int_0^\infty x^{2l+3}e^{-x}[L_{n-l-1}^{2l+1}(x)]^2dx$$

Which I cannot solve by using the known relations for Laguerre polynomials because the exponent on the x is neither the same or by one greater than Laguerre polynomial (2l+1).

$$\int_0^\infty x^{k+1}e^{-x}[L_{n}^{k}(x)]^2dx=\frac{(n+k)!}{n!}(2n+k+1)$$

Am I doing sth wrong? I've checked over and over and cannot find the flaw :\

Mathematica doesn't do these kind of calculations, and explanation in book by Zettili with deriving by l or electric charge and using Kramers recurrence relations aren't helping much :(

2. Jul 24, 2011

### dextercioby

I understand where you're coming from, but there;s little you can do. Some integrals can be done, some not. In your place, I'd concentrate on understanding how the Kramers recurrings can be derived without resorting to complicated integrals. Do you have a source which treats this thoroughly ?

3. Jul 25, 2011

### dingo_d

Well the Zettili's book gives solved problem, but I still have problems understanding as to why she derived the radial equation with orbital quantum number, or electric charge :\

I thought that once you have the recurrence relation the obtaining the formula should be straight forward :\

I tried with partial integration but it doesn't give good answer...

4. Jul 25, 2011

### dextercioby

I will look in Zettili;s book then and see what input I can give.

5. Jul 25, 2011

### dingo_d

Thank you for putting time and effort into this ^^

6. Jul 25, 2011

### dextercioby

I saw it only now. It was just a trick. Admittedly, I haven't seen it in any other source, it's very interesting, indeed. So this trick with the 'l' differentiation is neat, you have to take it for granted. It gives you <1/r^2> and the one next with the <e> differentiation gives you the <1/r>.

They do, it's just a recurring relation, so it suffices to know the average for 2 powers of r and you'll get it for all other, negative or positive.

For the <1/r> one could also use the virial theorem.

So the problems 6.2 & 6.3 of Zettili indeed solve a lot of issues, if you're able to prove Kramer's relation all by its own.

Zettili of course doesn't do it, which is the only drawback of this chapter 6.

7. Jul 25, 2011

### dextercioby

I just checked Constantinescu & Magyari problem book (Pergamon Press, 1971) on QM and saw the Kramers equation and the trick with the differentiation wrt l in problems 6 and 7 of chapter 3. Indeed the proof of Kramers one is difficult using integrals.

8. Jul 26, 2011

### dingo_d

Recursion!

Well thank you very much, I'll have it in mind. Altho I don't think my professor will ask me that on the exam, but I still wanted to know how did we get that :)

Thank you again ^^

9. Jul 26, 2011

### Dickfore

The mean value $\langle r^{-1}\rangle$ can be evaluated using the virial theorem. The potential energy is:
$$V = -\frac{\alpha}{r}$$
which gives a force:
$$F = -\frac{d V}{d r} = -\frac{\alpha}{r^{2}}$$
According to the Bohr model, the centripetal accleration is:
$$\frac{m v^{2}}{r} = \frac{\alpha}{r^{2}}$$
From here it follows that:
$$T = \frac{m \, v^{2}}{2} = \frac{\alpha}{2 \, r} = -\frac{V}{2}$$
This equality, derived under the assumptions of the Bohr model (a circular orbit and a simultaneous knowledge of the speed and position of the particle) is still correct when considered as a relation between the mean values (Virial Theorem):
$$\langle T \rangle = -\frac{\langle V \rangle}{2}$$
On the other hand, the energy is conserved:
$$E = T + V = \langle T \rangle + \langle V \rangle \Rightarrow E = \frac{\langle V \rangle}{2}$$
We can use the Bohr model to evalute the radii of the Bohr orbits and the stationary energies, which coincide with the solutions obtained by the more rigorous approach of solving the Schroedinger equation:
$$v^{2} \, r = \frac{\alpha}{m}$$
$$v \, r = \frac{n \, \hbar}{m}$$
$$r_{n} = \frac{\hbar^{2}}{m \, \alpha} \, n^{2} = a_{0} \, n^{2}$$
$$E_{n} = -\frac{\alpha}{2 \, r_{n}} = -\frac{m \, \alpha^{2}}{2 \, \hbar^{2} \, n^{2}}$$
Therefore:
$$\langle V \rangle = -\alpha \, \langle \frac{1}{r} \rangle = 2 E = -\frac{m \, \alpha^{2}}{\hbar^{2} \, n^{2}}$$
$$\langle \frac{1}{r} \rangle = \frac{m \, \alpha}{\hbar^{2} \, n^{2}} = \frac{1}{a_{0} \, n^{2}}$$

10. Jul 26, 2011

### Dickfore

If you want a more rigorous derivation, you can use the Hellmann-Feynman Theorem:
$$\left\langle \frac{\partial H}{\partial \lambda} \right\rangle = \frac{\partial E_{n}}{\partial \lambda}$$
where the expectation value is taken in the stationary state and $\lambda$ is some parameter entering in the expression for the Hamiltonian explicitly:
For the H-atom:
$$H = \frac{p^{2}}{2 \, m} - \frac{\alpha}{r}$$
Therefore:
$$\frac{\partial H}{\partial \alpha} = -\frac{1}{r}$$
The stationary energies are:
$$E_{n} = -\frac{m \, \alpha^{2}}{2 \, \hbar^{2} \, n^{2}}$$
Therefore, the partial derivative is:
$$\frac{\partial E_{n}}{\partial \alpha} = -\frac{m \, \alpha}{\hbar^{2} \, n^{2}}$$
Combining everything togehter and using the definition of the Bohr radius, you get the quoted relation.

Last edited: Jul 26, 2011