# Expectation value of Coulomb potential depends on relative spin

1. Sep 21, 2010

### Bill Foster

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

Show that the expectation value of the Coulomb potential $$v(\vec{r_1},\vec{r_2})=\frac{e^2}{|\vec{r_1}-\vec{r_2}|}$$, between two electrons depends on the relative orientation of spin of the two electrons. Assume each electron is in the product state form $$\phi(\vec{r})\chi_{\frac{1}{2}m}(s)$$.

2. Relevant equations

$$<A>_\phi = <\phi|A|\phi>$$

3. The attempt at a solution

No attempt as I don't even know where to start.

I'm guessing that somehow I have to write the Coulomb potential as an operator. And the two electrons as a state, and then use the equation for expectation value above.

2. Sep 22, 2010

### Thaakisfox

You just have to integrate. phi_1*x phi_2*x v x phi_1 x phi_2

3. Sep 22, 2010

### Bill Foster

If I'm looking for the expectation value of x of a wave function...

$$\langle x \rangle = \int x|\Psi\left(x\right)|^2dx$$

In this case, "x" is "v". And the wave function is a function of two electrons, one at $$r_1$$ with spin $$s_1$$ and one at $$r_2$$ with spin $$s_2$$.

To integrate over all v, I would have to integrate over all space, $$r_1$$ and $$r_2$$.

So there's something like this: $$dv \rightarrow d^3r$$.

So I set up the integral as...
$$\int d^3r \left(\phi(\vec{r_1})\chi_{\frac{1}{2}m}(s_1)\right)^\dagger\left(\phi(\vec{r_2})\chi_{\frac{1}{2}m}(s_2)\right)^\dagger \frac{e^2}{|\vec{r_1}-\vec{r_2}|} \left(\phi(\vec{r_1})\chi_{\frac{1}{2}m}(s_1)\right)\left(\phi(\vec{r_2})\chi_{\frac{1}{2}m}(s_2)\right)$$
$$=e^2\int d^3r \frac{1}{|\vec{r_1}-\vec{r_2}|} |\phi(\vec{r_1})\chi_{\frac{1}{2}m}(s_1)\phi(\vec{r_2})\chi_{\frac{1}{2}m}(s_2)|^2$$
$$=e^2\int d^3r \frac{1}{|\vec{r_1}-\vec{r_2}|} |\chi_{\frac{1}{2}m}(s_1)\chi_{\frac{1}{2}m}(s_2)|^2\delta_{r_1r_2}$$

It looks like the expectation value is zero unless r1 = r2. That doesn't seem right to me.

4. Sep 23, 2010

### Thaakisfox

You cannot eliminate the phi wavefunctions like that. They are a functions of r_1 and r_2.

5. Sep 23, 2010

### Bill Foster

$$=e^2\int d^3r \frac{1}{|\vec{r_1}-\vec{r_2}|} |\chi_{\frac{1}{2}m}(s_1)\chi_{\frac{1}{2}m}(s_2)| ^2\delta\left(r_1-r_2\right)$$

6. Sep 23, 2010

### Thaakisfox

No. You cannot do that. You cant just eliminate the spatial part of the wavefunctions as i said.

7. Sep 23, 2010

### Thaakisfox

The integral is

$$\int d^3 r ... \to\int d^3 r_1 d^3 r_2 ...$$

Since the r_1 and r_2 are in the coulomb potential itself you cannot simply eliminate the spatial wave functions.

8. Sep 23, 2010

### Bill Foster

Assuming the wave functions are normalized, then if we integrate them over all space, don't they become 1?

9. Sep 23, 2010

### Thaakisfox

Yes but we cant integrate them, as the variable they depend on is also in another term in the integrand i.e. in the coulomb potential, so we cant simply integrate them. Thats what i was talking about for the last few posts...

10. Sep 23, 2010

### Bill Foster

Then what's the next step?

11. Sep 23, 2010

### Thaakisfox

Ok. So then first of all you will have to write up those single electron wavefunctions as the product.
Best is to change to spherical coordinates and write them as the product of a function depending on r and another function depending on the angles.
Assuming that the potential in which the wavefunctions were calculated are spherically symmetric (why not..), the angle part will be the spherical harmonics.

So u will have something like this: Phi(vector r) = R(r)*Y_lm(theta,phi)

Next use the multipole expansion to express the inverse distance 1/r as the sum of the product of the spherical harmonics.

Now plug all this back into the integral.

12. Sep 27, 2010

### Bill Foster

$$\phi(\vec{r_1})\chi_{\frac{1}{2}m}(s_1)\phi(\vec{ r_2})\chi_{\frac{1}{2}m}(s_2) = \psi\left(\vec{r_1},\vec{r_2}\right)\chi\left(s_1,s_2\right)$$

$$\psi\left(\vec{r_1},\vec{r_2}\right)\chi_{\frac{1}{2}m}\left(s_1,s_2\right)=R\left(r\right)Y_l^m\left(\theta,\phi\right)\chi_{\frac{1}{2}m}\left(s_1,s_2\right)$$

Let one electron be at the origin, so the potential is:

$$\frac{e^2}{|r|} = e^2\sum_{j=1}^\infty \sum_{l=0}^\infty \sum_{m=-l}^l \frac{D_{l,j}^m}{r^j}Y_l^m\left(\theta,\phi\right)$$

$$\int d^3r e^2\sum_{j=1}^\infty \sum_{l=0}^\infty \sum_{m=-l}^l \frac{D_{l,j}^m}{r^j}Y_l^m\left(\theta,\phi\right)|R\left(r\right)Y_l^m\left(\theta,\phi\right)\chi_{\frac{1}{2}m}\left(s_1,s_2\right)|^2$$

13. Sep 27, 2010

### Bill Foster

Well, we're going about this all wrong. We don't need to evaluate this integral. We need to calculate the matrix elements. They will show that the Coulomb interaction depends on spin.

So...how about some help with evaluating the matrix elements?

14. Sep 28, 2010

### Bill Foster

I'm not sure how to evaluate the matrix elements, but suppose I put one electron at the origin, so instead of r_1 and r_2, I have one variable r.

Then I have:

$$V\left(\vec{r_1},\vec{r_2}\right) = \frac{e^2}{|\vec{r_1}-\vec{r_2}|}$$

this becomes:

$$V\left(\vec{r}\right) = \frac{e^2}{|\vec{r}|}$$

Then I can write the two wave functions as:

$$|r,s_1,s_2\rangle$$

So then I need to evaluate the matrix elements of:

$$\langle r,s_1,s_2|\frac{e^2}{|r|}|r,s_1,s_2 \rangle$$

Am I on the right track here?

15. Sep 30, 2010

### Bill Foster

Now I don't feel so bad. Evidently, nobody else knows how to do this either.

16. Oct 3, 2010

### fzero

(Note, the argument below is essentially the same as found in Sakurai's discussion of the Helium atom.)

Since the Coulomb potential commutes with the spin operators, we can almost neglect the spin computations in the matrix element. However, Fermi-Dirac statistic comes into play when we consider the wavefunctions.

Let the spatial wavefunctions be $$\phi_0(\vec{r})$$, where n refers to some quantum numbers. We'll take $$\phi_n(\vec{r})$$ to be the ground state.

We first consider the case when both electrons are in the ground state. Then by Fermi statistics, the spins must be opposite, so the spin state is the singlet. The 2-particle wavefunction is

$$\phi(\vec{r}_1,\vec{r}_2) = \phi_0(\vec{r}_1)\phi_0(\vec{r}_2) \chi_s(s_1,s_2).$$

There's no spin dependence of

$$\langle v(\vec{r_1},\vec{r_2}) \rangle$$

for this configuration.

Let's then consider the case where one electron is in the ground state and the other is in some excited state. The spins can either be opposite (singlet) or the same (triplet). For the singlet state the spatial part is symmetric:

$$\phi_+(\vec{r}_1,\vec{r}_2) =\frac{1}{\sqrt{2} }(\phi_0(\vec{r}_1)\phi_n(\vec{r}_2) + \phi_0(\vec{r}_2)\phi_n(\vec{r}_1) )\chi_s(s_1,s_2).$$

For the spin triplet, since the spins are symmetric, the spatial part must be antisymmetric:

$$\phi_-(\vec{r}_1,\vec{r}_2) = \frac{1}{\sqrt{2}}(\phi_0(\vec{r}_1)\phi_n(\vec{r}_2) -\phi_0(\vec{r}_2)\phi_n(\vec{r}_1) )\chi_t(s_1,s_2).$$

Now we can compute

$$\langle v(\vec{r_1},\vec{r_2}) \rangle_+ = \int d\vec{r}_1d\vec{r}_2 \phi_+^*(\vec{r}_1,\vec{r}_2) v(\vec{r_1},\vec{r_2}) \phi_+(\vec{r}_1,\vec{r}_2) = I + J$$

where

$$I = \int d\vec{r}_1d\vec{r}_2 v(\vec{r_1},\vec{r_2}) |\phi_0(\vec{r}_1)|^2|\phi_n(\vec{r}_2)|^2,$$

$$J = \int d\vec{r}_1d\vec{r}_2 v(\vec{r_1},\vec{r_2}) \phi_0^*(\vec{r}_1)\phi_0(\vec{r}_2) \phi_n^*(\vec{r}_2)\phi_n(\vec{r}_1).$$

However, for the triplet state, we find

$$\langle v(\vec{r_1},\vec{r_2}) \rangle_- = I - J.$$

So the spin dependence follows from spin statistics and not from any interaction term. Physically the reason is that when the electrons are in the triplet state, the spatial dependence is antisymmetric and the probability of finding the electrons close to one another is small. The electrostatic repulsion is minimized and the energy of the triplet state is lower than that of the singlet state. In the singlet state, the spatial function is symmetric and there is a high probability to find the electrons close to one another. Therefore the electrons see a higher electrostatic repulsion than in the triplet state.