Spin-orbit interaction

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In physics, in the area of quantum mechanics, the spin-orbit interaction is a shift in energy levels due to the potential energy of the spin magnetic moment of the electron in the magnetic field it feels as it moves through the electric field of the nucleus.

1 Spin Orbit Interaction
2 See also
3 References

[edit] Spin Orbit Interaction

The energy of a magnetic moment in a magnetic field is given by:

$\Delta H=-\boldsymbol{\mu}\cdot\boldsymbol{B}$

Where $μ$ is the magnetic moment of the particle and $B$ is the magnetic field it experiences.

[edit] Magnetic Field

We shall deal with the magnetic field first. For simplicity in many-electron systems we shall assume the nucleus to be stationary and not use the Biot-Savart Law. To calculate the magnetic field

$\boldsymbol{B} = - {\boldsymbol{v} \times \boldsymbol{E}\over c^2}$

Where v is the velocity of the particle and E the electric field it travels through. Now we know that E is radial so we can rewrite $\boldsymbol{E} =\left | {E\over r}\right| \boldsymbol{r}$ . Also we know that the momentum of the electron $\boldsymbol{p} =m_e \boldsymbol{v}$ . Substituting this in and changing the order of the cross product gives:

$\boldsymbol{B} = {\boldsymbol{r}\times\boldsymbol{p}\over m_ec^2} \left | {E\over r}\right|$

One then expresses the electric field as the gradient of the electric potential $\boldsymbol{E} = -\boldsymbol{\nabla}V$ . Here we make the central field approximation, that is, that the electrostatic potential is spherically symmetric so is only a function of radius. This approximation is exact for hydrogen, and indeed hydrogen-like systems. Now we can say

$\left | E\right| = {\partial V \over \partial r}={1\over e}{\partial U(r) \over \partial r}$

Where U is the potential energy in the central field; note that U is not the electrostatic potential, hence the factor of the electric charge e. Now we remember from classical mechanics that the angular momentum of a particle $\boldsymbol{l} = \boldsymbol{r}\times\boldsymbol{p}$ . Putting it all together we get

$\boldsymbol{B} = {1\over m_eec^2}{1\over r}{\partial U(r) \over \partial r} \boldsymbol{l}$

It is important to note at this point that B is a positive number multiplied by l, meaning that the magnetic field is parallel to the orbital angular momentum of the particle.

[edit] Magnetic Moment of the Electron

The magnetic moment of the electron is

$\boldsymbol{\mu} = -g_s\mu_B\boldsymbol{s}$

where $μ B$ is the Bohr magneton and $g_s\approx 2$ is the Landé g-factor. Here, $\boldsymbol{\mu}$ is a negative constant multiplied by the spin, so the magnetic moment is antiparallel to the spin angular momentum.

[edit] Interaction Energy

The interaction energy is

$\Delta H=-\boldsymbol{\mu}\cdot\boldsymbol{B}$

Let's substitute in the derived expressions.

$-\boldsymbol{\mu}\cdot\boldsymbol{B} = {2\mu_B\over m_eec^2}{1\over r}{\partial U(r) \over \partial r} \times \boldsymbol{l}\cdot\boldsymbol{s}$

We have not, thus far, taken into account the time dilation between the rest frames of the proton and electron; this effect is called Thomas precession and introduces a factor of $\frac{1}{2}$ . So

$-\boldsymbol{\mu}\cdot\boldsymbol{B} = {\mu_B\over m_eec^2}{1\over r}{\partial U(r) \over \partial r} \times \boldsymbol{l}\cdot\boldsymbol{s}$

[edit] Evaluating the energy shift

For this we will need degenerate perturbation theory. Basically

$\Delta E = \left\langle\Psi\right| \Delta H \left |\Psi\right\rangle$

From this we get two expectation values which must be evaluated;

$\left \langle {1\over r^3} \right \rangle = \frac{2}{a^3 n^3 l(l+1)(2l+1)}$

for hydrogenic wavefunctions (here $a = \hbar / Z \alpha m_e c$ is the Bohr radius divided by the nuclear charge Z) and

$\left \langle \boldsymbol{l}\cdot\boldsymbol{s} \right \rangle$

For this expectation value we must be careful due to the degeneracy of the states, so we need to choose a basis for the integral that allows perturbation theory to be used.

[edit] Evaluating $\left \langle \boldsymbol{l}\cdot\boldsymbol{s} \right \rangle$

The total angular momentum operator is

$\boldsymbol{j_z} = \boldsymbol{l_z} + \boldsymbol{s_z}$

with

$\boldsymbol{j}^2 = (\boldsymbol{l} + \boldsymbol{s})^2$

Evaluating the second operator

$\boldsymbol{j}^2 = \boldsymbol{l}^2 + \boldsymbol{s}^2 + 2\boldsymbol{l}\cdot\boldsymbol{s}$

$\boldsymbol{l}\cdot\boldsymbol{s}= {1\over 2}(\boldsymbol{j}^2 - \boldsymbol{l}^2 - \boldsymbol{s}^2)$

This gives

$\left\langle\Psi\right\|\boldsymbol{l}\cdot\boldsymbol{s} \left \|\Psi\right\rangle$	$=\left\langle\Psi\right\|{1\over 2}(\boldsymbol{j}^2-\boldsymbol{l}^2-\boldsymbol{s}^2)\left \|\Psi\right\rangle$
	$={1\over 2}(j(j+1) - l(l+1) -s(s+1))$