# Spin in a magnetic field

### The Hamiltonian

Both classical and quantum formulations of Hamiltonians require assumptions and only experimental data can confirm them in the end.

#### Classical physics

In classical physics, the energy associated with the spin of an electron in a magnetic field is the inner product between the magnetic field, $\vec{B}$, and the magnetic moment, $\vec{\mu}$ of the spin, $\vec{\sigma}$.

The magnetic moment is defined to be proportional to the spin, $$\vec{\mu} = \frac{\mu}{2}\vec{\sigma}$$ where the constant of proportionality, $\mu$, depends on the particle under consideration, but is related to the electric charge of the particle, in the sense that, if there were no charge then the magnetic field in which the particle is placed would have nothing to interact with. The factor of $\frac{1}{2}$ is also included by convention.

The classical energy stored in a system of a spin in a magnetic field is then $$E = \vec{\mu} \cdot \vec{B} = \frac{\mu}{2} \vec{\sigma} \cdot \vec{B}$$

If we chose $\vec{B} = B\hat{z}$ for simplicity, the energy is just $$E = \frac{\mu B}{2} \sigma_z$$

#### Quantum physics

We need to convert the classical statement about energy into a quantum statement about the Hamiltonian, where the spin and the magnetic moment are represented by Hermitian operators, rather than spatial vectors. However, the magnetic field does not play any quantum mechanical role since, we could for example be using a huge $10$ Tesla field on the particle.

So, in our (simple) equation for energy, we must replace the $z$-component of the spatial spin vector with the spin operator in the $z$-direction to get the Hamiltonian, $$H = \frac{\mu B}{2} \sigma_z$$

### Time evolution of spin

We are assuming that a particle in a spatial magnetic field $B\hat{z}$ has Hamiltonian $$H = \frac{\mu B}{2} \sigma_z$$

We can measure the spin in an arbitrary direction, $\sigma_n$ say, and compute the time derivative of its average, \begin{align*}\frac{\partial}{\partial t} \left\langle \sigma_n \right\rangle&= \frac{\mathrm{i}}{\hbar} \left\langle \left[H, \sigma_n\right] \right\rangle\\&= \frac{\mathrm{i}}{\hbar} \left\langle \left[\frac{\mu B}{2} \sigma_z, \sigma_n\right] \right\rangle\\&= \mathrm{i}\frac{\mu B}{2\hbar} \left\langle \left[\sigma_z, \sigma_n\right] \right\rangle\end{align*}

The commutator between $\sigma_z$ and $\sigma_n$ is \begin{align*}\left[\sigma_z, \sigma_n\right]&= \sigma_z \sigma_n - \sigma_n \sigma_z\\&= \begin{bmatrix}1 & 0 \\ 0 & -1\end{bmatrix}\begin{bmatrix}n_z & n_- \\ n_+ & -n_z\end{bmatrix} - \begin{bmatrix}n_z & n_- \\ n_+ & -n_z\end{bmatrix}\begin{bmatrix}1 & 0 \\ 0 & -1\end{bmatrix}\\&= \begin{bmatrix}n_z & n_- \\ -n_+ & n_z\end{bmatrix} - \begin{bmatrix}n_z & -n_- \\ n_+ & n_z\end{bmatrix}\\&= \begin{bmatrix}0 & 2n_- \\ -2n_+ & 0\end{bmatrix}\\&= 2\begin{bmatrix}0 & n_- \\ -n_+ & 0\end{bmatrix}\end{align*}

In particular, \begin{align*}\left[\sigma_z, \sigma_x\right] &= 2\begin{bmatrix}0 & 1 \\ -1 & 0\end{bmatrix} = 2\mathrm{i}\begin{bmatrix}0 & -\mathrm{i} \\ \mathrm{i} & 0\end{bmatrix} = 2\mathrm{i}\sigma_y\\ \left[\sigma_z, \sigma_y\right] &= 2\begin{bmatrix}0 & \mathrm{i} \\ -\mathrm{i} & 0\end{bmatrix} = -2\mathrm{i}\begin{bmatrix}0 & -1 \\ 1 & 0\end{bmatrix} = -2\mathrm{i}\sigma_x\\ \left[\sigma_z, \sigma_z\right] &= 2\begin{bmatrix}0 & 0 \\ 0 & 0\end{bmatrix} = 0\end{align*}

Thus, the time derivatives of the (averages of) the spins in the various spatial axes are \begin{align*}\frac{\partial}{\partial t} \left\langle \sigma_x \right\rangle &= \mathrm{i}\frac{\mu B}{2\hbar} \left\langle 2\mathrm{i}\sigma_y \right\rangle = -\frac{\mu B}{\hbar} \left\langle \sigma_y \right\rangle\\\frac{\partial}{\partial t} \left\langle \sigma_y \right\rangle &= \mathrm{i}\frac{\mu B}{2\hbar} \left\langle -2\mathrm{i}\sigma_x \right\rangle = \frac{\mu B}{\hbar} \left\langle \sigma_x \right\rangle\\\frac{\partial}{\partial t} \left\langle \sigma_z \right\rangle &= 0\end{align*}

### General solutions to the differential equations

Figure 9.1 - Spin in a magnetic field

The spin in the $\hat{z}$ direction doesn't change, hence the spin vector (in classical terms) is at a constant angle ($\phi$ in figure 9.1) to the $z$-azis.

Now, if we also write $\omega = \frac{\mu B}{\hbar}$ then the equations in the other two directions become \begin{align*}\frac{\partial}{\partial t} \left\langle \sigma_x \right\rangle &= - \omega \left\langle \sigma_y \right\rangle\\\frac{\partial}{\partial t} \left\langle \sigma_y \right\rangle &= \omega \left\langle \sigma_x \right\rangle\end{align*} which from a pair of coupled first order differential equations. We can de-couple them by considering the second derivatives, \begin{align*}\frac{\partial^2}{\partial t^2} \left\langle \sigma_x \right\rangle &= - \omega \frac{\partial}{\partial t} \left\langle \sigma_y \right\rangle = - \omega^2 \left\langle \sigma_x \right\rangle\\\frac{\partial^2}{\partial t^2} \left\langle \sigma_y \right\rangle &= + \omega \frac{\partial}{\partial t} \left\langle \sigma_x \right\rangle = - \omega^2 \left\langle \sigma_y \right\rangle\end{align*}

These are second order differential equations, with the following general solutions, \begin{align*}\left\langle \sigma_x(t) \right\rangle &= a \cos(\omega t) + b \sin(\omega t)\\\left\langle \sigma_y(t) \right\rangle &= c \cos(\omega t) + d \sin(\omega t)\end{align*} for some constants $a,b,c,d$.

### Example solution

If we take, for simplicity, $a=d=1$, and $b=c=0$, then a simple particular solution would be \begin{align*}\left\langle \sigma_x(t) \right\rangle &= \cos(\frac{\mu B}{\hbar} t)\\\left\langle \sigma_y(t) \right\rangle &= \sin(\frac{\mu B}{\hbar} t)\end{align*} which is analogous to the known classical solution described in the second diagram of figure 9.1. To correctly describe this in quantum mechanical terms, we can consider various times.

For all $t$
\begin{align*}\left\langle \sigma_z \right\rangle = 0 &\Rightarrow \sigma_z \text{ will be measured up or down with equal probability}\end{align*}

At $t = 0$
\begin{align*}\left\langle \sigma_x \right\rangle = 1 &\Rightarrow \sigma_x \text{ will be measured up}\\\left\langle \sigma_y \right\rangle = 0 &\Rightarrow \sigma_y \text{ will be measured up or down with equal probability}\end{align*}

At $t = \frac{\pi \hbar}{2 \mu B}$
\begin{align*}\left\langle \sigma_x \right\rangle = 0 &\Rightarrow \sigma_x \text{ will be measured up or down with equal probability}\\\left\langle \sigma_y \right\rangle = 1 &\Rightarrow \sigma_y \text{ will be measured up}\end{align*}

At $t = \frac{\pi \hbar}{\mu B}$
\begin{align*}\left\langle \sigma_x \right\rangle = -1 &\Rightarrow \sigma_x \text{ will be measured down}\\\left\langle \sigma_y \right\rangle = 0 &\Rightarrow \sigma_y \text{ will be measured up or down with equal probability}\end{align*}

At $t = \frac{3\pi \hbar}{2 \mu B}$
\begin{align*}\left\langle \sigma_x \right\rangle = 0 &\Rightarrow \sigma_x \text{ will be measured up or down with equal probability}\\\left\langle \sigma_y \right\rangle = -1 &\Rightarrow \sigma_y \text{ will be measured down}\end{align*}