Spin operators for single electrons YouTube

State space

In classical mechanics, the space that represents the spin of an electron would be a three-dimensional set of real numbers. But in quantum mechanics, the space that represents the states of the spin of a single electron is a two-dimensional vector space of the complex numbers, with the property that the square of the states is always one. That is, the space $$\left\{ \begin{bmatrix} \alpha \\ \beta \end{bmatrix} \in \mathbb{C}^2: \alpha^*\alpha + \beta^*\beta = 1\right\}$$

We choose, arbitrarily, the basis of the state-space to be the pair of orthonormal states $$\left\{\ket{u} = \begin{bmatrix} 1 \\ 0 \end{bmatrix}, \ket{d} = \begin{bmatrix} 0 \\ 1 \end{bmatrix}\right\}$$ where the letters $u,d$ represent up- and down-states of an electron respectively. By up we mean the axis of rotation is pointing in the positive $z$-direction, and by down we mean the axis of rotation is pointing in the negative $z$-direction. Thus, we can write $$\begin{align*} \ket{+z} &= \ket{u} \\ \ket{-z} &= \ket{d} \end{align*}$$

Spin operators

The observables associated with the spin states, called spin operators, are given by $$\sigma_n = \begin{bmatrix}n_z & n_-\\n_+ & -n_z\end{bmatrix}$$ where the unit vector $$\hat{n} = \begin{bmatrix} n_x \\ n_y \\ n_z \end{bmatrix}$$ is a direction of normal space and $$\begin{align*} n_+ &= n_x + \mathrm{i} n_y \\ n_- &= n_x - \mathrm{i} n_y \end{align*}$$

The spin operators in the three coordinate directions are $$\begin{align*} \sigma_x &= \begin{bmatrix}0 & 1\\1 & 0\end{bmatrix} \\ \sigma_y &= \begin{bmatrix}0 &-\mathrm{i}\\\mathrm{i} & 0\end{bmatrix} \\ \sigma_z &= \begin{bmatrix}1 & 0\\0 &-1\end{bmatrix} \end{align*}$$

An equivalent way to write the spin operators is to write down their actions on the basis states, in terms of the basis states,
$$\begin{matrix}\sigma_x\ket{u} = \ket{d} & \sigma_x\ket{d} = \ket{u}\\\sigma_y\ket{u} = \mathrm{i}\ket{d} & \sigma_y\ket{d} = -\mathrm{i}\ket{u}\\\sigma_z\ket{u} = \ket{u} & \sigma_z\ket{d} = -\ket{d}\\\sigma_n\ket{u} = n_z\ket{u} + n_+\ket{d} & \sigma_n\ket{d} = n_-\ket{u} - n_z\ket{d}\end{matrix}$$

Eigenvalues and eigenstates

The eigenvalues associated with $\sigma_n$ are $$\lambda_\pm = \pm 1$$ since $$\sigma_n^2 = \mathbf{I} = \begin{bmatrix}1 & 0\\0 & 1\end{bmatrix}$$

Nb. This makes sense, since we require that applying an observable to a state results in another state, and the square of the resulting state must also be one. Only unit-length eigenvalues satisfy this condition.

The eigenvectors of a particular spin operator represent those spin-states of an electron that would be measured to be $\pm 1$ with 100% certainty by the spin operator.

For all $\hat{n} \neq \hat{z}$ (where $n_z \neq 1$), the eigenstates of the spin operator are given by $$\ket{\pm n} = \sqrt{\frac{1 \pm n_z}{2}} \: \begin{bmatrix}1 \\ \frac{\pm 1 - n_z}{n_-} \end{bmatrix} = \sqrt{\frac{1 \pm n_z}{2}} \: \left( \ket{u} + \frac{\pm 1 - n_z}{n_-} \ket{d} \right)$$

For $\hat{n} = \hat{z}$ the eigenstates are just the basis states $$\begin{align*} \ket{+z} &= \begin{bmatrix}1\\ 0 \end{bmatrix} = \ket{u} \\ \ket{-z} &= \begin{bmatrix}0 \\ 1 \end{bmatrix} = \ket{d}\end{align*}$$

Trace, the determinant and averages

All Hermitian operators have real traces, determinants and averages, so this is true for spin operators.

The traces and the determinants are the same for all spin operators $$\begin{align*}\mathrm{Tr}(\sigma_n) &= 0\\\mathrm{Det}(\sigma_n) &= -1\end{align*}$$

The averages of the spin operator in the basis states are $$\begin{align*}\bra{u}\sigma_n \ket{u} &= n_z\\\bra{d}\sigma_n \ket{d} &= -n_z\end{align*}$$

So, for example $$\bra{u}\sigma_x \ket{u} = 0$$ which can be interpreted as saying that it is equally likely that an electron prepared with its spin in the $x$-direction of normal space will be $\pm 1$, if measured in the up direction. For completeness, $$\begin{align*}\bra{u}\sigma_x \ket{u} = 0 &\quad \bra{d}\sigma_x \ket{d} = 0\\\bra{u}\sigma_y \ket{u} = 0 &\quad \bra{d}\sigma_y \ket{d} = 0\\\bra{u}\sigma_z \ket{u} = 1 &\quad \bra{d}\sigma_z \ket{d} = -1\end{align*}$$