Spin in arbitrary directions YouTube


We can now define general spin operators (for a single electron) mathematically but, in fact, they have had to be confirmed experimentally.

Suppose we define a direction in normal space to be the unit vector, $$\hat{n} = \begin{bmatrix}n_x\\n_y\\n_z\end{bmatrix} \in \mathbb{R}^3$$

Then the spin operator that measures the spin of an electron in the $\hat{n}$ direction is given by $$\sigma_n = \sigma_x n_x + \sigma_y n_y + \sigma_z n_z$$

We can define a triple, $$\vec{\sigma} = (\sigma_x, \sigma_y, \sigma_z)$$ so that we can write $$\vec{\sigma} \cdot \hat{n} = \sigma_n$$ which is analogous to the definition of the spin vector in classical mechanics.

Nb. Professor Susskind uses slightly different notation, and usually uses numeric subscripts, $\sigma_1, \sigma_2, \sigma_3$, rather than latin $\sigma_x, \sigma_y, \sigma_z$.

We can write the operator as a matrix $$\begin{align*}\sigma_n &= \sigma_x n_x + \sigma_y n_y + \sigma_z n_z\\&= \begin{bmatrix}0 & 1\\1 & 0\end{bmatrix} n_x + \begin{bmatrix}0 & -\mathrm{i}\\\mathrm{i} & 0\end{bmatrix} n_y + \begin{bmatrix}1 & 0\\0 & -1\end{bmatrix} n_z\\&= \begin{bmatrix}n_z & n_x - \mathrm{i} n_y \\ n_x + \mathrm{i} n_y & -n_z\end{bmatrix}\\&= \begin{bmatrix}n_z & n_- \\ n_+ & -n_z\end{bmatrix}\end{align*}$$ where $n_{\pm} = n_x \pm \mathrm{i}n_y$ are complex conjugates.

So, by definition, $\sigma_n$ is a Hermitian matrix (diagonal values are real, cross-diagonal are complex conjugates).

Also, because $\hat{n}$ is a unit vector, we have the following identities: $$\begin{align*}n_x^2 + n_y^2 + n_z^2 &= 1\\n_+ n_- &= (n_x + \mathrm{i}n_y)(n_x - \mathrm{i} n_y)\\&= n_x^2 + n_y^2\\&= 1 - n_z^2\end{align*}$$


We have already seen in an earlier lecture that matrices of the form $$M = \begin{bmatrix}a & b^* \\ b & -a\end{bmatrix}$$ where $a \in \mathbb{R}$ and $b \in \mathbb{C}$, are Hermitian matrices with the property that if $$M^2 = \mathbf{I}$$ then its eigenvalues satisfy $$\lambda_{\pm} = \pm 1$$

The spin operator, $$\sigma_n = \begin{bmatrix}n_z & n_- \\ n_+ & -n_z\end{bmatrix}$$ satisfies this condition, so we want to find eigenstates, $\ket{\pm n}$, that satisfy $$\begin{align*}\sigma_n \ket{+n} &= \ket{+n}\\\sigma_n \ket{-n} &= -\ket{-n}\end{align*}$$

Nb. Professor Susskind uses the more explicit notation $\ket{\vec{\sigma} \cdot \hat{n} = \pm 1}$ instead of $\ket{\pm n}$.


The eigenstates of the spin operator are given by $$\ket{\pm n} = \ket{\pm z} \text{ if } n_z = 1$$ $$\ket{\pm n} = \sqrt{\frac{1 \pm n_z}{2}} \: \begin{bmatrix}1 \\ \frac{\pm 1 - n_z}{n_-} \end{bmatrix} \text{ if } n_z \neq 1$$

Trivially, if $n_z = 1$ then we must have that $$n_+ = n_- = 0$$ which implies that $$\sigma_n = \begin{bmatrix}1 & 0 \\ 0 & -1 \end{bmatrix} = \sigma_z$$ and so the eigenstates are simply $$\ket{\pm n} = \ket{\pm z}$$

Now suppose that $n_z \neq 1$ and write $$\ket{+n} = \begin{bmatrix}\alpha \\ \beta\end{bmatrix}$$ where $\alpha, \beta \in \mathbb{C}$ and $\alpha^*\alpha + \beta^*\beta = 1$.

Case 1: $\lambda = +1$

We want to find $\ket{+n}$ such that $$\sigma_n \ket{+n} = \ket{+n}$$ which leads to the following equation $$n_z \alpha + n_- \beta = \alpha$$ (there are actually two equations, but both lead to the same conclusion). Now, $$n_z \neq 1 \Rightarrow n_+ \neq 0 \Rightarrow n_- \neq 0 $$ so we can write $$\beta = \alpha \frac{1 - n_z}{n_-}$$

We now use the fact that states are unit vectors hence we want that $$\begin{align*}1 &= \braket{+n}{+n}\\&= \alpha^* \alpha + \beta^* \beta\\&= \alpha^* \alpha + \left( \alpha\frac{1 - n_z}{n_-} \right)^* \left( \alpha \frac{1 - n_z}{n_-} \right)\\&= \alpha^* \alpha \left( 1 + \left( \frac{\left( 1 - n_z \right)^2}{n_+ n_-} \right) \right)\\&= \alpha^* \alpha \left( \frac{n_+ n_- + \left( 1 - n_z \right)^2}{n_+ n_-} \right)\\&= \alpha^* \alpha \left( \frac{1 - n_z^2 + \left( 1 - n_z \right)^2}{1 - n_z^2} \right)\\&= \alpha^* \alpha \left( \frac{2 \left(1 - n_z\right)}{1 - n_z^2} \right)\\&= \alpha^* \alpha \left( \frac{2}{1 + n_z} \right)\end{align*}$$ which leads to $$\alpha^* \alpha = \frac{1 + n_z}{2}$$

The general solution is $$\alpha = \sqrt{\frac{1 + n_z}{2}}e^{i\theta}$$ of which the simplest - and the one derived by experiment - is $$\alpha = \sqrt{\frac{1 + n_z}{2}}$$

Thus, the eigenstate corresponding to the $\lambda_+ = +1$ eigenvalue is $$\ket{+n} = \sqrt{\frac{1 + n_z}{2}} \: \begin{bmatrix}1 \\ \frac{1 - n_z}{n_-} \end{bmatrix}$$

Case 2: $\lambda = -1$

A similar calculation for the other eigenvalue leads to $$\ket{-n} = \sqrt{\frac{1 - n_z}{2}} \: \begin{bmatrix}1 \\ \frac{-1 - n_z}{n_-} \end{bmatrix}$$

Given this more general definition of the spin operator and its eigenstates, we can recover the special cases of the three coordinate directions. For example, $$\hat{n} = \hat{x} = \begin{bmatrix}1 \\ 0 \\ 0 \end{bmatrix} \quad \Rightarrow \quad \left\{\begin{matrix}\sigma_n = \sigma_x = \begin{bmatrix}0 & 1\\1 & 0 \end{bmatrix}\\\ket{\pm n} = \ket{\pm x} = \tfrac{1}{\sqrt{2}} \begin{bmatrix}1\\\pm 1\end{bmatrix}\end{matrix}\right.$$

Average value

Given an arbitrary spin state and a direction in normal space, $$\ket{a} = \begin{bmatrix}\alpha \\ \beta \end{bmatrix} \:,\quad \hat{n} = \begin{bmatrix}n_x\\n_y\\n_z\end{bmatrix}$$ it is useful to calculate the average value of the general spin operator in that state, $$\begin{align*}\bra{a} \sigma_n \ket{a} &= \begin{bmatrix} \alpha^* & \beta^* \end{bmatrix} \begin{bmatrix} n_z & n_- \\ n_+ & -n_z \end{bmatrix} \begin{bmatrix} \alpha \\ \beta \end{bmatrix}\\&= \begin{bmatrix} \alpha^* & \beta^* \end{bmatrix} \begin{bmatrix} n_z \alpha + n_- \beta \\ n_+ \alpha - n_z \beta \end{bmatrix}\\&= \alpha^* \left( n_z \alpha + n_- \beta \right) + \beta^* \left( n_+ \alpha - n_z \beta \right)\\&= \alpha^* \left( n_z \alpha + n_x \beta -\mathrm{i} n_y \beta \right) + \beta^* \left( n_x \alpha + \mathrm{i} n_y \alpha - n_z \beta \right)\\&= \left( \beta^*\alpha + \alpha^* \beta \right)n_x + \mathrm{i}\left( \beta^*\alpha - \alpha^* \beta \right)n_y + \left( \alpha^*\alpha - \beta^* \beta \right)n_z\end{align*}$$

We can also express the average in terms of the coordinate spin operators, $$\begin{align*}\bra{a} \sigma_n \ket{a} &= \bra{a} \left( \sigma_x n_x + \sigma_y n_y + \sigma_z n_z \right) \ket{a}\\&= \bra{a} \sigma_x \ket{a} n_x + \bra{a} \sigma_y \ket{a} n_y + \bra{a} \sigma_z \ket{a} n_z\end{align*}$$

Thus, we can write $$\begin{align*}\bra{a} \sigma_x \ket{a} &= \beta^*\alpha + \alpha^* \beta\\\bra{a} \sigma_y \ket{a} &= \mathrm{i} \left( \beta^*\alpha - \alpha^* \beta \right)\\\bra{a} \sigma_z \ket{a} &= \alpha^*\alpha - \beta^* \beta\end{align*}$$

The last of these averages is clearly real, and it is simple to show that the first two are real as well, hence the average is real in general.

To see why the first two are real, note that for any $z = a + \mathrm{i}b \in \mathbb{C}$, we have $$\begin{matrix}z + z^* &=& a + \mathrm{i}b + a -\mathrm{i} b &=& 2a &\in \mathbb{R}\\i\left(z - z^*\right) &=& i\left(a + \mathrm{i}b - a + \mathrm{i}b\right) &=& -2b &\in \mathbb{R}\end{matrix}$$

Then set $$z = \beta^*\alpha$$ so that $$z^* = \alpha^*\beta$$ and the result follows.