### Spin states

Recall we defined three pairs of two-dimensional complex vectors, $$\ket{\pm x}, \ket{\pm y}, \ket{\pm z}$$ in terms of the two basis vectors $\ket{u}, \ket{d}$ of the state-space of a single electron, $$\begin{align*}\ket{\pm x} &= \tfrac{1}{\sqrt{2}}\begin{bmatrix} 1 \\ \pm 1 \end{bmatrix} = \tfrac{1}{\sqrt{2}}\left( \ket{u} \pm \ket{d} \right)\\\ket{\pm y} &= \tfrac{1}{\sqrt{2}}\begin{bmatrix} 1 \\ \pm \mathrm{i} \end{bmatrix} = \tfrac{1}{\sqrt{2}}\left( \ket{u} \pm \mathrm{i} \ket{d} \right)\\\ket{+z} &= \begin{bmatrix} 1 \\ 0 \end{bmatrix} = \ket{u}\\\ket{-z} &= \begin{bmatrix} 0 \\ 1 \end{bmatrix} = \ket{d}\end{align*}$$

These vectors represented states that the spin of an electron could be in. For example, an electron with its spin in the $+x$-direction of normal space would be in the state $\ket{+x}$.

An electron prepared in an arbitrary state can be made up from any (unit) linear combination of the $\ket{u}, \ket{d}$ state vectors, and the *squares* of components of that state are the probabilities that the given electron would be detected as being in either the up state or down state as measured in the *vertical* direction of normal space.

We would like to generalise this measuring the spin of an electron in *any* direction of space, not just the vertical.

### Measuring spin

We would like to define observables - Hermitian matrices - called **spin operators** which measure the spin of an electron in arbitrary directions of normal space. That is, suppose we are given vectors in state-space, $\ket{\pm n}$, that correspond to an arbitrary direction, $\hat n$, in normal space, then we want an operator $\sigma_n$, whose eigenvectors are exactly $\ket{\pm n}$, and the corresponding eigenvalues are $\pm 1$.

We start with the three coordinate directions and in the next lecture move on to the general case.

#### z-direction

We want to find an Hermitian operator, $\sigma_z$, such that $$\begin{align*}\sigma_z \ket{+z} &= \lambda_{+z} \ket{+z}\\\sigma_z \ket{-z} &= \lambda_{-z} \ket{-z}\end{align*}$$

For simplicity, we can insist that $\sigma_z^2 = 1$ so that the eigenvalues are guaranteed to be $\lambda_{\pm z} = \pm 1$, which means we can write $$\sigma_z = \begin{bmatrix}a & b^*\\ b & -a \end{bmatrix}$$ where $a \in \mathbb{R}$ and $b \in \mathbb{C}$.

Thus, we want the first equation is $$\begin{align*}\begin{bmatrix}a & b^*\\b & -a\end{bmatrix}\begin{bmatrix}1\\0\end{bmatrix}&=\begin{bmatrix}1\\0\end{bmatrix}\end{align*}$$ or $$\left.\begin{matrix}a \cdot 1 + b^* \cdot 0 = 1 \\ b \cdot 1 - a \cdot 0 = 0 \end{matrix}\right\} \Rightarrow \left\{\begin{matrix} a = 1 \\ b = 0 \end{matrix}\right.$$

Hence, $$\sigma_z = \begin{bmatrix}1 & 0\\ 0 & -1 \end{bmatrix}$$

We can check that this definition of $\sigma_z$ satifies the second equation, $$\begin{bmatrix}1 & 0\\ 0 & -1 \end{bmatrix}\begin{bmatrix}0\\1\end{bmatrix}= \begin{bmatrix}1 \cdot 0 + 0 \cdot 1\\0 \cdot 0 - 1 \cdot 0\end{bmatrix}= \begin{bmatrix}0\\-1\end{bmatrix}= -\begin{bmatrix}0\\1\end{bmatrix}$$

Ultimately, all spin operators have been derived empirically (by experiment) and by brilliant guesswork. But, given $\sigma_z$ and the probability postulate, we can calculate the spin operators in the other two coordinate directions.

#### x-direction

We can even derive the states, $\ket{\pm x}$, from $\sigma_z$ and the probability postulate. In the description of how the spin-state is measured, we expected that if an electron has been prepared in the state $$\ket{+z} = \begin{bmatrix}1\\0\end{bmatrix}$$ then the probability that it would be detected in the state $$\ket{+x} = \begin{bmatrix} \alpha \\ \beta \end{bmatrix}$$ should be exactly $1/2$. That is, $$\begin{align*}\tfrac{1}{2} &=\left | \braket{+x}{+z} \right |^2\\&= \left | \begin{bmatrix} \alpha^*& \beta^*\end{bmatrix} \begin{bmatrix}1\\0\end{bmatrix}\right |^2\\&= \left | \alpha^* \right |^2\\&= \alpha^* \alpha\end{align*}$$

The simplest value of $\alpha$ that satisfies this equation is $$\alpha = \tfrac{1}{\sqrt{2}}$$

Since all states are unit vectors, we can immediately choose $$\beta = \tfrac{1}{\sqrt{2}}$$ since $$\begin{align*}1 &= \alpha^* \alpha + \beta^* \beta\\&= \tfrac{1}{2} + \beta^* \beta\end{align*}$$

So, as previously defined (derived from experiment), $$\ket{+x} = \tfrac{1}{\sqrt{2}}\begin{bmatrix}1\\1\end{bmatrix}$$

We could now use this result to find $$\ket{-x} = \begin{bmatrix} \gamma \\ \delta \end{bmatrix}$$ since we argued from principle, that $$\begin{align*}0 &= \braket{-x}{+x}\\&= \tfrac{1}{\sqrt{2}}\begin{bmatrix}\gamma^* & \delta^*\end{bmatrix} \begin{bmatrix}1\\1\end{bmatrix}\\&= \tfrac{1}{\sqrt{2}}\left( \gamma^* + \delta^* \right)\end{align*}$$

Thus, $$\gamma^* + \delta^* = 0$$ or $$ \delta = -\gamma$$

We also know that $\ket{-x}$ is a unit vector, so $$\begin{align*}1 &= \braket{-x}{-x}\\&= \gamma^*\gamma + \delta^*\delta\\&= \gamma^*\gamma + \left( -\gamma^* \right)\left( -\gamma \right)\\&= 2\gamma^*\gamma\end{align*}$$

The simplest value of $\gamma$ that satisfies this equation is $$\gamma = \tfrac{1}{\sqrt{2}}$$ and this also means $$\delta = -\gamma = -\tfrac{1}{\sqrt{2}}$$

So, $$\ket{-x} = \tfrac{1}{\sqrt{2}}\begin{bmatrix}1\\-1\end{bmatrix}$$

Now we consider the **spin operator**, we can write $$\sigma_x = \begin{bmatrix}r & c\\c^* & -r\end{bmatrix}$$ where $r \in \mathbb{R}$ and $c \in \mathbb{C}$, since we want eigenvalues $\pm 1$. That is, we want $$\begin{align*}\sigma_x \ket{+x} &= \ket{+x}\\\sigma_x \ket{-x} &= -\ket{-x}\end{align*}$$

We can write the first equation as $$\begin{bmatrix}r & c\\c^* & -r\end{bmatrix} \tfrac{1}{\sqrt{2}}\begin{bmatrix}1\\1\end{bmatrix} = \tfrac{1}{\sqrt{2}}\begin{bmatrix}1\\1\end{bmatrix}$$ from which we get $$\begin{matrix}r + c = 1\\ c^* - r = 1\end{matrix}$$

The simplest solution (and the one defined by experiment) is to take $$\begin{matrix}r = 0\\ c = 1\end{matrix}$$ so that $$\sigma_x = \begin{bmatrix}0 & 1\\1 & 0\end{bmatrix}$$

We can check the second equation, $$\begin{align*}\sigma_x \ket{-x} &=\begin{bmatrix}0 & 1\\1 & 0\end{bmatrix} \tfrac{1}{\sqrt{2}}\begin{bmatrix}1\\-1\end{bmatrix}\\&= \tfrac{1}{\sqrt{2}}\begin{bmatrix}-1\\1\end{bmatrix}\\&= -\tfrac{1}{\sqrt{2}}\begin{bmatrix}1\\-1\end{bmatrix}\\&= -\ket{-x}\end{align*}$$

#### y-direction

We can get the values for the $\ket{\pm y}$ states by similar arguments to the above - or by experiment, of course. The **spin operator** turns out to be $$\sigma_y = \begin{bmatrix}0 & -\mathrm{i}\\\mathrm{i} & 0\end{bmatrix}$$

### Summary

#### Spin operators

The spin operators for the three coordinate directions of normal space, along with their corresponding eigenstates are $$\begin{matrix}\sigma_x = \begin{bmatrix}0 & 1\\1 & 0\end{bmatrix} & \ket{\pm x} = \tfrac{1}{\sqrt{2}}\begin{bmatrix}1\\ \pm 1\end{bmatrix}\\\sigma_y = \begin{bmatrix}0 & -\mathrm{i}\\\mathrm{i} & 0\end{bmatrix} & \ket{\pm y} = \tfrac{1}{\sqrt{2}}\begin{bmatrix}1\\ \pm \mathrm{i}\end{bmatrix}\\\sigma_z = \begin{bmatrix}1 & 0\\0 & -1\end{bmatrix}& \ket{+z} = \begin{bmatrix}1\\ 0\end{bmatrix}, \quad \ket{-z} = \begin{bmatrix}0\\ 1\end{bmatrix}\end{matrix}$$

#### Eigenvalues and eigenstates

The eigenvalues (or measurable values) of each spin operator are $\pm 1$, corresponding to the only two mutually exclusive outcomes that could occur after a measurement of an electron's spin has been made. The mutual exclusivity is an experimental fact, but follows from the definition of the eigenstates above.

#### Probability

The squares of the inner products between eigenstates of perpendicular directions are all equal to $1/2$, which is the same as saying that the probability of being measured in, say, the $\ket{\pm z}$ state, having been prepared in either of the $\ket{\pm x}$ or $\ket{\pm y}$ states, is $1/2$.

So, given the electron is prepared in the $\ket{+x}$, and is measured in the $z$-direction, then it is equally likely that the observable will be found to be $\pm 1$. We would thus expect that the average value of the observable is zero, $$\bra{+z} \sigma_x \ket{+z} = \begin{bmatrix}1 & 0\end{bmatrix} \begin{bmatrix}0 & 1\\1 & 0\end{bmatrix} \begin{bmatrix}1\\ 0\end{bmatrix}= \begin{bmatrix}1 & 0\end{bmatrix} \begin{bmatrix}0\\ 1\end{bmatrix} = 0$$

#### Measurement

In classical mechanics an electron has a magnetic moment, or spin vector, associated with it which, classically, can be measured continuously along each of its three spatial directions.

In quantum mechanics, however, only one of the three components can be measured at a time, and there are only two possible values that can be measured.