# Projection operators

Before we consider Bell's theorem, we need to know how to calculate the probability that an electron in a given state is measured in a given direction. We do this using projection operators.

### Normal space

A linear span of a set of vectors is the set of all linear combinations of those vectors. Thus, $$\mathbf{LS}\left\{\vec{x}_1, \dots, \vec{x}_N \right\} = \left\{\sum_{i=1}^N a_i \vec{x}_i: a_i \in \mathbb{R}\right\}$$

Figure 5.2 - Projection operator in normal space

A projection operator in $\mathbb{R}^3$ onto a direction (unit vector) $\hat{n}$ is a function \begin{align*}\Pi_\hat{n}: & \mathbb{R}^3 \rightarrow \mathbf{LS}\left\{\hat{n}\right\}\\& \vec{x} \mapsto \left(\hat{n} \cdot \vec{x}\right) \hat{n}\end{align*}

Further, we can extend this definition to projections in arbitrary vector spaces and along more than one direction.

Given a vector space $\mathbb{V}$, and a set of orthogonal directions, $\left\{\hat{n}_1, \dots, \hat{n}_N \right\}$, then the general definition of a projection operator is the function \begin{align*}\Pi_{\hat{n}_1, \dots, \hat{n}_N}: & \mathbb{V} \rightarrow \mathbf{LS}\left\{\hat{n}_1, \dots, \hat{n}_N\right\}\\& \vec{x} \mapsto \sum_{i=1}^N \left(\hat{n}_i \cdot \hat{x}\right) \hat{n}_i\end{align*}

Notice that the projection operator onto an orthonormal basis, $\left\{\hat{e}_1, \dots, \hat{e}_N \right\}$, for $\mathbb{V}$ is just the identity operator. That is, \begin{align*}\Pi_{\hat{e}_1, \dots, \hat{e}_N} \left(\vec{x}\right) &= \sum_{i=1}^N \left(\hat{e}_i \cdot \hat{x}\right) \hat{e}_i\\&= \sum_{i=1}^N x_i \hat{e}_i\\&= \vec{x}\end{align*}

### Spin projection operators

In our case, we will be projecting spin states onto the eigenspace of the spin operators. In the single-electron case, the eigenspace was only one eigenstate, but for the two-electron case, we have two pairs of eigenstates.

#### Single electron

There are two projection operators associated with the spin operator, $\sigma_n$, one for each of the eigenvalues, $\lambda_\pm = \pm 1$. They are \begin{align*}\Pi_{\pm n}: & \mathbb{V} \rightarrow \mathbf{LS}\left\{\ket{\pm n}\right\}\\& \ket{a} \mapsto \left(\braket{\pm n}{a}\right) \ket{\pm n}\end{align*} where $\mathbb{V}$ is just the state-space.

Note that is convenient to write $\left(\braket{\pm n}{a}\right) \ket{\pm n} = \ket{\pm n}\braket{\pm n}{a} = \left(\ket{\pm n}\bra{\pm n}\right)\ket{a}$ which is perfectly consistent because the product term is just a number and can be put either side of the state vector. Then, we can just say that $$\Pi_{\pm n} \equiv \ket{\pm n}\bra{\pm n}$$ and then \begin{align*}\Pi_{\pm n} \ket{a}&= \left(\ket{\pm n}\bra{\pm n}\right)\ket{a}\\ &= \ket{\pm n}\braket{\pm n}{a}\\ &= \left(\braket{\pm n}{a}\right) \ket{\pm n}\end{align*} as before.

Nb. Professor Susskind uses the more explicit notations: \begin{align*}\ket{\pm n} &\rightarrow \ket{\sigma_n = \pm 1}\\\Pi_{\pm n} &\rightarrow \Pi_{\sigma_n = \pm 1}\end{align*}

We can re-write the projection operators directly in terms of the spin operators. In fact, $$\Pi_{\pm n} = \frac{\mathbf{I} \pm \sigma_n}{2}$$

To see why this is true, consider any single-electron state $$\ket{a} = \begin{bmatrix} \alpha \\ \beta \end{bmatrix}$$

Then, projecting onto the $\ket{+n}$ eigenspace gives \begin{align*}\Pi_{+n} \ket{a} &= \ket{+n}\braket{+n}{a}\\ &= \left(\sqrt{\frac{1 + n_z}{2}} \: \begin{bmatrix}1 \\ \frac{1 - n_z}{n_-} \end{bmatrix}\right) \left(\sqrt{\frac{1 + n_z}{2}} \: \begin{bmatrix}1 & \frac{1 - n_z}{n_+} \end{bmatrix} \begin{bmatrix} \alpha \\ \beta \end{bmatrix}\right)\\ &= \frac{1 + n_z}{2}\begin{bmatrix}1 \\ \frac{1 - n_z}{n_-} \end{bmatrix}\left(\alpha + \frac{1 - n_z}{n_+} \beta\right)\\ &= \tfrac{1}{2}\begin{bmatrix} \left(1 + n_z\right) \alpha + \frac{1 - n_z^2}{n_+} \beta \\ \frac{1 - n_z^2}{n_-} \alpha + \frac{\left(1 - n_z^2\right)\left(1 - n_z\right)}{n_+ n_-} \beta \end{bmatrix}\\ &= \tfrac{1}{2}\begin{bmatrix} \left(1 + n_z\right) \alpha + \frac{n_+ n_-}{n_+} \beta \\ \frac{n_+ n_-}{n_-} \alpha + \frac{\left(n_+ n_-\right)\left(1 - n_z\right)}{n_+ n_-} \beta \end{bmatrix}\\ &= \tfrac{1}{2}\begin{bmatrix} \left(1 + n_z\right) \alpha + n_- \beta \\ n_+ \alpha + \left(1 - n_z\right) \beta \end{bmatrix}\\ &= \tfrac{1}{2}\begin{bmatrix} 1 + n_z & n_- \\ n_+ & 1 - n_z \end{bmatrix} \begin{bmatrix} \alpha \\ \beta \end{bmatrix}\\ &= \tfrac{1}{2}\left(\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} + \begin{bmatrix} n_z & n_- \\ n_+ & -n_z \end{bmatrix} \right) \begin{bmatrix} \alpha \\ \beta \end{bmatrix}\\ &= \frac{\mathbf{I} + \sigma_n}{2} \ket{a}\end{align*}

We can similarly show that $$\Pi_{-n} = \frac{\mathbf{I} - \sigma_n}{2}$$

#### Two electrons

Now there are four projection operators, two for each of the two spin operators, and for each projection, the eigenspace projected onto is the linear span of a pair of eigenvectors. Thus, \begin{align*}\Pi_{\pm n_\boldsymbol{\sigma}} &\equiv \sum_{i=1}^2 \ket{\pm n_\boldsymbol{\sigma}^i}\bra{\pm n_\boldsymbol{\sigma}^i}\\\Pi_{\pm n_\boldsymbol{\tau}} &\equiv \sum_{i=1}^2 \ket{\pm n_\boldsymbol{\tau}^i}\bra{\pm n_\boldsymbol{\tau}^i}\end{align*} recalling that, for example, \begin{align*}\ket{+n_\boldsymbol{\sigma}^1} &= \sqrt{\frac{1 + n_z}{2}}\left(\ket{uu} + \frac{1 - n_z}{n_-}\ket{du}\right)\\\ket{+n_\boldsymbol{\sigma}^2} &= \sqrt{\frac{1 + n_z}{2}}\left(\ket{ud} + \frac{1 - n_z}{n_-}\ket{dd}\right)\end{align*}

A very similar, but more complicated, argument to the single-electron case shows that we can write the projection operators in terms of their associated spin operators, \begin{align*}\Pi_{\pm n_\boldsymbol{\sigma}} &= \frac{\mathbf{I} \pm \boldsymbol{\sigma}_n}{2}\\\Pi_{\pm n_\boldsymbol{\tau}} &= \frac{\mathbf{I} \pm \boldsymbol{\tau}_n}{2}\end{align*}

### The probability postulate

#### Single electron

The postulate states that, for a given state $\ket{a}$, the probability that we will measure the spin, $\sigma_n$ to have the value $\pm 1$ is given by \begin{align*}\left|\braket{\pm n}{a}\right|^2 &= \braket{\pm n}{a}^*\braket{\pm n}{a}\\ &= \braket{a}{\pm n}\braket{\pm n}{a}\\ &= \bra{a}\left(\ket{\pm n}\bra{\pm n}\right)\ket{a}\\ &= \bra{a}\Pi_{\pm n}\ket{a}\end{align*} which is the average of the associated spin projection operator in the state $\ket{a}$.

#### Two electrons

For a pair of electrons in state $\ket{A}$ say, since their are two eigenvectors per spin operator, the probability that the first electron will be measured to be $\pm 1$ in a particular direction is a sum of inner products, \begin{align*}\sum_{i=1}^2 \left|\braket{\pm n_\boldsymbol{\sigma}^i}{A}\right|^2 &= \sum_{i=1}^2 \braket{\pm n_\boldsymbol{\sigma}^i}{A}^*\braket{\pm n_\boldsymbol{\sigma}^i}{A}\\ &= \sum_{i=1}^2 \braket{A}{\pm n_\boldsymbol{\sigma}^i}\braket{\pm n_\boldsymbol{\sigma}^i}{A}\\ &= \bra{A}\left(\sum_{i=1}^2 \ket{\pm n_\boldsymbol{\sigma}^i}\bra{\pm n_\boldsymbol{\sigma}^i}\right)\ket{A}\\ &= \bra{A}\Pi_{\pm n_\boldsymbol{\sigma}}\ket{A}\end{align*} which, again, is the average of the spin projection operator for the first electron. Similarly, for the second electron, $$\sum_{i=1}^2 \left|\braket{\pm n_\boldsymbol{\tau}^i}{A}\right|^2 = \bra{A}\Pi_{\pm n_\boldsymbol{\tau}}\ket{A}$$

#### Measuring spin of two electrons simultaneously

Recall that we can consider two-electron states as linear combinations of the basis product states, $$\ket{A} = \sum_{i,j=1}^2 \alpha_{ij} \ket{i} \otimes \ket{j}$$ where we have identified $\ket{u} \rightarrow \ket{1}, \ket{d} \rightarrow \ket{2}$. The two spin operators can also be written as products, \begin{align*}\boldsymbol{\sigma}_n &= \sigma_n \otimes \mathbf{I}\\\boldsymbol{\tau}_n &= \mathbf{I} \otimes \sigma_n\end{align*}

Hence, the projection operators are also products, \begin{align*}\Pi_{\pm n_\boldsymbol{\sigma}} &= \frac{\mathbf{I} \pm \boldsymbol{\sigma}_n}{2}\\&= \frac{\left(\mathbf{I} \otimes \mathbf{I}\right) \pm \left(\sigma_n \otimes \mathbf{I}\right)}{2}\\&= \frac{\mathbf{I} \pm \sigma_n}{2} \otimes \mathbf{I}\\&= \Pi_{\pm n} \otimes \mathbf{I}\\\Pi_{\pm n_\boldsymbol{\tau}} &= \cdots = \mathbf{I} \otimes \Pi_{\pm n}\end{align*}

Thus the spin projections for the two-electron states can be written as products of the single-electron projection, \begin{align*}\Pi_{\pm n_\boldsymbol{\sigma}} &= \Pi_{\pm n} \otimes \mathbf{I}\\\Pi_{\pm n_\boldsymbol{\tau}} &= \mathbf{I} \otimes \Pi_{\pm n}\end{align*}

This allows us to measure the spin of the two electrons simultaneously. Given two directions, $\hat{m}, \hat{n}$, the projection operator that simultaneously projects the state of the first electron onto the eigenspace of $\boldsymbol{\sigma}_m$ and the state of the second electron onto the eigenspace of $\boldsymbol{\tau}_n$ is given by \begin{align*}\Pi_{\pm m_\boldsymbol{\sigma}, \pm n_\boldsymbol{\tau}} &= \Pi_{\pm m_\boldsymbol{\sigma}} \Pi_{\pm n_\boldsymbol{\tau}}\\&= \left(\Pi_{\pm m} \otimes \mathbf{I}\right)\left(\mathbf{I} \otimes \Pi_{\pm n}\right)\\&= \Pi_{\pm m} \otimes \Pi_{\pm n}\end{align*}

We can of course express this in terms of the spins themselves, $$\Pi_{\pm m_\boldsymbol{\sigma}, \pm n_\boldsymbol{\tau}} = \frac{\mathbf{I} \pm \boldsymbol{\sigma}_m}{2} \frac{\mathbf{I} \pm \boldsymbol{\tau}_n}{2}$$

### Compatibility

We could have derived the probability of the combined statement simply because the two spin operators commute with each other, in which case we say that they are compatible.

Clearly, \begin{align*}\boldsymbol{\sigma}_m \boldsymbol{\tau}_n &= \left(\sigma_m \otimes \mathbf{I}\right)\left(\mathbf{I} \otimes \sigma_n\right)\\&= \sigma_m \otimes \sigma_n\\&= \left(\mathbf{I} \otimes \sigma_n\right) \left(\sigma_m \otimes \mathbf{I}\right)\\&= \boldsymbol{\tau}_n \boldsymbol{\sigma}_m\end{align*}

Hence, the projection operators are also compatible, $$\frac{\mathbf{I} \pm \boldsymbol{\sigma}_m}{2} \frac{\mathbf{I} \pm \boldsymbol{\tau}_n}{2} = \frac{\mathbf{I} \pm \boldsymbol{\tau}_n}{2} \frac{\mathbf{I} \pm \boldsymbol{\sigma}_m}{2}$$

Notice that, for a given state $\ket{A}$, $\Pi_{\pm n_\boldsymbol{\tau}}\ket{A}$ is in the eigenspace of $\boldsymbol{\tau}_n$. But, it's also just another state in the space of two electrons. That is, $$\frac{\mathbf{I} \pm \boldsymbol{\tau}_n}{2}\ket{A} \in \mathbf{LS}\left\{\ket{\pm n_\tau}\right\} \subset \mathbb{V}$$

Hence, $$\frac{\mathbf{I} \pm \boldsymbol{\sigma}_m}{2} \frac{\mathbf{I} \pm \boldsymbol{\tau}_n}{2} \ket{A} = \frac{\mathbf{I} \pm \boldsymbol{\sigma}_m}{2} \left(\frac{\mathbf{I} \pm \boldsymbol{\tau}_n}{2} \ket{A}\right) \in \mathbf{LS}\left\{\ket{\pm m_\sigma}\right\}$$ is in the eigenspace of $\boldsymbol{\sigma}_m$.

But, on the other hand, $$\frac{\mathbf{I} \pm \boldsymbol{\tau}_n}{2} \frac{\mathbf{I} \pm \boldsymbol{\sigma}_m}{2} \ket{A} = \frac{\mathbf{I} \pm \boldsymbol{\tau}_n}{2} \left(\frac{\mathbf{I} \pm \boldsymbol{\sigma}_m}{2} \ket{A}\right) \in \mathbf{LS}\left\{\ket{\pm n_\tau}\right\}$$ is in the eigenspace of $\boldsymbol{\tau}_n$.

Because the operators commute, the resulting state is the same for both results, so the compound operator projects a state such that the first electron lies in the eigenspace of $\boldsymbol{\sigma}_m$ and the second electron lies in the eigenspace of $\boldsymbol{\tau}_n$ simultaneously.

Nb. Notice that the compatibility of the double-electron projections does not depend on the single-electron projection being compatible with itself. In fact, for $\hat{m} \neq \hat{n}$, $\sigma_m \sigma_n \neq \sigma_n \sigma_m$.