Projection operators YouTube

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

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, $$\Pi_{\pm m_\boldsymbol{\sigma}, \pm n_\boldsymbol{\tau}} = \frac{\mathbf{I} \pm \boldsymbol{\sigma}_m}{2} \frac{\mathbf{I} \pm \boldsymbol{\tau}_n}{2}$$


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 $\hat{m} \neq \hat{n}$, $\sigma_m \sigma_n \neq \sigma_n \sigma_m$.