Spin operators for two electrons YouTube

Linear operators in general

The spin operator for the one electron state is a linear operator mapping states to states. We need to define similar operators for the two electron states. Suppose we define a state using the alternative notation, $$\ket{A} = \sum_{i,j = 1}^{2} \alpha_{ij} \ket{ij} = \sum_{i,j = 1}^{2} \alpha_{ij} \ket{i} \otimes \ket{j}$$ then we can show that a linear operator, $\mathbf{L}$ say, acting on $\ket{A}$ can have the form $$\mathbf{L} = L_1 \otimes L_2$$ where $L_1$ acts on the first state, and $L_2$ acts on the second. That is, $$\begin{align*}\mathbf{L} \ket{A} &= \mathbf{L} \left(\sum_{i,j = 1}^{2} \alpha_{ij} \ket{ij}\right)\\ &= \sum_{i,j = 1}^{2} \alpha_{ij} \mathbf{L} \ket{ij}\\ &= \sum_{i,j = 1}^{2} \alpha_{ij} \left(L_1 \otimes L_2\right) \left(\ket{i} \otimes \ket{j}\right)\\ &= \sum_{i,j = 1}^{2} \alpha_{ij} \left(L_1\ket{i}\right) \otimes \left(L_2\ket{j}\right)\end{align*}$$

Spin operators

Given a direction in normal space, $\hat{n}$, there exist two spin operators, $\boldsymbol{\sigma}_n, \boldsymbol{\tau}_n$ that can be written in terms of the single-electron spin operator, $\sigma_n$ such that $\boldsymbol{\sigma}_n$ measures the spin of the first electron and $\boldsymbol{\tau}_n$ measures the spin of the second electron.

Nb. The two-electron operators are rendered in a bold font, however the context in which these symbols appear will be clear.

We define $$\begin{align*}\boldsymbol{\sigma}_n &= \sigma_n \otimes \mathbf{I}\\\boldsymbol{\tau}_n &= \mathbf{I} \otimes \sigma_n\end{align*}$$ where $I$ is the identity operator (matrix).

We can now completely specify the new operators in terms of their actions on the basis states, for example $$\begin{align*}\boldsymbol{\sigma}_n \ket{uu} &= \left(\sigma_n \otimes \mathbf{I}\right)\left(\ket{u} \otimes \ket{u}\right)\\ &= \left(\sigma_n \ket{u} \right) \otimes \left(I \ket{u}\right)\\ &= \left(n_z\ket{u} + n_+\ket{d} \right) \otimes \ket{u}\\ &= n_z\ket{u} \otimes \ket{u} + n_+\ket{d} \otimes \ket{u}\\ &= n_z\ket{uu} + n_+\ket{du}\end{align*}$$ where we recall that $\sigma_n \ket{u} = n_z\ket{u} + n_+\ket{d}$.

Similarly, $$\begin{align*}\boldsymbol{\tau}_n \ket{uu} &= \left(I \otimes \sigma_n\right)\left(\ket{u} \otimes \ket{u}\right)\\ &= \left(I \ket{u} \right) \otimes \left(\sigma_n \ket{u}\right)\\ &= \ket{u} \otimes \left(n_z\ket{u} + n_+\ket{d} \right)\\ &= n_z\ket{u} \otimes \ket{u} + n_+\ket{u} \otimes \ket{d}\\ &= n_z\ket{uu} + n_+\ket{ud}\end{align*}$$

The complete specification is $$\begin{matrix}\boldsymbol{\sigma}_n \ket{uu} = n_z\ket{uu} + n_+\ket{du} &\quad \boldsymbol{\tau}_n \ket{uu} = n_z\ket{uu} + n_+\ket{ud}\\\boldsymbol{\sigma}_n \ket{ud} = n_z\ket{ud} + n_+\ket{dd} &\quad \boldsymbol{\tau}_n \ket{ud} = n_-\ket{uu} - n_z\ket{ud}\\\boldsymbol{\sigma}_n \ket{du} = n_-\ket{uu} - n_z\ket{du} &\quad \boldsymbol{\tau}_n \ket{du} = n_z\ket{du} + n_+\ket{dd}\\\boldsymbol{\sigma}_n \ket{dd} = n_-\ket{ud} - n_z\ket{dd} &\quad \boldsymbol{\tau}_n \ket{dd} = n_-\ket{du} - n_z\ket{dd}\end{matrix}$$

The spins corresponding to the three coordinate directions can then be derived, $$\begin{matrix}\boldsymbol{\sigma}_x \ket{uu} = \ket{du} &\quad \boldsymbol{\tau}_x \ket{uu} = \ket{ud}\\\boldsymbol{\sigma}_x \ket{ud} = \ket{dd} &\quad \boldsymbol{\tau}_x \ket{ud} = \ket{uu}\\\boldsymbol{\sigma}_x \ket{du} = \ket{uu} &\quad \boldsymbol{\tau}_x \ket{du} = \ket{dd}\\\boldsymbol{\sigma}_x \ket{dd} = \ket{ud} &\quad \boldsymbol{\tau}_x \ket{dd} = \ket{du}\end{matrix}$$

$$\begin{matrix}\boldsymbol{\sigma}_y \ket{uu} = \mathrm{i} \ket{du} &\quad \boldsymbol{\tau}_y \ket{uu} = \mathrm{i} \ket{ud}\\\boldsymbol{\sigma}_y \ket{ud} = \mathrm{i} \ket{dd} &\quad \boldsymbol{\tau}_y \ket{ud} = -\mathrm{i} \ket{uu}\\\boldsymbol{\sigma}_y \ket{du} = -\mathrm{i} \ket{uu} &\quad \boldsymbol{\tau}_y \ket{du} = \mathrm{i} \ket{dd}\\\boldsymbol{\sigma}_y \ket{dd} = -\mathrm{i} \ket{ud} &\quad \boldsymbol{\tau}_y \ket{dd} = -\mathrm{i} \ket{du}\end{matrix}$$

$$\begin{matrix}\boldsymbol{\sigma}_z \ket{uu} = \ket{uu} &\quad \boldsymbol{\tau}_z \ket{uu} = \ket{uu}\\\boldsymbol{\sigma}_z \ket{ud} = \ket{ud} &\quad \boldsymbol{\tau}_z \ket{ud} = -\ket{ud}\\\boldsymbol{\sigma}_z \ket{du} = -\ket{du} &\quad \boldsymbol{\tau}_z \ket{du} = \ket{du}\\\boldsymbol{\sigma}_z \ket{dd} = -\ket{dd} &\quad \boldsymbol{\tau}_z \ket{dd} = -\ket{dd}\end{matrix}$$

Matrix notation

We can use the above specification to define matrices representing the spin operators, where we consider states as four-dimensional (complex) vectors, $$\ket{A} = \alpha \ket{uu} + \beta \ket{ud} + \gamma \ket{du} + \delta \ket{dd} = \begin{bmatrix} \alpha \\ \beta \\ \gamma \\ \delta \end{bmatrix}$$

Then, $$\begin{align*}\boldsymbol{\sigma_n} &= \begin{bmatrix} n_z & 0 & n_- & 0 \\ 0 & n_z & 0 & n_-\\ n_+ & 0 &-n_z & 0 \\ 0 & n_+ & 0 &-n_z \end{bmatrix} = \begin{bmatrix} n_z \mathbf{I} & n_- \mathbf{I} \\ n_+ \mathbf{I} & -n_z \mathbf{I} \end{bmatrix}\\\boldsymbol{\tau_n} &= \begin{bmatrix} n_z & n_- & 0 & 0 \\ n_+ & -n_z & 0 & 0 \\ 0 & 0 & n_z & n_- \\ 0 & 0 & n_+ & -n_z \end{bmatrix} = \begin{bmatrix} \sigma_n & \mathbf{0} \\ \mathbf{0} & \sigma_n \end{bmatrix}\end{align*}$$ where I have also included mnemonic versions, involving the two-dimensional matrices $\sigma_n, \mathbf{I}, \mathbf{0}$ to highlight the pattern.

The spins corresponding to the three coordinate directions are then $$\begin{align*}\boldsymbol{\sigma_x} = \begin{bmatrix} \mathbf{0} & \mathbf{I} \\ \mathbf{I} & \mathbf{0} \end{bmatrix}&\quad\boldsymbol{\tau_x} = \begin{bmatrix} \sigma_x & \mathbf{0} \\ \mathbf{0} & \sigma_x \end{bmatrix}\end{align*}$$

$$\begin{align*}\boldsymbol{\sigma_y} = \begin{bmatrix} \mathbf{0} & -i \mathbf{I} \\ i \mathbf{I} & \mathbf{0} \end{bmatrix}&\quad\boldsymbol{\tau_y} = \begin{bmatrix} \sigma_y & \mathbf{0} \\ \mathbf{0} & \sigma_y \end{bmatrix}\end{align*}$$

$$\begin{align*}\boldsymbol{\sigma_z} = \begin{bmatrix} \mathbf{I} & \mathbf{0} \\ \mathbf{0} & -\mathbf{I} \end{bmatrix}&\quad\boldsymbol{\tau_z} = \begin{bmatrix} \sigma_z & \mathbf{0} \\ \mathbf{0} & \sigma_z \end{bmatrix}\end{align*}$$

Eigenvalues and eigenstates

Using the mnemonic version of the spin operator, we can see that $$\begin{align*}\boldsymbol{\sigma_n}^2 &= \begin{bmatrix} n_z \mathbf{I} & n_- \mathbf{I} \\ n_+ \mathbf{I} & -n_z \mathbf{I} \end{bmatrix}\begin{bmatrix} n_z \mathbf{I} & n_- \mathbf{I} \\ n_+ \mathbf{I} & -n_z \mathbf{I} \end{bmatrix}\\ &= \begin{bmatrix} \mathbf{I} & \mathbf{0} \\ \mathbf{0} & \mathbf{I} \end{bmatrix}\\ &= \mathbf{I}\end{align*}$$ where we have used $\mathbf{I}$ for both the two- and four-dimensional identity operators.

This implies, as for the two-dimensional case, that the eigenvalues of $\boldsymbol{\sigma_n}$ must be $\lambda = \pm 1$, but that, for each value, there are now two eigenstates. They are similar to the single-electron states, where for $\hat{n} \neq \hat{z}$, we have $$\ket{\pm n_\sigma} \in \left\{\begin{matrix}\sqrt{\frac{1 \pm n_z}{2}} \: \left( \ket{uu} + \frac{\pm 1 - n_z}{n_-} \ket{du} \right)\\\sqrt{\frac{1 \pm n_z}{2}} \: \left( \ket{ud} + \frac{\pm 1 - n_z}{n_-} \ket{dd} \right)\end{matrix}\right.$$

For $\hat{n} = \hat{z}$, we have $$\begin{align*}\ket{+z_\sigma} &\in \left\{\ket{uu}, \ket{ud}\right\}\\\ket{-z_\sigma} &\in \left\{\ket{du}, \ket{dd}\right\}\end{align*}$$

We can apply the same arguments for the $\boldsymbol{\tau_n}$ operator. Its four eigenstates are, for $\hat{n} \neq \hat{z}$, $$\ket{\pm n_\tau} \in \left\{\begin{matrix}\sqrt{\frac{1 \pm n_z}{2}} \: \left( \ket{uu} + \frac{\pm 1 - n_z}{n_-} \ket{ud} \right)\\\sqrt{\frac{1 \pm n_z}{2}} \: \left( \ket{du} + \frac{\pm 1 - n_z}{n_-} \ket{dd} \right)\end{matrix}\right.$$ and for $\hat{n} = \hat{z}$, $$\begin{align*}\ket{+z_\tau} &\in \left\{\ket{uu}, \ket{du}\right\}\\\ket{-z_\tau} &\in \left\{\ket{ud}, \ket{dd}\right\}\end{align*}$$