Pairs of electrons YouTube


We now consider a pair of electrons, idealised in the same way as for the single electron - we imagine each electron to be fixed in space and only the spins of the electrons are allowed to vary.

The state-space must now be described in terms of two quantum bits, which means four complex variables. Given the convention that up means in the $+z$-direction (and down means the $-z$-direction), the following state vectors form a basis on the state-space, $$\left\{ \ket{uu}, \ket{ud}, \ket{du}, \ket{dd} \right\} = \left\{ \begin{bmatrix} 1 \\ 0 \\ 0 \\ 0 \end{bmatrix},\begin{bmatrix} 0 \\ 1 \\ 0 \\ 0 \end{bmatrix},\begin{bmatrix} 0 \\ 0 \\ 1 \\ 0 \end{bmatrix},\begin{bmatrix} 0 \\ 0 \\ 0 \\ 1 \end{bmatrix} \right\}$$ where, for example, $\ket{ud}$ is the state where the spin of the first electron is up and the spin of the second is down. Then, in this basis, states have the form $$\ket{A} = \begin{bmatrix} \alpha \\ \beta \\ \gamma \\ \delta \end{bmatrix} = \alpha \ket{uu} + \beta \ket{ud} + \gamma \ket{du} + \delta \ket{dd}$$ where $\alpha, \beta, \gamma, \delta \in \mathbb{C}$

We still require that all (pure) states are unit length, that is $$\braket{A}{A} = \alpha^*\alpha + \beta^*\beta + \gamma^*\gamma + \delta^*\delta = 1$$

As before, the components of the vectors are probability amplitudes, and the squares of the components are probabilities. So, for example, $\alpha^*\alpha$ is the probability that a pair of electrons prepared in the state $\ket{A}$ will be measured to be in the $\ket{uu}$ state.

Nb. It is worth emphasising that there is no analagous idea of defining a state as a superposition of basis states in classical mechanics.

Product states

We can use tensor product notation to describe certain states. Given a pair of electrons, suppose the first electron is in state $\ket{a}$ and the second is in state $\ket{b}$. We could write the two-electron state as $$\ket{ab} = \ket{a} \otimes \ket{b}$$ where $\otimes$ is a bilinear operator, meaning it has the following properties, $$\begin{align*}\left( \ket{a} + \ket{b} \right) \otimes \ket{c} &= \left(\ket{a} \otimes \ket{c} \right) + \left(\ket{b} \otimes \ket{c} \right)\\\ket{a} \otimes \left( \ket{b} + \ket{c} \right) &= \left(\ket{a} \otimes \ket{b} \right) + \left(\ket{a} \otimes \ket{c} \right)\\\lambda \left(\ket{a} \otimes \ket{b} \right) &= \left(\lambda \ket{a} \right) \otimes \ket{b}\\&= \ket{a} \otimes \left(\lambda \ket{b} \right)\\&= \lambda \ket{a} \otimes \ket{b}\end{align*}$$

Now, not all states can be represented by a simple product of two states - in fact, entangled states are exactly those states that cannot be represented in this way. But, we can write the basis states in this way, $$\begin{align*} \ket{uu} &= \ket{u} \otimes \ket{u} \\ \ket{ud} &= \ket{u} \otimes \ket{d} \\ \ket{du} &= \ket{d} \otimes \ket{u} \\ \ket{dd} &= \ket{d} \otimes \ket{d} \end{align*}$$ and then, as stated above, all states can be written as a linear combination of these $$\ket{A} = \alpha \ket{u} \otimes \ket{u} + \beta \ket{u} \otimes \ket{d} + \gamma \ket{d} \otimes \ket{u} + \delta \ket{d} \otimes \ket{d}$$

We can express the bra states in a similar notation, $$\bra{A} = \bra{u} \otimes \bra{u} \alpha^* + \bra{u} \otimes \bra{d} \beta^* + \bra{d} \otimes \bra{u} \gamma^* + \bra{d} \otimes \bra{d} \delta^*$$

Inner product

Since $\ket{uu}, \ket{ud}, \ket{du}, \ket{dd}$ form an orthonormal basis, we expect that $$\begin{align*}\braket{uu}{uu} = 1 &\quad \braket{uu}{ud} = 0 & \braket{uu}{du} = 0 &\quad \braket{uu}{dd} = 0\\\braket{ud}{uu} = 0 &\quad \braket{ud}{ud} = 1 & \braket{ud}{du} = 0 &\quad \braket{ud}{dd} = 0\\\braket{du}{uu} = 0 &\quad \braket{du}{ud} = 0 & \braket{du}{du} = 1 &\quad \braket{du}{dd} = 0\\\braket{dd}{uu} = 0 &\quad \braket{dd}{ud} = 0 & \braket{dd}{du} = 0 &\quad \braket{dd}{dd} = 1\end{align*}$$

In order for this to be true, we define the inner product of two product states, $\ket{a} \otimes \ket{b}, \ket{c} \otimes \ket{d}$ to be $$\left( \bra{a} \otimes \bra{b} \right)\left( \ket{c} \otimes \ket{d} \right) = \braket{a}{c}\braket{b}{d}$$

Then, for some example basis states, $$\begin{align*}\braket{ud}{ud} &= \left( \bra{u} \otimes \bra{d} \right)\left( \ket{u} \otimes \ket{d} \right)\\&= \braket{u}{u}\braket{d}{d}\\&= 1 \cdot 1\\&= 1\end{align*}$$ and $$\begin{align*}\braket{uu}{du} &= \left( \bra{u} \otimes \bra{u} \right)\left( \ket{d} \otimes \ket{u} \right)\\&= \braket{u}{d}\braket{u}{u}\\&= 0 \cdot 1\\&= 0\end{align*}$$ as required.

The inner product between two states $$\begin{align*}\ket{A} &= \alpha_{uu} \ket{uu} + \alpha_{ud} \ket{ud} + \alpha_{du} \ket{du} + \alpha_{dd} \ket{dd}\\\ket{B} &= \beta_{uu} \ket{uu} + \beta_{ud} \ket{ud} + \beta_{du} \ket{du} + \beta_{dd} \ket{dd}\end{align*}$$ is then $$\begin{align*}\braket{B}{A} &= \left( \bra{uu} \beta_{uu}^* + \bra{ud} \beta_{ud}^* + \bra{du} \beta_{du}^* + \bra{dd} \beta_{dd}^* \right) \left( \alpha_{uu} \ket{uu} + \alpha_{ud} \ket{ud} + \alpha_{du} \ket{du} + \alpha_{dd} \ket{dd} \right)\\&= \beta_{uu}^*\alpha_{uu} + \beta_{ud}^*\alpha_{ud} + \beta_{du}^*\alpha_{du} + \beta_{dd}^*\alpha_{dd}\end{align*}$$ as expected.


Recall the discussion about how many real variables are needed to describe a single-electron state. There are four real numbers in two complex numbers, but we could reduce that to two real numbers because the state is a unit vector and it is phase invariant.

Using the same analysis to the two electron system, there are eight real variables in four complex numbers, which we can reduce by two to six real numbers because the state is a unit vector and it is phase invariant.

Classically, we expect we would need twice the number of real variables in a single-electron system to describe a double-electron system, that is four, but the general quantum state requires six. This shows that, in general, a system of two electrons cannot be thought of as a pair of independent one electron systems, although, as we shall see, such independent systems exist.

The space of states describing a pair of electrons contains states other than independent systems - a phenomenon called entanglement.

Alternative notation

We could use a more succinct notation for states of two electron systems. Write the two basis vectors as $$\begin{align*}\ket{u} &\rightarrow \ket{1}\\\ket{d} &\rightarrow \ket{2}\end{align*}$$

Then, we can express a state as a double sum, $$\ket{A} = \sum_{i,j = 1}^{2} \alpha_{ij} \ket{i} \otimes \ket{j}$$

Inner products then become, $$\begin{align*}\braket{B}{A} &= \left(\sum_{i,j} \bra{i} \otimes \bra{j} \beta_{ij}^*\right)\left(\sum_{k,l} \alpha_{kl} \ket{k} \otimes \ket{l}\right)\\ &= \sum_{i,j} \beta_{ij}^* \sum_{k,l} \alpha_{kl} \left(\bra{i} \otimes \bra{j}\right) \left(\ket{k} \otimes \ket{l}\right)\\ &= \sum_{i,j} \beta_{ij}^* \sum_{k,l} \alpha_{kl} \braket{i}{k} \braket{j}{l}\\ &= \sum_{i,j} \beta_{ij}^* \sum_{k,l} \alpha_{kl} \delta_{ik} \delta_{jl}\\ &= \sum_{i,j} \beta_{ij}^* \alpha_{ij}\end{align*}$$