### State-space

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.

### Entanglement

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