# Composite systems

### Composite states

Entanglement can only occur in systems composed of more than one subsystem.

We can use tensor products to describe the states such systems can be in, as we did for two-electron systems such as the singlet state.

In the most general terms, we could define a basis $$\left\{\ket{e_i} \otimes \ket{e_j}: i=1,2,\cdots,N_1 \text{ and } j=1,2,\cdots,N_2\right\}$$ and then the most general pure state can be wrtten as $$\ket{\Psi} = \sum_{i=1}^{N_1} \sum_{j=1}^{N_2}\psi_{ij}\ket{e_i} \otimes \ket{e_j}$$ where $\psi_{ij} \in \mathbb{C}$. As usual, we require this to be normalised state, so \begin{align*}1 &= \braket{\Psi}{\Psi}\\&= \left(\sum_{i=1}^{N_1} \sum_{j=1}^{N_2}\bra{e_i} \otimes \bra{e_j} \psi_{ij}^*\right) \left(\sum_{k=1}^{N_1} \sum_{l=1}^{N_2}\psi_{kl}\ket{e_k} \otimes \ket{e_l}\right)\\&= \sum_{i=1}^{N_1} \sum_{j=1}^{N_2} \sum_{k=1}^{N_1} \sum_{l=1}^{N_2} \psi_{ij}^* \psi_{kl}\braket{e_i}{e_k}\braket{e_j}{e_l}\\&= \sum_{i=1}^{N_1} \sum_{j=1}^{N_2} \psi_{ij}^* \psi_{ij}\end{align*}

### Probability density

Composite linear operators, $\boldsymbol{M}$, are defined to be products of linear operators, $M, M'$ say, $$\boldsymbol{M} = M \otimes M'$$ such that \begin{align*}\boldsymbol{M}\ket{\Psi} &= \left(M \otimes M'\right)\left(\sum_{i=1}^{N_1} \sum_{j=1}^{N_2}\psi_{ij}\ket{e_i} \otimes \ket{e_j}\right)\\&= \sum_{i=1}^{N_1} \sum_{j=1}^{N_2}\psi_{ij}\left(M\ket{e_i} \otimes M'\ket{e_j}\right)\end{align*}

Thus, the probability density associated with a system is also a product, $$\boldsymbol{\rho} = \rho \otimes \rho'$$

The average density of a Hernitian operator $\boldsymbol{M}$ is then \begin{align*}\left\langle \boldsymbol{M} \right\rangle &= \mathrm{Tr}\left(\boldsymbol{\rho}\boldsymbol{M}\right)\\&= \sum_{i=1}^{N_1} \sum_{j=1}^{N_2} \bra{e_i} \otimes \bra{e_j}\left(\rho M \otimes \rho' M' \right)\ket{e_i} \otimes \ket{e_j}\\&= \left(\sum_{i=1}^{N_1} \bra{e_i}\rho M\ket{e_i}\right)\left(\sum_{j=1}^{N_2} \bra{e_j}\rho' M'\ket{e_j}\right)\\&= \mathrm{Tr}\left(\rho M\right) \mathrm{Tr}\left(\rho' M'\right)\\&= \left\langle M \right\rangle \left\langle M' \right\rangle\end{align*}

### Pure composite systems do not necessarily have pure subsystems

Suppose we know that a composite system as described above is in the pure state $\ket{\Psi}$.

We show that, in general, the first subsystem has a non-trivial probability density and so is in a mixed state. We do this by ignoring the second subsystem altogether, not observing it and assuming no knowledge of it.

Not observing it means that the Hermitian operator has the form $\boldsymbol{M} = M \otimes \mathbf{I}$, so the average density becomes $$\left\langle \boldsymbol{M} \right\rangle = \left\langle M \right\rangle \left\langle \mathbf{I} \right\rangle$$

Assuming no knowledge of it means that we must assign equal probabilities to each of the eigenstates. Thus, in the basis in which the probability density of the second subsystem is diagonal, we must have $$\rho'_{ij} = \left\{\begin{matrix}\tfrac{1}{N_2} & i = j\\0 & i \neq j\end{matrix}\right.$$

Since trace is invariant, we can calculate $\left\langle \mathbf{I} \right\rangle$ in this basis, $$\left\langle \mathbf{I} \right\rangle = \mathrm{Tr}\left(\rho' \mathbf{I}\right) = \sum_{j=1}^{N_2} \rho'_{jj} \delta_{jj} = \sum_{j=1}^{N_2} \tfrac{1}{N_2} = 1$$

Hence, the average density of the composite system is equal to the average density of the first subsystem. We can calculate in terms of its probability density (on some arbitrary basis), \begin{align*}\left\langle \boldsymbol{M} \right\rangle &= \left\langle M \right\rangle\\&= \sum_{i=1}^{N_1} \bra{e_i}\rho M \ket{e_i}\\&= \sum_{i=1}^{N_1} \bra{e_i}\rho \left(\sum_{k=1}^{N_1}\ket{e_k}\bra{e_k}\right)M \ket{e_i}\\&= \sum_{i=1}^{N_1} \sum_{k=1}^{N_1} \bra{e_i}\rho \ket{e_k} \bra{e_k} M \ket{e_i}\\&= \sum_{i=1}^{N_1} \sum_{k=1}^{N_1} \rho_{ik} M_{ki}\end{align*}

However, we can also calculate the average of $\boldsymbol{M}$ using the pure state $\ket{\Psi}$ since we know the composite system is in this state, \begin{align*}\left\langle \boldsymbol{M} \right\rangle &= \bra{\Psi}\boldsymbol{M}\ket{\Psi}\\&= \left(\sum_{i=1}^{N_1} \sum_{j=1}^{N_2}\bra{e_i} \otimes \bra{e_j} \psi_{ij}^*\right) M \otimes \mathbf{I} \left(\sum_{k=1}^{N_1} \sum_{l=1}^{N_2}\psi_{kl}\ket{e_k} \otimes \ket{e_l}\right)\\&= \sum_{i=1}^{N_1} \sum_{j=1}^{N_2}\sum_{k=1}^{N_1} \sum_{l=1}^{N_2} \psi_{ij}^*\psi_{kl} \bra{e_i}M\ket{e_k} \braket{e_j}{e_l}\\&= \sum_{i=1}^{N_1} \sum_{j=1}^{N_2}\sum_{k=1}^{N_1} \psi_{ij}^*\psi_{kj} \bra{e_i}M\ket{e_k}\\&= \sum_{i=1}^{N_1} \sum_{k=1}^{N_1} \left(\sum_{j=1}^{N_2} \psi_{ij}^*\psi_{kj}\right) M_{ik}\\&= \sum_{i=1}^{N_1} \sum_{k=1}^{N_1} \left(\sum_{j=1}^{N_2} \psi_{kj}^*\psi_{ij}\right) M_{ki}\end{align*} where we have swapped the dummy variables in the last step, so that we can equate the term in brackets with the probability density of the first subsystem, $$\rho_{ik} = \sum_{j=1}^{N_2} \psi_{kj}^*\psi_{ij}$$

So, the probability density of the first subsystem is a summation over all the possible states of the second subsystem. As we have seen, if $\rho$ did represent a pure state of the first subsystem then we would have, for some $j$, $$\rho_{ik} = \left\{\begin{matrix}1 & i = k = j\\0 & \text{ otherwise}\end{matrix}\right.$$

Clearly, this is not true in general. In general, the first subsystem is in a mixed state, even though the composite system is in a pure state.

We note that no analogue of such behaviour exists in classical mechanics, where complete knowledge of the composite system implies complete knowledge of each of the subsystems.