### Product states

We can find pure composite systems made up of pure subsystems, namely those that are know to be in *product* states.

To see why this is true, suppose that the composite system is known to be in a product state $$\ket{\Psi} = \ket{\phi} \otimes \ket{\chi}$$ where $\ket{\phi}, \ket{\chi}$ are pure states.

Equivalently, we can write in terms of components on some basis, $$\psi_{ij} = \phi_i \chi_j$$ where, for product states, we are able to *separate the variables*.

Now if we measure the first subsystem and ignore the second, we find that the probability density for the first subsystem is $$\begin{align*}\rho_{ik} &= \sum_{j=1}^{N_2} \psi_{kj}^*\psi_{ij}\\&= \sum_{j=1}^{N_2} \phi_k^* \chi_j^* \phi_i \chi_j\\&= \phi_k^* \phi_i \left(\sum_{j=1}^{N_2} \chi_j^* \chi_j\right)\\&= \phi_k^* \phi_i \left(1\right)\\&= \phi_k^* \phi_i\end{align*}$$ (using the fact that the sub-states are normalised).

Ignoring the second system means a composite operator $\boldsymbol{M}$ has the form $$\boldsymbol{M} = M \otimes \mathbf{I}$$ and the average of the *first* subsystem operator $M$ is then $$\begin{align*}\left\langle M \right\rangle &= \sum_{i=1}^{N_1} \sum_{k=1}^{N_1} \rho_{ik} M_{ki}\\&= \sum_{i=1}^{N_1} \sum_{k=1}^{N_1} \phi_k^* \phi_i M_{ki}\\&= \sum_{i=1}^{N_1} \sum_{k=1}^{N_1} \phi_i^* \phi_k M_{ik} & \text{ swapping dummy variables}\\&= \sum_{i=1}^{N_1} \sum_{k=1}^{N_1} \phi_i^* M_{ik} \phi_k\\&= \bra{\phi}M\ket{\phi}\end{align*}$$ which implies that the first subsystem is in the pure state $\ket{\phi}$.

We can make the same argument to show that the second subsystem is also in a pure state, namely $\ket{\chi}$.

The entropy for the composite system in this state is *zero*, since we have full knowledge about each of the subsystems.

In general, we might say that the smaller the entropy of a composite system, then the more like a product state it is. Similarly, the larger the entropy, the more the system is entangled and less like a product state.

### Singlet state

Suppose a composite system is known to be in the singlet state, $$\ket{\Psi} = \frac{\ket{ud}-\ket{du}}{\sqrt{2}}$$ the components of which are $$\begin{matrix}\psi_{uu} = 0 & \psi_{ud} = \tfrac{1}{\sqrt{2}}\\\psi_{du} = -\tfrac{1}{\sqrt{2}} & \psi_{dd} = 0\end{matrix}$$

The expression for the components of the probability density of the first subsystem of a two-bit system was, $$\begin{align*}\rho_{ik} &= \sum_{j=1}^{N_2} \psi_{kj}^*\psi_{ij}\\& \text{where } i,k=1,\cdots,N_1\end{align*}$$

In our case, we let $i,k \in \left\{u, d\right\}$ and the components of the probability density associated with the first electron is $$\rho_{ik} = \sum_{j \in \left\{u, d\right\}} \psi_{kj}^*\psi_{ij} = \psi_{ku}^*\psi_{iu} + \psi_{kd}^*\psi_{id}$$ and so $$\begin{align*}\rho_{uu} &= \psi_{uu}^*\psi_{uu} + \psi_{ud}^*\psi_{ud} = 0 \cdot 0 + \tfrac{1}{\sqrt{2}} \cdot \tfrac{1}{\sqrt{2}} = \tfrac{1}{2}\\\rho_{ud} &= \psi_{du}^*\psi_{uu} + \psi_{dd}^*\psi_{ud} = -\tfrac{1}{\sqrt{2}} \cdot 0 + 0 \cdot \tfrac{1}{\sqrt{2}} = 0\\\rho_{du} &= \psi_{uu}^*\psi_{du} + \psi_{ud}^*\psi_{dd} = 0 \cdot \left(-\tfrac{1}{\sqrt{2}}\right) + \tfrac{1}{\sqrt{2}} \cdot 0 = 0\\\rho_{dd} &= \psi_{du}^*\psi_{du} + \psi_{dd}^*\psi_{dd} = \left(-\tfrac{1}{\sqrt{2}}\right)\left(-\tfrac{1}{\sqrt{2}}\right) + 0 \cdot 0 = \tfrac{1}{2}\end{align*}$$

Thus the (first) probability density is already a diagonal matrix $$\rho = \begin{bmatrix}\tfrac{1}{2} & 0\\0 & \tfrac{1}{2}\end{bmatrix} = \tfrac{1}{2}\mathbf{I}$$

As required, the trace of the probability density is *one* and the entropy associated with the first electron is the maximum it could be $$S = - \tfrac{1}{2} \log \tfrac{1}{2} - \tfrac{1}{2} \log \tfrac{1}{2} = \log 2$$ hence we say the subsystem is **maximally entangled**.

So, even though the composite pair of electrons are in a pure state, the first electron, at least, is maximally *mixed*.

Nb. The same is true for the second electron, simply by symmetry.

We can illustrate this entanglement (entropy) if we consider the average of the spin operator (of the first electron) in the general $\hat{n}$ direction on the singlet state $\rho$, $$\begin{align*}\left\langle \sigma_n \right\rangle &= \mathrm{Tr}\left(\rho \sigma_n\right)\\&= \mathrm{Tr}\left(\tfrac{1}{2}\mathbf{I}\begin{bmatrix}n_z & n_- \\ n_+ & n_z \end{bmatrix}\right)\\&= \mathrm{Tr}\left(\tfrac{1}{2}\begin{bmatrix}n_z & n_- \\ n_+ & n_z \end{bmatrix}\right)\\&= \tfrac{1}{2} \left(n_z - n_z\right)\\&= 0\end{align*}$$

So, as discovered before, the average spin of the singlet state - in any direction - is *zero*, meaning that whichever direction you measure the spin of the singlet state, it will equally likely to be *up* or *down* in that direction.

### States with equal coefficients

Suppose now that a composite system is known to be in the state with equal coefficients, $$\ket{\Psi} = \frac{\ket{uu} + \ket{ud} + \ket{du} + \ket{dd}}{2}$$

So, the coefficients are $$\begin{matrix}\psi_{uu} = \tfrac{1}{2} & \psi_{ud} = \tfrac{1}{2}\\\psi_{du} = \tfrac{1}{2} & \psi_{dd} = \tfrac{1}{2}\end{matrix}$$ and $$\braket{\Psi}{\Psi} = 4\cdot\tfrac{1}{2}\cdot\tfrac{1}{2} = 1$$ as required.

The probability density of the first electron is easily calculated in the standard basis, with each entry equal to $$\tfrac{1}{2}\cdot\tfrac{1}{2} + \tfrac{1}{2}\cdot\tfrac{1}{2} = \tfrac{1}{2}$$

So the probability density, $$\rho = \begin{bmatrix}\tfrac{1}{2} & \tfrac{1}{2} \\ \tfrac{1}{2} & \tfrac{1}{2} \end{bmatrix}$$ is not a diagonal matrix in the standard basis. We could go throught the normal method of calculating the eigenvalues, which we can denote $\rho_1, \rho_2$ and the associated eigenvectors but we can also use the two results from the last lecture, $$\mathrm{Tr}\left(M\right) = \sum_{i=1}^{N} \lambda_i$$ and $$\mathrm{Det}\left(M\right) = \prod_{i=1}^{N} \lambda_i$$ where $\lambda_i$ are the eigenvalues of $M$.

In our case, since both the trace and determinant are invariant we can calculate them from the density in the standard basis, $$\begin{align*}\mathrm{Tr}\left(\rho\right) &= \tfrac{1}{2} + \tfrac{1}{2} = 1\\\mathrm{Det}\left(\rho\right) &= \tfrac{1}{2}\tfrac{1}{2} - \tfrac{1}{2}\tfrac{1}{2} = 0\end{align*}$$

Then, in the basis in which $\rho$ is diagonal, the eigenvalues $\rho_1, \rho_2$ must satisfy $$\begin{align*}\rho_1 + \rho_2 &= \mathrm{Tr}\left(\rho\right) = 1\\\rho_1 \cdot \rho_2 &= \mathrm{Det}\left(\rho\right) = 0\end{align*}$$

Thus, there are two equivalent cases, $$\rho_1 = 1 \text{ and } \rho_2 = 0$$ or $$\rho_1 = 0 \text{ and } \rho_2 = 1$$

The physics is the same for both cases (easy to show). In particular the entropy is *zero*, since we have complete knowledge about the first electron - that is, the first electron is in a *pure* state.

We can find out what this state is by determining the eigenspace of $\rho$. Write $\ket{\rho\!=\!1}, \ket{\rho\!=\!0}$ be the eigenvectors corresponding to $\rho_1 = 1, \rho_2 = 0$. Then, $$\begin{align*}\rho \ket{\rho = 1} = 1 \cdot \ket{\rho = 1} &\Rightarrow \ket{\rho = 1} = \begin{bmatrix} \tfrac{1}{\sqrt{2}} \\ +\tfrac{1}{\sqrt{2}} \end{bmatrix} = \ket{\sigma_x = +1}\\\rho \ket{\rho = 0} = 0 \cdot \ket{\rho = 0} &\Rightarrow \ket{\rho = 0} = \begin{bmatrix} \tfrac{1}{\sqrt{2}} \\ -\tfrac{1}{\sqrt{2}} \end{bmatrix} = \ket{\sigma_x = -1}\end{align*}$$ where we have indicated that these are the same eigenvectors as that of the spin operator, $\sigma_x$, corresponding to the $\hat{x}$ direction.

We can show this in a more general way, by calculating the average of the arbitrary spin operator, $\sigma_n$, in the state $\rho$, $$\begin{align*}\left\langle \sigma_n \right\rangle &= \mathrm{Tr}\left(\rho \sigma_n\right)\\&= \mathrm{Tr}\left(\begin{bmatrix}\tfrac{1}{2} & \tfrac{1}{2} \\ \tfrac{1}{2} & \tfrac{1}{2} \end{bmatrix}\begin{bmatrix}n_z & n_- \\ n_+ & -n_z \end{bmatrix}\right)\\&= \mathrm{Tr}\begin{bmatrix}\tfrac{n_z + n_+}{2} & \tfrac{n_- - n_z}{2} \\ \tfrac{n_z + n_+}{2} & \tfrac{n_- - n_z}{2} \end{bmatrix}\\&= \frac{n_z + n_+}{2} + \frac{n_- - n_z}{2}\\&= \frac{n_+ + n_-}{2}\\&= n_x\end{align*}$$

For example, $$\begin{align*}\hat{n} = \hat{x} &\Rightarrow \left\langle \sigma_n \right\rangle = 1\\\hat{n} \perp \hat{x} &\Rightarrow \left\langle \sigma_n \right\rangle = 0\end{align*}$$

That is, in the $\hat{x}$ direction, the spin will definitely be *up*, and in the plane perpendicular to $\hat{x}$, the spin will measure *up* or *down* with equal likelihood.

Thus, the state with equal coefficients is a pure state in the eigenspace of the spin operator in the $\hat{x}$ direction.

### Entanglement entropy

We could adjust the coefficients in the above example so that the density *doesn't* correspond to a pure state. As a rule, though, small changes of this kind will result in a correspondingly small amount of entropy, or, more precisely, a small amount of **entanglement entropy**.

In general, even if the composite system is in a pure state, with *zero* entropy, the subsystems are entangled and do have entropy.

As we have noted before, there is no analogue of this behaviour in classical sysytems.

Also, entropy isn't additive - that is, the entropy of the composite system is not, in general, the sum of the entropies of the subsystems, even though, in some special cases, it is.