## 5.9 *Remarks and exercises*

### 5.9.1 Entangled or not?

Let a joint state of

Show that, if

|\psi\rangle is a product state, then\det (c_{ij}) = 0 .Show that the converse (

\det(c_{ij})=0 implies that|\psi\rangle is a product state) holds only for qubits. Explain why.Deduce that the state

\frac12\big(|00\rangle + |01\rangle + |10\rangle + (-1)^k|11\rangle\big) is entangled fork=1 and unentangled fork=0 . Express the latter case explicitly as a product state.

There is a lot of interesting physics behind the previous innocuous-looking mathematical statement.
For example, think again about the state ^{82}
But can you actually use this effect to send a message faster than light?
What would happen if you tried?

Hopefully you can see that it would not work, since the result of the measurement is random — you cannot choose the bit value you want to send. We shall return to this, and other related phenomena, later on.

### 5.9.2 SWAP circuit

Show that, for any states

(Swapping).

### 5.9.3 Controlled-NOT circuit

Show that the circuit below implements the controlled-

(Controlled-

### 5.9.4 Measuring with controlled-NOT

The controlled-

Take a look at the circuit below, where

(?).

Now assume that the top two qubits are in the state

### 5.9.5 Arbitrary controlled-U on two qubits

Any unitary operation

### 5.9.6 Entangled qubits

Two entangled qubits in the state

What is the probability that Alice and Bob will register identical results? Can any correlations they observe be used for instantaneous communication?

Prior to the measurements in the computational basis, Alice and Bob apply unitary operations

R_\alpha andR_\beta (respectively) to their respective qubits:The gate

R_\theta is defined by its action on the basis states:\begin{aligned} |0\rangle &\longmapsto \cos\theta|0\rangle + \sin\theta|1\rangle \\|1\rangle &\longmapsto -\sin\theta|0\rangle + \cos\theta|1\rangle. \end{aligned} Show that the state of the two qubits prior to the measurements is\begin{aligned} &\frac{1}{\sqrt2}\cos(\alpha-\beta)\big( |00\rangle + |11\rangle \big) \\- &\frac{1}{\sqrt2}\sin(\alpha-\beta)\big( |01\rangle - |10\rangle \big). \end{aligned} What is the probability that Alice and Bob’s outcomes are identical?

### 5.9.8 Playing with conditional unitaries

The swap gate

Show that the four Bell states

\frac{1}{\sqrt2}(|00\rangle\pm|11\rangle) and\frac{1}{\sqrt2}(|01\rangle\pm|10\rangle) are eigenvectors ofS that form an orthonormal basis in the Hilbert space associated to two qubits. Which Bell states span the*symmetric subspace*(i.e. the space spanned by all eigenvectors with eigenvalue1 ), and which the*antisymmetric*one (i.e. that spanned by eigenvectors with eigenvalue-1 )? CanS have any eigenvalues apart from\pm1 ?Show that

P_\pm = \frac12(\mathbf{1}\pm S) are two orthogonal projectors which form the decomposition of the identity and project on the symmetric and antisymmetric subspaces. Decompose the state vector|a\rangle|b\rangle of two qubits into symmetric and antisymmetric components.Consider the quantum circuit below, composed of two Hadamard gates, one controlled-

S operation (also known as the**controlled-swap**, or**Fredkin**gate), and the measurementM in the computational basis. Suppose that the state vectors|a\rangle and|b\rangle are normalised but*not*orthogonal to one another. Step through the execution of this network, writing down the quantum states of the three qubits after each computational step. What are the probabilities of observing0 or1 when the measurementM is performed?

(Symmetric and antisymmetric projection).

Explain why this quantum network implements projections on the symmetric and antisymmetric subspaces of the two qubits.

Two qubits are transmitted through a quantum channel which applies the same, randomly chosen, unitary operation

U to each of them. Show that the symmetric and antisymmetric subspaces are invariant underU\otimes U .Polarised photons are transmitted through an optical fibre. Due to the variation of the refractive index along the fibre, the polarisation of each photon is rotated by the same unknown angle. This makes communication based on polarisation encoding unreliable. However, if you can prepare any polarisation state of the two photons then you can still use the channel to communicate without any errors. How can this be achieved?

“Spooky action at a distance” is a loose translation of the German “spukhafte Fernwirkung”, which is the phrase that Albert Einstein used in his 1947 letter to Max Born.↩︎