## 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|\psi\rangle=|a\rangle|b\rangle ) holds only for qubits. Explain why.Deduce that the state

\frac{1}{2}\big(|00\rangle + |01\rangle + |10\rangle + (-1)^k|11\rangle\big) is entangled for odd values ofk and unentangled for even values ofk . 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

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 — it is not at all a lost cause!

### 5.9.2 SWAP circuit

Show that, for any states

(Swapping).

### 5.9.3 Controlled-NOT circuit

Show that the circuit below gives another implementation of 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

What is this circuit actually measuring?

### 5.9.5 Arbitrary controlled-U on two qubits

Recall Section 3.5: any unitary operation

Suppose that you can implement any single-qubit gate, and that you have a bunch of controlled-

### 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), for some fixed values\alpha,\beta\in\mathbb{R} , to their respective qubits:where 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}{\sqrt{2}}\cos(\alpha-\beta)\big( |00\rangle + |11\rangle \big) \\- &\frac{1}{\sqrt{2}}\sin(\alpha-\beta)\big( |01\rangle - |10\rangle \big). \end{aligned} What is the probability that Alice and Bob’s outcomes are identical?

What is the geometric interpretation of the operator

R_\theta ?

### 5.9.8 Playing with conditional unitaries

The swap gate

Show that the four Bell states

\frac{1}{\sqrt{2}}(|00\rangle\pm|11\rangle) and\frac{1}{\sqrt{2}}(|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 = \frac{1}{2}(\mathbf{1}\pm S) are two orthogonal projectors which form the decomposition of the identity and project onto 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 of the four computational steps. What are the probabilities of observing0 or1 when the measurementM is finally 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, i.e.U\otimes U . Show that the symmetric and antisymmetric subspaces are invariant under this operation.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 are able to prepare

*any*polarisation state of the two photons then you can still use the channel to communicate without any errors — how?

### 5.9.9 Tensor products in components

In our discussion of tensor products we have so far taken a rather abstract approach.
There are, however, situations in which we have to put numbers in, and write tensor products of vectors and matrices explicitly.
For example, here is the standard basis of two qubits written explicitly as column vectors:^{102}

The matrix elements of the tensor product operation

are given by

This product **Kronecker product** of matrices, which generalises the **outer product** of two vectors that we saw in 0.8.

The tensor product induces a natural partition of matrices into blocks. Multiplication of block matrices works pretty much the same as regular matrix multiplication (assuming the dimensions of the sub-matrices are appropriate), except that the entries are now matrices rather than numbers, and so may not commute.

Evaluate the following matrix product of

(4\times 4) block matrices:\left[ \, \begin{array}{c|c} \mathbf{1}& X \\\hline Y & Z \end{array} \, \right] \left[ \, \begin{array}{c|c} \mathbf{1}& Y \\\hline X & Z \end{array} \, \right] (whereX ,Y , andZ are the Pauli matrices).Using the block matrix form of

A\otimes B expressed in terms ofA_{ij} andB_{ij} (as described above), explain how the following operations are performed on the block matrix:- transposition
(A\otimes B)^T ; **partial transpositions**A^T\otimes B andA\otimes B^T ;- trace
\operatorname{tr}(A\otimes B) ; **partial traces**(\operatorname{tr}A)\otimes B andA\otimes(\operatorname{tr}B) .

- transposition

### 5.9.10 Hadamard transforms in components

Consider the Hadamard transform

- The output state is a superposition of all binary strings:
\sum_{x\in\{0,1\}^3} c_x|x\rangle . Where in theH^{\otimes 3} matrix will you find the coefficientsc_x ?

Do you want to write down

Consider again the single-qubit Hadamard gate matrix

Using the fact that

(A\otimes B)_{ik,jl} = A_{ij}B_{kl} , or any other method, analyse the pattern of the\pm1 in the tensor product of Hadamard matrices. What is the entryH^{\otimes 4}_{0101,1110} ?For any two binary strings

a=(a_1,\ldots, a_n) andb =(b_1,\ldots , b_n) of the same length we can define their “scalar” product asa\cdot b = (a_1b_1\oplus \ldots \oplus a_n b_n) . Show that, up to the constant(1/\sqrt{2})^n , the entryH^{\otimes n}_{a,b} is(-1)^{a\cdot b} for anyn and for any binary stringsa andb of lengthn .Show that

H^{\otimes n} acts as|a\rangle \longmapsto \left(\frac{1}{\sqrt{2}}\right)^n \sum_{b\in\{0,1\}^n} (-1)^{a\cdot b}|b\rangle A quantum register of

10 qubits holds the binary string0110101001 . The Hadamard Transform is then applied to this register yielding a superposition of all binary strings of length10 . What is the sign in front of the|0101010101\rangle term?

### 5.9.11 The Schmidt decomposition

An arbitrary vector in the Hilbert space *non-negative* numbers.
This is known as the **Schmidt decomposition** of

Any bipartite state can be expressed in this form, but remember that *the bases used depend on the state being expanded*.
Indeed, given two bipartite states *cannot* perform the Schmidt decomposition using the *same* orthonormal bases in

The Schmidt decomposition follows from the **singular value decomposition** (often abbreviated to **SVD**): *any* **singular values** of

You can visualize the SVD by thinking of

Using the index notation **rank** of **Schmidt number**.
You should be able to see that all bipartite states of rank-one are separable.

The Schmidt decomposition is *almost* unique.
The ambiguity arises when we have two or more identical singular values, as, for example, in the case of the Bell states.
Then any unitary transformation of the basis vectors corresponding to a degenerate singular value, both in

We always use the lexicographic order

00<01<10<11 .↩︎