Quantum Computation and Quantum Information - Exercise 4.16

Question

Exercise Description

In the book "Quantum Computation and Quantum Information - 10th Anniversary Edition", there is the following exercise ($H$ being the Hadamard matrix and $x_1$,$x_2$ qubits):

Questions

I think I managed to solve the exercise for the first case (the second one is analogous), but it would be great if someone could check to see if it is indeed correct. (I could not find a solution online.) Concretely:

• Is my solution correct (also with respect to notation)?
• Is there a better / more straightforward approach?

Proposed Solution

The computational basis vectors are $$ |00\rangle = (1,0,0,0)^T\\ |01\rangle = (0,1,0,0)^T\\ |10\rangle = (0,0,1,0)^T\\ |11\rangle = (0,0,0,1)^T $$ and we know that $$ H |0\rangle = \frac{|0\rangle + |1\rangle}{\sqrt{2}}\\ H |1\rangle = \frac{|0\rangle - |1\rangle}{\sqrt{2}} $$ The idea is to build a (unimodular) $4 \times 4$ matrix $A$ that works as follows: $$ A|00\rangle = |0\rangle\frac{|0\rangle + |1\rangle}{\sqrt{2}} = \frac{|00\rangle}{\sqrt{2}} + \frac{|01\rangle}{\sqrt{2}}\\ A|01\rangle = |0\rangle\frac{|0\rangle - |1\rangle}{\sqrt{2}} = \frac{|00\rangle}{\sqrt{2}} - \frac{|01\rangle}{\sqrt{2}}\\ A|10\rangle = |1\rangle\frac{|0\rangle + |1\rangle}{\sqrt{2}} = \frac{|10\rangle}{\sqrt{2}} + \frac{|11\rangle}{\sqrt{2}}\\ A|11\rangle = |1\rangle\frac{|0\rangle - |1\rangle}{\sqrt{2}} = \frac{|10\rangle}{\sqrt{2}} - \frac{|11\rangle}{\sqrt{2}} $$ We now determine the columns of $A$ using each basis vector and expected result. For $|00\rangle$, this looks as follows: $$ \begin{bmatrix} * & * & * & *\\ * & * & * & *\\ * & * & * & *\\ * & * & * & * \end{bmatrix} \begin{bmatrix} 1\\0\\0\\0 \end{bmatrix} = \begin{bmatrix} 1/\sqrt{2}\\1/\sqrt{2}\\0\\0 \end{bmatrix}, $$ which gives the first column of $A$: $$ \begin{bmatrix} 1/\sqrt{2} & * & * & *\\ 1/\sqrt{2} & * & * & *\\ 0 & * & * & *\\ 0 & * & * & * \end{bmatrix} $$ We can then proceed analogously with the remaining basis vectors to find all columns of $A$. Eventually, we end up with $$ A = \frac{1}{\sqrt{2}} \begin{bmatrix} 1 & 1 & 0 & 0\\ 1 & -1 & 0 & 0\\ 0 & 0 & 1 & 1\\ 0 & 0 & 1 & -1 \end{bmatrix} $$ It's easy to verify that $\det A = 1$ (i.e. $A$ is unimodular) and that multiplying $A$ with any of the basis vectors yields the desired result.

Answer 1

• Yes, your answer is correct.
• A more straightforward approach would take into account the tensor product structure of the two-qubit wavefunction. In fact, the Hilbert space $\mathcal{H}$ for joint states of your two qubits, spanned by the four computational basis vectors you listed above, is nothing else than the tensor product of the two-dimensional Hilbert spaces of each qubit $\mathcal{H}_{1}$ and $\mathcal{H}_{2}$, i.e. $\mathcal{H}=\mathcal{H}_{1}\otimes\mathcal{H}_{2}$. This mathematical relation can be exploited to describe unitaries acting on $\mathcal{H}_{1}$ and $\mathcal{H}_{2}$ (in your example the unitaries $H$ and $I$, respectively) by means of such a tensor product: the unitary operation acting on $\mathcal{H}$ is $U=H\otimes I$. You know the matrix representation of both $H$ and $I$ in the computational basis, tensoring them yields the $A$ you constructed step by step.