2
$\begingroup$

So in the single-qubit case, we can write any unitary operation as an instance of the following parametrized unitary:

$$U(\theta, \phi, \lambda) = \begin{bmatrix} \cos(\theta) & -e^{i\lambda}\sin(\theta) \\ e^{i\phi}\sin(\theta) & e^{i(\lambda+\phi)}\cos(\theta) \\ \end{bmatrix}$$

What's the extension of this idea into 2-qubit operations? Can we write any 2-qubit unitary operation as an instance of some parametrized unitary?

$\endgroup$

2 Answers 2

2
$\begingroup$

There is some parametrized matrix form for a 2-qubit unitary, but it would be extremely inconvenient to work with. An $n$-qubit gate is an element of the group $SU(2^n)$, which has dimension $2^{2n} - 1$ (in other words, the number of required parameters). This is manageable for 1-qubit unitaries because there are only 3 parameters to manage. Going to 2 qubits, this number balloons to 15; no matter how the unitary is parametrized, this is going to be quite unwieldy. This is probably why you won't find such a gate actually implemented.

$\endgroup$
4
  • $\begingroup$ Is there a formula for how many of those parameters control norms of the elements in the matrix and how many control the phases? So in the case of $U(2)$ only $\theta$ defines the norms of the elements and the others define the phases? $\endgroup$
    – Dani007
    Jun 25 at 21:41
  • $\begingroup$ I'm honestly not sure that there even is a fixed formula for all parametrizations $\endgroup$
    – Cody Wang
    Jun 25 at 22:05
  • $\begingroup$ @CodyWang Why do you say that an $n$ qubit gate is an element of $SU(2^n)$? For me it is simply an element of $U(2^n)$. Why do you add the extra condition that the determinant of the matrix is exactly equal to $1$? $\endgroup$ Jun 26 at 8:32
  • $\begingroup$ That's true; I guess I'm really talking about equivalence classes of gates that differ only by a global phase. $\endgroup$
    – Cody Wang
    Jun 28 at 4:13
1
$\begingroup$

What I propose here could possibly be simplified further, but that's at least a first direction.

First, any unitary can be written as $U=e^{-it/\hbar \widetilde{H}}$ for some $t$ and $\widetilde{H}$ (the Hamiltonian) hermitian which we can rewrite $U=e^{-i H}$ where $H$ is also Hermitian.

Then, the $n$-Pauli matrices form a basis for any operator acting on $n$ qubits.

For this reason, $H=\sum_{i_1...i_n} \alpha_{i_1...i_n} \sigma_{i_1} \otimes ... \otimes \sigma_{i_n}$ with $\alpha_{i_1...i_n}$ a real coefficient, and $\sigma_k$ being equal to the identity matrix in two dimensions for $k=0$, or $X$ for $k=1$, $Y$ for $k=2$, $Z$ for $k=3$.

In general, an $n$-qubit unitary matrix can then be written as:

$$U=e^{-i \sum_{i_1...i_n} \alpha_{i_1...i_n} \sigma_{i_1} \otimes ... \otimes \sigma_{i_n}}$$

Your question is a particular case for $n=2$.

Now it might not be exactly the question you ask as you would probably like to have it in an non exponentiated form. Also, I guess that some symmetries could be exploited to simplify my answer (for instance a global phase doesn't matter).

But maybe that's still somewhat usefull for you, I don't know.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.