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?


2 Answers 2


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.

  • $\begingroup$ 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
    Commented Jun 25, 2022 at 21:41
  • $\begingroup$ I'm honestly not sure that there even is a fixed formula for all parametrizations $\endgroup$
    – Cody Wang
    Commented Jun 25, 2022 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 $1$? $\endgroup$ Commented Jun 26, 2022 at 8:32
  • 1
    $\begingroup$ That's true; I guess I'm really talking about equivalence classes of gates that differ only by a global phase. $\endgroup$
    – Cody Wang
    Commented Jun 28, 2022 at 4:13

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.


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