3
$\begingroup$

I am trying to find a way to implement a unitary transformation in Q# that implements e^(iA) where A is a square matrix. However, I only found ways to do this in Q# if A can be represented as a tensors of Paulis (the Exp operation: https://docs.microsoft.com/en-us/qsharp/api/qsharp/microsoft.quantum.primitive.exp or using a PauliEvolutionSet when trying to do Hamiltonian simulation: https://docs.microsoft.com/en-us/qsharp/api/qsharp/microsoft.quantum.simulation.paulievolutionset).

Does anyone know how to do this for a generic matrix A? If for example you want to use a 8x8 matrix, you can't express it in terms of tensors of Paulis...

Thanks in advance for the help!

$\endgroup$

1 Answer 1

3
$\begingroup$

One thing to point out is that because the Paulis form a basis, you can actually represent any $2^n \times 2^n$ matrix in terms of a sum of tensors of Paulis, i.e., members of the $n$-qubit Pauli group. That is, you can write any $8 \times 8$ matrix $A$ as a sum of the form $$A =\sum_{i,j,k}h_{ijk}\ \sigma_i\otimes\sigma_j\otimes\sigma_k$$ where $h_{ijk}$ are the coefficients in the Pauli basis. The answer to this related question describes how to solve for these coefficients.

Now, once you have expressed $A$ in the Pauli basis, you can use the Q# operations you mention to implement the evolution. Some more background on implementing this in Q# is available at this link. The general idea is that once you have expressed the matrix $A$ in the Pauli basis, you can now use something like a Trotter–Suzuki expansion to approximately express the exponential $e^{iA}$ as a product of exponentials of Paulis, which can then in turn be implemented on a quantum computer (and also via built-in Q# tools such as Exp and PauliEvolutionSet).

Why doesn't a language like Q# include a built-in library for implementing a matrix exponential $e^{iA}$ for some general matrix $A$? Essentially because such an operation is, in general, extremely inefficient to implement on a quantum computer. To understand why, note that for a general $n$-qubit unitary, there are $4^n$ coefficients required to represent it in a basis like the Pauli basis, which means that your resulting circuit depth will be on the order of $4^n$ -- far too deep to be practical for anything beyond very small systems.

The exception is the case where the matrix $A$ has "sparsity" in some representation -- for example, if only a constant number of the $4^n$ coefficients in the Pauli basis are non-zero. In that case, the circuit resulting from a Trotter-Suzuki decomposition would have only constant depth, rather than going as $4^n$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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