I was wondering if there’s any way to implement adjoints for the cubic phase gate or the Kerr gate? I’m quite new to CV so I’m unsure as to whether there is a work-around for it. I would imagine that if the matrix is truncated, then there is a possibility to simply take the adjoint of that? How accurate is something like this?
Thanks!
Ieva
Hi @ievutec!
At the moment, there is no built-in method to take the Hermitian conjugate of a quantum operation. However, in the case of these two gates, it is simply sufficient to negate the parameter.
For example, for the Kerr interaction:
K(\kappa)^\dagger = \left(e^{i\kappa \hat{n}^2}\right)^\dagger = e^{-i\kappa \hat{n}^2} = K(-\kappa)
since the number operator is self-adjoint (\hat{n}^\dagger = \hat{n}). The same result holds true for the cubic phase gate, as \hat{x}^\dagger = \hat{x}:
V(\gamma)^\dagger = \left(e^{i\gamma\hat{x}^3/3\hbar}\right)^\dagger = e^{-i\gamma\hat{x}^3/3\hbar} = V(-\gamma)
With regards to your second question about the accuracy of truncation; if choosing between the two non-linear CV operations (Kerr and cubic phase), in general, it is always better to use the Kerr interaction.
This is simply due to numerical accuracy — since the number operator \hat{n} is diagonal the Fock basis, computing the matrix exponential and therefore the action of the Kerr interaction is relatively easy and accurate computationally.
The position operator x is not diagonal, and so we need to compute a truncated matrix exponential, resulting in the additional numerical error.
Ah brilliant thanks! I’d worked that out for myself but something was going wrong in the computation so at least now I know it wasn’t the conjugate 
Ieva