Kerr gate matrix

Photonic Software Strawberry Fields
1

Hey I was wondering if I’ve understood the matrix implementation for the Kerr gate correctly (having trawled through the repo code). As far as I can see it’s a diagonal matrix with exp(i x kappa x n**2) for n in the range of 0 to cutoff_dim. So something like:

1 0 0 0…
0 exp(i x kappa) 0 0 …
0 0 exp(4i x kappa) 0 0 …
…

Thanks in advance!

2

Hi @ievutec. Yes, that’s exactly right!

1 Like
3

Hi @josh For the diagonal matrix representation of the Squeezing gate image , how would I find the value of a hat? Thank you.

4

Hi @sophchoe, are you looking for the Fock representation of the squeezing operation?

Unfortunately, the Fock representation is not as simple as the Kerr gate, since the squeezing operation is not diagonal in the Fock basis. In the Fock basis, the creation and annihilation operators are given by:

\hat{a}^\dagger = \begin{bmatrix} 0 & 0 & 0 & 0 & \cdots & 0 & \cdots \\ \sqrt{1} & 0 & 0 & 0 & \cdots & 0 & \cdots \\ 0 & \sqrt{2} & 0 & 0 & \cdots & 0 & \cdots \\ 0 & 0 & \sqrt{3} & 0 & \cdots & 0 & \cdots \\ \vdots & \vdots & \vdots & \ddots & \ddots & \cdots & \cdots \\ 0 & 0 & 0 & \cdots & \sqrt{n} & 0 & \cdots & \\ \vdots & \vdots & \vdots & \vdots & \vdots & \ddots & \ddots \end{bmatrix}

and

a =\begin{bmatrix} 0 & \sqrt{1} & 0 & 0 & \cdots & 0 & \cdots \\ 0 & 0 & \sqrt{2} & 0 & \cdots & 0 & \cdots \\ 0 & 0 & 0 & \sqrt{3} & \cdots & 0 & \cdots \\ 0 & 0 & 0 & 0 & \ddots & \vdots & \cdots \\ \vdots & \vdots & \vdots & \vdots & \ddots & \sqrt{n} & \cdots \\ 0 & 0 & 0 & 0 & \cdots & 0 & \ddots \\ \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \ddots \end{bmatrix}

We can create this in NumPy, up to a certain Fock cutoff or truncation, using np.diag:

>>> cutoff = 8
>>> a = np.diag(np.sqrt(np.arange(1, cutoff)), k=1)
>>> adag = np.diag(np.sqrt(np.arange(1, cutoff)), k=-1)

We can then use scipy.linalg.expm to compute the matrix exponential:

>>> z = 0.76
>>> sp.linalg.expm(0.5 * (np.conj(z) * a @ a - z * adag @ adag))

However, this is quite inefficient, and the matrix exponential can introduce numerical errors for large Fock basis states.

Instead, we can use The Walrus to compute the Fock representation of the squeezing operation:

>>> from thewalrus.fock_gradients import squeezing
>>> squeezing(r=z, theta=0, cutoff=cutoff)
5

Hi @josh, thank you so much for your reply. It gave me a much better understanding. I can see how the Wigner projection onto the phase space would make the standard CV gates alter their forms substantially.

I made an example calculation with cutoff dim = 4 and with the matrix exponential. I see how the same would apply to the other gates.

Thank you so much!!!

1 Like
6

no worries @sophchoe!