1
$\begingroup$

I am familiar with the quantum gates of qiskit, but I become very confused when looking at the continous variable (CV) operations in the pennylane documentation. I am especially interested in the Karr, Squeeze and Displacement operators as I have seen them used in Quantum Neural networks. See: https://pennylane.readthedocs.io/en/stable/introduction/operations.html

I dont understand the mathematical representations of these operators. How would one implement them on for example qiskit?

$\endgroup$
2
$\begingroup$

Short, sort-of right answer: you can't

This is in essence due to the superconducting qubits that e.g. IBM use being, well, qubits, while continuous variable (CV) operations don't act on qubits. Well, sort of.

These are two fundamentally different ways of going about making a quantum computer, so let's start from first principles:

When you take a state $\left|\psi\right\rangle$, such as the state of a qubit, you're using first quantisation, where the wave function is used as the state of the system to ask 'what is the state of the particle?' (in quantum computing, usually in terms of the basis states $\left|0\right>$ and $\left|1\right>$). However, CV operaations instead use second quantisation. That is, as an example, we have a physical waveguide/optical cable that light can travel down, known as a 'mode' and we ask 'how many particles are there in each state and mode?'.

To do this, we use the idea of creation and annihilation operators where each creation operator $a^\dagger$ creates a particle and each annihilation operator $a$ destroys a particle. We can further define $n = a^\dagger a$ as the number of particles. These are the operators that are used in defining the CV operations you linked to.

Worth noting is that one can be related to the other using $\left|1\right> = a^\dagger\left|0\right>$ and therefore $\left|0\right> = a\left|1\right>$. This then gives a more meaningful (at some level) understanding of those basis states - $\left|0\right>$ is the state with no excitations and $\left|1\right>$ is the state with one excitation. This can be further extended to refer to $\left|n\right>$ (defined by $a\left|n\right> = \sqrt{n}\left|n-1\right>$ and $a\left|0\right> = 0$) as the state with n excitations, where the states with 0 and 1 excitation are used as the basis for your qubit.

Now, just because you're only accessing those bottom two excitation levels ('ground' and 'first excited' states) doesn't mean that the other levels don't exist, so there is no fundamental reason those operation you linked to can't be performed on superconducting qubits (after including higher energy levels). Indeed, this appears to be possible.

This gives the full answer:

This is something that's not completely unfeasible but either the specific way that IBM implements its operations on its qubits, what they haven't tried or just don't allow the users to do mean that it's not something we can do on IBM's quperconducting qubit devices.

$\endgroup$

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.