I would like to understand what is a continuous quantum register. I know the direct definition is a quantum register that stores a real number defined by an observable with a spectrum consisting of $\mathbb{R}$ but that seems really abstract to me.

Also how it relates to qubits? Is a set of qubits used for simulating discretely a continuous quantum register? If yes how?


1 Answer 1


direct definition is a quantum register that stores a real number defined by an observable with a spectrum consisting of ℝ

Yes, qubits can be used to discretize a continuous quantum system.


Lets say $\Phi $ is an operator on the continuous system which we want to simulate using qubits. We need a discrete register ($D$) to store the qubit values. Now the discretization is done by measuring the shifts between the eigenstates of $\Phi$ denoted by operator say $M$. This is done by DFT. $$ M = F^*\Phi F $$

Where $F$ is the DFT operator.

Now all this is considering that you already have a register which is storing continuous quantum output of a continuous system. "What is that register?" is I believe to be your original question. For that, you can take an example of cavity QED.

A typical quantum system with a continuous degree of freedom is the quantum field and normally, quantum simulations of quantum field theories rely on discretisation of this field.

Whereas a cavity QED output is a continuous quantum field and this acts as a continuous quantum register. This thesis has even proposed an algorithm to be directly implemented on this continuous quantum field and only discretizing after the implementation, thus utilizing the underlying properties of continuous system.

  • $\begingroup$ Would it be possible to have a numerical example ? I am still lost about it. Like how is the real number encoded in this continuous register and what is the shift you mention? $\endgroup$
    – cnada
    Aug 22, 2018 at 17:52
  • $\begingroup$ So $\Phi$ is an operator on an infinite dimensional $\mathcal{H}$ and $F$ is the DFT on a $d^n$ dimensional Hilbert space. Then your composition doesn't make sense. Say your types. $\endgroup$
    – AHusain
    Aug 22, 2018 at 18:45
  • $\begingroup$ @cnada and AHusain, I dont remember the complete mathematical proof of the descretization. I recollected that from my memory of reading a recent paper by a team at IQC. I will link that paper when I get home. $\endgroup$
    – artha
    Aug 23, 2018 at 4:49

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