When people generally speak of quantum information in the context of continuous variables, what is generally meant is that observables, like position/momentum or the field quadratures of quantum optics, can take a continuous range of values and that information processing tasks are performed using operators on these variables.

However, most work in this area restricts attention to a countable number of modes of the radiation field (in quantum optics). In this case, the state of the system can be described using a countably infinite number of variables, like the number eigenstates in the field. But this is still called 'continuous variable' quantum information because the relevant observables still take continuous values. Is that so?

If so, is the boson sampling problem in the domain of continuous variables? I don't know what exactly the relevant variables are in this case. A preliminary reading seems to suggest to me that the observable is the photon number, since what is observed is the probability after measuring in some basis. This would now mean that in the sense I think of continuous variables, boson sampling is not a problem in this domain.

Am I right/wrong? Which of my statements here are false?


1 Answer 1


I'd say, your statements are all more or less correct. To be sure, let me just restate essentially what you said in my own words:

Well, I'd say that the theoretical underpinning of Boson Sampling is not a continuous variable application - however it arises from a continuous variable setting. That said, what do we actually mean by CV quantum information?

As you say, the term "continuous variable" refers to the fact that there is something nondiscrete in the setting of the problem. Most prominently, in quantum optics, the canonical commutation relations cannot be represented by bounded (let alone compact) operators on a Hilbert space, which nearly immediately results in unbounded continuous spectra. In doing quantum optics, we have to work with these continuities - for example, the Gaussian states have a Gaussian (hence continuous) distribution in some representation of phase space. Of course, we will very often work (at least as far as I got, I'm still learning) with the discrete covariance matrix of the Gaussian state, etc. but underlying our efforts is the continuous Wigner function and we're forced to come back to it once in a while. Mathematically, this implies we suddenly also have to care about convergences, etc.

However, if we now investigate Boson sampling, which was posed in the context of quantum optics, the original form is devoid of continuous variables. We only have to work with the discrete set of modes/particle numbers. The quantities we want to consider are finite dimensional matrices (and finite dimensional representations thereof), the result hopes to calculate the expectation value of a finite dimensional quantitiy which happens to be connected to the permanent - the boson equivalent of the determinant, etc.

In short, the computer science/theoretical physics perspective on Boson sampling seems to me inherently discrete - at no point does it seem necessary to revert to the underlying wave-packets or the like. Hence, from a purely theoretical point of view, I'd consider Boson sampling as discrete as most models for doing computations.

However, if you want to implement this in your lab (as an experimental physicist), I would assume you'd also have to deal with the continuous variable background of the problem - namely quantum optics - so from this point of view, it might actually look continuous. In this sense, the lines could be blurred...

  • $\begingroup$ Thanks for the answer! I'll hold from accepting it for now because I would like to see some more discussion on the precise meaning of continuous variables in CV quantum information. $\endgroup$
    – Abhinav
    Jan 22, 2014 at 18:23
  • $\begingroup$ Since I read it this morning, I though you might also want to have a look at arxiv-web3.library.cornell.edu/abs/1401.4679, where a definition along these lines is given (pg. 3) $\endgroup$
    – Martin
    Jan 23, 2014 at 10:28

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