Physics

New paper: Error models for mode-mismatch in linear optics quantum computing

5 Comments

I’ve just submitted my latest paper with Tim Ralph to Physical Review A, entitled Error models for mode-mismatch in linear optics quantum computing. Previously I’ve done a lot of work into the effects of mode-mismatch in linear optics quantum computing (LOQC) (see previous physics posts for explanations of both these terms). This paper focusses on a basic building block of LOQC, called the parity gate, which is fundamental to many recent LOQC proposals, most notably the cluster- and parity-state approaches. Both these approaches are very promising and have completely changed the nature of the LOQC playing field. It is important to understand how errors are introduced in these models, of which mode-mismatch is one of the most dominant.

5 thoughts on “New paper: Error models for mode-mismatch in linear optics quantum computing”

  1. Hi Peter, your latest paper reads very well and since I am new to the idea of cluster states for computing I found that section very illuminating.

    I did have a question about the parity operation. When initially citing that operator, you mentioned Refs. [9] and [10], which I understand to describe a Bell-state projection operator, using both a PBS and a beam splitter (and detectors). Is the parity operator identical, or is it just the action of the PBS and detectors that embodies the parity measurement? I suspect the latter, but I hoped you could clarify the connection.

    Many thanks,
    Travis.

    Reply

  2. Hi Travis,
    Thanks for you comments. The terminology surrounding Bell measurements and parity measurements are a little confusing, especially in the optical contexxt. Formally, a projector onto one of the Bell states could be expressed
    P_B = (|HH>+|VV>)(< HH|+< VV|)
    whereas the even-parity projector is
    P_P = |HH>< VV|
    Technicially these are different projectors. However, in optics these measurements are destructive. Once you include this fact the operators are in fact the same.
    Alternately, acting P_B and P_P on an arbitrary state will, in general, be different. However, if you trace out the measured modes following application of the projectors, you will be left with the same state in both cases.
    I hope this answers your question.
    Regards,
    Peter.

    Reply

  3. Thanks Peter, your example and the explanation in terms of tracing out the measured modes helps me to understand the equivalence of these two projections.

    I am still thinking about whether there are any consequences to this equivalence in practice, as well as in theory, e.g., in the case that the incident state is not (polarization) entangled, it would seem P_B might yield different results than P_P, such as

    P_B | HH > = | HH > + | VV >

    P_P | HH > = | HH >

    Perhaps this exception is to be expected for separable states. If you get a chance, I would interested to hear your perspective.

    Thanks,
    Travis.

    Reply

  4. Hi Travis. You raise some good points. There are a few subtleties associated with these projectors, that I hope I can clarify. You’re quite right that, in the form I wrote them, these projectors are apparently not equivalent in general. If fact they are…

    Consider the completely general state

    |psi> = alpha |HH>|W> + beta |HV>|X> + gamma |VH>|Y> + delta |VV>|Z>

    Here the first ket in each term corresponds to the modes being acted on, and the second ket is the remainder of the state.

    Applying the Bell projector we get

    P_B |psi> = (|HH> + |VV>)(alpha |W> + delta |Z>)

    Applying the parity projector we get

    P_P |psi> = alpha |HH>|W> + delta|VV>|Z>

    It’s clear looking at these two states that if we now trace out the first two modes the former case will be left in a coherent superposition of |W> and |Z> while the later will be left in the incoherent mixture of |W> and |Z>.

    However, (here’s the sublety) you’ll notice in the paper that the parity gate places Hadamard gates after the PBS. This destroys H/V information so we are actually left with

    P_P |psi> = alpha |DD>|W> + delta|DD>|Z> = |DD> (alpha |W> + delta |Z>)

    where |D> is the diagonally polarized state.

    Now tracing out the detected modes

    tr(P_B |psi>) = alpha |W> + delta |Z>
    tr(P_P |psi>) = alpha |W> + delta |Z>

    And both are equivalent.

    Hope this clarifies things.

    Reply

  5. Thanks again Peter, I had overlooked the part about measuring in the diagonal basis. It all make more sense now.

    Reply

Leave a Reply