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PBR Paper, Calculation of Outcome Probabilities

  1. Jun 11, 2013 #1
    I'm trying to understand the PBR proof as explained in their paper on arXiv (http://arxiv.org/abs/1111.3328). I'm having trouble following the calculation of the probability of measurement outcomes on page 6 (A7). Specifically, going from the first line to the second line of A7, they seem to equate [itex]\langle x_1...x_n|H^{\otimes n}[/itex] with [itex](\Sigma_z (-1)^{x\cdot z}\langle z |)/\sqrt{2^n}[/itex]. I don't understand that equality, probably because I don't understand what the [itex]z[/itex]'s are or where they came from. Doesn't seem to be defined in the paper -- later in the paper it says [itex]z[/itex] is the sum over [itex]i[/itex] of [itex]z_i[/itex], but the [itex]z_i[/itex]'s don't appear to be defined.

    If someone could clarify it would be much appreciated.
  2. jcsd
  3. Jun 12, 2013 #2
    I think I might see what it is now. It looks like they are treating tensor states like [itex]|\psi_{x_1}\rangle \otimes \ldots\otimes |\psi_{x_n}\rangle[/itex] as vectors in [itex]\mathbb{R}^n[/itex] with coordinates [itex](x_1,...,x_n)[/itex] (which works out here despite the fact that the tensor space is really [itex]2^n[/itex] dimensional because all the states of the component systems are either (complex) multiples of |0> or |1>, no combinations thereof). Then the z's are the typical orthonormal basis elements of this n-dimensional space, and similarly the quantum state is "encoded" as an element of this basis denoted x, and the usual n-dimensional dot product is used. The math seems to work out if you assume this.
  4. Aug 7, 2013 #3
    I'm catching up on a huge backlog of papers I've been meaning to read for ages and only reached the PBR paper yesterday. I'm stuck on Appendix A, the stuff labelled A7. Maybe I don't understand how the inner product on the tensor product space works, but I can't see how they got from line 4 to line 5. According to my understanding it looks like a plus and some parentheses have gone missing.
  5. Aug 7, 2013 #4
    Scratch that, I googled a bit and now it makes sense :) It also helped when I bothered to read a few lines lower that they are using |z| for what most people would have written as |z|^2.
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