Chris Pacejo
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 Mar 17 awarded Nice Question Jun 18 comment Are not the field equations proof of the holographic principle? I'm 100% clear that time is no different from space. See my comment about arbitrarily slicing the 4-space along a hyperplane. I've already "read up on the topic". I'll wait for another answer as it seems we're talking past each other. Jun 18 comment Are not the field equations proof of the holographic principle? It seems you are implying that the holographic principle states that the universe can be described by equations relating three dimensions deterministically (as opposed to three dimensions unconstrained by any equations), and hence the universe contains only two dimensions of information? Jun 18 comment Are not the field equations proof of the holographic principle? Yes I know. They deterministically specify the time evolution of a three-dimensional system. Given a 3-dimensional initial condition (let's say the hyperplane at t=0, but x=0 or y=z work equally well), the equations describe a unique four-dimensional field; i.e. they project from 3 to 4 dimensions. Is this not the very definition of a hologram? Jun 18 comment Are not the field equations proof of the holographic principle? I'm not sure what you're referring to? My second paragraph talks about 2+1, not 3+1. Jun 17 asked Are not the field equations proof of the holographic principle? Jan 19 accepted Controlled-measurement of a quantum register Jan 17 comment Controlled-measurement of a quantum register I have revised the question, hopefully this clears things up. Jan 17 revised Controlled-measurement of a quantum register big clarification Jan 17 awarded Commentator Jan 17 comment Controlled-measurement of a quantum register OK, I see what your method is doing now -- it is effectively saving the value of $b$ to $c$ so that eventually, when I measure the entire system, if $a=1$, I can ferret out what $b$ would have been had I measured it at the point when I saved its value. What's missing is that I wanted to know what $b$ would be if $a=1$ before continuing with the rest of the operation; but as Peter Shor pointed out this is impossible. Jan 17 comment Controlled-measurement of a quantum register Your understanding is correct, but since such an operation isn't reversible, obviously there's no such unitary transform. Hence why I'm fishing from the "measurement" angle (since measurement is a "special" operation). Note also my edit; what I'm actually after is controlled-initialization (another non-reversible operation); I had just figured controlled-measurement was the most expedient means of achieving this (though impossible as Peter Shor points out). Jan 17 comment Controlled-measurement of a quantum register I edited the last paragraph to generalize it for any 2-qubit state (though I left the resultant phase of the square root unspecified; what it is is inconsequential for my use). Your first example would remain unchanged (i.e. keep the square root). Your second example would become $\left[1/2,\sqrt{1/2},1/2,0\right]$, though it's not clear to me what the phase of the second entry ($\left|01\right>$) "should" be. Jan 17 revised Controlled-measurement of a quantum register clarified controlled-initialization Jan 17 revised Controlled-measurement of a quantum register expanded on controlled-initialization Jan 17 comment Controlled-measurement of a quantum register Thank you, this makes a lot of sense. However I should have been clearer as to my goal: I actually wish to conditionally initialize some qubits. Since obviously initialization is not a unitary operation, I have been modeling it as a measurement + an XOR based on the measurement. Assuming instead that initialization is an elementary operation, it seems that controlled initialization would not violate the rule you state above (since no measurement is being made). Is this correct? Would controlled-initialization run into a different problem? (I've updated the question accordingly.) Jan 17 comment Controlled-measurement of a quantum register While that evolution works for the example, it doesn't perform what I want in the general case, which is to collapse $b$, which I don't think is possible with a unitary transform in the general case since it is a lossy operation. I understand that $b$ may measure differently than in such an uncontrolled measurement, but that would only be the case if $a$ measures differently that what it was controlled as. i.e. I only want to know, if $a$ were eventually to measure as 1, what will $b$ measure as? Jan 16 revised Controlled-measurement of a quantum register added tag Jan 16 comment Controlled-measurement of a quantum register No, I want to measure what $b$ is conditioned on $a$ being 0; I don't want to measure $b$ when $a$ is 0 and $0$ when $a$ is 1. Jan 16 comment Controlled-measurement of a quantum register That question pertains to the intersection of control theory and quantum mechanics. My question does not concern control theory (a branch of mathematics), but rather controlled gates (a concept in quantum computation).