network profile
| bio | website | |
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| location | ||
| age | ||
| visits | member for | 4 months |
| seen | Jan 23 at 21:11 | |
| stats | profile views | 3 |
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Jan 19 |
accepted | Controlled-measurement of a quantum register |
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Jan 17 |
comment |
Controlled-measurement of a quantum register I have revised the question, hopefully this clears things up. |
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Jan 17 |
revised |
Controlled-measurement of a quantum register big clarification |
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Jan 17 |
awarded | Commentator |
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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. |
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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). |
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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. |
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Jan 17 |
revised |
Controlled-measurement of a quantum register clarified controlled-initialization |
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Jan 17 |
revised |
Controlled-measurement of a quantum register expanded on controlled-initialization |
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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.) |
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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? |
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Jan 16 |
revised |
Controlled-measurement of a quantum register added tag |
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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. |
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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). |
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Jan 16 |
awarded | Editor |
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Jan 16 |
revised |
Controlled-measurement of a quantum register added 715 characters in body |
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Jan 16 |
comment |
Controlled-measurement of a quantum register Maybe I'm missing something, but I'm not sure this effects a controlled-measurement as I described it. Consider the example where $a$ and $b$ are both in the first Hadamard state, and $c$ is 0, so the state vector (ignoring phase) is $\left[\sqrt{1/4},\sqrt{1/4},\sqrt{1/4},\sqrt{1/4},0,0,0,0\right]$. After CCNOT, this becomes $\left[\sqrt{1/4},\sqrt{1/4},\sqrt{1/4},0,0,0,0,\sqrt{1/4}\right]$, which means that $c$ is 0 with 75% probability, and 1 with 25% probability, when it should be 50% for both. |
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Jan 16 |
asked | Controlled-measurement of a quantum register |
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Jan 16 |
awarded | Scholar |
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Jan 16 |
accepted | If quantum computation is reversible, what is the point of Grover's search algorithm? |
