New answers tagged hamiltonian-simulation
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The remark that you state is absolutely critical. Let's try and introduce a notation that takes the spaces into account better. So, we're going to have a set of $a$ qubits denoted $C$, a set of $b$ qubits denoted $D$ and a set of $s$ qubits denoted $S$. Now I can use $U_{CS}$ to mean apply $U$ on qubits in sets $C$ (the ancillas) and $S$, and act as identity ...
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We can't implement $e^{iZ_1 \otimes Z_2 \otimes Z_3 \theta}$ with three separate rotations. In other words:
$$e^{iZ_1 \otimes Z_2 \otimes Z_3 \theta} \ne e^{i Z_1 \theta} \otimes e^{i Z_2 \theta} \otimes e^{i Z_3 \theta}$$
The implementation of this gate can be found in this answer. The $e^{-iI \otimes I \otimes I\theta} = e^{-i\theta} I \otimes I \otimes I$ ...
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You're right, that this factor can be separated as an operation that applies a global phase factor. Matrix exponentiation is distributive over an additive argument if and and only if the additive terms commute. The identity operator and the scalar $-i\theta$, which can incidentally be though of as a constant of the gate design, commute with all unitary gates....
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For all we know, it seems that fully-fledged quantum computations indeed require gates (and thus Hamiltonians generating those gates) which don't commute. (But this does not mean that we know this for sure!)
However, it is not true that otherwise, the system can be classically simulated (at least, again, there is strong evidence for that). That is, such a ...
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First, a minor point: it doesn't make any sense to say that "a Hamiltonian commutes". You mean that the different terms in the Hamilonian commute. When it comes to commutation, it takes two (or more) to tango.
It is indeed often said that if all the terms of a Hamiltonian commute, then the Hamiltonian is classical. But that isn't really true in my opinion. ...
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