I Quantum Circuit Confusion On Time Evolution

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The discussion centers on the challenges of visualizing time evolution in quantum circuits, particularly when using the Heisenberg picture with non-commuting Hamiltonians H1 and H2. The author questions the validity of stacking unitary gates vertically to represent the conjugation of an operator O, as it suggests an elapsed time of 2t, which may not accurately reflect the physical process. There is a debate on whether the amount of time spent in the evolution is significant, impacting the stacking method's appropriateness. The conversation also highlights the need for a suitable Hamiltonian H that can simplify the relationship between the two Hamiltonians using the Baker-Campbell-Hausdorff formula. Ultimately, the discussion emphasizes the complexities involved in representing quantum time evolution accurately in circuit form.
thatboi
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Hi all,
When working in the Heisenberg picture, we can represent implementing time evolution on an operator via a Hamiltonian H through a quantum circuit type picture like the following:
1686890731927.png

where time is on the vertical axis and increases going up and the block represents the unitary gate ##e^{-iHt}##. However, I am struggling to picture how this would look on a circuit if instead, we wanted to conjugate some operator ##O## via the unitaries ##e^{-iH_{1}t},e^{-iH_{2}t}## where ##[H_{1},H_{2}] \neq 0##. That is, ##O(t) = e^{-iH_{2}t}e^{-iH_{1}t}Oe^{iH_{1}t}e^{iH_{2}t}##. Vertically stacking the gates on top of each other doesn't seem to make much sense to me since it would then seem to imply that we have elapsed a time ##2t## through this time evolution.
Any thoughts?
 
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Is it really important how much time was spent? If not, then vertically stacking gates is OK. If yes, then you need to find an operator ##H## such that
$$e^{iH_1t}e^{iH_2t}=e^{iHt}$$
For that purpose, you need to use some version of Baker-Campbell-Hausdorff formula.
 
For the quantum state ##|l,m\rangle= |2,0\rangle## the z-component of angular momentum is zero and ##|L^2|=6 \hbar^2##. According to uncertainty it is impossible to determine the values of ##L_x, L_y, L_z## simultaneously. However, we know that ##L_x## and ## L_y##, like ##L_z##, get the values ##(-2,-1,0,1,2) \hbar##. In other words, for the state ##|2,0\rangle## we have ##\vec{L}=(L_x, L_y,0)## with ##L_x## and ## L_y## one of the values ##(-2,-1,0,1,2) \hbar##. But none of these...

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