New answers tagged matrix-representation
1
If you set $P_y(t)=0$ for a particular length of time, then you just get an $X$ rotation. Similarly, if you set $P_x(t)=0$ for a certain length of time, you just get a $Y$ rotation. So, you do a sequence that looks something like
\begin{align*}
P_X(t)=\left\{\begin{array}{cc}
J & 0\leq t\leq \frac{d}{2J} \\
0 & 0<t-\frac{d}{2J}\leq \frac{c}{2J} \\
...
1
First of all, note that if you want to know the action of a quantum circuit of a given quantum state, you essentially have two options:
Writing done the matrix representation of this circuit
Feeding the state you're interested through the circuit to get the outcome
You simply have to pick the option that is faster. In this answer, I will describe the first ...
1
You can build a matrix by knowing the action of the operator on each basis state... encoding as:
We need the following action
The other states are left unchanged (or mapped to themselves).
In decimal 10000 and 10001 is 16 and 17... so defining the final operator in decimal we have:
(This is the operator for the final operation in the circuit given)
A ...
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