Physics Asked on February 10, 2021
I stumbled upon such notations: $X=|0ranglelangle1|+|1ranglelangle0|$ for of logical NOT gate, or $|00ranglelangle00|+|01ranglelangle01|+|10ranglelangle10|+|11ranglelangle11|$ for 2-qubit indentity.
I am not sure I understand it correctly: does it mean that gate $|0ranglelangle0|+i|1ranglelangle1|= begin{bmatrix}1&0&iend{bmatrix}$?
How would you write SWAP gate in such notation?
$newcommand{ket}[1]{left|#1rightrangle}$ $newcommand{bra}[1]{leftlangle#1right|}$
Lets say, we define our states as $ket0 = begin{bmatrix}1 end{bmatrix}$ and $ket{1} = begin{bmatrix} 0 1end{bmatrix}$. Then, the NOT operation does, $Xket0 = ket1$ and $Xket1 =ket0$, which can be described by the matrix $$X = begin{bmatrix}0&11&0end{bmatrix} = ket0 bra1 + ket1 bra0.$$
Now, if we consider two qubits, the states can be defined as, $$ket{00} = begin{bmatrix}1 end{bmatrix},quad ket{01} = begin{bmatrix}0 1 end{bmatrix},quad ket{10} = begin{bmatrix}0 1 end{bmatrix},quad ket{11} = begin{bmatrix}0 1 end{bmatrix}$$
Now, the SWAP operation $S$ does the following:
$$
Sket{00} = ket{00}
Sket{01} = ket{10}
Sket{10} = ket{01}
,Sket{11} = ket{11}.$$
This is represented by the matrix
$$S = begin{bmatrix}0&0&0&1 &0&1&0 0&1&0&1 0&0&0&1end{bmatrix}.$$
And if you check that this matrix can be written as,
$$S = |00ranglelangle00|+|01ranglelangle10|+|10ranglelangle01|+|11ranglelangle11|.$$
So, the answer to your questions are yes.
Answered by sss_1983 on February 10, 2021
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