$\require{cancel} \newcommand{\Ket}[1]{\left|{#1}\right\rangle} \newcommand{\Bra}[1]{\left\langle{#1}\right|} \newcommand{\Braket}[1]{\left\langle{#1}\right\rangle} \newcommand{\Rsr}[1]{\frac{1}{\sqrt{#1}}} \newcommand{\RSR}[1]{1/\sqrt{#1}} \newcommand{\Verti}{\rvert} \newcommand{\HAT}[1]{\hat{\,#1~}} \DeclareMathOperator{\Tr}{Tr}$
Hadamard on Multi-Qubits¶
First created in September 2018
It is useful to express multi-qubit quantum states in terms of a sumation of all possible states.
Unbiased Superposition¶
For an $n$-qubit system, there are $N=2^n$ states. By applying Hadamard transform to all $n$ qubits, an unbiased superposition is obtained.
$\displaystyle H^{\otimes n}\Ket0^{\otimes n}=\Rsr{2^n}\sum_{k=0}^{N-1}\Ket k .$
The simplest form when $n=1$ is $H\Ket0=\Rsr2(\Ket0+\Ket1).$
When $n=2,~H\otimes H\Ket{00}={1\over2}(\Ket{00}+\Ket{01}+\Ket{10}+\Ket{11}).$
Phase¶
Instead of an all-zero $\Ket0^{\otimes n}$, let us examine bit-patterns of $\Ket{x_k}^{\otimes n}$ where $x_k\in\{0,1\}$.
Notation: Pattern $\Ket x=\Ket{x_1x_2\ldots x_n}~~\text{with iteration variable }\Ket z=\Ket{z_1z_2\ldots z_n},$
When $n=1,~\Ket x=\Ket{x_1},~\Ket z=\Ket{z_1},$
$H\Ket{x_1} =\Rsr2(\Ket0+(-1)^{x_1}\Ket1) =\Rsr2((-1)^{x_1\cdot 0}\Ket0+(-1)^{x_1\cdot 1}\Ket1) =\Rsr2\sum_{z=0}^1(-1)^{x_1z_1}\Ket z .$
When $n=2,~\Ket{x_1x_2},~\Ket z=\Ket{z_1z_2}$
$H\otimes H\Ket{x_1x_2}\\ ={1\over2}~(\Ket0+(-1)^{x_1}\Ket1)~(\Ket0+(-1)^{x_2}\Ket1)\\ ={1\over2}(\Ket{00}+(-1)^{x_2}\Ket{01}+(-1)^{x_1}\Ket{10}+(-1)^{x_1+x_2}\Ket{11}) .$
It appears that $(-1)^{x_k}$ appears when and only when $z_k=1$, so we can generally apply $(-1)^{x_kz_k}$ to all values of $z$.
$H\otimes H\Ket{x_1x_2}={1\over2}\sum_{z=0}^3(-1)^{x_1z_1+x_2z_2}\Ket{z_1z_2}.$
Generally, for $\Ket x=\Ket{x_1x_2\ldots x_n},~\Ket z=\Ket{z_1z_2\ldots z_n},$
$\displaystyle H^{\otimes n}\Ket{x_1x_2\ldots x_n} =\Rsr{2^n}\sum_{z_1,z_2,\ldots,z_n}(-1)^{x_1z_1+x_2z_2+\ldots+x_nz_n}\Ket{z_1z_2\ldots z_n} .$
$\boxed{H^{\otimes n}\Ket x=\Rsr{2^n}\sum_z(-1)^{x\cdot z}\Ket z}$ where $x\cdot z$ is the sum of the bit-wise AND result of the two patterns $x$ and $z$.
Code Section Template¶
# Initialisation
import sys
sys.path.append('../')
from qtol import *
# Iteration
# Number of qubits
qbNum = 1
# Define the Quantum and Classical Registers
q = QuantumRegister(qbNum)
c = ClassicalRegister(qbNum)
qc = QuantumCircuit(q, c)
# Preparation
qc.iden(q)
# Circuit building
# ...
# Finalisation
# ...
show_me(qc, q, c, show_latex=True, show_bloch_vector=True, show_histogram=True)
circuit_drawer(qc)
