title: Qutuam Computation

#+STARTUP: overview

Basic states and operations

ketbra

• Ket: $$|a⟩=\left(\begin{array}{c}{a}_1\\ a_2\end{array}\right)$$

• Bra: $$⟨b|=|b{⟩}^{+}={\left(\begin{array}{c}{b}_1\\ b_2\end{array}\right)}^{+}=\left(\begin{array}{cc}{b}_1^{\ast }& b_2^{\ast }\end{array}\right)$$

• Production of Ket and Bra:

$$⟨b|a⟩=\left(\begin{array}{cc}{b}_1^{\ast }& b_2^{\ast }\end{array}\right)\cdot \left(\begin{array}{c}{a}_1\\ a_2\end{array}\right)=a_1{b}_1^{\ast }+a_2{b}_2^{\ast }$$

states

$$|0⟩:=\left(\begin{array}{c}1\\ 0\end{array}\right)$$

$$|1⟩:=\left(\begin{array}{c}0\\ 1\end{array}\right)$$ If a measurement will collapse either on |0⟩ or |1⟩, which means |0⟩ or |1⟩ is its eigentstate, we call it as Z-measurement, and denote as ${\sigma }_z$.

$$|+⟩:=\frac{1}{\sqrt{2}}(|0⟩+|1⟩)$$

$$|-⟩:=\frac{1}{\sqrt{2}}(|0⟩-|1⟩)$$ If a measurement will collapse either on |+⟩ or |-⟩, which means |+⟩ or |-⟩ is its eigentstate, we call it as X-measurement, and denote as ${\sigma }_x$.

$$|+i⟩:=\frac{1}{\sqrt{2}}(|0⟩+i|1⟩)$$

$$|-i⟩:=\frac{1}{\sqrt{2}}(|0⟩-i|1⟩)$$ If a measurement will collapse either on |+i⟩ or |-i⟩, which means |+i⟩ or |-i⟩ is its eigentstate, we call it as Y-measurement, and denote as ${\sigma }_y$.

The coefficient has to be $\frac{1}{\sqrt{2}}$ because of normalisation$⟨\varphi |\varphi ⟩=1$.

Measurements

for states basis of{ $x_1$ } and { $x_2$ }, the measurement of state $\varphi$ on { $x_1$ } is $⟨x_1|\varphi {⟩}^2$.

Single Qutantum circuits

• bit flip: ${\sigma }_x|0⟩=|1⟩$ :

  $${\sigma }_x=\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)$$

• phase flip: ${\sigma }_z|+⟩=|-⟩$ : $${\sigma }_z=\left(\begin{array}{cc}1& 0\\ 0& -1\end{array}\right)$$

• bit & phase filp: ${\sigma }_y|+i⟩=|-i⟩$ : $${\sigma }_y=\left(\begin{array}{cc}0& -i\\ i& 0\end{array}\right)$$

• Hadamard gate: $H|0⟩=|+⟩$, $H|1⟩=|-⟩$, $$H=\frac{1}{\sqrt{2}}\left(\begin{array}{cc}1& 1\\ 1& -1\end{array}\right)$$

• Phase gate: $S|+⟩=|+i⟩$, $S|-⟩=|-i⟩$ $$S=\left(\begin{array}{cc}1& 0\\ 0& i\end{array}\right)$$

Multipartite quantum states

we use tensor products to describe multiple states $$|a⟩\oplus |b⟩=\left(\begin{array}{c}{a}_1\\ a_2\end{array}\right)\oplus \left(\begin{array}{c}{b}_1\\ b_2\end{array}\right)=\left(\begin{array}{c}{a}_1{b}_1\\ a_1{b}_2\\ a_2{b}_1\\ a_2{b}_2\end{array}\right)$$ only such can to described by $\oplus$ of other states are called uncorrelated, otherweise it's correlated, and when some fully correlated are called entangled.

XOR gate

$$CNOT=\left(\begin{array}{cccc}1& 0& 0& 0\\ 0& 1& 0& 0\\ 0& 0& 1& 0\\ 0& 0& 0& 1\end{array}\right)$$

#+header: :headers '("\\usepackage{qcircuit}")

\Qcircuit @C=1em @R=1em {
\lstick{\ket{x}} & \ctrl{1} & \rstick{\ket{x}} \qw \\
\lstick{\ket{y}} & \targ   & \rstick{\ket{x \oplus y }} \qw
}

Photo Link here

Bell states

Example of fully correlated states (maximally entangled),

$$|{\phi }^{00}⟩=\frac{1}{\sqrt{2}}(|00{⟩}_{AB}+|11{⟩}_{AB})$$ $$|{\phi }^{01}⟩=\frac{1}{\sqrt{2}}(|01{⟩}_{AB}+|10{⟩}_{AB})$$ $$|{\phi }^{10}⟩=\frac{1}{\sqrt{2}}(|00{⟩}_{AB}-|11{⟩}_{AB})$$ $$|{\phi }^{11}⟩=\frac{1}{\sqrt{2}}(|01{⟩}_{AB}-|10{⟩}_{AB})$$

Create Bell states

#+header: :headers '("\\usepackage{qcircuit}")

\Qcircuit @C=1em @R=1em {
\lstick{\ket{i}_A} & \gate{H} & \ctrl{1} &  \qw \\
\lstick{\ket{j}_B} &  \qw    &   \targ   &  \qw
  }

Photo Link here

Teleportion

If Alise and Bob share the same bell states $|{\phi }_{AB}^{00}⟩=\frac{1}{\sqrt{2}}(|00{⟩}_{AB}+|11{⟩}_{AB})$, Now Alias want to send stetas $|\varphi {⟩}_S=\alpha |0{⟩}_S+\beta |1{⟩}_S$ to Bob,

$$\Phi =|\varphi {⟩}_S\oplus {\phi }_{AB}^{00}⟩=(\alpha |0{⟩}_S+\beta |1{⟩}_S)\oplus \frac{1}{\sqrt{2}}(|00{⟩}_{AB}+|11{⟩}_{AB})$$

$$\Phi =\frac{1}{\sqrt{2}}(\alpha |000{⟩}_{SAB}+\alpha |011{⟩}_{SAB}+\beta |100{⟩}_{SAB}+\beta |111{⟩}_{SAB})$$ $$\Phi =\frac{1}{2\sqrt{2}}(\alpha |000{⟩}_{SAB}+\beta |001{⟩}_{SAB}+\alpha |110{⟩}_{SAB}+\beta |111{⟩}_{SAB}+\alpha |011{⟩}_{SAB}+\alpha |101{⟩}_{SAB}+\beta |010{⟩}_{SAB}+\beta |100{⟩}_{SAB}+\alpha |000{⟩}_{SAB}-\beta |001{⟩}_{SAB}-\alpha |110{⟩}_{SAB}+\beta |111{⟩}_{SAB}+\alpha |011{⟩}_{SAB}-\alpha |101{⟩}_{SAB}-\beta |010{⟩}_{SAB}+\beta |100{⟩}_{SAB})$$ $$\begin{gathered} \Phi =\frac{1}{2\sqrt{2}}[(|00{⟩}_{SA}+|11{⟩}_{SA})]\oplus (\alpha |0{⟩}_B+\beta |1{⟩}_B)+(|01{⟩}_{SA}+|10{⟩}_{SA})]\oplus (\alpha |1{⟩}_B+\beta |0{⟩}_B) \\ +(|00{⟩}_{SA}-|11{⟩}_{SA})]\oplus (\alpha |0{⟩}_B-\beta |1{⟩}_B)+(|01{⟩}_{SA}-|10{⟩}_{SA})]\oplus (\alpha |1{⟩}_B-\beta |0{⟩}_B)] \end{gathered}$$ $$\Phi =\frac{1}{2\sqrt{2}}[|{\phi }_{AB}^{00}⟩\oplus |\varphi {⟩}_B+|{\phi }_{AB}^{01}⟩\oplus {\sigma }_x|\varphi {⟩}_B+|{\phi }_{AB}^{10}⟩\oplus {\sigma }_z|\varphi {⟩}_B+|{\phi }_{AB}^{11}⟩\oplus {\sigma }_x{\sigma }_z|\varphi {⟩}_B]$$

• Alice preforms a measurement in the Bell basis
• she send her classical output (i,j) to Bob
• Bob apply ${\sigma }_z^i{\sigma }_x^j$ to get the orignal $|\varphi ⟩$.

#+header: :headers '("\\usepackage{qcircuit}")

\Qcircuit @C=1em @R=.7em {
 \lstick{\ket{\phi_s}_A}          & \multigate{1}{Bell Meas} & \cw & \cw    & \cwx[2] \\
 \lstick{\ket{\varphi^{00}_A}}    & \ghost{Bell Meas}        & \cw & \cwx[1]   \\
 \lstick{\ket{\varphi^{00}_B}}    &  \qw                    & \qw & \gate{\sigma_x^j}& \gate{\sigma_x^j }& \qw &  \rstick{\ket{\phi_s}_B} 
}

Photo Link Here

Deutsche-Jose

Bit oracle

#+header: :headers '("\\usepackage{qcircuit}")

Photo Link here

$O_f|x⟩|y⟩=|x⟩|y\oplus f(x)⟩$, $U_f$ : phase oracle, which is independt of y, $$U_f|x⟩=(-1{)}^{f(x)}|x⟩$$.

Hadamard on n qubits

for $x\in \{0,1\}$, $$|x⟩->H->|y⟩$$ , $$|y⟩=\frac{1}{\sqrt{2}}\sum _{k\in \{0,1\}}(-1{)}^{k\cdot x}|k⟩$$

for for $x\in \{0,1{\}}^n$, $$|x_0⟩->H->|y_0⟩$$ , $$|x_1⟩->H->|y_1⟩$$ , $$|x_2⟩->H->|y_2⟩$$ , $$|y⟩=H^{\otimes n}=\frac{1}{\sqrt{2^n}}\sum _{k\in \{0,1{\}}^n}(-1{)}^{k\cdot x}|k⟩$$

Deutsche Jose algorithm

#+header: :headers '("\\usepackage{qcircuit}")

\Qcircuit @C=1em @R=1em {
\lstick{\ket{0}}  & \gate{H} & \multigate{2}{U_f} & \gate{H} & \meter & \cw  & \rstick{\ket{y_0}} \\
\lstick{\ket{0}} & \gate{H} & \ghost{U_f} & \gate{H} & \meter & \cw  & \rstick{\ket{y_1}} \\
\lstick{\ket{0}} & \gate{H} & \ghost{U_f} & \gate{H} & \meter & \cw  & \rstick{\ket{y_n}} 
}

Photo Link here

Proof: $$|{\phi }_0⟩=|0000.....0{⟩}^{\otimes n}$$

$$|{\phi }_1⟩=H^{\otimes n}|{\phi }_0⟩=\frac{1}{\sqrt{2^n}}\sum _{x\in \{0,1{\}}^n}(-1{)}^{x\cdot {\phi }_0}|x⟩$$ because $x\cdot {\phi }_0=0$, so , $$|{\phi }_1⟩=\frac{1}{\sqrt{2^n}}\sum _{x\in \{0,1{\}}^n}|x⟩$$

$$|{\phi }_2⟩=U_f|{\phi }_1⟩=\frac{1}{\sqrt{2^n}}\sum _{x\in \{0,1{\}}^n}{U}_f|x⟩=\frac{1}{\sqrt{2^n}}\sum _{x\in \{0,1{\}}^n}(-1{)}^{f(x)}|x⟩$$

$$|{\phi }_3⟩=H^{\otimes n}|{\phi }_2⟩=\frac{1}{\sqrt{2^n}}\sum _{x\in \{0,1{\}}^n}(-1{)}^{f(x)}{H}^{\otimes n}|x⟩$$ $$|{\phi }_3⟩=\frac{1}{2^n}\sum _{x\in \{0,1{\}}^n}(-1{)}^{f(x)}\sum _{k\in \{0,1{\}}^n}(-1{)}^{k\cdot x}|k⟩$$ $$|{\phi }_3⟩==\sum _{k\in \{0,1{\}}^n}[\frac{1}{2^n}\sum _{x\in \{0,1{\}}^n}(-1{)}^{f(x)+k\cdot x}]\cdot |k⟩=\sum _{k\in \{0,1{\}}^n}{C}_k|k⟩$$

The Probability to measure the Zero( ${\phi }_0$) string is: $$P[y=|0000...0⟩]=|⟨00000...0|{\phi }_3⟩{|}^2=|\sum _{k\in \{0,1{\}}^n}{C}_k⟨0000..0|k⟩{|}^2=|C_{\mathrm{000..0}}{|}^2$$ $$P=|\frac{1}{2^n}\sum _{x\in \{0,1{\}}^n}(-1{)}^{f(x)}{|}^2$$ $P=1$ if f is constant, and $P=0$ if f is balanced

Gover's Algrithmus