$z = x + iy = re^{iθ}$
$r = \sqrt{x^2 + y^2}$
$θ = tan^{-1}({y \over x})$
$x = rcosθ$
$y = rsinθ$
${y \over x} = tan(θ)$
$e^{iθ} = cosθ + isinθ$
$e^{iθ} + e^{-iθ}= 2cosθ$
$x^2 + y^2 = r^2$
$z^* = x - iy = re^{-iθ}$
| $ | z | = r$\ |
| $ | z | ^2 = r^2 = zz^*$ |
$e^{iπ} = -1 = i^2$
$\sqrt{e^{iπ}} = i = e^{iπ \over 2} = e^{iπ \over 6}$
$e^{i2π} = 1$
$e^{i3π \over 2} = -i$
$sin(u ± v) = sin(u)cos(v)±cos(u)sin(v)$
$cos(u ± v) = cos(u)cos(v)±sin(u)sin(v)$
$\begin{bmatrix}
α
β
\end{bmatrix}^T =
\begin{bmatrix}
α
β
\end{bmatrix}$
$\begin{bmatrix}
α
β
\end{bmatrix}^{\dagger} =
\begin{bmatrix}
α^*
β^*
\end{bmatrix}$
$|ψ⟩ = \begin{bmatrix}
α
β
\end{bmatrix}$
it’s called ket
$⟨ψ| =
\begin{bmatrix}
α^*
β^*
\end{bmatrix} =
\begin{bmatrix}
α^*
0
\end{bmatrix} +
\begin{bmatrix}
0
β^*
\end{bmatrix} =
α^⟨0| + β^⟨1|$
it’s called bra
$⟨ψ|Φ⟩ =
\begin{bmatrix}
α^*
β^*
\end{bmatrix}
\begin{bmatrix}
γ
δ
\end{bmatrix} =
α^γ + β^δ
$
it’s a inner product also called as bra-ket
| $⟨Φ | ψ⟩ = ⟨ψ | Φ⟩^*$ |
$⟨ψ|ψ⟩ =
\begin{bmatrix}
α^*
β^*
\end{bmatrix}
\begin{bmatrix}
α
β
\end{bmatrix} =
|α^2| + |β^2| = 1
$
$⟨Uψ|Uψ⟩ = ⟨ψ|U^{\dagger}U|ψ⟩ = ⟨ψ|ψ⟩ = 1 $
if the result of $⟨ψ|ψ⟩$ is $1$ so the state is normalized
$⟨+|-⟩ = ⟨0|1⟩ = ⟨i|-i⟩ = 0$
if the result is 0, so these states are orthogonal
if you test the states and they are normalized and orthogonal, so you have a orthonormal basis
| $ | ψ⟩ = ⟨0 | ψ⟩ | 0⟩ + ⟨1 | ψ⟩ | 1⟩$ |
| $ | ψ⟩ = α | a⟩ + β | b⟩$\ |
| $α = ⟨a | ψ⟩$\ | ||
| $β = ⟨b | ψ⟩$\ | ||
| $⟨a | ψ⟩$ is a projection of | ψ⟩ on | a⟩ |
$(AB)^{\dagger} = A^{\dagger}B^{\dagger}$
$|ψ⟩^{\dagger} = ⟨ψ|$
$⟨Uψ| = (|Uψ⟩)^{\dagger} = U^{\dagger}|ψ⟩^{\dagger} = ⟨ψ|U^{\dagger}$
$U^{\dagger}U = I$
If a matrix $U$ times $U^{\dagger}$ result in $I$, so it’s a unitary matrix
$M^{-1}M = I$
$U^{-1} = U^{\dagger}$
$|ψ⟩⟨Φ| =
\begin{bmatrix}
α
β
\end{bmatrix}
\begin{bmatrix}
γ^*
δ^*
\end{bmatrix} =
\begin{bmatrix}
αγ^* \space αδ^
βγ^ \space βδ^*
\end{bmatrix}$
$U = \begin{bmatrix}
0 \space 1
1 \space 0
\end{bmatrix}$
$U = 0|0⟩⟨0| + 1|0⟩⟨1| + 1|1⟩⟨0| + 0|1⟩⟨1| $
$U = |0⟩⟨1| + |1⟩⟨0| = X$
| $ | a⟩⟨a | + | b⟩⟨b | = I$ |
$|ψ⟩ = α|a⟩ + β|b⟩$
$|ψ⟩ = ⟨a|ψ⟩|a⟩ + ⟨b|ψ⟩|b⟩$
$|ψ⟩ = |ψ⟩|a⟩⟨a| + |ψ⟩|b⟩⟨b|$
$|ψ⟩ = (|a⟩⟨a| + |b⟩⟨b|)|ψ⟩$
$|ψ⟩ = I|ψ⟩$
If you can do this with an orthonormal basis, this is an complete orthonormal basis
| $ | 0⟩ ⊗ | 0⟩ = | 00⟩$\ |
| $⟨0 | ⊗ ⟨0 | = ⟨00 | $ |
| $c_{0} | 00⟩ + c_{1} | 01⟩ + c_{2} | 10⟩ + c_{3} | 11⟩$\ |
| The probability of each state is $ | c_{x} | ^2$ |
| $ | 0⟩^{⊗n} = | 0⟩ ⊗ | 0⟩ ⊗\dots⊗ | 0⟩ = | 0^{n}⟩$ |
| $⟨01 | 00⟩ = ⟨0 | 0⟩⟨1 | 0⟩$ |
Kronecker product, it’s the tensor product for linear algebra
$
\begin{bmatrix}
0
1
\end{bmatrix}⊗
\begin{bmatrix}
1
0
\end{bmatrix} =
\begin{bmatrix}
0
0
1
0
\end{bmatrix}
$
$c_{0}|00⟩ + c_{1}|01⟩ + c_{2}|10⟩ + c_{3}|11⟩ = \begin{bmatrix}
c_{0}
c_{1}
c_{2}
c_{3}
\end{bmatrix}$
$⟨00| = \begin{bmatrix} 1 \space 0 \end{bmatrix} ⊗ \begin{bmatrix} 1 \space 0 \end{bmatrix} = \begin{bmatrix} 1 \space 0 \space 0 \space 0 \end{bmatrix} $
$|ψ⟩ = \sum_{j=0}^{n-1} c_{j}|j⟩ = c_{0}|0⟩ + c_{1}|1⟩ + c_{2}|2⟩ + \dots + c_{n-1}|n-1⟩$
$⟨ψ| = \sum_{j=0}^{n-1} c_{j}⟨j| = c_{0}⟨0| + c_{1}⟨1| + c_{2}⟨2| + \dots + c_{n-1}⟨n-1|$
where $n$ is the number of qubits
In a 3 qubits circuit, with the collapsed state for the last and middle qubits are both 0, the state will be $c_{0}|000⟩ + c_{1}|001⟩ \over \sqrt{|c_{0}|^2 + |c_{1}|^2}$
| $(H ⊗ I)( | 0⟩⊗ | 0⟩) = H | 0⟩ ⊗ I | 0⟩$\ |
| $H^{⊗n} | 0⟩^{⊗n} = H | 0⟩ ⊗ I | 0⟩$ |
| $(H ⊗ I) | 00⟩ = {1 \over \sqrt2}( | 00⟩ + | 10⟩)$\ |
| $(H ⊗ I) | 01⟩ = {1 \over \sqrt2}( | 01⟩ + | 11⟩)$\ |
| $(H ⊗ I) | 10⟩ = {1 \over \sqrt2}( | 00⟩ - | 10⟩)$\ |
| $(H ⊗ I) | 11⟩ = {1 \over \sqrt2}( | 01⟩ - | 11⟩)$ |
| $(H ⊗ X) | 00⟩ = {1 \over \sqrt2}( | 01⟩ + | 11⟩)$ |
$CNOT(I ⊗ X) = (I ⊗ X)CNOT$ this is represented in a circuit from right to left
| $ | ψ_{0}⟩ | ψ_{1}⟩ = (α_{0} | 0⟩ + β_{0} | 1⟩)(α_{1} | 0⟩ + β_{1} | 1⟩) = α_{0}α_{1} | 00⟩ + α_{0}β_{1} | 01⟩ + β_{0}α_{1} | 10⟩ + β_{0}β_{1} | 11⟩$ |
An entangled state can’t be factored. Because of that, for classical computers, it’s difficult to map all the amplitudes, once in a entangled state you need to map all $2^n$ possible states, but with no entangled states you only need to map $2n$
the $CNOT$ with the $H$ gate, can create a entangled state
$CNOT|+⟩|0⟩ = {1\over\sqrt2}(|00⟩ + |11⟩) = |Φ^+⟩$
$CNOT|-⟩|0⟩ = {1\over\sqrt2}(|00⟩ - |11⟩) = |Φ^-⟩$
$CNOT|+⟩|1⟩ = {1\over\sqrt2}(|01⟩ + |10⟩) = |ψ^+⟩$
$CNOT|-⟩|1⟩ = {1\over\sqrt2}(|01⟩ - |10⟩) = |ψ^-⟩$
These are the bell states
$A(α|0⟩ + β|1⟩) = 1$
$(Aα)(Aα)^* + (Aβ)(Aβ)^* = 1$
Measurements in X
$|0⟩ = {1 \over \sqrt{2}}(|+⟩ + |-⟩)$
$|1⟩ = {1 \over \sqrt{2}}(|+⟩ - |-⟩)$
Measurements in Y
$|0⟩= {1 \over \sqrt{2}}(|i⟩ + |-i⟩)$
$|1⟩= {-i \over \sqrt{2}}(|i⟩ - |-i⟩)$
Measurements in other basis
$|a⟩ = α|0⟩ + β|1⟩$
$|b⟩ = γ|0⟩ + δ|1⟩$
$|0⟩ = α^|a⟩ + β^|b⟩$
$|1⟩ = γ^|a⟩ + δ^|b⟩$
consecutive measurements in different basis the result is always 1/2 for both states
| ${1 \over \sqrt{2}} | 00⟩ + {1 \over 2} | 01⟩ + {\sqrt{3} \over 4} | 10⟩ + {1 \over 4} | 11⟩$\ | |
| The probability of $ | x0⟩$ is $ | {1 \over \sqrt{2}} | ^2 + | {\sqrt{3} \over 4} | ^2$\ |
| If we measured the first qubit(rightmost in little endian system) and the result is zero, so the state of the circuit is $A({1 \over \sqrt{2}} | 00⟩ + {1 \over 2} | 01⟩)$ |
The states can also be measured sequentially $Prob(|00⟩) = Prob(|0⟩)Prob(|0⟩)$
global phase can be ignored at measurements
| ${1 \over \sqrt{2}}( | 0⟩+i | 1⟩) = {1 \over \sqrt{2}}( | 0⟩ + e^{iπ \over 2}) = {1 \over \sqrt{2}}( | 0⟩+ | 1⟩) = {1 \over \sqrt{2}}( | +⟩ + | -⟩)$ |
| $ | α | ^2 + | β | ^2 = 1$ |
α need to be a positive real number and β a complex number
$α = cos({θ \over 2})$
$β = e^{iθ}sin({θ \over 2})$
Ancillas need to be reset.
θ is the angle between the poles (z-axis)
Φ is the angle between x and y axis (this is present in $e^{iΦ}$)
to know if two states are in opposite sides in the bloch sphere
$θb = π - θa$
$Φb = Φa + π$
cartesian coordinates in the bloch sphere
$z = cosθ$
$y = sinΦ sinθ$
$x = cosΦ sinθ$
${\displaystyle \mathbf {\hat {n}} } = n _{x}{\displaystyle \mathbf {\hat {x}} } + n _{y}{\displaystyle \mathbf {\hat {y}} } + n _{z}{\displaystyle \mathbf {\hat {z}} }$
${\displaystyle \mathbf {\hat {n}} }$ is a unit vector that represents a rotation on bloch sphere
$n _{(x,y,z)}$ represents the rotation number for each axis
${\displaystyle \mathbf {\hat {x}} }$,${\displaystyle \mathbf {\hat {y}} }$ and ${\displaystyle \mathbf {\hat {z}} }$ represents points on the sphere
also $|n _{x}|^2 + |n _{y}|^2 +|n _{z}|^2 = 1$
A gate, need to be always unitary and reversible
The number of outputs of a gate needs to be the same of the inputs to be reversible and map all possible results
$X^{100} = I$
$X^{101} = X$
| $X | 0⟩ = | 1⟩$\ |
| $X | 1⟩ = | 0⟩$ |
$XY = iZ$
$Y|0⟩ = i|1⟩$
$Y|1⟩ = -i|0⟩$
Y gate is not a classical gate, since it has a $i$ and $-i$ phase
Y, X, Z gates rotate 180° on its axis(x, y, z respectively)
$S|0⟩ = |0⟩$
$S|1⟩ = i|1⟩$
$S = \sqrt{Z}$ or $S^2 = Z$
S gate rotates 90° on the Z axis
$T|0⟩ = |0⟩$
$T|1⟩ = e^{iπ \over 4}|1⟩$
$T^2 = S$
$T^4 = Z$
T gate is also called $π \over 8$ gate
T gate rotates 45° on the Z axis\
$H|0⟩ = {1 \over \sqrt{2}}(|0⟩ + |1⟩) = |+⟩$
$H|1⟩ = {1 \over \sqrt{2}}(|0⟩ - |1⟩) = |-⟩$
H gate is also called Hadamard gate
H gate rotates 180° between X and Y axis
$X^2 = Y^2 = S^4 = T^8 = H^2 = I$
$U = e^{iγ}[cos({θ \over 2})I - isin({θ \over 2})(n _{x}X + n _{y}Y + n _{z}Z)]$
Where $γ$ is the global phase
| $αU | 0⟩ + βU | 1⟩ = U(α | 0⟩ + β | 1⟩) = αe^{iγ}[cos({θ \over 2})I - isin({θ \over 2})(n _{x}X + n _{y}Y + n _{z}Z)] | 0⟩ + βe^{iγ}[cos({θ \over 2})I - isin({θ \over 2})(n _{x}X + n _{y}Y + n _{z}Z)] | 1⟩$ |
$U|0⟩ = a|0⟩ + b|1⟩ = \begin{bmatrix}
a
b
\end{bmatrix}$
$U|1⟩ = c|0⟩ + d|1⟩ = \begin{bmatrix}
c
d
\end{bmatrix}$
$U = \begin{bmatrix}
a \space c
b \space d
\end{bmatrix}$
$U|ψ⟩ = \begin{bmatrix}
a \space c
b \space d
\end{bmatrix} \begin{bmatrix}
α
β
\end{bmatrix} = \begin{bmatrix}
aα + cβ
bα + dβ
\end{bmatrix}$
$|aα + cβ|^2 + |bα + dβ|^2 = 1$
$U|ψ⟩ = |Uψ⟩$
$⟨Uψ| = \begin{bmatrix}
a^α^ + c^β^ \space\space\space
b^α^ + d^β^\end{bmatrix} =
\begin{bmatrix}
α^\space
β^
\end{bmatrix}
\begin{bmatrix}
a^* \space c^
b^ \space d^*
\end{bmatrix} =
\begin{bmatrix}
α^\space
β^
\end{bmatrix}
\begin{bmatrix}
a \space c
b \space d
\end{bmatrix}^{\dagger} = ⟨ψ|U^{\dagger}$
CNOT (CX) inverts the right qubit(target) if the left qubit(control) is 1
$CNOT(|10⟩) = |11⟩$
the control qubit is not changed because it has a XOR
$CNOT|a⟩|b⟩ = |a⟩|a ⊗ b⟩$
$CNOT(c_{0}|00⟩ + c_{1}|01⟩ + c_{2}|10⟩ + c_{3}|11⟩) = c_{0}|00⟩ + c_{1}|01⟩ + c_{2}|11⟩ + c_{3}|10⟩$ it inverts the $c_{2}$ with the $c_{3}$ amplitude
$CNOT = CNOT_{ij} = CNOT_{10}$ where $i$ is the control and $j$ the target
$(H ⊗ H) CNOT_{01} (H ⊗ H) = CNOT_{10}$ it inverts the control with the target
$(X ⊗ I) CNOT (X ⊗ I)$ it is called $anti \space CNOT$
Controlled U (CU), it applies U in the right qubit, when the left qubit is 1
$CU|00⟩ = |00⟩$
$CU|01⟩ = |01⟩$
$CU|10⟩ = |1⟩ ⊗ U|0⟩$
$CU|11⟩ = |1⟩ ⊗ U|1⟩$
SWAP, the swap inverts two qubits positions, but differently from CNOT, it can’t create entangled states
$SWAP|00⟩ = |00⟩$
$SWAP|01⟩ = |10⟩$
$SWAP|10⟩ = |01⟩$
$SWAP|11⟩ = |11⟩$
it also invert amplitudes
$SWAP(c_{0}|00⟩ + c_{1}|01⟩ + c_{2}|10⟩ + c_{3}|11⟩) = c_{0}|00⟩ + c_{1}|10⟩ + c_{2}|01⟩ + c_{3}|11⟩$
$SWAP = CNOT \space CNOT_{01} \space CNOT$
Toffoli gate (CCX), inverts the right qubit if the other two are 1
$Toffoli|111⟩ = |110⟩$
$Toffoli|110⟩ = |111⟩$
$R{_r} = {360 \degree \over 2^r} \space on \space Z axis$
$R{_1} = Z$
$R{_2} = S$
$R{_3} = T$
$R{_4} = 22.5 \degree$
$R{_r}^{\dagger} = -{360 \degree \over 2^r}$
An unknown quantum state can’t be reproduced(replicated) if we don’t know each amplitude
$|ψ⟩|0⟩ = |ψ⟩|ψ⟩$
$U|ψ⟩|0⟩ = |ψ⟩|ψ⟩$ there’s no operator $U$ that can pass $|ψ⟩$ state to the right qubit, once the result is $|ψ⟩^2$ which isn’t linear
But for $(I ⊗ H)|+⟩|0⟩ = |+⟩|+⟩$ we can clone, $|+⟩$, once we know the amplitudes
Generally $U$ can only copy states that are orthogonal
$(I ⊗ H)|+⟩|0⟩ = |+⟩|+⟩$
$(x ⊗ H)|-⟩|0⟩ = |-⟩|-⟩$
$(I ⊗ I)|0⟩|0⟩ = |0⟩|0⟩$
$(I ⊗ X)|1⟩|0⟩ = |1⟩|1⟩$
$(I ⊗ Y)|i⟩|0⟩ = |i⟩|i⟩$
$(X ⊗ Y)|-i⟩|0⟩ = |-i⟩|-i⟩$