# Quantum Teleportation
###### tags:`quantum computing`

## Unitary Matrix
- $U^*U = UU^* = I\quad(U^* = U^H)$
- $|\lambda_n| = 1$

>$$
\begin{aligned}
Ux &= \lambda x
\end{aligned}
$$
>
>$$
\begin{aligned}
||x||^2 = \langle Ux,Ux \rangle &= x^TU^TUx = ||x||^2
\end{aligned}
\tag{1}
$$
>
>$$
\begin{aligned}
||Ux||^2 = ||\lambda x||^2 &= |\lambda|^2||x||^2
\end{aligned}
\tag{2}
$$
>
>$$
\begin{aligned}
(1) = (2)
\end{aligned}
$$
>
>$\therefore |\lambda| = 1 \rightarrow$ this means eigenvalue lies on the unit circle of palane, which means $\lambda =a+ib = e^{i\theta}$
- $|det(U)| = 1$

## Phase Kickback
Let $\psi$ be a eigenvector of unitary matrix.
以上電路代表 :
$$
\begin{aligned}
CU(|\psi\rangle\otimes H|0\rangle) = CU(|\psi\rangle\otimes H|+\rangle)
\end{aligned}
$$
繼續推倒:
$$
\begin{aligned}
CU(|\psi\rangle\otimes H|+\rangle) &= CU(|\psi\rangle\otimes (\frac{1}{\sqrt{2}}|0\rangle + \frac{1}{\sqrt{2}}|1\rangle))\\
&= CU(\frac{1}{\sqrt{2}}|\psi\rangle \otimes|0\rangle + \frac{1}{\sqrt{2}}|\psi\rangle \otimes|1\rangle)\\
&= \frac{1}{\sqrt{2}}CU(|\psi\rangle\otimes|0\rangle)+\frac{1}{\sqrt{2}}CU(|\psi\rangle\otimes|1\rangle)\\
&= \frac{1}{\sqrt{2}}|\psi\rangle\otimes|0\rangle+\frac{1}{\sqrt{2}}U|\psi\rangle\otimes|1\rangle\\
&= \frac{1}{\sqrt{2}}|\psi\rangle\otimes|0\rangle+\frac{1}{\sqrt{2}}\lambda|\psi\rangle\otimes|1\rangle\\
&= \frac{1}{\sqrt{2}}|\psi\rangle\otimes|0\rangle+\frac{1}{\sqrt{2}}e^{i\phi}|\psi\rangle\otimes|1\rangle\\
&= |\psi\rangle(\frac{1}{\sqrt{2}}|0\rangle + \frac{1}{\sqrt{2}}e^{i\phi}|1\rangle)
\end{aligned}
$$
我們可以看到 phase 被踢到 first qubit 的 $|1\rangle$ 上

## Circuit of Quantum Teleportation
- Q1 : 是輸入要傳輸的 $\psi$ signal，由 Alice 傳給 Bob
- Q2 : Alice 的專用 channel
- Q3 : Bob 的專用 channel
### Fist state
$$
\begin{aligned}
|\psi_0\rangle &= |\psi\rangle \otimes(\frac{1}{\sqrt{2}}|00\rangle + |11\rangle)\\
&= (\alpha|0\rangle+\beta|1\rangle)\otimes(\frac{1}{\sqrt{2}}|00\rangle + \frac{1}{\sqrt{2}}|11\rangle) \\
& = \frac{\alpha}{\sqrt{2}}|000\rangle + \frac{\alpha}{\sqrt{2}}|011\rangle + \frac{\beta}{\sqrt{2}}|100\rangle+\frac{\beta}{\sqrt{2}}|111\rangle
\end{aligned}
$$

### Entanglement

<div style="text-align: center;">
<img src="https://hackmd.io/_uploads/B1oixuxzlg.png" alt="image">
</div>

$$
\begin{aligned}
|\psi\rangle &= \alpha|0\rangle + \beta|1\rangle\\
\beta_{00} &= \frac{1}{\sqrt{2}}|00\rangle + \frac{1}{\sqrt{2}}|11\rangle
\end{aligned}
$$
我們想證明 $|\psi\rangle$ 和 $\beta_{00}$ entanglement 在一起，像是以下形式

$$
|\psi\rangle\otimes\beta_{00} =
\begin{bmatrix}
\frac{\alpha}{\sqrt{2}}\\
0\\
0\\
\frac{\alpha}{\sqrt{2}}\\
\frac{\beta}{\sqrt{2}}\\
0\\
0\\
\frac{\beta}{\sqrt{2}}
\end{bmatrix} =
\begin{bmatrix}
a\\
b\\
c\\
d\\
\end{bmatrix} \otimes
\begin{bmatrix}
e\\
f\\
\end{bmatrix} =
\begin{bmatrix}
ae\\
af\\
be\\
bf\\
ce\\
cf\\
de\\
df
\end{bmatrix}
$$
我們可以得到一下的關係式
$$
\begin{cases}
ae = \frac{\alpha}{\sqrt{2}}\\
af = 0\\
be = 0\\
bf = \frac{\alpha}{\sqrt{2}}\\
ce = \frac{\beta}{\sqrt{2}}\\
cf = 0\\
de = 0\\
df = \frac{\beta}{\sqrt{2}}
\end{cases}
\Rightarrow
\begin{cases}
a\neq0, e\neq0\\
a = 0\space or \space f = 0\\
b = 0\space or \space e =0\\
b \neq0, f\neq 0\\
c \neq0, e\neq0\\
c =0 \space or \space f = 0\\
d =0 \space or \space e=0\\
d \neq 0, f\neq 0
\end{cases}
$$
因為 $(a\neq0, e\neq) \wedge (b\neq0, f\neq 0)$ 但跟 ($a = 0\space or \space f = 0$) 矛盾，所以$|\psi\rangle$ 和 $\beta_{00}$ entanglement。
### general form
- 懶得寫

### Second state
$$
|\psi_1\rangle = \begin{cases}
CNOT(Q_1,Q_2) = \alpha|0\rangle\otimes(\frac{1}{\sqrt{2}}|00\rangle + \frac{1}{\sqrt{2}}|11\rangle) = \alpha|0\rangle\otimes(\frac{1}{\sqrt{2}}|00\rangle+\frac{1}{\sqrt{2}}|11\rangle)\\
CNOT(Q_1,Q_2) = \beta|1\rangle\otimes X(\frac{1}{\sqrt{2}}|00\rangle + \frac{1}{\sqrt{2}}|11\rangle) = \beta|1\rangle\otimes(\frac{1}{\sqrt{2}}|10\rangle+\frac{1}{\sqrt{2}}|01\rangle)
\end{cases}
$$
$$
\begin{aligned}
|\psi_1\rangle = (\frac{\alpha}{\sqrt{2}}|000\rangle+\frac{\alpha}{\sqrt{2}}|011\rangle) + (\frac{\beta}{\sqrt{2}}|110\rangle+\frac{\beta}{\sqrt{2}}|101\rangle)
\end{aligned}
$$

### Third state

$$
\begin{aligned}
|\psi_2\rangle &= \alpha H|0\rangle\otimes(\frac{1}{\sqrt{2}}|00\rangle + \frac{1}{\sqrt{2}}|11\rangle) + \beta H|1\rangle\otimes(\frac{1}{\sqrt{2}}|10\rangle + \frac{1}{\sqrt{2}}|01\rangle)\\
&= \alpha|+\rangle\otimes(\frac{1}{\sqrt{2}}|00\rangle + \frac{1}{\sqrt{2}}|11\rangle) + \beta |-\rangle\otimes(\frac{1}{\sqrt{2}}|10\rangle + \frac{1}{\sqrt{2}}|01\rangle)\\
&= \alpha(\frac{1}{\sqrt{2}}|000\rangle + \frac{1}{\sqrt{2}}|011\rangle + \frac{1}{\sqrt{2}}|100\rangle + \frac{1}{\sqrt{2}}|111\rangle) + \beta(\frac{1}{\sqrt{2}}|010\rangle + \frac{1}{\sqrt{2}}|001\rangle + \frac{1}{\sqrt{2}}|110\rangle + \frac{1}{\sqrt{2}}|101\rangle)\\
&=\frac{1}{2}|00\rangle(\alpha|0\rangle+\beta|1\rangle) + \frac{1}{2}|01\rangle(\alpha|1\rangle+\beta|0\rangle) + \frac{1}{2}|10\rangle(\alpha|0\rangle-\beta|1\rangle) + \frac{1}{2}|11\rangle(\alpha|1\rangle-\beta|0\rangle)
\end{aligned}
$$

### Measurement
- 當我們觀測到哪一個 state 時，因為特定的 state 會跟特定的 state entanglement 在一起
- 我們就做對應的修正，就可以得到 Alice 傳輸的 $\psi$

## phase kick back 在哪?
:face_with_diagonal_mouth: 當初看完 Quantum Teleportation 時，想說靠北，phase kickback 是用在哪裡
- 首先想說的是當 channel Q2 和 Q3 的 qubit 經過 H 和 CNOT Gate，變成 bell type $I$。
- bell type $I$ 是 CNOT gate 的 eigenvector
$$
\begin{aligned}
(X \otimes X)\frac{1}{\sqrt{2}}(|00\rangle + |11\rangle)
&= \frac{1}{\sqrt{2}}(X|0\rangle\otimes X|0\rangle + X|1\rangle\otimes X|1\rangle)\\
&= \frac{1}{\sqrt{2}}(|00\rangle + |11\rangle)\\
&= \lambda\frac{1}{\sqrt{2}}(|00\rangle + |11\rangle)
\end{aligned}
$$
- 可以對應到一開始介紹 Phase kick back 的電路。
- 從框框可以看到我們把channel Q1 phasae kickback 到 Q3，只是需要修正。
$$
\begin{aligned}
\frac{1}{2}|00\rangle\boxed{(\alpha|0\rangle+\beta|1\rangle)} + \frac{1}{2}|01\rangle\boxed{(\alpha|1\rangle+\beta|0\rangle)} + \frac{1}{2}|10\rangle\boxed{(\alpha|0\rangle-\beta|1\rangle)} + \frac{1}{2}|11\rangle\boxed{(\alpha|1\rangle-\beta|0\rangle)}
\end{aligned}
$$
### 求 $X\otimes X$ 的 eigenvalue
$$
X\otimes X=
\begin{bmatrix}
0&0&0&1 \\
0&0&1&0 \\
0&1&0&0 \\
1&0&0&0 \\
\end{bmatrix}
$$
- $det(X-\lambda I) = 0$
$$
A-\lambda I =
\begin{vmatrix}
-\lambda&0&0&1\\
0&-\lambda&1&0\\
0&1&-\lambda&0\\
1&0&0&-\lambda\\
\end{vmatrix}
$$
- X 的 $\lambda = 1(重根), -1(重根)$
### Bell type 是 $X\otimes X$ 的 eigenvalue
- 懶得求
## Referece
- Prof. Pao-Ta Yu. 2025. 量子電腦程式設計概論. 中正大學, Chiayi, Taiwan。