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Homework III
exercise 3.1
Q：U U U U ρ U † U\rho U^\dagger U ρ U †
A：
证明厄密性
( U ρ U † ) † = ( U † ) † ( U ρ ) † = U ρ † U † = U ρ U † (1) (U\rho U^\dagger) ^\dagger=(U^\dagger)^\dagger(U\rho)^\dagger=U\rho^\dagger U^\dagger=U\rho U^\dagger
\tag{1}
( U ρ U † ) † = ( U † ) † ( U ρ ) † = U ρ † U † = U ρ U † ( 1 )
U ρ U † U\rho U^\dagger U ρ U †
考虑迹的轮换性（详情见 ep.1 的 exercise 1.1）
T r ( ( U ρ U † ) = T r ( U † U ρ ) = T r ( ρ ) = 1 (2) Tr((U\rho U^\dagger)=Tr(U^\dagger U\rho)=Tr(\rho)=1
\tag{2}
T r (( U ρ U † ) = T r ( U † U ρ ) = T r ( ρ ) = 1 ( 2 )
证明半正定性（已知 ρ \rho ρ T r ( ρ 2 ) ≤ 1 Tr(\rho^2)\le1 T r ( ρ 2 ) ≤ 1
T r ( U ρ U † U ρ U † ) = T r ( U ρ 2 U † ) = T r ( ρ 2 ) ≤ 1 (3) Tr(U\rho U^\dagger U\rho U^\dagger)=Tr(U\rho^2 U^\dagger)=Tr(\rho^2)\le1
\tag{3}
T r ( U ρ U † U ρ U † ) = T r ( U ρ 2 U † ) = T r ( ρ 2 ) ≤ 1 ( 3 )
本题通过利用密度矩阵的性质来证明 U ρ U † U\rho U^\dagger U ρ U †
exercise 3.2
Q：试计算下述量子态的 Bloch 矢量：
∣ ψ ⟩ = 1 2 ( ∣ 0 ⟩ + i ∣ 1 ⟩ ) \ket{\psi}=\frac{1}{\sqrt{2}}(\ket{0}+i\ket{1}) ∣ ψ ⟩ = 2 1 ( ∣ 0 ⟩ + i ∣ 1 ⟩ ) ∣ ψ ⟩ = 1 3 ∣ 0 ⟩ − 2 i 3 ∣ 1 ⟩ \ket{\psi}=\frac{1}{\sqrt{3}}\ket{0}-\frac{\sqrt{2}i}{\sqrt{3}}\ket{1} ∣ ψ ⟩ = 3 1 ∣ 0 ⟩ − 3 2 i ∣ 1 ⟩ ρ = [ 1 / 4 i 2 / 5 − i 2 / 5 3 / 4 ] \rho=\begin{bmatrix}1/4 &i2/5\\-i2/5 &3/4\end{bmatrix} ρ = [ 1/4 − i 2/5 i 2/5 3/4 ]
A：
求密度矩阵：
ρ = ∣ ψ ⟩ ⟨ ψ ∣ = 1 2 ( ∣ 0 ⟩ ⟨ 0 ∣ − i ∣ 0 ⟩ ⟨ 1 ∣ + i ∣ 1 ⟩ ⟨ 0 ∣ + ∣ 1 ⟩ ⟨ 1 ∣ ) = 1 2 [ 1 − i i 1 ] = 1 2 ( I + n ⃗ ⋅ σ ⃗ ) (4) \begin{align}
\rho=\ket{\psi}\bra{\psi}&=\frac12(\ket{0}\bra{0}-i\ket{0}\bra{1}+i\ket{1}\bra{0}+\ket{1}\bra{1})\\
&=\frac12\begin{bmatrix}1 &-i\\i &1\end{bmatrix}=\frac12(\mathbb{I}+\vec{n}\cdot\vec{\sigma})
\end{align}
\tag{4}
ρ = ∣ ψ ⟩ ⟨ ψ ∣ = 2 1 ( ∣ 0 ⟩ ⟨ 0 ∣ − i ∣ 0 ⟩ ⟨ 1 ∣ + i ∣ 1 ⟩ ⟨ 0 ∣ + ∣ 1 ⟩ ⟨ 1 ∣ ) = 2 1 [ 1 i − i 1 ] = 2 1 ( I + n ⋅ σ ) ( 4 )
对 xyz 三个方向投影：
n x = T r ( ρ σ x ) = T r ( 1 2 [ 1 − i i 1 ] [ 0 1 1 0 ] ) = 0 n y = T r ( ρ σ y ) = T r ( 1 2 [ 1 − i i 1 ] [ 0 − i i 0 ] ) = 1 n z = T r ( ρ σ z ) = T r ( 1 2 [ 1 − i i 1 ] [ 1 0 0 − 1 ] ) = 0 (5) \begin{align}
n_x&=Tr(\rho \sigma_x)=Tr(\frac12\begin{bmatrix}1 &-i\\i &1\end{bmatrix}\begin{bmatrix}0 &1\\1 &0\end{bmatrix})=0\\
n_y&=Tr(\rho \sigma_y)=Tr(\frac12\begin{bmatrix}1 &-i\\i &1\end{bmatrix}\begin{bmatrix}0 &-i\\i &0\end{bmatrix})=1\\
n_z&=Tr(\rho \sigma_z)=Tr(\frac12\begin{bmatrix}1 &-i\\i &1\end{bmatrix}\begin{bmatrix}1 &0\\0 &-1\end{bmatrix})=0
\end{align}
\tag{5}
n x n y n z = T r ( ρ σ x ) = T r ( 2 1 [ 1 i − i 1 ] [ 0 1 1 0 ] ) = 0 = T r ( ρ σ y ) = T r ( 2 1 [ 1 i − i 1 ] [ 0 i − i 0 ] ) = 1 = T r ( ρ σ z ) = T r ( 2 1 [ 1 i − i 1 ] [ 1 0 0 − 1 ] ) = 0 ( 5 )
因此得到 Bloch 矢量 n ⃗ = ( 0 , 1 , 0 ) \vec{n}=(0,1,0) n = ( 0 , 1 , 0 )
同理：
ρ = ∣ ψ ⟩ ⟨ ψ ∣ = 1 3 ∣ 0 ⟩ ⟨ 0 ∣ + 2 i 3 ∣ 0 ⟩ ⟨ 1 ∣ − 2 i 3 ∣ 1 ⟩ ⟨ 0 ∣ + 2 3 ∣ 1 ⟩ ⟨ 1 ∣ = [ 1 3 2 3 i − 2 3 i 2 3 ] = 1 2 ( I + n ⃗ ⋅ σ ⃗ ) (6) \begin{align}
\rho=\ket{\psi}\bra{\psi}&=\frac13\ket{0}\bra{0}+\frac{\sqrt2i}{3}\ket{0}\bra{1}-\frac{\sqrt2i}{3}\ket{1}\bra{0}+\frac23\ket{1}\bra{1}\\
&=\begin{bmatrix}\frac13 &\frac{\sqrt2}3i\\-\frac{\sqrt2}3i &\frac23\end{bmatrix}=\frac12(\mathbb{I}+\vec{n}\cdot\vec{\sigma})
\end{align}
\tag{6}
ρ = ∣ ψ ⟩ ⟨ ψ ∣ = 3 1 ∣ 0 ⟩ ⟨ 0 ∣ + 3 2 i ∣ 0 ⟩ ⟨ 1 ∣ − 3 2 i ∣ 1 ⟩ ⟨ 0 ∣ + 3 2 ∣ 1 ⟩ ⟨ 1 ∣ = [ 3 1 − 3 2 i 3 2 i 3 2 ] = 2 1 ( I + n ⋅ σ ) ( 6 )
对 xyz 三个方向投影：
n x = T r ( ρ σ x ) = T r ( 1 3 [ 1 2 i − 2 i 2 ] [ 0 1 1 0 ] ) = 0 n y = T r ( ρ σ y ) = T r ( 1 3 [ 1 2 i − 2 i 2 ] [ 0 − i i 0 ] ) = − 2 2 3 n z = T r ( ρ σ z ) = T r ( 1 3 [ 1 2 i − 2 i 2 ] [ 1 0 0 − 1 ] ) = − 1 3 (7) \begin{align}
n_x&=Tr(\rho \sigma_x)=Tr(\frac13\begin{bmatrix}1 &\sqrt2i\\-\sqrt2i &2\end{bmatrix}\begin{bmatrix}0 &1\\1 &0\end{bmatrix})=0\\
n_y&=Tr(\rho \sigma_y)=Tr(\frac13\begin{bmatrix}1 &\sqrt2i\\-\sqrt2i &2\end{bmatrix}\begin{bmatrix}0 &-i\\i &0\end{bmatrix})=-\frac{2\sqrt2}{3}\\
n_z&=Tr(\rho \sigma_z)=Tr(\frac13\begin{bmatrix}1 &\sqrt2i\\-\sqrt2i &2\end{bmatrix}\begin{bmatrix}1 &0\\0 &-1\end{bmatrix})=-\frac13
\end{align}
\tag{7}
n x n y n z = T r ( ρ σ x ) = T r ( 3 1 [ 1 − 2 i 2 i 2 ] [ 0 1 1 0 ] ) = 0 = T r ( ρ σ y ) = T r ( 3 1 [ 1 − 2 i 2 i 2 ] [ 0 i − i 0 ] ) = − 3 2 2 = T r ( ρ σ z ) = T r ( 3 1 [ 1 − 2 i 2 i 2 ] [ 1 0 0 − 1 ] ) = − 3 1 ( 7 )
得到 Bloch 矢量 n ⃗ = ( 0 , − 2 2 3 , − 1 3 ) \vec{n}=(0,-\frac{2\sqrt2}{3},-\frac13) n = ( 0 , − 3 2 2 , − 3 1 )
已经得到密度矩阵，投影：
n x = T r ( ρ σ x ) = T r ( [ 1 / 4 i 2 / 5 − i 2 / 5 3 / 4 ] [ 0 1 1 0 ] ) = 0 n y = T r ( ρ σ y ) = T r ( [ 1 / 4 i 2 / 5 − i 2 / 5 3 / 4 ] [ 0 − i i 0 ] ) = − 4 5 n z = T r ( ρ σ z ) = T r ( [ 1 / 4 i 2 / 5 − i 2 / 5 3 / 4 ] [ 1 0 0 − 1 ] ) = − 1 2 (8) \begin{align}
n_x&=Tr(\rho \sigma_x)=Tr(\begin{bmatrix}1/4 &i2/5\\-i2/5 &3/4\end{bmatrix}\begin{bmatrix}0 &1\\1 &0\end{bmatrix})=0\\
n_y&=Tr(\rho \sigma_y)=Tr(\begin{bmatrix}1/4 &i2/5\\-i2/5 &3/4\end{bmatrix}\begin{bmatrix}0 &-i\\i &0\end{bmatrix})=-\frac45\\
n_z&=Tr(\rho \sigma_z)=Tr(\begin{bmatrix}1/4 &i2/5\\-i2/5 &3/4\end{bmatrix}\begin{bmatrix}1 &0\\0 &-1\end{bmatrix})=-\frac12
\end{align}
\tag{8}
n x n y n z = T r ( ρ σ x ) = T r ( [ 1/4 − i 2/5 i 2/5 3/4 ] [ 0 1 1 0 ] ) = 0 = T r ( ρ σ y ) = T r ( [ 1/4 − i 2/5 i 2/5 3/4 ] [ 0 i − i 0 ] ) = − 5 4 = T r ( ρ σ z ) = T r ( [ 1/4 − i 2/5 i 2/5 3/4 ] [ 1 0 0 − 1 ] ) = − 2 1 ( 8 )
得到 Bloch 矢量 n ⃗ = ( 0 , − 4 5 , − 1 2 ) \vec{n}=(0,-\frac45,-\frac12) n = ( 0 , − 5 4 , − 2 1 )
作为一个可以将单个 qubit 态非常直观地展现出来的工具，Bloch 球包含了许多值得挖掘的信息和特性。在 exercise 2.3 中我大概介绍了 Bloch 球的引入原因和一些帮我们构建 Bloch 球和量子态之间关系的元素。在本题中我们可以看到，由 Bloch 球衍生出的 Bloch 矢量和量子态之间是可以相互转化的。类比欧拉几何空间中我们将一个普通的矢量做内积投影到各个基矢量上，我们把量子态和二维希尔伯特空间的基——Pauli 矩阵做矩阵乘法然后再求迹，由此可得到某量子态在各个方向上的投影。同时，利用 Bloch 矢量可以判断是否为纯态，如本题的1、2问得到的 n ⃗ \vec{n} n n ⃗ < 1 \vec{n}<1 n < 1
除此之外，Bloch 球还有一些有意思的性质如：Bloch球面上中心对称的两个矢量对应的态是正交的；Bloch 球面上任意两点 A A A B B B C C C A A A B B B ρ C = C B A B ρ A + ( 1 − C B A B ρ B ) \rho_C=\frac{CB}{AB}\rho_A+(1-\frac{CB}{AB}\rho_B) ρ C = A B CB ρ A + ( 1 − A B CB ρ B ) 東雲正樹大佬的文章 。
另，补充 Bloch 矢量转化为量子态的公式，对应点 ψ ( c o s φ sin θ , s i n φ sin θ , cos θ ) \psi\left(cos\varphi\sin\theta,sin\varphi\sin\theta,\cos\theta\right) ψ ( cos φ sin θ , s in φ sin θ , cos θ )
∣ ψ ⟩ = cos θ 2 ∣ 0 ⟩ + e i φ sin θ 2 ∣ 1 ⟩ (9) |\psi\rangle=\cos\frac\theta2|0\rangle+e^{i\varphi}\sin\frac\theta2|1\rangle
\tag{9}
∣ ψ ⟩ = cos 2 θ ∣0 ⟩ + e i φ sin 2 θ ∣1 ⟩ ( 9 )
exercise 3.3
Q：给定 U = e − i ω t 2 σ z U=e^{-i\frac{\omega t}2\sigma_z} U = e − i 2 ω t σ z ρ = U ρ 0 U † \rho=U\rho_0U^\dagger ρ = U ρ 0 U †
A：有：
ρ ( t ) = U ρ 0 U † = e − i ω t 2 σ z ρ 0 e i ω t 2 σ z (10) \rho(t)=U\rho_0U^\dagger=e^{-i\frac{\omega t}2\sigma_z}\rho_0 e^{i\frac{\omega t}2\sigma_z}
\tag{10}
ρ ( t ) = U ρ 0 U † = e − i 2 ω t σ z ρ 0 e i 2 ω t σ z ( 10 )
对等式两边同时求导：
d ρ ( t ) d t = d U d t ρ 0 U † + U ρ 0 d U † d t = ( − i ω 2 σ z ) U ρ 0 U † + U ρ 0 U † ( i ω 2 σ z ) = − i [ ω 2 σ z ρ ( t ) − ρ ( t ) ω 2 σ z ] = − i [ ω 2 σ z , ρ ( t ) ] (11) \begin{align}
\frac{\mathrm{d}\rho(t)}{\mathrm{d}t}&=\frac{\mathrm{d}U}{\mathrm{d}t}\rho_0U^\dagger+U\rho_0\frac{\mathrm{d}U^\dagger}{\mathrm{d}t}\\
&=(-i\frac{\omega}2\sigma_z)U\rho_0U^\dagger+U\rho_0U^\dagger(i\frac{\omega}2\sigma_z)\\
&=-i[\frac{\omega}2\sigma_z\rho(t)-\rho(t)\frac{\omega}2\sigma_z]\\
&=-i[\frac{\omega}2\sigma_z,\rho(t)]
\end{align}
\tag{11}
d t d ρ ( t ) = d t d U ρ 0 U † + U ρ 0 d t d U † = ( − i 2 ω σ z ) U ρ 0 U † + U ρ 0 U † ( i 2 ω σ z ) = − i [ 2 ω σ z ρ ( t ) − ρ ( t ) 2 ω σ z ] = − i [ 2 ω σ z , ρ ( t )] ( 11 )
我认为此题主要展示了一下含矩阵的微分方程是怎么求的。与此相比更重要的是3.3和3.4共同的主题，即量子态的演化。两题中都出现了一个算符 U = e − i ω t 2 σ z U=e^{-i\frac{\omega t}2\sigma_z} U = e − i 2 ω t σ z t t t ω \omega ω ω t \omega t ω t
d ρ ( t ) d t = − i [ ω 2 σ z , ρ ( t ) ] \frac{\mathrm{d}\rho(t)}{\mathrm{d}t}=-i[\frac{\omega}2\sigma_z,\rho(t)] d t d ρ ( t ) = − i [ 2 ω σ z , ρ ( t )] i ℏ ∂ ∂ t ∣ ψ ⟩ = H ^ ∣ ψ ⟩ i\hbar\frac{\partial}{\partial t}\ket{\psi}=\hat H\ket{\psi} i ℏ ∂ t ∂ ∣ ψ ⟩ = H ^ ∣ ψ ⟩
exercise 3.4
Q：假定初始态对应的 Bloch 向量为 ( r , θ , ϕ ) (r,\theta,\phi) ( r , θ , ϕ ) ρ = U ρ 0 U † \rho=U\rho_0U^\dagger ρ = U ρ 0 U † ( r , θ , ϕ + ω ) (r,\theta,\phi+\omega) ( r , θ , ϕ + ω ) U = e − i ω 2 σ z U=e^{-i\frac{\omega}2\sigma_z} U = e − i 2 ω σ z
A：有：
ρ 0 = 1 2 ( I + n ⃗ ⋅ σ ⃗ ) = 1 2 ( I + r sin θ cos ϕ σ x + r sin θ sin ϕ σ y + r cos θ σ z ) (12) \begin{align}
\rho_0&=\frac12(\mathbb{I}+\vec{n}\cdot\vec{\sigma})\\
&=\frac12(\mathbb{I}+r\sin\theta\cos\phi\sigma_x+r\sin\theta\sin\phi\sigma_y+r\cos\theta\sigma_z)
\end{align}
\tag{12}
ρ 0 = 2 1 ( I + n ⋅ σ ) = 2 1 ( I + r sin θ cos ϕ σ x + r sin θ sin ϕ σ y + r cos θ σ z ) ( 12 )
把 U = e − i ω 2 σ z U=e^{-i\frac{\omega}2\sigma_z} U = e − i 2 ω σ z ρ = U ρ 0 U † \rho=U\rho_0U^\dagger ρ = U ρ 0 U †
ρ = U ρ 0 U † = e − i ω t 2 σ z ρ 0 e i ω t 2 σ z = ( I cos ω 2 − i sin ω 2 σ z ) 1 2 ( I + r sin θ cos ϕ σ x + r sin θ sin ϕ σ y + r cos θ σ z ) ( I cos ω 2 + i sin ω 2 σ z ) = 1 2 [ I ( cos 2 ω 2 + sin 2 ω 2 − i r sin ω 2 cos θ cos ω 2 + i r cos ω 2 cos θ sin ω 2 ) + σ x ( r cos ω 2 sin θ cos ϕ cos ω 2 − r sin ω 2 sin θ cos ϕ sin ω 2 − r sin ω 2 sin θ sin ϕ cos ω 2 − r cos ω 2 sin θ sin ϕ cos ω 2 ) + σ y ( r cos ω 2 sin θ sin ϕ cos ω 2 − r sin ω 2 sin θ sin ϕ sin ω 2 + r sin ω 2 sin θ cos ϕ cos ω 2 + r cos ω 2 sin θ cos ϕ sin ω 2 ) + σ z ( r cos ω 2 cos θ cos ω 2 + r sin ω 2 cos θ sin ω 2 − i sin ω 2 cos ω 2 + i cos ω 2 sin ω 2 ) ] = 1 2 [ I + r sin θ cos ( ϕ + ω ) σ x + r sin θ sin ( ϕ + ω ) σ y + r cos θ σ x ] (13) \begin{align}
\rho &= U\rho_0U^\dagger=e^{-i\frac{\omega t}2\sigma_z}\rho_0e^{i\frac{\omega t}2\sigma_z}\\
&=(\mathbb{I}\cos\frac\omega2-i\sin\frac\omega2\sigma_z)\frac12(\mathbb{I}+r\sin\theta\cos\phi\sigma_x+r\sin\theta\sin\phi\sigma_y+r\cos\theta\sigma_z)(\mathbb{I}\cos\frac\omega2+i\sin\frac\omega2\sigma_z)\\
&=\frac12\big[\mathbb{I}(\cos^2\frac\omega2+\sin^2\frac\omega2-ir\sin\frac\omega2\cos\theta\cos\frac\omega2+ir\cos\frac\omega2\cos\theta\sin\frac\omega2)\\
&+\sigma_x(r\cos\frac\omega2\sin\theta\cos\phi\cos\frac\omega2-r\sin\frac\omega2\sin\theta\cos\phi\sin\frac\omega2\\
&-r\sin\frac\omega2\sin\theta\sin\phi\cos\frac\omega2-r\cos\frac\omega2\sin\theta\sin\phi\cos\frac\omega2)\\
&+\sigma_y(r\cos\frac\omega2\sin\theta\sin\phi\cos\frac\omega2-r\sin\frac\omega2\sin\theta\sin\phi\sin\frac\omega2\\
&+r\sin\frac\omega2\sin\theta\cos\phi\cos\frac\omega2+r\cos\frac\omega2\sin\theta\cos\phi\sin\frac\omega2)\\
&+\sigma_z(r\cos\frac\omega2\cos\theta\cos\frac\omega2+r\sin\frac\omega2\cos\theta\sin\frac\omega2-i\sin\frac\omega2\cos\frac\omega2+i\cos\frac\omega2\sin\frac\omega2)\big]\\
&=\frac12\big[\mathbb{I}+r\sin\theta\cos(\phi+\omega)\sigma_x+r\sin\theta\sin(\phi+\omega)\sigma_y+r\cos\theta\sigma_x\big]
\end{align}
\tag{13}
ρ = U ρ 0 U † = e − i 2 ω t σ z ρ 0 e i 2 ω t σ z = ( I cos 2 ω − i sin 2 ω σ z ) 2 1 ( I + r sin θ cos ϕ σ x + r sin θ sin ϕ σ y + r cos θ σ z ) ( I cos 2 ω + i sin 2 ω σ z ) = 2 1 [ I ( cos 2 2 ω + sin 2 2 ω − i r sin 2 ω cos θ cos 2 ω + i r cos 2 ω cos θ sin 2 ω ) + σ x ( r cos 2 ω sin θ cos ϕ cos 2 ω − r sin 2 ω sin θ cos ϕ sin 2 ω − r sin 2 ω sin θ sin ϕ cos 2 ω − r cos 2 ω sin θ sin ϕ cos 2 ω ) + σ y ( r cos 2 ω sin θ sin ϕ cos 2 ω − r sin 2 ω sin θ sin ϕ sin 2 ω + r sin 2 ω sin θ cos ϕ cos 2 ω + r cos 2 ω sin θ cos ϕ sin 2 ω ) + σ z ( r cos 2 ω cos θ cos 2 ω + r sin 2 ω cos θ sin 2 ω − i sin 2 ω cos 2 ω + i cos 2 ω sin 2 ω ) ] = 2 1 [ I + r sin θ cos ( ϕ + ω ) σ x + r sin θ sin ( ϕ + ω ) σ y + r cos θ σ x ] ( 13 )
则 ρ \rho ρ ( r , θ , ϕ + ω ) (r,\theta,\phi+\omega) ( r , θ , ϕ + ω )
exercise 3.5
Q：Suppose a 2-qubit pure state is of the form ∣ Φ ⟩ A B = 1 2 ∣ 0 ⟩ ( 1 2 ∣ 0 ⟩ + 3 2 ∣ 1 ⟩ ) + 1 2 ∣ 1 ⟩ ( 3 2 ∣ 0 ⟩ + 1 2 ∣ 1 ⟩ ) \ket{\Phi}_{AB}=\frac{1}{\sqrt{2}}\ket{0}(\frac12\ket{0}+\frac{\sqrt3}{2}\ket{1})+\frac{1}{\sqrt{2}}\ket{1}(\frac{\sqrt3}2\ket{0}+\frac12\ket{1}) ∣ Φ ⟩ A B = 2 1 ∣ 0 ⟩ ( 2 1 ∣ 0 ⟩ + 2 3 ∣ 1 ⟩ ) + 2 1 ∣ 1 ⟩ ( 2 3 ∣ 0 ⟩ + 2 1 ∣ 1 ⟩ ) ρ A \rho_A ρ A ρ B \rho_B ρ B
A：有 ∣ Φ ⟩ A B = 1 2 ∣ 0 ⟩ ( 1 2 ∣ 0 ⟩ + 3 2 ∣ 1 ⟩ ) + 1 2 ∣ 1 ⟩ ( 3 2 ∣ 0 ⟩ + 1 2 ∣ 1 ⟩ ) \ket{\Phi}_{AB}=\frac{1}{\sqrt{2}}\ket{0}(\frac12\ket{0}+\frac{\sqrt3}{2}\ket{1})+\frac{1}{\sqrt{2}}\ket{1}(\frac{\sqrt3}2\ket{0}+\frac12\ket{1}) ∣ Φ ⟩ A B = 2 1 ∣ 0 ⟩ ( 2 1 ∣ 0 ⟩ + 2 3 ∣ 1 ⟩ ) + 2 1 ∣ 1 ⟩ ( 2 3 ∣ 0 ⟩ + 2 1 ∣ 1 ⟩ )
ρ A = T r B ( ∣ Φ ⟩ A B ⟨ Φ ∣ A B ) = T r B [ 1 2 ∣ 0 ⟩ A ( 1 2 ∣ 0 ⟩ B + 3 2 ∣ 1 ⟩ B ) + 1 2 ∣ 1 ⟩ A ( 3 2 ∣ 0 ⟩ B + 1 2 ∣ 1 ⟩ B ) ] [ 1 2 ⟨ 0 ∣ A ( 1 2 ⟨ 0 ∣ B + 3 2 ⟨ 1 ∣ B ) + 1 2 ⟨ 1 ∣ A ( 3 2 ⟨ 0 ∣ B + 1 2 ⟨ 1 ∣ B ) ] = 1 8 ∣ 0 ⟩ A ⟨ 0 ∣ A T r ( ∣ 0 ⟩ B ⟨ 0 ∣ B ) + 3 8 ∣ 0 ⟩ A ⟨ 0 ∣ A T r ( ∣ 0 ⟩ B ⟨ 1 ∣ B ) + 3 8 ∣ 0 ⟩ A ⟨ 1 ∣ A T r ( ∣ 0 ⟩ B ⟨ 0 ∣ B ) + ⋯ + 1 8 ∣ 1 ⟩ A ⟨ 1 ∣ A T r ( ∣ 1 ⟩ B ⟨ 1 ∣ B ) = 1 8 ∣ 0 ⟩ A ⟨ 0 ∣ A + 3 8 ∣ 0 ⟩ A ⟨ 1 ∣ A + 3 8 ∣ 0 ⟩ A ⟨ 0 ∣ A + 3 8 ∣ 0 ⟩ A ⟨ 1 ∣ A + 3 8 ∣ 1 ⟩ A ⟨ 0 ∣ A + 3 8 ∣ 1 ⟩ A ⟨ 1 ∣ A + 3 8 ∣ 1 ⟩ A ⟨ 0 ∣ A + 1 8 ∣ 1 ⟩ A ⟨ 1 ∣ A = 1 2 ∣ 0 ⟩ A ⟨ 0 ∣ A + 3 4 ∣ 0 ⟩ A ⟨ 1 ∣ A + 3 4 ∣ 1 ⟩ A ⟨ 0 ∣ A + 1 2 ∣ 1 ⟩ A ⟨ 1 ∣ A (14) \begin{align}
\rho_A&=Tr_B(\ket{\Phi}_{AB}\bra{\Phi}_{AB})\\
&=Tr_B[\frac1{\sqrt2}\ket{0}_A(\frac12\ket{0}_B+\frac{\sqrt3}2\ket{1}_B)+\frac1{\sqrt2}\ket{1}_A(\frac{\sqrt3}2\ket{0}_B+\frac12\ket{1}_B)]\\
&[\frac1{\sqrt2}\bra{0}_A(\frac12\bra{0}_B+\frac{\sqrt3}2\bra{1}_B)+\frac1{\sqrt2}\bra{1}_A(\frac{\sqrt3}2\bra{0}_B+\frac12\bra{1}_B)]\\
&=\frac18\ket{0}_A\bra{0}_ATr(\ket{0}_B\bra{0}_B)+\frac{\sqrt3}8\ket{0}_A\bra{0}_ATr(\ket{0}_B\bra{1}_B)\\
&+\frac{\sqrt3}8\ket{0}_A\bra{1}_ATr(\ket{0}_B\bra{0}_B)+\cdots+\frac18\ket{1}_A\bra{1}_ATr(\ket{1}_B\bra{1}_B)\\
&=\frac18\ket{0}_A\bra{0}_A+\frac{\sqrt3}8\ket{0}_A\bra{1}_A+\frac38\ket{0}_A\bra{0}_A+\frac{\sqrt3}8\ket{0}_A\bra{1}_A+\frac{\sqrt3}8\ket{1}_A\bra{0}_A\\
&+\frac38\ket{1}_A\bra{1}_A+\frac{\sqrt3}8\ket{1}_A\bra{0}_A+\frac18\ket{1}_A\bra{1}_A\\
&=\frac12\ket{0}_A\bra{0}_A+\frac{\sqrt3}4\ket{0}_A\bra{1}_A+\frac{\sqrt3}4\ket{1}_A\bra{0}_A+\frac12\ket{1}_A\bra{1}_A
\end{align}
\tag{14}
ρ A = T r B ( ∣ Φ ⟩ A B ⟨ Φ ∣ A B ) = T r B [ 2 1 ∣ 0 ⟩ A ( 2 1 ∣ 0 ⟩ B + 2 3 ∣ 1 ⟩ B ) + 2 1 ∣ 1 ⟩ A ( 2 3 ∣ 0 ⟩ B + 2 1 ∣ 1 ⟩ B )] [ 2 1 ⟨ 0 ∣ A ( 2 1 ⟨ 0 ∣ B + 2 3 ⟨ 1 ∣ B ) + 2 1 ⟨ 1 ∣ A ( 2 3 ⟨ 0 ∣ B + 2 1 ⟨ 1 ∣ B )] = 8 1 ∣ 0 ⟩ A ⟨ 0 ∣ A T r ( ∣ 0 ⟩ B ⟨ 0 ∣ B ) + 8 3 ∣ 0 ⟩ A ⟨ 0 ∣ A T r ( ∣ 0 ⟩ B ⟨ 1 ∣ B ) + 8 3 ∣ 0 ⟩ A ⟨ 1 ∣ A T r ( ∣ 0 ⟩ B ⟨ 0 ∣ B ) + ⋯ + 8 1 ∣ 1 ⟩ A ⟨ 1 ∣ A T r ( ∣ 1 ⟩ B ⟨ 1 ∣ B ) = 8 1 ∣ 0 ⟩ A ⟨ 0 ∣ A + 8 3 ∣ 0 ⟩ A ⟨ 1 ∣ A + 8 3 ∣ 0 ⟩ A ⟨ 0 ∣ A + 8 3 ∣ 0 ⟩ A ⟨ 1 ∣ A + 8 3 ∣ 1 ⟩ A ⟨ 0 ∣ A + 8 3 ∣ 1 ⟩ A ⟨ 1 ∣ A + 8 3 ∣ 1 ⟩ A ⟨ 0 ∣ A + 8 1 ∣ 1 ⟩ A ⟨ 1 ∣ A = 2 1 ∣ 0 ⟩ A ⟨ 0 ∣ A + 4 3 ∣ 0 ⟩ A ⟨ 1 ∣ A + 4 3 ∣ 1 ⟩ A ⟨ 0 ∣ A + 2 1 ∣ 1 ⟩ A ⟨ 1 ∣ A ( 14 )
同理：
ρ B = T r A ( ∣ Φ ⟩ A B ⟨ Φ ∣ A B ) = 1 8 ∣ 0 ⟩ B ⟨ 0 ∣ B + 3 8 ∣ 0 ⟩ B ⟨ 1 ∣ B + 3 8 ∣ 0 ⟩ B ⟨ 0 ∣ B + 3 8 ∣ 0 ⟩ B ⟨ 1 ∣ B + 3 8 ∣ 1 ⟩ B ⟨ 0 ∣ B + 3 8 ∣ 1 ⟩ B ⟨ 1 ∣ B + 3 8 ∣ 1 ⟩ B ⟨ 0 ∣ B + 1 8 ∣ 1 ⟩ B ⟨ 1 ∣ B = 1 2 ∣ 0 ⟩ B ⟨ 0 ∣ B + 3 4 ∣ 0 ⟩ B ⟨ 1 ∣ B + 3 4 ∣ 1 ⟩ B ⟨ 0 ∣ B + 1 2 ∣ 1 ⟩ B ⟨ 1 ∣ B (15) \begin{align}
\rho_B&=Tr_A(\ket{\Phi}_{AB}\bra{\Phi}_{AB})\\
&=\frac18\ket{0}_B\bra{0}_B+\frac{\sqrt3}8\ket{0}_B\bra{1}_B+\frac38\ket{0}_B\bra{0}_B+\frac{\sqrt3}8\ket{0}_B\bra{1}_B+\frac{\sqrt3}8\ket{1}_B\bra{0}_B\\
&+\frac38\ket{1}_B\bra{1}_B+\frac{\sqrt3}8\ket{1}_B\bra{0}_B+\frac18\ket{1}_B\bra{1}_B\\
&=\frac12\ket{0}_B\bra{0}_B+\frac{\sqrt3}4\ket{0}_B\bra{1}_B+\frac{\sqrt3}4\ket{1}_B\bra{0}_B+\frac12\ket{1}_B\bra{1}_B
\end{align}
\tag{15}
ρ B = T r A ( ∣ Φ ⟩ A B ⟨ Φ ∣ A B ) = 8 1 ∣ 0 ⟩ B ⟨ 0 ∣ B + 8 3 ∣ 0 ⟩ B ⟨ 1 ∣ B + 8 3 ∣ 0 ⟩ B ⟨ 0 ∣ B + 8 3 ∣ 0 ⟩ B ⟨ 1 ∣ B + 8 3 ∣ 1 ⟩ B ⟨ 0 ∣ B + 8 3 ∣ 1 ⟩ B ⟨ 1 ∣ B + 8 3 ∣ 1 ⟩ B ⟨ 0 ∣ B + 8 1 ∣ 1 ⟩ B ⟨ 1 ∣ B = 2 1 ∣ 0 ⟩ B ⟨ 0 ∣ B + 4 3 ∣ 0 ⟩ B ⟨ 1 ∣ B + 4 3 ∣ 1 ⟩ B ⟨ 0 ∣ B + 2 1 ∣ 1 ⟩ B ⟨ 1 ∣ B ( 15 )
这道题用到了一个操作叫 Partial Trace，正如它的名字一样，该操作只对大于一个 qubit 的量子态的一部分求迹，举个例子：
有量子态 ∣ Φ + ⟩ = 1 2 ( ∣ 0 A 0 B ⟩ + ∣ 1 A 1 B ⟩ ) |\Phi^+\rangle=\frac{1}{\sqrt{2}}(|0_A0_B\rangle+|1_A1_B\rangle) ∣ Φ + ⟩ = 2 1 ( ∣ 0 A 0 B ⟩ + ∣ 1 A 1 B ⟩)
T r B ( ∣ Φ + ⟩ ⟨ Φ + ∣ ) = T r B [ 1 2 ( ∣ 0 A 0 B ⟩ + ∣ 1 A 1 B ⟩ ) ( ⟨ 0 A 0 B ∣ + ⟨ 1 A 1 B ∣ ) ] = 1 2 ( ∣ 0 A ⟩ ⟨ 0 A ∣ T r ( ∣ 0 B ⟩ ⟨ 0 B ∣ ) + ∣ 1 A ⟩ ⟨ 0 A ∣ T r ( ∣ 1 B ⟩ ⟨ 0 B ∣ ) + ∣ 0 A ⟩ ⟨ 1 A ∣ T r ( ∣ 0 B ⟩ ⟨ 1 B ∣ ) + ∣ 1 A ⟩ ⟨ 1 A ∣ T r ( ∣ 1 B ⟩ ⟨ 1 B ∣ ) = 1 2 ( ∣ 0 A ⟩ ⟨ 0 A ∣ + ∣ 1 A ⟩ ⟨ 1 A ∣ ) (16) \begin{aligned}
&Tr_{B}(\left|\Phi^{+}\right\rangle\left\langle\Phi^{+}\right|)=Tr_{B}\left[\frac{1}{2}(\left|0_{A}0_{B}\right\rangle+\left|1_{A}1_{B}\right\rangle)(\left\langle0_{A}0_{B}\right|+\left\langle1_{A}1_{B}\right|)\right] \\
&=\frac{1}{2}(|0_A\rangle\langle0_A| Tr(|0_B\rangle\langle0_B|)+|1_A\rangle\langle0_A| Tr(|1_B\rangle\langle0_B|)+|0_A\rangle\langle1_A| Tr(|0_B\rangle\langle1_B|)\\
&+|1_A\rangle\langle1_A| Tr(|1_B\rangle\langle1_B|) \\
&=\frac{1}{2}(|0_{A}\rangle\langle0_{A}|+|1_{A}\rangle\langle1_{A}|)
\end{aligned}
\tag{16}
T r B ( Φ + ⟩ ⟨ Φ + ) = T r B [ 2 1 ( ∣ 0 A 0 B ⟩ + ∣ 1 A 1 B ⟩ ) ( ⟨ 0 A 0 B ∣ + ⟨ 1 A 1 B ∣ ) ] = 2 1 ( ∣ 0 A ⟩ ⟨ 0 A ∣ T r ( ∣ 0 B ⟩ ⟨ 0 B ∣ ) + ∣ 1 A ⟩ ⟨ 0 A ∣ T r ( ∣ 1 B ⟩ ⟨ 0 B ∣ ) + ∣ 0 A ⟩ ⟨ 1 A ∣ T r ( ∣ 0 B ⟩ ⟨ 1 B ∣ ) + ∣ 1 A ⟩ ⟨ 1 A ∣ T r ( ∣ 1 B ⟩ ⟨ 1 B ∣ ) = 2 1 ( ∣ 0 A ⟩ ⟨ 0 A ∣ + ∣ 1 A ⟩ ⟨ 1 A ∣ ) ( 16 )
Homework IV
exercise 4.1
Q：试得到下述线路的酉矩阵：
A：先看左图，列出真值表：
设该线路的酉矩阵为 U 1 U_1 U 1
U 1 [ α 1 α 2 α 1 β 2 β 1 α 2 β 1 β 2 ] = 1 2 [ α 1 ( α 2 + β 2 ) α 1 ( α 2 − β 2 ) β 1 ( α 2 + β 2 ) β 1 ( α 2 − β 2 ) ] (17) U_1\begin{bmatrix}
\alpha_1\alpha_2\\
\alpha_1\beta_2\\
\beta_1\alpha_2\\
\beta_1\beta_2
\end{bmatrix}=\frac1{\sqrt2}\begin{bmatrix}
\alpha_1(\alpha_2+\beta_2)\\
\alpha_1(\alpha_2-\beta_2)\\
\beta_1(\alpha_2+\beta_2)\\
\beta_1(\alpha_2-\beta_2)
\end{bmatrix}
\tag{17}
U 1 α 1 α 2 α 1 β 2 β 1 α 2 β 1 β 2 = 2 1 α 1 ( α 2 + β 2 ) α 1 ( α 2 − β 2 ) β 1 ( α 2 + β 2 ) β 1 ( α 2 − β 2 ) ( 17 )
由上式可以反推出 U 1 U_1 U 1
U 1 = 1 2 [ 1 1 0 0 1 − 1 0 0 0 0 1 1 0 0 1 − 1 ] = I ⊗ H (18) U_1=\frac1{\sqrt2}\begin{bmatrix}
1 & 1 & 0& 0\\
1 & -1 & 0& 0\\
0& 0 & 1 & 1\\
0& 0 & 1 &-1
\end{bmatrix}=\mathbb{I}\otimes H
\tag{18}
U 1 = 2 1 1 1 0 0 1 − 1 0 0 0 0 1 1 0 0 1 − 1 = I ⊗ H ( 18 )
同理，我们也可以得到右图的酉矩阵 U 2 U_2 U 2
U 2 = 1 2 [ 1 0 1 0 0 1 0 1 1 0 − 1 0 0 1 0 − 1 ] = H ⊗ I (19) U_2=\frac1{\sqrt2}\begin{bmatrix}
1 & 0 & 1 & 0\\
0 & 1 & 0 & 1\\
1 & 0 & -1 & 0\\
0 & 1 & 0 & -1
\end{bmatrix}=H\otimes\mathbb{I}
\tag{19}
U 2 = 2 1 1 0 1 0 0 1 0 1 1 0 − 1 0 0 1 0 − 1 = H ⊗ I ( 19 )
此题看似简单，实则是在向我们介绍量子门的工作原理。本题中出现的仅为两个量子比特的门，如果是有 n n n 2 n 2^n 2 n
那我们要如何处理这类问题呢？答案是找规律！仔细观察 U 1 U 2 U_1U_2 U 1 U 2 ∣ A B ⟩ = ∣ A ⟩ ⊗ ∣ B ⟩ \ket{AB}=\ket{A}\otimes\ket{B} ∣ A B ⟩ = ∣ A ⟩ ⊗ ∣ B ⟩ U 1 U_1 U 1 I \mathbb{I} I H = 1 2 [ 1 1 1 − 1 ] H=\frac1{\sqrt2}\begin{bmatrix}1 & 1\\1 & -1\end{bmatrix} H = 2 1 [ 1 1 1 − 1 ] U 2 U_2 U 2 I ⊗ H \mathbb{I}\otimes H I ⊗ H H ⊗ I H\otimes\mathbb{I} H ⊗ I
exercise 4.2
Q：从交换门的定义由来得到其矩阵表达式：
A：已知两种 C-NOT 门的矩阵：
写成解析式形式为：U 1 = ∣ 0 ⟩ ⟨ 0 ∣ ⊗ I + ∣ 1 ⟩ ⟨ 1 ∣ ⊗ X U_1=\ket{0}\bra{0}\otimes\mathbb{I}+\ket{1}\bra{1}\otimes X U 1 = ∣ 0 ⟩ ⟨ 0 ∣ ⊗ I + ∣ 1 ⟩ ⟨ 1 ∣ ⊗ X U 2 = I ⊗ ∣ 0 ⟩ ⟨ 0 ∣ + X ⊗ ∣ 1 ⟩ ⟨ 1 ∣ U_2=\mathbb{I}\otimes\ket{0}\bra{0}+X\otimes\ket{1}\bra{1} U 2 = I ⊗ ∣ 0 ⟩ ⟨ 0 ∣ + X ⊗ ∣ 1 ⟩ ⟨ 1 ∣
U C N O T = U 1 U 2 U 1 = [ 1 0 0 0 0 0 1 0 0 1 0 0 0 0 0 1 ] (20) U_{CNOT}=U_1U_2U_1=
\begin{bmatrix}
1 & 0 & 0 & 0\\
0 & 0 & 1 & 0\\
0 & 1 & 0 & 0\\
0 & 0 & 0 & 1
\end{bmatrix}
\tag{20}
U CNOT = U 1 U 2 U 1 = 1 0 0 0 0 0 1 0 0 1 0 0 0 0 0 1 ( 20 )
相比于上一题，本题提供的信息更多了，首先是多个门连续作用后表达式的计算顺序。在矩阵乘法的规则中，没多进行一次映射，就要将映射的矩阵乘到原矩阵的左边去，因此（20）的具体顺序为：U C N O T = U 1 3 U 2 2 U 1 1 U_{CNOT}=U_1^3U_2^2U_1^1 U CNOT = U 1 3 U 2 2 U 1 1
接着我们就要来找规律了，由于 C-NOT 门要考虑到一个比特对另一个比特的影响，所以它的表达式 U 1 = ∣ 0 ⟩ ⟨ 0 ∣ ⊗ I + ∣ 1 ⟩ ⟨ 1 ∣ ⊗ X U_1=\ket{0}\bra{0}\otimes\mathbb{I}+\ket{1}\bra{1}\otimes X U 1 = ∣ 0 ⟩ ⟨ 0 ∣ ⊗ I + ∣ 1 ⟩ ⟨ 1 ∣ ⊗ X ∣ 1 ⟩ \ket{1} ∣ 1 ⟩ ∣ 0 ⟩ ⟨ 0 ∣ \ket{0}\bra{0} ∣ 0 ⟩ ⟨ 0 ∣ ∣ 1 ⟩ \ket{1} ∣ 1 ⟩ ∣ 0 ⟩ \ket{0} ∣ 0 ⟩ ∣ 0 ⟩ \ket{0} ∣ 0 ⟩ ∣ 1 ⟩ \ket{1} ∣ 1 ⟩ ∣ 1 ⟩ \ket{1} ∣ 1 ⟩ ∣ 1 ⟩ ⟨ 1 ∣ \ket{1}\bra{1} ∣ 1 ⟩ ⟨ 1 ∣ σ x \sigma_x σ x
结合上面的结论再对比 U 1 U 2 U_1U_2 U 1 U 2
延申一点，刚刚我们讨论了 U 1 = ∣ 0 ⟩ ⟨ 0 ∣ ⊗ I + ∣ 1 ⟩ ⟨ 1 ∣ ⊗ X U_1=\ket{0}\bra{0}\otimes\mathbb{I}+\ket{1}\bra{1}\otimes X U 1 = ∣ 0 ⟩ ⟨ 0 ∣ ⊗ I + ∣ 1 ⟩ ⟨ 1 ∣ ⊗ X U 2 = I ⊗ ∣ 0 ⟩ ⟨ 0 ∣ + X ⊗ ∣ 1 ⟩ ⟨ 1 ∣ U_2=\mathbb{I}\otimes\ket{0}\bra{0}+X\otimes\ket{1}\bra{1} U 2 = I ⊗ ∣ 0 ⟩ ⟨ 0 ∣ + X ⊗ ∣ 1 ⟩ ⟨ 1 ∣ ∣ 0 ⟩ ⟨ 0 ∣ \ket{0}\bra{0} ∣ 0 ⟩ ⟨ 0 ∣ ∣ 1 ⟩ ⟨ 1 ∣ \ket{1}\bra{1} ∣ 1 ⟩ ⟨ 1 ∣ U 3 = ∣ 1 ⟩ ⟨ 1 ∣ ⊗ I + ∣ 0 ⟩ ⟨ 0 ∣ ⊗ X U_3=\ket{1}\bra{1}\otimes\mathbb{I}+\ket{0}\bra{0}\otimes X U 3 = ∣ 1 ⟩ ⟨ 1 ∣ ⊗ I + ∣ 0 ⟩ ⟨ 0 ∣ ⊗ X
结果在这里！
如图，这就是 U 3 = ∣ 1 ⟩ ⟨ 1 ∣ ⊗ I + ∣ 0 ⟩ ⟨ 0 ∣ ⊗ X U_3=\ket{1}\bra{1}\otimes\mathbb{I}+\ket{0}\bra{0}\otimes X U 3 = ∣ 1 ⟩ ⟨ 1 ∣ ⊗ I + ∣ 0 ⟩ ⟨ 0 ∣ ⊗ X ∣ 0 ⟩ \ket{0} ∣ 0 ⟩ ∣ 1 ⟩ \ket{1} ∣ 1 ⟩
exercise 4.3
Q：试说明下述量子线路的等价性：
A：先求下面这个门的矩阵
按照我们发现的规律，易得表达式：
U H = H ⊗ H = 1 2 [ 1 1 1 1 1 − 1 1 − 1 1 1 − 1 − 1 1 − 1 − 1 1 ] (21) U_H=H\otimes H=\frac12
\begin{bmatrix}
1 & 1 & 1 & 1\\
1 & -1 & 1 & -1\\
1 & 1 & -1 & -1\\
1 & -1 & -1 & 1
\end{bmatrix}
\tag{21}
U H = H ⊗ H = 2 1 1 1 1 1 1 − 1 1 − 1 1 1 − 1 − 1 1 − 1 − 1 1 ( 21 )
已知：
则：
U = U H U C N O T U H = 1 4 [ 1 1 1 1 1 − 1 1 − 1 1 1 − 1 − 1 1 − 1 − 1 1 ] [ 1 0 0 0 0 1 0 0 0 0 0 1 0 0 1 0 ] [ 1 1 1 1 1 − 1 1 − 1 1 1 − 1 − 1 1 − 1 − 1 1 ] = [ 1 0 0 0 0 0 0 1 0 0 1 0 0 1 0 0 ] (22) \begin{align}
U&=U_HU_{CNOT}U_H=\frac14
\begin{bmatrix}
1 & 1 & 1 & 1\\
1 & -1 & 1 & -1\\
1 & 1 & -1 & -1\\
1 & -1 & -1 & 1
\end{bmatrix}
\begin{bmatrix}
1 & 0 & 0 & 0\\
0 & 1 & 0 & 0\\
0 & 0 & 0 & 1\\
0 & 0 & 1 & 0
\end{bmatrix}
\begin{bmatrix}
1 & 1 & 1 & 1\\
1 & -1 & 1 & -1\\
1 & 1 & -1 & -1\\
1 & -1 & -1 & 1
\end{bmatrix}\\
&=\begin{bmatrix}
1 & 0 & 0 & 0\\
0 & 0 & 0 & 1\\
0 & 0 & 1 & 0\\
0 & 1 & 0 & 0
\end{bmatrix}
\end{align}
\tag{22}
U = U H U CNOT U H = 4 1 1 1 1 1 1 − 1 1 − 1 1 1 − 1 − 1 1 − 1 − 1 1 1 0 0 0 0 1 0 0 0 0 0 1 0 0 1 0 1 1 1 1 1 − 1 1 − 1 1 1 − 1 − 1 1 − 1 − 1 1 = 1 0 0 0 0 0 0 1 0 0 1 0 0 1 0 0 ( 22 )
得证。
只要搞明白了前两题，这题就没啥技术含量了，可能就是想让你惊讶一下这个等价吧。
exercise 4.4
Q：试得到下述量子线路的输出量子态：
A：设 T = exp ( i π σ y 4 ) = cos π 4 I − i sin π 4 σ y T=\exp(\frac{i\pi\sigma_y}{4})=\cos\frac\pi4\mathbb{I}-i\sin\frac\pi4\sigma_y T = exp ( 4 iπ σ y ) = cos 4 π I − i sin 4 π σ y U U U
U = U C N O T ( T ⊗ I ) = [ cos π 4 0 − sin π 4 0 0 cos π 4 0 − sin π 4 0 sin π 4 0 cos π 4 sin π 4 0 cos π 4 0 ] (23) \begin{align}
U&=U_{CNOT}(T\otimes\mathbb{I})\\
&=\begin{bmatrix}
\cos\frac\pi4 & 0 & -\sin\frac\pi4 & 0\\
0 & \cos\frac\pi4 & 0 & -\sin\frac\pi4\\
0 & \sin\frac\pi4 & 0 & \cos\frac\pi4\\
\sin\frac\pi4 & 0 & \cos\frac\pi4 & 0
\end{bmatrix}
\end{align}
\tag{23}
U = U CNOT ( T ⊗ I ) = cos 4 π 0 0 sin 4 π 0 cos 4 π sin 4 π 0 − sin 4 π 0 0 cos 4 π 0 − sin 4 π cos 4 π 0 ( 23 )
由于输入为 ∣ 00 ⟩ \ket{00} ∣ 00 ⟩
U ∣ 00 ⟩ = U [ 1 0 0 0 ] = [ cos π 4 0 0 sin π 4 ] (24) U\ket{00}=U\begin{bmatrix}
1\\
0\\
0\\
0
\end{bmatrix}
=\begin{bmatrix}
\cos\frac\pi4\\
0\\
0\\
\sin\frac\pi4
\end{bmatrix}
\tag{24}
U ∣ 00 ⟩ = U 1 0 0 0 = cos 4 π 0 0 sin 4 π ( 24 )
这道题结合了 exercise 2.5 的带矩阵的欧拉公式，但是只要算出 T = exp ( i π σ y 4 ) = cos π 4 I − i sin π 4 σ y T=\exp(\frac{i\pi\sigma_y}{4})=\cos\frac\pi4\mathbb{I}-i\sin\frac\pi4\sigma_y T = exp ( 4 iπ σ y ) = cos 4 π I − i sin 4 π σ y
exercise 4.5
Q：试得到下述线路的输出量子态，并说明此量子线路的设计目的：
A：对这种题，直接列出真值表：
∣ 000 ⟩ → ∣ 000 ⟩ ∣ 010 ⟩ → ∣ 011 ⟩ ∣ 100 ⟩ → ∣ 101 ⟩ ∣ 110 ⟩ → ∣ 110 ⟩ (25) \begin{align}
\ket{000} & \rightarrow & \ket{000}\\
\ket{010} & \rightarrow & \ket{011}\\
\ket{100} & \rightarrow & \ket{101}\\
\ket{110} & \rightarrow & \ket{110}
\end{align}
\tag{25}
∣ 000 ⟩ ∣ 010 ⟩ ∣ 100 ⟩ ∣ 110 ⟩ → → → → ∣ 000 ⟩ ∣ 011 ⟩ ∣ 101 ⟩ ∣ 110 ⟩ ( 25 )
从真值表的结果来看，该量子线路可以用来判断 x , y x,y x , y ∣ 0 ⟩ \ket{0} ∣ 0 ⟩ ∣ 1 ⟩ \ket{1} ∣ 1 ⟩
