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8.1 Schmidt Decomposition and Partial Trace

We've known \(\text{Herm}(A)\subset \mathcal{L}(A)\) in real and \(\dim(\text{Herm}(A))=|A|^2\)​

\(\text{Herm}(\mathbb{C^2})\) we have basis of Pauli matrices
\(\sigma_{0},I： \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}\) \(\sigma_{1},X： \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}\) \(\sigma_{2},Y: \begin{pmatrix} 0 & -i \\ i & 0 \end{pmatrix}\) \(\sigma_{3},Z: \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}\)

Here they are not pure since the eigenvalues are negative and \(\text{Tr}\neq 1\). Clearly \(\sigma_j^2=I_2\), \(\forall j\in \{0,1,2,3\}\)

For example

  • \(I = |0\rangle\langle0| + |1\rangle\langle1|\)
  • \(\sigma_{1} = 2 \cdot |+\rangle\langle+| - I\)
  • \(\sigma_{2} = 2 \cdot |+i\rangle\langle+i| - I\)
  • \(\sigma_{3} = 2 \cdot |0\rangle\langle0| - I\)

where \(|+\rangle = \frac{1}{\sqrt{2}}(|0\rangle+|1\rangle) = \frac{1}{\sqrt{2}} \begin{pmatrix} 1 \\ 1 \end{pmatrix}\) and \(|+i\rangle=\frac{1}{\sqrt{2}}(|0\rangle+i|1\rangle)=\frac{1}{\sqrt{2}} \begin{pmatrix} 1 \\ i \end{pmatrix}\)

Then \(\left\{|0\rangle\langle0|, |1\rangle\langle1|, |+\rangle\langle+|, |+i\rangle\langle+i|\right\}\) is a basis with pure states

Homework 1: Find a basis of \(\text{Herm}(A)\) consisting of pure states of the form \(|\psi\rang\lang\psi|\)

Solution

The matrices \(H_{a,b}\), with \(1 \le a,b \le n\), form an orthogonal basis for \(\text{Herm}(A)\)

\[ H_{a,b}= \begin{cases} E_{a,a} & \text{if }a = b \\ E_{a,b}+ E_{b,a} & \text{if }a < b \\ i(E_{a,b}- E_{b,a}) & \text{if }a > b \end{cases} \]

where \(E_{a,b} = |e_a\rangle\langle e_b|\) and \(|e_a\rangle\) a state with 1 in the \(a\)-th entry and all other entries zeros.

Now define the states: \(|\psi_{a,b}\rangle = \begin{cases}|e_a\rangle & \text{if } a = b \\\frac{1}{\sqrt{2}}(|e_a\rangle + |e_b\rangle) & \text{if } a < b \\\frac{1}{\sqrt{2}}(i|e_a\rangle + |e_b\rangle) & \text{if } a > b\end{cases}\)
and the pure density matrices \(\rho_{a,b}= |\psi_{a,b}\rangle\langle\psi_{a,b}|\).

After some calculations we get \(H_{a,a} = \rho_{a,a}\) and \(H_{a,b} = 2\rho_{a,b} - \rho_{a,a} - \rho_{b,b}\)

So \(\rho_{a,b}\) form a basis.

Schatten p-Norm

For \(p\geq 1\), given \(M\in\mathbb{C}^{n\times m}\) and \(||M||_p:=(\text{Tr}|M|^p)^{\frac{1}{p}}\) and \(|M|=\sqrt{M^*M}\)

The trace norm: \(||M||_1=\text{Tr}|M|=\text{Tr}\sqrt{M^*M}\)

The operator norm: \(||M||_\infty=\lambda_{\max}(|M|)\) where \(\lambda_{\max}(|M|)\) is the largest eigenvalue of \(|M|\)​

Property

Invariance: For any two Hilbert spaces \(A', B'\) with \(|A'| \ge |A|\) and \(|B'| \ge |B|\), and any isometries \(V \in \mathcal{L}(B, B')\) and \(U \in \mathcal{L}(A, A')\)

\[ ||VMU^*||_{p}=||M||_{p} \]

Hölder Inequality:

\[ ||MN||_{1}\leq ||M||_{p}||N||_{q}\text{ for }p,q\in[1,\infty),\frac{1}{p}+\frac{1}{q}=1 \]

Monotonicity:

\[ ||M||_{p}\geq ||M||_{p'}\text{ if }p\leq p' \]

Linear Operators as Bipartite Vector

\(|\psi\rangle \in A \otimes B\) and \(A \otimes B = AB\). Def: \(M|x\rangle:=\sum_{y}M_{xy}|y\rangle\) where \(M:\tilde{A}\to B\)

\(|\psi\rangle=\sum_{x=1}^{|A|}\sum_{y=1}^{|B|}M_{xy}|x\rangle^{A}\otimes|y\rangle ^{B}=\sum_{x=1}^{|A|}|x\rangle^{A}\otimes\underbrace{\sum_{y=1}^{B}M_{xy}|y\rangle^{B}} _{M|x\rang}=(I^{A}\otimes M)\sum_{x=1}^{|A|}|x\rangle^{A}|x\rangle^{\tilde{A}}\) since tensor product is linear

Then we define \(|\Omega^{A\tilde{A}}\rangle := \sum_{x=1}^{|A|}|x\rangle^{A}\otimes |x\rangle ^{\tilde{A}}\)

Then \(\langle\Omega^{A\tilde{A}}|\Omega^{A\tilde{A}}\rangle = \sum_{x'=1}^{|A|}\langle x'|\langle x'| \sum_{x=1}^{|A|}|x\rangle|x\rangle = \sum_{x,x'}^{|A|}\underbrace{\langle x|x' \rangle}_{\delta_{xx'}} \langle x|x' \rangle = \sum_{x=1}^{|A|}1 = |A|\) since \(\delta_{xx'} = \begin{cases} 1 & \text{if } x=x' \\ 0 & \text{if } x\neq x' \end{cases}\)

Now \(|\psi^{AB}\rang =I^{A}\otimes M|\Omega^{A\tilde{A}}\rang\) where \(|\Omega^{A\tilde{A}}\rangle := \sum_{x=1}^{|A|}|x\rangle^{A}\otimes |x\rangle^{\tilde{A}}\)

Homework 2. Suppose \(|A|=|B|\)

  1. \(\langle\Omega^{A\tilde{A}}|I^{A}\otimes M|\Omega^{A\tilde{A}}\rangle = \text{Tr} [M]\)

    Proof

    \(\langle\Omega^{A\tilde{A}}|I^{A}\otimes M|\Omega^{A\tilde{A}}\rangle=\langle\Omega ^{A\tilde{A}}|I^{A}\otimes M\sum_{x=1}^{|A|}|x\rangle^{A}\otimes|x\rangle^{\tilde{A}}\)

    \(=\langle\Omega^{A\tilde{A}}|\sum_{x=1}^{|A|}\left(I^{A}\otimes M\right)\left(|x\rangle ^{A}\otimes|x\rangle^{\tilde{A}}\right)=\langle\Omega^{A\tilde{A}}|\sum_{x=1}^{|A|} I^{A}|x\rangle^{A}\otimes M|x\rangle^{\tilde{A}}\)

    \(=\left(\sum_{x^{\prime}=1}^{|A|}\lang x^{\prime}|^{A}\otimes\lang x^{\prime}|^{\tilde{A}} \right)\left(\sum_{x=1}^{|A|}|x\rangle^{A}\otimes M|x\rangle^{\tilde{A}}\right)=\left (\sum_{x,x^{\prime}=1}^{|A|}\lang x^{\prime}|^{A}\otimes\lang x^{\prime}|^{\tilde{A}} \right)\left(|x\rangle^{A}\otimes M|x\rangle^{\tilde{A}}\right)\)

    \(=\sum_{x,x^{\prime}=1}^{|A|}\langle x^{\prime}|x\rangle\otimes\langle x^{\prime} |M|x\rangle^{\tilde{A}}=\sum_{x=1}^{|A|}\langle x|x\rangle\otimes\langle x|M|x\rangle ^{\tilde{A}}=\sum_{x=1}^{|A|}\langle x|M|x\rangle^{\tilde{A}}=\text{Tr}[M]\)

  2. \(I \otimes M|\Omega^{A\tilde{A}}\rangle = M^{T}\otimes I^{\tilde{A}}|\Omega^{A\tilde{A}} \rangle\) Generalized version: \(|\psi^{AB}\rangle=I^{A}\otimes M^{\tilde{A}\to B}|\Omega^{A\tilde{A}}\rangle=\left (M^{T}\right)^{\tilde{B}\to A}\otimes I^{B}|\Omega^{\tilde{B}B}\rangle\)

    Proof

    \(I\otimes M|\Omega^{A\tilde{A}}\rangle=\sum_{x=1}^{|A|}|x\rangle^{A}\otimes M|x\rangle ^{\tilde{A}}=\sum_{x=1}^{|A|}\sum_{y}M_{xy}|x\rangle\otimes|y\rangle\)

    \(M^{T}\otimes I^{\tilde{A}}|\Omega^{A\tilde{A}}\rangle=\sum_{x=1}^{|A|}M^{T}|x\rangle ^{A}\otimes|x\rangle^{\tilde{A}}=\sum_{x=1}^{|A|}\sum_{y}M_{yx}|y\rangle\otimes|x \rangle\)

    Then we can change the index in the second line: \(x\Leftrightarrow y\), we will get the same result

  3. \(M \otimes N|\Omega^{A\tilde{A}}\rangle = |\Omega^{A\tilde{A}}\rangle \iff M = (N ^{-1})^{T}\)

    Proof

    \(\Leftarrow\)) \(M\otimes N|\Omega^{A\tilde{A}}\rangle=(N^{-1})^{T}\otimes N|\Omega^{A\tilde{A}}\rangle =\left(\left(N^{-1}\right)^{T}\otimes N\right)\left(\sum_{x=1}^{|A|}|x\rangle^{A} \otimes|x\rangle^{\tilde{A}}\right)\)

    \(=\sum_{x=1}^{|A|}\left(\left(N^{-1}\right)^{T}|x\rangle^{A}\otimes N|x\rangle^{\tilde{A}} \right)=\sum_{x=1}^{|A|}\left(\sum_{y}(N^{-1})_{xy}^{T}|y\rangle\otimes\sum_{z}N_{xz} |z\rangle\right)\)

    \(=\sum_{x=1}^{|A|}\sum_{y,z}(N^{-1})_{xy}^{T}N_{xz}\left(|y\rangle\otimes|z\rangle \right)=\sum_{y,z}\sum_{x=1}^{|A|}(N^{-1})_{yx}N_{xz}\left(|y\rangle\otimes|z\rangle \right)\)

    \(=\sum_{y,z}\delta_{yz}\left(|y\rangle\otimes|z\rangle\right)=\sum_{y}|y\rangle^{A} \otimes|y\rangle^{\tilde{A}}=|\Omega^{A\tilde{A}}\rangle\)

    \(\Rightarrow\)) \(M\otimes N|\Omega^{A\tilde{A}}\rangle=|\Omega^{A\tilde{A}}\rangle\)

    LHS \(=\) \(\sum_{x,y,z} M_{xy} N_{xz} |y\rangle \otimes |z\rangle\) and RHS = \(\sum_k |k\rangle \otimes |k\rangle\).

    Then the coefficients should be same: \(\sum_{x}M_{xy}N_{xz}= \delta_{yz}\) and we know \(MN^T=I\) iff \(\sum\)

    Then \(M=(N^{-1})^{T}\)

Schmidt Decomposition

Theorem

Let \(|\psi\rangle \in AB\). Suppose \(\langle\psi|\psi\rangle=1\) and \(||\psi||=1\). Then, there exists K-orthonormal Vectors \(\{|u_x^A\rangle\}_{x \in [K]}\) of \(A\), and K-orthonormal Vectors of \(B\) \(\{|v_x^B\rangle\}_{x \in [K]}\) s.t.

\[ |\psi^{AB}\rangle=\sum_{x=1}^{K}\sqrt{p_{x}}|u_{x}^{A}\rangle|v_{x}^{B}\rangle \]

Here \(\sqrt{p_x}\) is Schmidt-Coefficient where \(K \le \min\{|A|, |B|\}\)
Probability vector: \(\vec{p}=(p_{1}, ..., p_{K})\)

To prove it, we use the following theorem

Singular Valued Decomposition

Given \(M\in\mathbb{C}^{n\times m}\), there exists unitary matrices \(U\in \mathbb{C}^{n\times n},V\in \mathbb{C}^{m\times m}\) s.t.

\[ M=UDV\quad D\text{ is diagonal but not necessary square} \]

\(D = \begin{pmatrix} \lambda_{1} & 0 & \cdots & 0 \\ 0 & \lambda_{2} & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & \lambda_{k} \\ 0&0&\cdots&0\\ \vdots&\vdots&\cdots&\vdots\\ 0&0&\cdots&0\\ \end{pmatrix}\) and \(K = \min\{n, m\}\) where \(\lambda_1 \ge \lambda_2 \ge \dots \ge \lambda_k > 0\)
And each \(\lambda\)s are singular values of \(M\)​

Suppose \(m=n\), then \(|M|=\sqrt{M^*M}=\sqrt{V^*D^*U^*UDV}=\sqrt{V^*D^2V}=\sqrt{(V^*DV)^2}=V^*DV\) since \(V^*DV\) is Hermitian

Polar decomposition

Then \(|M|=V^*DV=V^*U^*UDV=V^*U^*M\)

Since \(V,U\) is unitary, then \(M=UV|M|\), then \(M=U'|M|\)

Proof of the first theorem

We assume \(m=n\), then \(M=UDV\), then \(|\psi^{AB}\rang=I\otimes M|\Omega\rang=(I\otimes UDV)|\Omega\rangle\)

By tensor product \(=(I\otimes UD)(I\otimes V)|\Omega\rangle\)

Then use Homework 2.2: \(=(I\otimes UD)(V^{T}\otimes I)|\Omega\rangle\)

Thus \(|\psi^{AB}\rang = (V^{T}\otimes U)(I\otimes D)|\Omega\rang\)

Since \(D= \begin{pmatrix} \lambda_{1} & \cdots & 0 \\ \vdots & \lambda_{i} & \vdots \\ 0 & \cdots & \lambda_{n} \end{pmatrix}\), then \((I\otimes D)|\Omega\rangle=\sum^{n}_{x=1}\lambda_{x}|xx\rang\)

Then \(|\psi^{AB}\rang =\sum^n_{x=1}\lambda_x(V^T\otimes U)|x\rang|x\rang=\sum^n_{x=1}\lambda_x(V^T|x\rang)(U|x\rang)=\sum^n_{x=1}\lambda_x|v_x\rang|u_x\rang\)

Then \(|\psi^{AB}\rang =\sum_{x\in [n]}\lambda_x|v_x\rang|u_x\rang\)

Then by \(1=\langle\psi^{AB}|\psi^{AB}\rangle=\sum_{y\in[n]}\lambda_{y}\langle v_{y}|\langle u_{y}|\sum_{x\in[n]}\lambda_{x}|v_{x}\rangle|u_{x}\rangle=\sum_{x,y}\lambda_{x}\lambda _{y}\underbrace{\langle v_{y}|v_{x}\rangle}_{\delta_{xy}}\underbrace{\langle u_{y}|u_{x}\rangle} _{\delta_{xy}}=\sum_{x=1}^{n}\lambda_{x}^{2}=1\)

Partial Trace

\(\text{Tr}_B:\mathcal{L}(AB)\to \mathcal{L}(A)\) where \(x,x'\in \{1,...,|A|\},y,y'\in\{1,...,|B|\}\)

Orthonormal Basis of \(\mathcal{L}(AB)\) is \(\{|x\rang\lang x'|\otimes |y\rang\lang y'|\}\)

We define \(\text{Tr}_{B}[|x\rangle\langle x^{\prime}|^{A}\otimes|y\rangle\langle y^{\prime} |^{B}]=|x\rangle\langle x^{\prime}|^{A}\underbrace{\text{Tr}[|y\rangle\langle y^{\prime}|]}_{\delta_{yy'}}\)

Let \(|\psi\rang\in AB,\psi^{AB}:=|\psi^{AB}\rang\lang\psi^{AB}|\)

\(=\sum_{x}\sqrt{p_{x}}|u_{x}\rangle |v_{x}\rangle \sum_{y}\sqrt{p_{y}}\langle u_{y} | \langle v_{y}|\) \(=\sum_{x,y}\sqrt{p_{x} p_{y}}|u_{x} \rangle \langle u_{y} | \otimes |v_{x} \rangle \langle v_{y} |\)

Then \(\text{Tr}[\psi^{AB}]=\sum_{x,y}\sqrt{p_{x}p_{y}}|u_{x}\rang\lang u_{y}|\cdot \text{Tr} [|v_{x}\rang\lang v_{y}|] =\sum_{x,y}\sqrt{p_{x}p_{y}}|u_{x}\rang\lang u_{y}|\cdot \lang v_y|v_x\rang\)

\(=\sum_xp_x|u_x\rang\lang u_x|\)

Then \(\rho^{A}:=\text{Tr}_{B}[\psi^{AB}]=\sum^{|A|}_{x=1}p_{x}|u_{x}\rang\lang u_{x}|^{A}\). This is the reduced density matrix of \(\psi^{AB}\)

If we have a reduced density matrix, can we know the \(|\phi\rang\) where \(\phi:=|\phi\rang\lang\phi|\)?

We say \(|\phi^{AB}\rang\) is a purification of \(\rho^{A}\) if \(\rho^{A}=\text{Tr}_{B}[\phi^{AB}]\)

Actually \(\sqrt{\rho}\otimes I|\Omega\rangle\) where \(\Omega=|\Omega\rangle\langle\Omega|\) is a purification of \(\rho\) why?

Proof

First \(\sqrt{\rho}\otimes I|\Omega\rangle=(\sqrt{\rho}\otimes I)(\sum_{x=1}^{|A|}|x\rangle|x\rangle ))=\sum_{x=1}^{|A|}\sqrt{\rho}|x\rangle\otimes|x\rangle=|\psi^{AB}\rangle\)

Then \(\text{Tr}_{B}(|\psi^{AB}\rangle\langle\psi^{AB}|) = \text{Tr}_{B}((\sum_{x=1}^{|A|} \sqrt{\rho}| x\rangle \otimes |x\rangle) (\sum_{y=1}^{|A|}\langle y|\sqrt{\rho}\otimes \langle y |))\) since \(\rho\) is hermitian
\(= \text{Tr}_{B} (\sum_{x,y}\sqrt{\rho}|x\rangle \langle y| \sqrt{\rho}|x\rangle \langle y|) = \sum_{x,y}^{|A|}\sqrt{\rho}|x\rangle \langle y| \sqrt{\rho}\,\text{Tr}_{B} (|x\rangle \langle y|)\)
\(= \sum_{x,y} \sqrt{\rho} |x\rangle \langle y| \sqrt{\rho} \otimes \delta_{x,y}\)
\(= \sum_{x} \sqrt{\rho}|x\rangle \langle x| \sqrt{\rho}= \sqrt{\rho}\sum_{x} |x\rangle \langle x| \sqrt{\rho} = \rho\)

Any purification can be written as \(|\psi^{AB}\rangle=\sqrt{\rho}\otimes V^{A\to B}|\Omega^{A\tilde{A}}\rangle\) where \(\rho^{A}:=\text{Tr}_{B}[|\psi^{AB}\rang\lang\psi^{AB}|]\)

Proof

\(\text{Tr}_{B}(|\psi^{AB}\rangle\langle\psi^{AB}|)=\text{Tr}_{B}((\sum_{x=1}^{|A|} \sqrt{\rho}|x\rangle\otimes V^{A\to B}|x\rangle)(\sum_{y=1}^{|A|}\langle y|\sqrt{\rho} \otimes\langle y|{V^{A\to B}}^*))\) since \(\rho\) is hermitian
​\(=\text{Tr}_{B}(\sum_{x,y}\sqrt{\rho}|x\rangle\langle y|\sqrt{\rho}V^{A\to B}|x\rangle \langle y|{V^{A\to B}}^{*})=\sum_{x,y}^{|A|}\sqrt{\rho}|x\rangle\langle y|\sqrt{\rho} \,\text{Tr}_{B}(V^{A\to B}|x\rangle\langle y|{V^{A\to B}}^{*})\)
​\(= \sum_{x,y} \sqrt{\rho} |x\rangle \langle y| \sqrt{\rho} \otimes \delta_{x,y}\)
​\(= \sum_{x} \sqrt{\rho}|x\rangle \langle x| \sqrt{\rho}= \sqrt{\rho}\sum_{x} |x\rangle \langle x| \sqrt{\rho} = \rho\)

\(\text{Tr}_{\tilde{A}}[\Omega^{A\tilde{A}}] = I\)

Proof

\(\Omega^{A\tilde{A}}=\left(\sum_{x=1}^{|A|}|x\rangle^{A}|x\rangle^{\tilde{A}}\right )\left(\sum_{y=1}^{|A|}\langle y|^{A}\langle y|^{\tilde{A}}\right)=\sum_{x=1}^{|A|} \sum_{y=1}^{|A|}|x\rangle^{A}\langle y|^{A}\otimes|x\rangle^{\tilde{A}}\langle y| ^{\tilde{A}}\)

Then \(\text{Tr}_{\tilde{A}}[\Omega^{A\tilde{A}}]=\text{Tr}_{\tilde{A}}\left[\sum_{x=1} ^{|A|}\sum_{y=1}^{|A|}|x\rangle^{A}\langle y|^{A}\otimes|x\rangle^{\tilde{A}}\langle y|^{\tilde{A}}\right]=\sum_{x=1}^{|A|}\sum_{y=1}^{|A|}|x\rangle^{A}\langle y|^{A} \text{Tr}\left\lbrack|x\rangle^{\tilde{A}}\langle y|^{\tilde{A}}\right\rbrack=\sum_{x=1}^{|A|}|x\rang\lang x|=I\)

Homework 3: Show that if both \(|\psi^{AB}\rangle\) and \(|\phi^{AC}\rangle\) are purifications of \(\rho^A\).

Then, \(|\phi^{AC}\rangle = (I^{A} \otimes U)|\psi^{AB}\rangle\) where \(U:B\rightarrow C\) is a isometry

Proof

Since \(|\psi^{AB}\rangle=\sum_{x=1}^{|A|}\sqrt{p_{x}}|u_{x}^{A}\rangle|v_{x}^{B}\rangle\) and \(|\phi^{AC}\rangle=\sum_{x=1}^{|A|}\sqrt{q_{x}}|u_{x}^{A}\rangle|z_{x}^{C}\rangle\)

Also \(\rho^{A}=\text{Tr}_{B}[\psi^{AB}]=\sum_{x=1}^{|A|}p_{x}|u_{x}\rangle\langle u_{x} |^{A}\) and \(\rho^{A}=\text{Tr}_{C}[\phi^{AC}]=\sum_{x=1}^{|A|}q_{x}|u_{x}\rangle\langle u_{x} |^{A}\)

Then \(\sum_{x=1}^{|A|}p_{x}|u_{x}\rangle\langle u_{x}|^{A}=\sum_{x=1}^{|A|}q_{x}|u_{x} \rangle\langle u_{x}|^{A}\), then \(p_x=q_x\)

Then \(|\phi^{AC}\rangle=\sum_{x=1}^{|A|}\sqrt{p_{x}}|u_{x}^{A}\rangle|z_{x}^{C}\rangle\)​

Also \((I^{A}\otimes U)|\psi^{AB}\rangle=(I^{A}\otimes U)\sum_{x=1}^{|A|}\sqrt{p_{x}}|u _{x}^{A}\rangle|v_{x}^{B}\rangle=\sum_{x=1}^{|A|}\sqrt{p_{x}}|u_{x}^{A}\rangle\otimes U|v_{x}^{B}\rangle\)

We define \(U:B\to C\) s.t. \(U|v_{x}^{B}\rangle=|z_{x}^{C}\rangle\) for all \(x\), let show this is a isometry.

Since \(\langle Uv|Uv\rangle=\langle\sum_{x}Uc_{x}v_{x}|\sum_{x}Uc_{x}v_{x}\rangle=\sum_{x} \langle Uc_{x}v_{x}|Uc_{x}v_{x}\rangle+\sum_{x\neq y}\langle Uc_{x}v_{x}|Uc_{y}v_{y} \rangle=\sum_{x}||c_x||^2\langle z_{x}|z_{x}\rangle+\sum_{x\neq y}\bar c_xc_y\langle z_{x}|z_{y}\rangle =\sum_x||c_x||^2+0=\sum_x||c_x||^2=\langle v|v\rangle\), thus \(U\) is a isometry.

Thus we complete the proof

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