[Note] Quantum Computation and Quantum Information - Chapter 2: Introduction to quantum mechanics

上課筆記，原文書:

Quantum Computation and Quantum Information, Michael A. Nielsen & Isaac L. Chuang

2.1 Linear algebra

Notation	Description
$z^*$	Complex conjugate of the complex number $z$. $(1+i)^*=1-i$
$\vert\psi\rangle$	Column vector. Also known as a ket
$\langle\psi\vert$	Row vector dual to $\vert\psi\rangle$. Also known as bra
$\langle \varphi\vert\psi\rangle$	Inner product between the vectors $\vert\varphi\rangle$ and $\vert\psi\rangle$
$\vert\varphi\rangle \otimes\vert\psi\rangle$	Tensor product of $\vert\varphi\rangle$ and $\vert\psi\rangle$
$A^*$	Complex conjugate of the $A$ matrix
$A^T$	Transpose of the $A$ matrix
$A^\dagger$	Hermitian conjugate or adjoint of the $A$ matrix, $A^\dagger=(A^T)^*$. $\left[\begin{matrix}a & b\ c & d\end{matrix}\right]^\dagger = \left[\begin{matrix} a^* & c^*\\b^* & d^*\end{matrix}\right]$
$\langle \varphi\vert A\vert\psi \rangle$	Inner product between $\vert\varphi\rangle$ and $A\vert\psi\rangle$. Equivalently, inner product between $A^\dagger\vert\varphi\rangle$ and $\vert\psi\rangle$.

2.1.1 Bases and linear independence

• Spanning set：a set of vectors $\vert v_1\rangle,\dots,\vert v_n\rangle$ such that any vector $\vert v\rangle$ in the vector space can be written as a linear combination $\vert v \rangle=\sum_i a_i\vert v_i\rangle$ of vectors in that set.
• A set of non-zero vectors $\vert v_1 \rangle, \dots, \vert v_n\rangle$ are lineraly dependent if there exists a set of complex numbers $a_1,\dots, a_n$ with $a_i\ne 0$ for at least one value of $i$, such that $$a_i\vert v_1\rangle + a_2\vert v_2\rangle + \cdots + a_n\vert v_n\rangle = 0$$

2.1.2 Linear operators and matrics

• $A: V\to W$ is said to be a linear operator between vector spaces $V$ and $W$ if $$A\left(\sum_i a_i\vert v_i\rangle\right)=\sum_i a_i A\vert v_i\rangle$$
• Suppose $A:V\to W$ is a linear operator between vector spaces $V$ and $W$. Suppose $\vert v_1\rangle,\dots,\vert v_m\rangle$ is a basis for $V$ and $\vert w_1\rangle,\dots, \vert w_n\rangle$ is a basis for $W$. Then for each $j$ in the range $1,\dots,m$, there exist complex numbers $A_{1j}$ through $A_{nj}$ such that $$A\vert v_j \rangle=\sum_i A_{ij} \vert w_i\rangle$$

2.1.3 The Pauli matrics

$$\begin{eqnarray*} \sigma_0 \equiv I \equiv \left[\begin{matrix} 1 & 0 \\0 & 1\end{matrix}\right] \quad & \sigma_1\equiv\sigma_x\equiv X\equiv \left[\begin{matrix} 0 & 1\\ 1 & 0 \end{matrix}\right]\\ \sigma_2\equiv\sigma_y\equiv Y\equiv \left[\begin{matrix} 0 & -i\\ i & 0 \end{matrix}\right] \quad & \sigma_3\equiv\sigma_z\equiv Z\equiv\left[\begin{matrix}1&0\\0&-1\end{matrix}\right] \end{eqnarray*}$$

The $X$ matrix is often known as the quantum NOT gate, by analogy to the classical NOT gate:

• $X\vert 0\rangle=\vert 1\rangle$, $X\vert 1\rangle=\vert 0\rangle$
• $X\vert +\rangle=\vert +\rangle$, $X\vert -\rangle=(-1)\vert -\rangle$
  
  where $\vert+\rangle=\frac{1}{\sqrt{2}}(\vert 0\rangle + \vert 1\rangle)$, and $\vert-\rangle=\frac{1}{\sqrt{2}}(\vert 0\rangle - \vert 1\rangle)$

The $X$ and $Z$ are also sometimes referred toas the bit flip and phase flip matrices. $Z\vert 1\rangle=(-1)\vert 1\rangle$, $-1$ is known as a phase factor.

2.1.4 Inner products

1. $(\cdot,\cdot)$ is linear in the second argument, $$\left(\vert v\rangle, \sum_i\lambda_i\vert w_i\rangle\right)=\sum_i\lambda_i\left(\vert v\rangle, \vert w_i\rangle\right)$$
2. $\left(\vert v\rangle, \vert w\rangle\right)=\left(\vert w\rangle, \vert v\rangle\right)^*$
3. $\left(\vert v\rangle, \vert v\rangle\right)\ge 0$ with equality if and only if $\vert v\rangle=0$

Let $\vert w\rangle=\sum_i w_i\vert i \rangle$ and $\vert v\rangle=\sum_j v_j\vert j \rangle$ be representations of vectprs $\vert w\rangle$ and $\vert v \rangle$ with respect to some orthonormal basis $\vert i\rangle$. Then, since $\langle i\vert j\rangle=\delta_{ij}$,

$$\begin{align*} \langle v\vert w\rangle &= \left(\sum_j v_i\vert i \rangle, \sum_i w_j\vert j \rangle\right) = \sum_{ij}v^*_iw_j\delta_{ij}=\sum_iv^*_iw_i\\ &= \left[\begin{matrix}v^*_1 \dots v^*_n\end{matrix}\right]\left[\begin{matrix}w_1\\ \vdots\\w_n\end{matrix}\right] \end{align*}$$

2.1.5 Eigenvectors and eigenvalues

• An eigenvector of a linear operator $A$ on a vector space is a non-zero vector $\vert v \rangle$ such that $A\vert v \rangle=\lambda\vert v\rangle$, where $\lambda$ is a complex number knwon as the eigenvalue of $A$ corresponding to $\vert v \rangle$.
• The characteristic function is defined to be $c(\lambda)\equiv\det\vert A-\lambda I\vert$, where $\det$ is the determinant function for matrics.

2.1.6 Adjoints and Hermitian operators

• Suppose $A$ is any linear operator on a Hilbert space, $V$. It turns out that there exists a unique linear operator $A^\dagger$ on $V$ such that for all vectors $\vert v\rangle, \vert w \rangle\in V$, $$(\vert v \rangle, A\vert w\rangle)=(A^\dagger\vert v\rangle, \vert w\rangle)$$
• $(AB)^\dagger=B^\dagger A^\dagger$
• $\vert v \rangle^\dagger \equiv \langle v \vert$
• A matrix $U$ is said to be unitary if $U^\dagger U=I$.
• Unitary operators preserve inner products between vectors. Let $\vert v\rangle$ and $\vert w \rangle$ be any two vectors. $$(U\vert v \rangle, U\vert w \rangle)=\langle v\vert U^\dagger U\vert w\rangle=\langle v \vert I\vert w\rangle=\langle v\vert w \rangle$$
• A positive operator $A$ is defined to be an operator such that for any vector $\vert v\rangle$, $(\vert v\rangle, A\vert v\rangle)$ is a real, non-negative number. If $(\vert v\rangle, A\vert v\rangle)>0, ~~\forall \vert v\rangle \ne 0$，we say that $A$ is positive definite.

2.1.7 Tensor products

• Suppose $V$ and $W$ are vector spaces of dimension $m$ and $n$ respectively, then $V\otimes W$ (read V tensor W) is a $mn$ dimensional vector space.
• The elements of $V\otimes W$ are linear combinations of tensor products $\vert v\rangle \otimes \vert w \rangle, ~~\text{where}~\vert v\rangle\in V, \vert w\rangle\in W$
• $\vert i\rangle\otimes \vert j\rangle$ is a basis for $V\otimes W$ if $\vert i\rangle$ and $\vert j\rangle$ are orthonormal bases for the spaces $V$ and $W$
• Sometimes abbreviate notations to $\vert v \rangle\vert w\rangle$, $\vert v, w\rangle$, or even $\vert vw\rangle$

Properties:

• For an arbitrary scalar $z$ and elements $\vert v \rangle$ of $V$ and $\vert w\rangle$ of W, $$z(\vert v\rangle\otimes\vert w\rangle)=(z\vert v\rangle)\otimes \vert w\rangle=\vert v\rangle \otimes (z\vert w\rangle)$$
• For arbitrary $\vert v_1\rangle$ and $\vert v_2\rangle$ in $V$ and $\vert w\rangle$ in $W$, $$(\vert v_1\rangle + \vert v_2\rangle)\otimes\vert w \rangle=\vert v_1\rangle\otimes\vert w\rangle + \vert v_2\rangle \otimes\vert w\rangle$$
• For arbitrary $\vert v\rangle$ in $V$ and $\vert w_1\rangle$ and $\vert w_2\rangle$ in $W$, $$\vert v\rangle \otimes (\vert w_1\rangle + \vert w_2\rangle)=\vert v\rangle \otimes \vert w_1\rangle + \vert v\rangle \otimes \vert w_2\rangle$$

Suppose $\vert v\rangle$ and $\vert w \rangle$ are vectors in $V$ and $W$, and $A$ and $B$ are linear operators on $V$ and $W$, respectively. Then we can define a linear operator $A\otimes B$ on $V\otimes W$ by the equation

$$(A\otimes B)(\vert v \rangle\otimes \vert w\rangle)\equiv A\vert v\rangle \otimes B\vert w\rangle$$

Suppose A is an $m$ by $n$ matrix, and $B$ is a $p$ by $q$ matrix. Then we have the matrix representation:

$$A\otimes B = \underbrace{\left.\left[ \vphantom{\begin{matrix}{c}1\\1\\1\\1\end{matrix}} \begin{matrix} A_{11}B&A_{12}B&\cdots &A_{1n}B\\ A_{21}B&A_{22}B&\cdots &A_{2n}B\\ \vdots&\vdots&\ddots&\vdots&\\ A_{m1}B&A_{m2}B&\cdots &A_{mn}B \end{matrix} \right]\right\}}_{nq\text{ columns}} \,mp\text{ rows}$$

• $\vert \psi\rangle^{\otimes k}$ means $\vert \psi\rangle$ tensored with itself $k$ times.

2.1.8 Operator functions

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2.1.9 The commutator and anti-commutator

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2.1.10 The polar and singular value decompositions

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2.2 The postulates of quantum mechanics

2.2.1 State space

Postulate 1: Associated to any isolated physical system is a complex vector space with inner product (that is, a Hilbert space) known as the state space of the system. The system is completely described by its state vector, which is a unit vector in the system’s state space.

• A qubit has a two-dimensional state space.
• Suppose $\vert 0 \rangle$ and $\vert 1 \rangle$ form an orthonormal basis for that state space. Then an arbitrary state vector in the state space can be written $$\vert \psi\rangle=a\vert 0\rangle + b \vert 1\rangle$$ where, $a$ and $b$ are complex numbers.
• $\vert \psi\rangle$ is a unit vector implies that $\langle \psi \vert \psi \rangle=1$, and also equivalent to $\vert a\vert^2 + \vert b\vert^2=1$.
• We say that any linear combination $\sum_i\alpha_i\vert \psi_i\rangle$ is a superposition of the states $\vert \psi_i\rangle$ with amplitude $\alpha_i$ for the state $\vert \psi_i\rangle$.

2.2.2 Evolution

Postulate 2: The evolution of a closed quantum system is described by a unitary transformation. That is, the state $\vert \psi \rangle$ of the system at time $t_1$ is related to the state $\vert \psi’\rangle$ of the system at time $t_2$ by unitary operator $U$ which depends only on the times $t_1$ and $t_2$,

$$\vert \psi'\rangle=U\vert \psi\rangle$$

Hadamard gate $H$ is an unitary operator：

$$H=\frac{1}{\sqrt2}\left[\begin{matrix}1&1\\1&-1\end{matrix}\right]$$

This has the action：

• $H\vert 0\rangle\equiv\frac{1}{\sqrt2}(\vert 0\rangle + \vert 1\rangle)$
• $H\vert 1\rangle\equiv\frac{1}{\sqrt2}(\vert 0\rangle - \vert 1\rangle)$

Postulate 2’: The time evolution of the state of a closed quantum system is described by the Schrödinger equation,

$$i\hbar\frac{d}{dt}\vert \psi(t)\rangle=\hat{H}\vert\psi(t)\rangle$$

• $\hbar$ is known as Planck’s constant
• $\hat{H}$ is referred to Hamiltonian operator.

2.2.3 Quantum measurement

Postulate 3: Quantum measurements are described by a collection $\{M_m\}$ of measurement operators. These are opaertors acting on the state space of the system being measured. The index $m$ refers to the measurement outcomes that may occur in the experiment. If the state of the quantum system is $\vert\psi\rangle$ immediately before the measurement then the probability that result $m$ occurs is given by

$$p(m)=\langle\psi\vert M^\dagger_m M_m\vert\psi\rangle$$

and the state of the system after the measurement (post-measurement state) is

$$\frac{M_m\vert\psi\rangle}{\sqrt{\langle\psi\vert M^\dagger_m M_m\vert\psi\rangle}}$$

The measurement operators satisfy the completeness equation,

$$\sum_m M^\dagger_m M_m=I$$

The completeness equation expresses the fact that probabilities sum to one:

$$1=\sum_m p(m)=\sum_m\langle\psi\vert M^\dagger_m M_m\vert \psi\rangle$$

• measurement of a qubit in the computational basis. $M_0=\vert0\rangle\langle0\vert$, $M_1=\vert1\rangle\langle1\vert$.
  • Hermitian
  • $M_0^2=M_0, M_1^2=M_1$
  • completeness relation is obeyed

• Pauli measurements:

  $$X\equiv\left[\begin{matrix}0& 1\\1&0\end{matrix}\right], \quad Y\equiv\left[\begin{matrix}0& -i\\-i&0\end{matrix}\right], \quad Z\equiv\left[\begin{matrix}1& 0\\0&-1\end{matrix}\right]$$

• Pauli-X (Classical NOT gate):

  $$\begin{align*} X\vert 0\rangle = \vert 1\rangle, &\quad X\vert 1\rangle = \vert 0\rangle\\ X\vert +\rangle = \vert +\rangle, &\quad X\vert -\rangle = (-1)\vert -\rangle \end{align*}$$

2.2.4 Distinguishing quantum states

• Non-orthogonal quantum states cannot be distinguished.
• Orthogonal quantum states are perfectly distinguishable

2.2.5 Projective measurement

• The special class of measurements (Postulate 3) is known as projective measurements.
• Primarily concerned measurements with projective measurements.
• $P_i^2=P_i$, where $P_i$ is a measurement operator, and is called projector.

Projective measurements: A projective measurement is described by an observable, $M$, a Hermitian operator on the state space of the system being observed. The observable has a spectral decomposition,

$$M=\sum_m mP_m$$

where $P_m$ is the projector onto the eigenspace of $M$ with eigenvalue $m$. The possible outcomes of the measurement correspond to the eigenvalues, $m$, of the observable. Upon measuring the state $\vert\psi\rangle$, the probability of getting result $m$ is given by

$$p(m)=\langle\psi\vert P_m\vert\psi\rangle$$

Given that outcome $m$ occurred, the state of the quantum system immediately after the measurement is

$$\frac{P_m\vert\psi\rangle}{\sqrt{p(m)}}$$

• The average value of the measurement:

  $$\begin{align*} \mathbb{E}(M) =\langle M\rangle &= \sum_m m~p(m)\\ &= \sum_m m\langle \psi\vert P_m\vert\psi\rangle\\ &= \langle \psi \vert \left(\sum_m mP_m\right)\vert\psi\rangle\\ &= \langle \psi\vert M\vert\psi\rangle \end{align*}$$

• The square of the standard deviation:

  $$\begin{align*} [\Delta(M)]^2 &= \langle (M-\langle M \rangle)^2\rangle\\ &= \langle M^2 \rangle - \langle M\rangle ^2 \end{align*}$$

• Standard deviation $\Delta(M)=\sqrt{\langle M^2 \rangle - \langle M\rangle ^2}$

• Every hermitian matrix $M=\sum_\lambda \lambda\vert\psi_\lambda\rangle\langle\psi_\lambda\vert$ can be viewed as a measurement in physics. They are called observables and most physical quantities are observables e.g. position, momentum, electric field, strength, number of photon

2.2.6 POVM measurements

• POVM (Positive Operator-Valued Measure) $$E_m\equiv M^\dagger_mM_m$$
• $E_m$ is a positive operator such that $\sum_mE_m=I$ and $p(m)=\langle\psi\vert E_m\vert\psi\rangle$
  • Positive operator is a positive semi-definite matrix.
  • $E_m$ is also known as POVM element
• A complete set $\{E_m\}$ is known as POVM

Example: If we want to distinguish $\{\vert 0\rangle, \vert + \rangle\}$. We can set measurement operations to:

$$\begin{align*} E_1 &= \frac{\sqrt{2}}{1+\sqrt{2}}\vert 1\rangle\langle 1\vert\\ E_2 &= \frac{\sqrt{2}}{1+\sqrt{2}}\vert -\rangle\langle -\vert\\ E_3 &= Id-E_1-E_2 \quad \text{(Should check if $E_3$ is a positive operator)} \end{align*}$$

Since $\vert 0\rangle$ is orthogonal to the measurement $E_1$, that is $\langle 0\vert E_1\vert 0\rangle=0$. Also, $\vert+\rangle$ is orthogonal to the measurement $E_2$, that is $\langle +\vert E_2\vert +\rangle=0$:

1. If we see outcome 1, it must be $\vert+\rangle$.
2. If we see outcome 2, it must be $\vert 0\rangle$.
3. If we see outcome 3, we cannot distinguish it.

2.2.7 Phase

Global state:
Consider the state $e^{i\theta}\vert\psi\rangle$

• $\vert\psi\rangle$ is a state vector
• $e^{i\theta}$ is the global phase factor
• Global phase do not matter. Let $\vert\psi’\rangle=e^{i\theta}\vert\psi\rangle$ measured with $\{M_i\}$: $$\begin{align*} p'_i &= \langle \psi'\vert M^\dagger_iM_i\vert\psi'\rangle\\ &= \langle \psi\vert M^\dagger_i e^{-i\theta}e^{i\theta}M_i\vert\psi\rangle\\ &= p_i \end{align*}$$

Relative state:
Consider $\vert+\rangle=\frac{1}{\sqrt{2}}\vert 0\rangle+\frac{1}{\sqrt{2}}\vert 1\rangle$, $\vert-\rangle=\frac{1}{\sqrt{2}}\vert0\rangle-\frac{1}{\sqrt{2}}\vert1\rangle$. The relative phase of the state $\vert 1\rangle$ in $\vert+\rangle$ and $\vert-\rangle$ are orthogonal.

2.2.8 Composite systems

Postulate 4: The state space of a composite physical system is the tensor product of the state spaces of the component physical systems. Moreover, if we have systems numbered $1$ through $n$, and system number $i$ is prepared in the state $\vert \psi_i\rangle$, then the joint state of the total system is $\vert \psi_1\rangle\otimes\vert\psi_2\rangle\otimes\cdots\otimes\vert\psi_n\rangle$

• For a single quantum system, we can discribe the composite state of two quantum states $\vert x\rangle$ and $\vert y\rangle$ as $\alpha\vert x\rangle+\beta\vert y\rangle$, where $\vert\alpha\vert^2+\vert\beta\vert^2=1$. We call this superposition principle of quantum mechanics.
• Similary, for two quantum systems $A$ and $B$, we use tensor product to define the joint system $AB$, and we might denote the state as $\vert A\rangle\vert B\rangle$ for the joint system, where each $\vert A\rangle$ and $\vert B\rangle$ are a state of the system $A$ and $B$, respectively.

Letting $\vert 0\rangle$ be any fixed state of an ancilla system (附屬系統) $M$, having an orthonormal basis $\vert m\rangle$ in one-to-one correspondence with the possible outcomes of the measurement $\{M_m\}$, define an operator $U$ on products $\vert\psi\rangle\vert0\rangle$ of states $\vert\psi\rangle$ from a system $Q$ with the state $\vert0\rangle$ by

$$U\vert\psi\rangle\vert0\rangle\equiv\sum_m M_m\vert\psi\rangle\vert m\rangle$$

we can see that $U$ preserves inner products between states of the form $\vert\psi\rangle\vert 0\rangle$:

$$\begin{align*} \langle\varphi\vert\langle0\vert U^\dagger U\vert\psi\rangle\vert 0\rangle &= \sum_{m,m'}\langle\varphi\vert M_m^\dagger M_{m'}\vert\psi\rangle\langle m\vert m'\rangle\\ &= \sum_{m}\langle\varphi\vert M^\dagger_m M_m\vert\phi\rangle\\ &= \langle \varphi\vert\psi\rangle \end{align*}$$

We have proved that $U: Q\to Q\otimes M$ is an unitary operator If we can prove Exercise 2.67, then we can say $U$ can be extended to a unitary operator on the space $Q\otimes M$.

Exercise 2.67: Suppose $V$ is a Hilbert space with a subspace $W$. Suppose $U:W\to V$ is a linear operator which preserves inner products, that is, for any $\vert w_1\rangle$ and $\vert w_2\rangle$ in W,

$$\langle w_1\vert U^\dagger U\vert w_2\rangle = \langle w_1\vert w_2\rangle$$

Prove that there exists a unitary operator $U’:V\to V$ which extends $U$. That is, $U’\vert w\rangle=U\vert w\rangle$ for all $\vert w\rangle$ in $W$, but $U’$ is defined on the entire space $V$.

Proof: In order to complete the proof, we first show that $U’$ is an unitary operator, then we show that $U’ \vert w\rangle=U\vert w\rangle$ is hold for all $\vert w\rangle$ in $W$:

Let $\vert v\rangle=U\vert w\rangle$, where $\vert v\rangle \in V$, $\vert w\rangle \ne \vert w_2\rangle$ is in $W$. It is reasonable to define $U=\vert v\rangle\langle w\vert$ and the conjugate transpose $U^\dagger=\vert w\rangle\langle v\vert$, since $\vert v\rangle=\vert v\rangle\langle w\vert w\rangle=U\vert w\rangle$. However, we discovered that $U^\dagger U=\vert w\rangle\langle v\vert v\rangle\langle w\vert =\vert w\rangle\langle w\vert \ne I$, which means $\langle w_1\vert U^\dagger U\vert w_2\rangle = \langle w_1\vert w_2\rangle$ is not hold unless we can show that $U^\dagger U=\sum_i\vert w_i\rangle\langle w_i\vert=I$ for any orthonormal basis $\{\vert w_i\rangle\}$ in $W$.

Hense, we let $\{\vert w_i\rangle\}$ and $\{\vert v_k\rangle\}$ be an orthonormal basis that span $W$ and $V$, where $i\in\dim(W)$ and $k\in\dim(V)$. We redefine $U=\sum_i\vert v_i\rangle\langle w_i\vert$ and the conjugate transpose $U^\dagger=\sum_i\vert w_i\rangle\langle v_i\vert$. Now the equation holds, since

$$\begin{align*}U^\dagger U =\left(\sum_i\vert w_i\rangle\langle v_i\vert\right)\left(\sum_i\vert v_i\rangle\langle w_i\vert\right) =\sum_i \vert w_i\rangle\langle v_i\vert v_i\rangle\langle w_i\vert =\sum_i\vert w_i\rangle\langle w_i\vert=I\end{align*}$$

Let $\{\vert x_j\rangle\}$ be an orthonormal basis in complement space $X=V\setminus W$, where $j\in\dim(X)$. The mutually orthogonal set $\{\vert w_i\rangle, \vert x_j\rangle~\vert~ \forall i, j\}$, which can span $V$, is also an orthonormal basis in $V$. We then define $U’=\sum_i\vert v_i\rangle\langle w_i\vert + \sum_j\vert v_j\rangle\langle x_j\vert$, and the conjugate transpose $(U’)^\dagger = \sum_i\vert w_i\rangle\langle v_i\vert + \sum_j\vert x_j\rangle\langle v_j\vert$. As such, we can show that $U’$ is an unitary operator by

$$\begin{align*}(U')^\dagger U' &= \left(\sum_i\vert w_i\rangle\langle v_i\vert + \sum_j\vert x_j\rangle\langle v_j\vert\right)\left(\sum_i\vert v_i\rangle\langle w_i\vert + \sum_j\vert v_j\rangle\langle x_j\vert\right)\\ &= \left(\sum_i\vert w_i\rangle\langle v_i\vert\right)\left(\sum_i\vert v_i\rangle\langle w_i\vert\right)+\left(\sum_j\vert x_j\rangle\langle v_j\vert\right)\left(\sum_j\vert v_j\rangle\langle x_j\vert\right)\\ &= \sum_i\vert w_i\rangle\langle w_i\vert + \sum_j \vert x_j\rangle\langle x_j \vert = I\end{align*}$$

Same for $U’(U’)^\dagger = I$. Next, we show that $U’$ extends $U$

$$\begin{align*}U'\vert w\rangle &= \left(\sum_i\vert v_i\rangle\langle w_i\vert + \sum_j\vert v_j\rangle\langle x_j\vert\right)\vert w\rangle\\ &= \left(\sum_i\vert v_i\rangle\langle w_i\vert\right)\vert w\rangle \quad\quad(\because \vert x_j\rangle \bot \vert w\rangle, ~\forall~\vert x_j\rangle\in X\\ &= U\vert w\rangle\end{align*}$$

2.3 Application: superdense coding

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2.4 The density operator

2.4.1 Ensembles of quantum states

• Density operator, or density matrix:

  $$\rho\equiv\sum_i p_i\vert \psi_i\rangle\langle \psi_i\vert$$

  where $\vert \psi_i\rangle$ is an unknown state, $p_i$ is its probability. Then we call $\{p_i, \vert\psi_i\rangle\}$ an ensemble of pure states.

• The evolution descried by the unitary operator $U$ of the density operator:

  $$\rho=\sum_i p_i\vert \psi_i\rangle\langle \psi_i\vert \xrightarrow{U} \sum_i p_i U\vert \psi_i\rangle\langle\psi_i\vert U^\dagger=U\rho U^\dagger$$

• Measurement, if the initial state is $\vert \psi_i\rangle$, the probability of getting result $m$ is

  $$p(m\vert i)=\langle \psi_i\vert M^\dagger_m M_m \vert \psi_i\rangle=\text{tr}(M^\dagger_m M_m\vert \psi_i\rangle\langle \psi_i\vert)$$

• The total probability of getting result $m$ is:

  $$\begin{align*} p(m) &= \sum_i p(m\vert i)p_i\\ &=\sum_i p_i \text{tr}(M^\dagger_mM_m\vert \psi_i\rangle\langle \psi_i\vert)\\ &= \text{tr}(M^\dagger_m M_m \rho) \end{align*}$$

• The post-measurement state of $\vert \psi_i\rangle$:

  $$\vert \psi^m_i\rangle = \frac{M_m\vert \psi_i\rangle}{\sqrt{\langle \psi_i\vert M^\dagger_mM_m\vert \psi_i\rangle}}$$

• The density matrix after measurement:

  $$\rho_m=\sum_i p(i\vert m)\vert\psi^m_i\rangle\langle \psi^m_i\vert=\sum_i p(i\vert m)\frac{M_m\vert\psi_i\rangle\langle\psi_i\vert M^\dagger_m}{\langle \psi_i\vert M^\dagger_mM_m\vert \psi_i\rangle}$$

  since $p(i\vert m)=p(m, i)/p(m)=p(m\vert i)p_i/p(m)$, we obtain

  $$\begin{align*} \rho_m &= \sum_i p_i\frac{M_m\vert \psi_i\rangle\langle\psi_i\vert M^\dagger_m}{\text{tr}(M^\dagger_m M_m \rho)}\\ &= \frac{M_m\rho M^\dagger_m}{\text{tr}(M^\dagger_m M_m\rho)} \end{align*}$$

What we have shown is that the basic postulates of quantum mechanics related to
unitary evolution and measurement can be rephrased in the language of density operators.

• A quantum systen whose state $\vert \psi\rangle$ is known exactly is said to be in a pure state, the density operator is $\rho=\vert\psi\rangle\langle \psi \vert$. Otherwise, $\rho$ is in mixed state (a mixture of the different pure state).
• A pure state $\text{tr}(\rho^2)=1$, mixed state $\text{tr}(\rho^2)<1$
• A quantum system in mixed state $\rho_i$ arising from some ensemble $\{p_{i, j}, \vert \psi_{ij}\rangle\}$ of pure states with probability $p_i$. The density matrix for the system is $$\rho = \sum_{ij}p_ip_{ij}\vert \psi_{ij}\rangle\langle\psi_{ij}\vert=\sum_i p_i\rho_i$$ We say $\rho$ is a mixture of the state $\rho_i$ with probabilities $p_i$.
• A quantum system is in the state $\rho_m$ with probability $p(m)$ after the measurement $m$. The state of the quantum system would be described by the density operator $$\begin{align*} \rho' &=\sum_m p(m)\rho_m\\ &= \sum_m \text{tr}(M^\dagger_m M_m\rho)\frac{M_m\rho M^\dagger_m}{\text{tr}(M^\dagger_mM_m\rho)}\\ &= \sum_m M_m\rho M^\dagger_m \end{align*}$$

2.4.2 General properties of the density operator

Theorem 2.5 (Characterization of density operators) An operator $\rho$ is the density operator associated to some ensemble $\{p_i, \vert \psi_i\rangle\}$ if and only if satisfies the conditions:

1. (Trace condition) $\rho$ has trace equal to one.
2. (Positive condition) $\rho$ is a positive operator.

Proof
Suppose $\rho=\sum_ip_i\vert\psi_i\rangle\langle\psi_i\vert$ is a density operator. Then

$$\text{tr}(\rho)=\sum_ip_i\text{tr}(\vert\psi_i\rangle\langle\psi_i\vert)=\sum_ip_i=1$$

so the trace condition $\text{tr}(\rho)=1$ is satisfied. Suppose $\vert\varphi\rangle$ is an arbitrary vector in state space. Then

$$\begin{align*}\langle \varphi\vert\rho\vert\varphi\rangle &= \sum_ip_i\langle\varphi\vert\psi_i\rangle\langle\psi_i\vert\varphi\rangle\\ &= \sum_i p_i\vert\langle \varphi\vert\psi_i\rangle\vert^2\\ & \ge 0\end{align*}$$

Conversely, suppose $\rho$ is any operator satifsying the trace and positivity conditions. Since $\rho$ is positive, it must have a spectral decomposition

$$\rho = \sum_j\lambda_j\vert j\rangle\langle j\vert$$

where the vectors $\vert j\rangle$ are orthogonal, and $\lambda_j$ are real, non-negative eigenvalues of $\rho$.

From the trace condition we see that $\sum_j \lambda_j=1$. Therefore, a system in state $\vert j\rangle$ with probability $\lambda_j$ will have density opeartor $\rho$. That is, the ensemble $\{\lambda_j, \vert j\rangle\}$ is an ensemble of states giving rise to the density operator $\rho$.

We can reformulate the postulates:

Postulate 1: Associated to any isolated physical system is a complex vector space with inner product (that is, a Hilbert space) known as the state space of the system. The system is completely discribed by its density operator, which is a positive operator $\rho$ with trace one, acting on the state space of the system. If a quantum system is in the state $\rho_i$ with probability $p_i$, then the density operator for the system is $\sum_i p_i\rho_i$.

Postulate 2: The evolution of a closed quantum system is described by a unitary transformation That is, the state $\rho$ of the system at time $t_1$ is related to the state $\rho’$ of th esystem at time $t_2$ by a unitary operator $U$ which depends only on the times $t_1$ and $t_2$,

$$\rho' = U\rho U^\dagger$$

Postulate 3: Quantum measurements are described by a collection $\{M_m\}$ of measurement operators. These are operators acting on the state space of the system being measured. The index $m$ refers to the measurement outcomes that may occur in the experiment. If the state of the quantum system is $\rho$ immediately before the measurement then the probability that result $m$ occurs is given by

$$p(m)=\text{tr}(M^\dagger_mM_m\rho)$$

and the state of the system after the measurement is

$$\frac{M_m\rho M^\dagger_m}{\text{tr}(M^\dagger_mM_m\rho)}$$

The measurement operators satisfy the completeness equation,

$$\sum_m M^\dagger_mM_m=I$$

Postulate 4: The state space of a composite physical system is the tensor product of the state spaces of the compoenet physical systems. Moreover, if we have systems numbered $1$ through $n$, and system number $i$ is prepared in the state $\rho_i$, then the joint state of the total system is $\rho_1\otimes \rho_2\otimes \dots \otimes \rho_n$

Theorem 2.6 (Unitary freedom in the ensemble for density matrices) The sets $\vert\tilde{\psi}_i\rangle$ amd $\vert\tilde{\varphi}_j\rangle$ generate the same density matrix if and only if

$$\vert\tilde{\psi}_i\rangle=\sum_j u_{ij}\vert\tilde\varphi_j\rangle$$

where $u_{ij}$ is a unitary matrix of complex numbers, with indices $i$ and $j$, and we ‘pad’ whichever set of vectors $\vert\tilde\psi_i\rangle$ or $\vert\tilde\varphi_j\rangle$ is smaller with additional vectors $0$ so that the two sets have the same number of elements.

Also, $\rho=\sum_ip_i\vert\psi_i\rangle\langle\psi_i\vert=\sum_jq_j\vert\varphi_j\rangle\langle\varphi_j\vert$ if and only if

$$\sqrt{p_i}\vert\psi_i\rangle=\sum_j u_{ij}\sqrt{q_j}\vert\varphi_j\rangle$$

where $u_{ij}$ is the element of some unitary matrix $U$.

Proof
Suppose $\vert\tilde\psi_i\rangle=\sum_ju_{ij}\vert\tilde\varphi_j\rangle$ for some unitary $u_{ij}$. Then

$$\begin{align*}\sum_i\vert\tilde\psi_i\rangle\langle\tilde\psi_i\vert &= \sum_{ijk}u_{ij}u_{ik}^*\vert\tilde\varphi_j\rangle\langle\tilde\varphi_k\vert\\ &= \sum_{jk}\left(\sum_i u_{ki}^\dagger u_{ij}\right)\vert\tilde\varphi_j\rangle\langle\tilde\varphi_k\vert\\ &= \sum_{jk}\delta_{kj}\vert\tilde\varphi_j\rangle\langle\tilde\varphi_k\vert\\ &= \sum_j \vert\tilde\varphi_j\rangle\langle\tilde\varphi_j\vert\end{align*}$$

Conversely, suppose

$$A=\sum_i\vert\tilde\psi_i\rangle\langle\tilde\psi_i\vert=\sum_j\vert\tilde\varphi_j\rangle\langle\tilde\varphi_j\vert$$

Let $A=\sum_k\lambda_k\vert k\rangle\langle k\vert$ be a spectral decomposition for $A$ such that states $\vert k \rangle$ are orthonormal, and the $\lambda_k$ are strictly positive. Our strategy is to relate the states $\vert\tilde\psi_i\rangle$ to the states $\vert\tilde k\rangle\equiv\sqrt{\lambda_k}\vert k\rangle$, and similarly relate the states $\vert \tilde\varphi_j\rangle$ to the states $\vert\tilde k\rangle$. Combining the two relations will give the result. Let $\vert\psi\rangle$ be any vector orthonormal to the space spanned by the $\vert\tilde k\rangle$, so $\langle \psi\vert\tilde k\rangle\langle\tilde k\vert\psi\rangle=0$ for all $k$, abd thus we see that

$$0=\langle\psi\vert A\vert\psi\rangle=\sum_i\langle\psi\vert\tilde\psi_i\rangle\langle\tilde\psi_i\vert\psi\rangle=\sum_i\vert\langle\psi\vert\tilde\psi_i\rangle\vert^2$$

Thus $\langle\psi\vert\tilde\psi_i\rangle=0$ for all $i$ and all $\vert\psi\rangle$ orthonormal to the space spanned by the $\vert\tilde k\rangle$. It follows that each $\vert\tilde\psi\rangle$ can be expressed as a linear combination of the $\vert\tilde k\rangle$, $\vert\tilde\psi_i\rangle=\sum_kc_{ik}\vert\tilde k\rangle$. Since $A=\sum_k\vert \tilde k \rangle\langle\tilde k\vert=\sum_i\vert\tilde\psi_i\rangle\langle\tilde\psi_i\vert$ we see that

$$\sum_k\vert\tilde k\rangle\langle \tilde k\vert = \sum_{kl}\left(\sum_i c_{ik}c_{il}^*\right)\vert\tilde k\rangle\langle\tilde l\vert$$

The operator $\vert\tilde k\rangle\langle \tilde l\vert$ are easily seen to be linearly independent, and thus it must be that $\sum_i c_{ik}c_{il}^*=\delta_{kl}$. This ensures that we may append extra columns to $c$ to obtain a unitary matrix $v$ such that $\vert\tilde\psi_i\rangle=\sum_kv_{ik}\vert\tilde k\rangle$, where we have appended zero vectors to the list of $\vert\tilde k \rangle$. Similarly, we can find a unitary matrix $w$ such that $\vert\tilde\varphi_j\rangle=\sum_kw_{jk}\vert\tilde k\rangle$. Thus $\vert\tilde\psi_i\rangle=\sum_j u_{ij}\vert\tilde\varphi_j\rangle$
, where $u=vw^\dagger$ is unitary.

2.4.3 The reduced density operator

Suppose we have physical systems $A$ and $B$, whose state is described by a density operator $\rho^{AB}$.

• The reduced density operator for system $A$ is

  $$\rho^A\equiv \text{tr}_B(\rho^{AB})$$

  where $\text{tr}_B$ is called partial trace over system $B$, defined by

  $$\text{tr}_B(\vert a_1\rangle\langle a_2\vert\otimes\vert b_1\rangle\langle b_2\vert)\equiv\vert a_1\rangle\langle a_2\vert \text{tr}(\vert b_1\rangle\langle b_2\vert)$$

• Suppose a quantum system is in the product state $\rho^{AB}=\rho\otimes \sigma$, where $\rho$ is a density operator for system $A$, and $\sigma$ is a density operator for system $B$. Then

  $$\rho^A=\text{tr}_B(\rho\otimes\sigma)=\rho\text{tr}(\sigma) = \rho$$

2.5 The Schmidt decomposition and purification

Theorem 2.7 (Schmidt decomposition) Suppose $\vert\psi\rangle$ is a pure state of a composite system, $AB$. Then there exist orthonormal states $\vert i_A\rangle$ for system $A$, and orthonormal states $\vert i_B\rangle$ of system $B$ such that

$$\vert \psi\rangle = \sum_i \lambda_i \vert i_A\rangle\vert i_B\rangle$$

where $\lambda_i$ are non-negative real numbers satisfying $\sum_i \lambda_i^2=1$ known as Schmidt co-efficients.

Proof
Let $\vert j\rangle$ and $\vert k\rangle$ be any fixed orthonormal bases for systems $A$ and $B$, respectively. Then $\vert \psi\rangle$ can be written

$$\vert \psi\rangle = \sum_{jk} a_{jk}\vert j\rangle \vert k\rangle$$

for some matrix $A$ of complex numbers $a_{jk}$. By the singular value decomposition $A=UDV$, where $D$ is a positive, diagonal matrix. $U$ and $V$ are unitary matrices. Thus

$$\vert \psi\rangle = \sum_{ijk}u_{ji}d_{ii}v_{ik}\vert j\rangle\vert k\rangle$$

Defining $\vert i_A\rangle\equiv\sum_{j}u_{ji}\vert j\rangle$, $\vert i_B\rangle\equiv\sum_k v_{ik}\vert k\rangle$, and $\lambda_i\equiv d_{ii}$, this gives

$$\vert\psi\rangle\equiv\sum_i\lambda_i\vert i_A\rangle\vert i_B\rangle$$

It is easy to check that $\vert i_A\rangle$ forms an orthonormal set, from the unitarity of $u$ and the orthonormality of $\vert j\rangle$, and similarlt that the $\vert i_B\rangle$ form an orthonormal set.

• $\vert i_A\rangle$ and $\vert i_B\rangle$ are called Schmidt bases for $A$ and $B$, respectively.
• $\lambda_i$ is called Schmidt co-efficients and the number of $\lambda_i$ is called Schmidt number or Schmidt rank for the state $\vert\psi\rangle$.
  • The Schmidt number is preserved under unitary transformations on system $A$ or system $B$ alone.
  • $\vert\psi\rangle=\sum_i\lambda_i\vert i_A\rangle\vert i_B\rangle \xrightarrow{U_A} U_A\vert\psi\rangle=\sum_i\lambda_i\left(U_A\vert i_A\rangle\right)\vert i_B\rangle$

Suppose we are given a state $\rho^A$ of a quantum system $A$. It is possible to introduce another system, which we denote $R$, and define a pure state $\vert AR\rangle$ for the joint system $AR$ such that $\rho^A=\text{tr}_R(\vert AR\rangle\langle AR\vert)$. This procedure is called purification.

• We call the system $R$ a reference system.
• Only a mathematical procedure
• Usful in simplifying proofs because pure states are easier to manipulate

Theorem Every density matrix $\rho$ can be purified.

Proof
To prove that we can purify any state in system $A$, we show that we can construct a system $R$ and purification $\vert AR\rangle$ for $\rho^A$:

Suppose the orthonormal decomposition for $\rho^A=\sum_i p_i\vert i^A\rangle\langle i^A\vert$. We introduce a system $R$ which has the same state space as system $A$, with orthonormal basis states $\vert i^R\rangle$, and define a pure state for the combined system:

$$\vert AR\rangle\equiv\sum_i \sqrt{p_i}\vert i^A\rangle\vert i^R\rangle$$

Verify that $\vert AR\rangle$ is actually a purification of $\rho^A$

$$\begin{align*}\text{tr}_R(\vert AR\rangle\langle AR\vert) &= \sum_{ij}\sqrt{p_ip_j}\vert i^A\rangle\langle j^A\vert \text{tr}(\vert i^R\rangle\langle j^R\vert)\\ &= \sum_{ij}\sqrt{p_ip_j}\vert i^A\rangle\langle j^A\vert \delta_{ij}\\ &= \sum_i p_i\vert i^A\rangle\langle i^A\vert\\ &= \rho^A\end{align*}$$

2.6 EPR and the Bell inequality

待補