Math 401 Paper 1, Side note 1: Quantum information theory and Measure concentration

Typicality

The idea of typicality in high-dimensions is very important topic in understanding this paper and taking it to the next level of detail under language of mathematics. I’m trying to comprehend these material and write down my understanding in this note.

Let $X$ be the alphabet of our source of information.

Let $x^n=x_1,x_2,\cdots,x_n$ be a sequence with $x_i\in X$ .

We say that $x^n$ is $\epsilon$ -typical with respect to $p(x)$ if

For all $a\in X$ with $p(a)>0$ , we have

\|\frac{1}{n}N(a|x^n)-p(a)|\leq \frac{\epsilon}{\|X\|}

For all $a\in X$ with $p(a)=0$ , we have

N(a|x^n)=0

Here $N(a|x^n)$ is the number of times $a$ appears in $x^n$ . That’s basically saying that:

The difference between the probability of $a$ appearing in $x^n$ and the probability of $a$ appearing in the source of information $p(a)$ should be within $\epsilon$ divided by the size of the alphabet $X$ of the source of information.
The probability of $a$ not appearing in $x^n$ should be 0.

Here are two easy propositions that can be proved:

For $\epsilon>0$ , the probability of a sequence being $\epsilon$ -typical goes to 1 as $n$ goes to infinity.

If $x^n$ is $\epsilon$ -typical, then the probability of $x^n$ is produced is $2^{-n[H(X)+\epsilon]}\leq p(x^n)\leq 2^{-n[H(X)-\epsilon]}$ .

The number of $\epsilon$ -typical sequences is at least $2^{n[H(X)+\epsilon]}$ .

Recall that $H(X)=-\sum_{a\in X}p(a)\log_2 p(a)$ is the entropy of the source of information.

Shannon theory in Quantum information theory

Shannon theory provides a way to quantify the amount of information in a message.

Practically speaking:

A holy grail for error-correcting codes
Conceptually speaking:
An operationally-motivated way of thinking about correlations
What’s missing (for a quantum mechanic)?
- Features from linear structure:
  - Entanglement and non-orthogonality

Partial trace and purification

Partial trace

Recall that the bipartite state of a quantum system is a linear operator on $\mathscr{H}=\mathscr{A}\otimes \mathscr{B}$ , where $\mathscr{A}$ and $\mathscr{B}$ are finite-dimensional Hilbert spaces.

Definition of partial trace for arbitrary linear operators

Let $T$ be a linear operator on $\mathscr{H}=\mathscr{A}\otimes \mathscr{B}$ , where $\mathscr{A}$ and $\mathscr{B}$ are finite-dimensional Hilbert spaces.

An operator $T$ on $\mathscr{H}=\mathscr{A}\otimes \mathscr{B}$ can be written as (by the definition of tensor product of linear operators )

T=\sum_{i=1}^n a_i A_i\otimes B_i

where $A_i$ is a linear operator on $\mathscr{A}$ and $B_i$ is a linear operator on $\mathscr{B}$ .

The $\mathscr{B}$ -partial trace of $T$ ( $\operatorname{Tr}_{\mathscr{B}}(T):\mathcal{L}(\mathscr{A}\otimes \mathscr{B})\to \mathcal{L}(\mathscr{A})$ ) is the linear operator on $\mathscr{A}$ defined by

\operatorname{Tr}_{\mathscr{B}}(T)=\sum_{i=1}^n a_i \operatorname{Tr}(B_i) A_i

Partial trace for density operators

Let $\rho$ be a density operator in $\mathscr{H}_1\otimes\mathscr{H}_2$ , the partial trace of $\rho$ over $\mathscr{H}_2$ is the density operator in $\mathscr{H}_1$ (reduced density operator for the subsystem $\mathscr{H}_1$ ) given by:

\rho_1\coloneqq\operatorname{Tr}_2(\rho)

Examples

Let $\rho=\frac{1}{\sqrt{2}}(|01\rangle+|10\rangle)$ be a density operator on $\mathscr{H}=\mathbb{C}^2\otimes \mathbb{C}^2$ .

Expand the expression of $\rho$ in the basis of $\mathbb{C}^2\otimes\mathbb{C}^2$ using linear combination of basis vectors:

\rho=\frac{1}{2}(|01\rangle\langle 01|+|01\rangle\langle 10|+|10\rangle\langle 01|+|10\rangle\langle 10|)

Note $\operatorname{Tr}_2(|ab\rangle\langle cd|)=|a\rangle\langle c|\cdot \langle b|d\rangle$ .

Then the reduced density operator of the subsystem $\mathbb{C}^2$ in first qubit is, note the $\langle 0|0\rangle=\langle 1|1\rangle=1$ and $\langle 0|1\rangle=\langle 1|0\rangle=0$ :

\begin{aligned} \rho_1&=\operatorname{Tr}_2(\rho)\\ &=\frac{1}{2}(\langle 1|1\rangle |0\rangle\langle 0|+\langle 0|1\rangle |0\rangle\langle 1|+\langle 1|0\rangle |1\rangle\langle 0|+\langle 0|0\rangle |1\rangle\langle 1|)\\ &=\frac{1}{2}(|0\rangle\langle 0|+|1\rangle\langle 1|)\\ &=\frac{1}{2}I \end{aligned}

is a mixed state.

Purification

Let $\rho$ be any state (may not be pure) on the finite dimensional Hilbert space $\mathscr{H}$ . then there exists a unit vector $w\in \mathscr{H}\otimes \mathscr{H}$ such that $\rho=\operatorname{Tr}_2(|w\rangle\langle w|)$ is a pure state.

Proof

Let $(u_1,u_2,\cdots,u_n)$ be an orthonormal basis of $\mathscr{H}$ consisting of eigenvectors of $\rho$ for the eigenvalues $p_1,p_2,\cdots,p_n$ . As $\rho$ is a states, $p_i\geq 0$ for all $i$ and $\sum_{i=1}^n p_i=1$ .

We can write $\rho$ as

\rho=\sum_{i=1}^n p_i |u_i\rangle\langle u_i|

Let $w=\sum_{i=1}^n \sqrt{p_i} u_i\otimes u_i$ , note that $w$ is a unit vector (pure state). Then

\begin{aligned} \operatorname{Tr}_2(|w\rangle\langle w|)&=\operatorname{Tr}_2(\sum_{i=1}^n \sum_{j=1}^n \sqrt{p_ip_j} |u_i\otimes u_i\rangle \langle u_j\otimes u_j|)\\ &=\sum_{i=1}^n \sum_{j=1}^n \sqrt{p_ip_j} \operatorname{Tr}_2(|u_i\otimes u_i\rangle \langle u_j\otimes u_j|)\\ &=\sum_{i=1}^n \sum_{j=1}^n \sqrt{p_ip_j} \langle u_i|u_j\rangle |u_i\rangle\langle u_i|\\ &=\sum_{i=1}^n \sum_{j=1}^n \sqrt{p_ip_j} \delta_{ij} |u_i\rangle\langle u_i|\\ &=\sum_{i=1}^n p_i |u_i\rangle\langle u_i|\\ &=\rho \end{aligned}

is a pure state.

Drawing the connection between the space $S^{2n+1}$ , $CP^n$ , and $\mathbb{R}$

A pure quantum state of size $N$ can be identified with a Hopf circle on the sphere $S^{2N-1}$ .

A random pure state $|\psi\rangle$ of a bipartite $N\times K$ system such that $K\geq N\geq 3$ .

The partial trace of such system produces a mixed state $\rho(\psi)=\operatorname{Tr}_K(|\psi\rangle\langle \psi|)$ , with induced measure $\mu_K$ . When $K=N$ , the induced measure $\mu_K$ is the Hilbert-Schmidt measure.

Consider the function $f:S^{2N-1}\to \mathbb{R}$ defined by $f(x)=S(\rho(\psi))$ , where $S(\cdot)$ is the von Neumann entropy. The Lipschitz constant of $f$ is $\sim \ln N$ .