Formalism of quantum statistical mechanics

(Also see Kardar 6.4 (Quantum microstates) and Kardar 6.5 (Quantum macrostates))

Thus far, we dove into quantum stat-mech using the replacement recepie

\[ Z=\sum_{n} e^{-\beta E_{n}}, \hat{H}\left|E_{n}\right\rangle=E_{n}\left|E_{n}\right\rangle \]

This is correct, but its worth developing the general formalism.

Classically, we have a distribution over microstates ” \(\mu\) “, \(p(\mu)\), with \(\langle O\rangle=\sum_{\mu} p(\mu) O(\mu)\)

The distribution \(p(\mu)\) depends on ensemble,

\[ p(\mu)=\frac{1}{\Omega} \delta(H(\mu)-E), \qquad p(\mu)=\frac{1}{z} e^{-\beta H(\mu)} \text {, etc. } \]

The quantum analog is the “density matrix”

\[\begin{split} \begin{aligned} & \hat{\rho}=\sum_{\alpha} p_{\alpha}|\alpha\rangle\langle\alpha|, \\ & \langle\hat{O}\rangle=\sum_{\alpha} p_{\alpha}\langle a|\hat{\theta}| \alpha\rangle \\ & \operatorname{Tr}(\hat{j})=1 \quad \sum_{\alpha} \rho_{\alpha}=1 \\ & =\operatorname{Tr}(\hat{\rho} \hat{O}) \end{aligned} \end{split}\]

The various eq. ensembles are then taken diagonal in \(\hat{H}\) :

\[\begin{split} \begin{aligned} & \left.\hat{\rho}_{E}=\frac{1}{\Omega} \sum_{n} \delta\left(E_{n}-E\right)\left|E_{n}\right\rangle<E_{n} \right\rvert\, \\ & \hat{\rho}_{\beta}=\frac{1}{Z} \sum_{n} e^{-\beta E_{n}}\left|E_{n}\right\rangle\left\langle E_{n}\right| \\ & \hat{\rho}_{\beta, \mu}=\frac{1}{Q} \sum_{n} e^{-\beta\left(E_{n}-\mu N_{n}\right)}\left|E_{n}\right\rangle\left\langle E_{n}\right) \\ & \quad \tau \quad[\hat{H}, \hat{N}]=0 \end{aligned} \end{split}\]

N-conserved

QH analog of a distribution over mierostates is the “density matrix”:

Consider a system being in QH micratate \(\left|\psi_{\alpha}\right\rangle\) with probability \(p_{\alpha}\). This can be summarized by the density matrix

\[ \hat{\rho} \equiv \sum_{\alpha} \rho_{\alpha}\left|p_{\alpha}\right\rangle\left\langle\psi_{\alpha}\right|, \quad \sum \rho_{\alpha}=1=\operatorname{Tr} (\hat{\rho}) \]

Expectation value

\[ \begin{aligned} & \langle\hat{O}\rangle_{\hat{\rho}}=\sum_{\alpha} p_{\alpha}\left\langle\psi_{\alpha}|\hat{O}| \psi_{\alpha}\right\rangle \stackrel{(*)}{=} \operatorname{Tr}(\hat{O} \hat{\rho}) \end{aligned} \]

To see (*), note that

\[ \operatorname{Tr}(\hat{O} \hat{\rho})=\sum_{\alpha_{1}}\langle\alpha|\hat{O} \hat{\rho}| \alpha)=\sum_{\alpha \beta}\langle\alpha|\hat{O}| \beta\rangle \underbrace{\langle\beta|\hat{\rho}| \alpha\rangle}_{=p_{\alpha}\delta_{\alpha, \beta}} =\sum_{\alpha} p_{\alpha}\langle\alpha|\hat{O}| \alpha\rangle \]

\(\hat{\rho}\) can be expanded in any orthogonal basis

\[ \hat{\rho}=\sum_{i, j} \rho_{i, j}|i\rangle\langle j| \;, \qquad \rho_{i j}=\langle i|\hat{\rho}| j\rangle \]

Note that not all \(\rho_{i j}\) are valid density matrices. Requirements:

(1) \(\operatorname{Tr}(\hat{\rho})=1\)

(2) \(\hat{\rho}^{+}=\sum_{\alpha} p_{\alpha}^{*}\left(\left|\psi_{\alpha}\right\rangle\left\langle\psi_{\alpha}\right|\right)^{*}=\hat{\rho} \). (Hermitian)

(3) For any \(|v\rangle\),

\[\begin{split} \begin{gathered} \langle v|\hat{\rho}| v\rangle=\sum_{\alpha} p_{\alpha}\left\langle v \mid \psi_{\alpha}\right\rangle\left\langle\psi_{\alpha} \mid v\right\rangle \\ =\sum_{\alpha} p_{\alpha}\left|\left\langle v \mid \psi_{\alpha}\right\rangle\right|^{2} \geqslant 0 \end{gathered} \end{split}\]

I.e. \(\hat{\rho}\) is “positive semi-definite”

In a general basis, the density matrix typically has the following structure:

By positive semi-definiteness and the definition of the d.m.s., we have

\[ 0 \leq \rho_{i i} \leq 1, \sum \rho_{i i}=1 . \]

So the “populations”look like probabilities. However, this is basis dependent!

For example, for a single spin, we can convert matrix from the up-down basis to the left- right basis, we obtain

\[\begin{split} \frac{1}{2}\left[\begin{array}{ll} |\uparrow\rangle & |\downarrow\rangle \\ \frac{1}{2} & \frac{1}{2} \\ \frac{1}{2} & \frac{1}{2} \end{array}\right] \leftrightarrow\left[\begin{array}{cc} |\rightarrow\rangle & |\leftarrow\rangle \\ 1 & 0 \\ c & 0 \end{array}\right] \end{split}\]

There is always a special basis in which coherence part vanishes, e.g.

\[ \hat{\rho}=\operatorname{diag}\left(p_{i}\right) . \]

We will mostly assume that this basis is time -independent, so we can heuristically work with population \(p_{i}\).

\[ \operatorname{Tr}[A, B]=A_{i j} B_{j i}-B_{i j} \cdot A_{j i}=0 \text {. finite ! } \quad \mid \quad \underbrace{}_{-i \hbar}\langle\psi|=\langle\psi| \hat{H} \]

Dynamics:

Exercise: Show that the Schrödinger equation

\[i\hbar \partial_{t}|\varphi\rangle=\hat{H}|\psi\rangle\]

implies the von Neumann equation

\[i \hbar \partial_{t} \hat{\rho}=[\hat{H}, \hat{\rho}]\]

So

\[ i\hbar \partial_{t} \sum_{i} \rho_{i, i}=i \hbar \partial_{t}(\operatorname{Tr}(\hat{\rho}))=\operatorname{Tr}([\hat{H}, \hat{\rho}])=0 \text {. } \]

(since \(\operatorname{Tr}([A, B])=0\) )

Thus, \(\operatorname{Tr}(\hat{\rho})=1\) is respected under unitarian time evolution.

For time-independent Hamiltonian, we have \(\psi(t)=\hat{U}(t) \psi(0)\) with \( \hat{U}=e^{-i \hat{H} t(\hbar} \) and so

\[ \hat{\rho}(t)=\hat{U}(t) \hat{\rho}_{0} \hat{U}^{+} (t) \]

( For a unitarian evolution in a open systan, one obtaines the so-called Lindblad quation which has aspects of a Masters equation)

Density matrices in equilbrium.

The various equilibrium ensembles are then taken to be diagonal in the \(\hat{H}\) basis:

\[\begin{split} \begin{aligned} & \hat{\rho}_{E}=\frac{1}{\Omega} \sum_{n} \delta\left(E_{n}-E\right)\left|E_{n}\right\rangle\left\langle E_{n}\right| \\ & \hat{\rho}_{\beta}=\frac{1}{Z} \sum_{n} e^{-\beta E_{n}}\left|E_{n}\right\rangle\left\langle E_{n}\right| \\ & \hat{\rho}_{\beta, \mu}=\frac{1}{Q} \sum_{n} e^{-\beta\left(E_{n}-\mu N_{n}\right)} \left|E_{n}\right\rangle\left\langle E_{n}\right\rangle \end{aligned} \end{split}\]

(Note that \([\hat{H}, \hat{N}]=0\)). The density matrices can be conveniently summarized as operator expressions, eg.

\[ \hat{\rho}_{\beta}=\frac{1}{Z} e^{-\beta \hat{H}}, \quad Z=\operatorname{Tr}\left(e^{-\beta \hat{H}}\right) \]

This points to an interesting observation: with

\[ \beta=\frac{1}{K_{B} T} \rightarrow i t / A, \quad e^{-\beta \hat{H}} \rightarrow e^{-i t \hat{H} / h}=\text { Time Evolution! } \]

Thus the thermal density matrix is formally equivalent to “imaginary time evolution”,

\[ \hat{\rho}_{\beta}=\frac{1}{Z} \hat{U}(t=-i \beta) \]

This mathematical coincidence (??) gives rise to a number of technical tricks (Wick rotation | Euclidean path integral) important to quantum field and many body theory.

For an ensemble like \(\hat{H}-\mu \hat{N}\), where \([\hat{H}, \hat{N}]=0\), it is easy to show that

\[\begin{split} \begin{aligned} & \langle E\rangle=\langle\hat{H}\rangle=\frac{1}{Z} \partial_{(-\beta)} \operatorname{Tr}\left(e^{-\beta \hat{H}}\right) \\ & =\frac{1}{Z} \partial_{-\beta} Z=-\partial_{\beta} \ln (Z) \end{aligned} \end{split}\]

Note that \(Z\) is just a number (as in classical stat mech) in contrast to \(\hat{H}\) and \(\hat{\rho}\).

So, fortunately, all our thermo results like \(F=-\frac{1}{\beta} \ln (Z), \mu=-\frac{\partial F}{\partial N}\), etc and Maxwell’s relations / thermodynamic inequalities are unchanged by quantum effects. It’s just harder to compute \(Z\) because we need eigenspectrom \(\hat{H} \longrightarrow E_{n}\), which is itself hard!

The Many-Body Hilbert space

The sum \(\sum_{n} e^{-\beta E_{n}}\left|E_{n}\right\rangle\left\langle E_{n}\right|\) runs over the Hilbert space \(\left|E_{n}\right\rangle \in \mathcal{H}\), so its worth understanding the structure of \(\mathcal{H}\) in a many-body system. 3 common cases:

(1) Spins / quits \(|\uparrow \uparrow \downarrow \uparrow \cdots\rangle\)

(2) Bosons \(\left.\psi\left(x_{1}, x_{2}\right)=\psi\left(x_{2}, x_{1}\right)\right\}\) 1st/2nd

(3) Fermions \(\left.\psi\left(x_{1}, x_{2}\right)=-\psi\left(x_{2}, x_{1}\right)\right\}\) quantization

We start with (1) amd some rudiments of quantum information- (2, 3) are covered in future lectures.

Spins 1 “qu-dits”

The Hilbert space of \(s=1 / 2\) spin is

\[\begin{split} \begin{aligned} & H=\operatorname{span}\left(\{|\uparrow\rangle,|\downarrow\rangle \})=\mathbb{C}^{2}\right. \\ & |\psi\rangle=\psi_{\tau}|\lambda\rangle+\psi_{+}|\downarrow\rangle-\left(\begin{array}{l} \psi_{\uparrow} \\ \psi_{\downarrow} \end{array}\right) \end{aligned} \end{split}\]

In quant-info, call it “qubit”, \(\{|0\rangle,|1\rangle\}\)

Operators spanned by \(2 \times 2\) Paulies \(\sigma^{x / y / z}\), often denoted \(X, Y, Z\) in quant-info.

For spin S,

\[\begin{split} \begin{aligned} \mathcal{H}= & \operatorname{span}\left(\{|m\rangle \}\right)=\mathbb{C}^{2 S+1} \\ & -S \leq m \leq S \end{aligned} \end{split}\]

or “qudit”, \(d=2s+1\).

What about 2 spins?

Now,

\[\begin{split}H^{(2)}=\operatorname{span}(\{|\uparrow \uparrow\rangle,|\uparrow \downarrow\rangle,|\downarrow \uparrow\rangle,|\downarrow \downarrow\rangle\})\\ =\mathbb{C}^{2} \otimes \mathbb{C}^{2}=\mathbb{C}^{2*2=4} \end{split}\]

The 2-spin \(\mathcal{H}^{(2)}=\mathcal{H}^{(1)} \otimes \mathcal{H}^{(1)}\) is a “tensor product”,

\[\begin{split} |\psi\rangle=\sum_{\sigma_{1}, \sigma_{2}=\tau / \downarrow} \psi_{\sigma_{1}} \sigma_{2}\left|\sigma_{1} \sigma_{2}\right\rangle=\left(\begin{array}{l} \psi_{\uparrow \pi} \\ \psi_{\tau \downarrow} \\ \psi_{\downarrow \pi} \\ \psi_{\downarrow \downarrow} \end{array}\right) \end{split}\]

We can extend to \(N\)-spins:

\[\begin{split} \begin{aligned} \mathcal{H}^{(N)} & =3 \operatorname{span}\left(\left\{| \sigma_{1}, \sigma_{2}, \cdots, \sigma_{N}>\right\}\right) \\ & =\mathbb{C}^{2} \otimes \mathbb{C}^{2} \otimes \mathbb{C}^{2} \otimes \cdots \\ & =\mathbb{C}^{2^{N}} \end{aligned} \end{split}\]

Key: dimension of \(\mathcal{H}^{(N)}=2^{N}\), NOT \(2 * N\)

So to encode a generic microstate

\[ |\psi\rangle=\sum_{\sigma_{1}, \sigma_{2} \ldots \sigma_{N}} \psi_{\sigma_{1}, \sigma_{2},\cdots} \mid\left\{\sigma\}\right\rangle \]

we need \(2^{N}\) complex numbers \(\psi_{\{\sigma\}}\)

Exponentially large in N!!

In contrast, a microstate of the classical Ising model, \(\mu=\sigma_{1} \sigma_{2} \cdots \sigma_{N}\), requires \(N\) classical bits. So your Monte-Carlo simulation of \(20 \times 20=400\) spins needed \(<1\) K B RAM. If we store quantum \(\psi_{\{\sigma\}}\) using 128-bit precision floating point, need

\[\begin{split} \begin{aligned} 128 \cdot 2^{400} & =3 \cdot 10^{122} \mathrm{bits} \\ & =3 \cdot 10^{112} \mathrm{~GB} \end{aligned} \end{split}\]

So… simulating quantum w/ classical hard!

Many body operators

The tensor product structure also extends to operators:

\[\begin{split} \begin{array}{r} \hat{Z}_{i}\left|\sigma_{1}, \sigma_{2}, \cdots \sigma_{N}\right\rangle \equiv \sigma_{i}\left|\sigma_{1}, \cdots \sigma_{N}\right\rangle \\ \left(\hat{Z}_{i} \cdot \hat{Z}_{j}\right)\left|\sigma_{1}, \cdots, \sigma_{N}\right\rangle \equiv \sigma_{i} \cdot \sigma_{j}\left|\sigma_{1}, \cdots, \sigma_{N}\right\rangle \end{array} \end{split}\]

Since \(\hat{X}=\left[\begin{array}{ll}0 & 1 \\ 1 & 0\end{array}\right]\), with notation \(\sigma\in\{1 ,-1\}\),

\[\begin{split} \begin{aligned} & \hat{X}|\sigma\rangle=|-\sigma\rangle \quad \text{("Bitflip")} \\ & \hat{X}_{1}\left|\sigma_{1}, \sigma_{2}, \cdots \sigma_{N}\right\rangle=\left|-\sigma_{1}, \sigma_{2}, \cdots\right\rangle \end{aligned} \end{split}\]

The \(\hat{Z}_{i}, \hat{X}_{i}\) are \(2^{N} \times 2^{N}\) matrices:

Ex: for \(N=2, \quad\) Hilbert space\(=\{|1,1\rangle, \mid 1,-1\rangle,|-1,1\rangle,|-1,-1\rangle\}\)

\[\begin{split} \begin{array}{ll} \hat{Z}_{1}=\left[\begin{array}{llll} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & -1 \end{array}\right] & \hat{Z}_{2}=\left[\begin{array}{llll} 1 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & -1 \end{array}\right] \\ \\ \hat{X}_{1}=\left[\begin{array}{llll} 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \end{array}\right] & \hat{X}_{2}=\left[\begin{array}{llll} 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \end{array}\right] \end{array} \end{split}\]

Generic \(2^{N} \times 2^{N}\) operators can be built up by adding / multiplying \(\hat{Z}_{i}, \hat{X}_{i}\).

Ex: Transverse - Field Ising model

\[\begin{split} \begin{aligned} & \hat{H}_{T F I} \sum_{i=1}^{N}\left[-J \hat{Z}_{i} \cdot \hat{Z}_{i+1}-g \hat{X}_{i}\right] \\ & \hat{m}=\frac{1}{N} \sum_{i} \hat{Z}_{i} \quad \text { (magnetization) } \end{aligned} \end{split}\]

Typical question:

Given \(\hat{\rho}_{\beta}=e^{-\beta \hat{H} T F I} / Z\), what’s \(\operatorname{Tr}\left(\hat{m} \hat{\rho}_{\beta}\right)\) ?

Not easy, because to obtain \(\hat{\rho}_\beta\) we need to diagonalize \(2^{N} \times 2^{N}\) matrix \(\hat{H}_{T F I}\) ! The biggest computations of this form can handle \(N \sim 20\) or so (see discussions of “quantum supremacy”)

Reduced Density matrices / Entanglement

In classical probability theory, a distribution over two random variables, \(\mu=\{\mu_{1}, \mu_{2}\}, \quad p\left(\mu_{1}, \mu_{2}\right)\), allows us to define a marginal distribution over one:

\[ p_{1}\left(\mu_{1}\right) \equiv \sum_{\mu_{2}} p\left(\mu_{1}, \mu_{2}\right) \]

If we only care about observables \(O_{1}\left(\mu_{1}\right) \quad\left(\right.\) rather than \(\left.O\left(\mu_{1}, \mu_{2}\right)\right)\)

\[\begin{split} \begin{aligned} \left\langle O_{1}\right\rangle & =\sum_{\mu_{1}, \mu_{2}} O_{1}\left(\mu_{1}\right) p\left(\mu_{1}, \mu_{2}\right) \\ & =\sum_{\mu_{1}} O_{1}\left(\mu_{1}\right) p\left(\mu_{1}\right) \end{aligned} \end{split}\]

So, the marginal \(p_{1}\) is sufficient to recover \(\left\langle O_{1}\right\rangle\). Note that, in general,

\[ \rho\left(\mu_{1}, \mu_{2}\right) \neq \rho_{1}\left(\mu_{1}\right) \cdot \rho_{2}\left(\mu_{2}\right) \]

For example,

\[ \rho(\{\sigma \})=\frac{1}{Z} e^{-\beta \sum_{i} \sigma_{i} \cdot \sigma_{i+1}} \]

does not factorize.

In contrast, for

\[\begin{split} \rho(\sigma)=\frac{1}{Z} e^{-\beta \sum_{i} h \sigma_{i}}\\ =\frac{1}{Z} \prod_{i} e^{-\beta h \sigma_{i}} \\ =\rho_{1}\left(\sigma_{1}\right) \rho_{2}\left(\sigma_{2}\right) \ldots \end{split}\]

\(p\left(\mu_{1}, \mu_{2}\right) \neq p_{1}\left(\mu_{1}\right) \cdot p_{2}\left(\mu_{2}\right)\) implies \(\mu_{1}, \mu_{2}\) are “correlated”, and \(\left\langle O_{1} O_{2}\right\rangle \neq\left\langle O_{1}\right\rangle\left\langle O_{2}\right\rangle\).

For a quantum system on \(\mathcal{H}=\mathcal{H}_{1} \bigotimes\mathcal{H}_{2}\), there is an equivalent notion of marginal: the “Reduced Density Matrix”

Let \(\operatorname{span}\left\{\left|i_{1}\right\rangle\right\}=H_{1}=\mathbb{C}^{D_{1}}\), \(i_{1} =1, \cdots, D_{1}\) and \(\operatorname{span}\left(\left\{\left|i_{2}\right\rangle\right\}\right)=H_{2}=\mathbb{C}^{D_{2}}\), \(i_{2} =1, \cdots, D_{2}\)

Then \(\quad \mathcal{H}=\operatorname{span}\left(\left\{\left|i_{1}, i_{2}\right\rangle\right\}\right)=\mathbb{C}^{D_{1} \cdot D_{2}}\)

A generic state is

\[\begin{split} |\psi\rangle=\sum_{i_{1}, i_{2}} \psi_{i_{1}, i_{2}}\left|i_{1}, i_{2}\right\rangle \\ \hat{\rho}=\sum \rho_{\alpha} |\psi_{\alpha}\rangle\langle\psi_{\alpha}|=\sum \rho_{i_{1}, i_{2} ; j_{1}, j_{2}}| i_{1}, i_{2}\rangle\langle j_{1}, j_{2}| \end{split}\]

Consider an observable \(\hat{O}=\hat{O} \otimes \hat{\mathbb{I}}\), which only makes a measurement on subsystem ,’

\[ \hat{O}_{1}=\sum_{i_{1}, i_{2}, j_{1}} \hat{O}_{i_{1}, j_1}\left|i_{1}, i_{2}\right\rangle \left\langle j_{1}, i_{2}\right| \]

For example, if we have 2 spins, \(\mathcal{H}=\mathbb{C}^{2} \otimes C^{2}\), we might physically separate them to different regions of the lab, and then use Stern-Gerlach to measure only \(\operatorname{spin} 1: \hat{G}=\hat{Z}_{1} \otimes \hat{\mathbb{1}}_{2}\)

We have

\[\begin{split} \left\langle\hat{O}\right\rangle=\operatorname{Tr}\left(\hat{O} \hat{\rho}\right)\\ =\sum O_{i_{1}, j_{1}}\left\langle j_{1}, i_{2}|\hat{\rho}| i_{1}, i_{2}\right\rangle \\ =\sum_{i_{1}, i_{2}, j_{1}} O_{i_{1}, j_{1}} \rho_{j_{1}, i_{2} ; i_{1}, i_{2}} \end{split}\]

If we define the \(D_{1} \times D_{1}\) matrix

\[ \rho_{j_{1} ; i_{1}}^{(1)}\equiv\sum_{i_{2}} \rho_{j, i_{2} ; i_{1}, i_{2}} \]

and the corresponding operator

\[ \hat{\rho}^{(1)}=\sum \rho_{i_{1}, j,}^{(1)}\left|i_{1}\right\rangle\left\langle j_{1}\right| \]

we obtain

\[ \left\langle\hat{O}_{1}\right\rangle=\operatorname{Tr}\left(\hat{O}_{1} \hat{\rho}^{(1)}\right) \]

So, the “partial trace” \(\hat{\rho}^{(1)}=\operatorname{Tr}_{2}(\hat{\rho})\) is the quantum analog of marginal; called “reduced density matrix” for subsystem 1.

Ex: EPR state \(|E P R\rangle=\frac{1}{\sqrt{2}}(|\uparrow \downarrow\rangle+|\downarrow \uparrow\rangle)\)

\[\begin{split} \begin{aligned} \hat{\rho} & =|E P R\rangle\langle E P R| \quad \text { "Pure state" } \\ & =\frac{1}{2}(|\uparrow \downarrow \rangle+|\downarrow \uparrow\rangle)(|\uparrow \downarrow \rangle+|\downarrow \uparrow\rangle) \\ & =\frac{1}{2}\left[\begin{array}{llll} 0 & 0 & 0 & 0 \\ 0 & 1 & 1 & 0 \\ 0 & 1 & 1 & 0 \\ 0 & 0 & 0 & 0 \end{array}\right] \end{aligned} \end{split}\]

So

\[\begin{split} \\ \hat{\rho}_{1} & =\left[\begin{array}{cc} \frac{1}{2} & 0 \\ 0 & \frac{1}{2} \end{array}\right] \end{split}\]

Note \(\hat{\rho}_{1}\) is mixed even though \(\hat{\rho}\) is pure! So

\[ \left\langle X_{1}\right\rangle=\left\langle Y_{1}\right\rangle=\left\langle Z_{1}\right\rangle=0 \]

even though \(\left\langle Z_{1} Z_{2}\right\rangle=-1\).

The two EPR spins are correlated even though the system (as a whole) is in a definite (pure) state.

Entropy:

\[\quad S=-\operatorname{Tr}(\hat{\rho} \ln (\hat{\rho}))\]

For \(\hat{\rho}=\sum p_{\alpha}|\alpha\rangle\langle\alpha|, \quad\langle\alpha \mid \beta\rangle=\delta_{\alpha \beta}\).

\[ S=-\sum_\alpha p_{\alpha} \ln p_{\alpha} \]

We call a state \(\hat{\rho}\) “pure” if

\[ S[\hat{\rho}]=0 \Longleftrightarrow \hat{\rho}=|\psi\rangle\langle\psi| \]

Otherwise, \(\quad S[\hat{\rho}]>0 \Rightarrow \hat{\rho}\) “mixed

Entanglement:

Let \(\mathcal{H}=\mathcal{H}_{1} \otimes \mathcal{H}_{2}\), and \(\quad|\psi\rangle \in \mathcal{H}\).

Clearly, \(\hat{\rho}=|\psi\rangle\langle\psi|\) has \(S[\hat{\rho}]=0\).

However, the RDM \(\hat{\rho}_{1}=\operatorname{Tr}_{2}(\hat{\rho})\) can be mixed. We call \(S\left[\hat{\rho}_{1}\right]\) the “entanglement entropy”.

Even though, the system is in a single fixed pure state \(|\psi\rangle\), measurements between \(1\) and \(2\) are correlated:

\[ \left\langle O_{1} O_{2}\right\rangle \neq\left\langle O_{1}\right\rangle\left\langle O_{2}\right\rangle \]

Our earlier example was \(|\psi\rangle=|E P R\rangle\), where \(\hat{\rho}_{1}=\text{diag}\left[\begin{array}{ll}1 / 2 & 1 / 2\end{array}\right] \rightarrow S=\ln (2)\) : one bit of entanglement.

Note

Check out recent colloquium by Matthew Fisher on phase transitions in entanglement entropy.

Measures of quantum information

Many but not all results of classical information theory transfer to the quantum case.

For \(\quad H_{A} \otimes \mathcal{H}_{B}=\mathcal{H}_{A B}, \quad \hat{\rho}_{A}=\operatorname{Tr}_{B}\left(\hat{\rho}_{A B}\right)\), one defines the “mutual information”, which is a measure of total correlation

\[ I(A: B)=S\left[\hat{\rho}_{A}\right]+S\left[\hat{\rho}_{B}\right]-S\left[\hat{\rho}_{A B}\right] \geq 0 \]

\[ I(A: B)=0 \quad \Longleftrightarrow \hat{\rho}_{A B}=\hat{\rho}_{a} \otimes \hat{\rho}_{B} \]

Important property:

\[\begin{split} \begin{aligned} & I(A: B C) \geq I(A: B) \\ & I(A: B C)-I(A: B)=S_{A}+S_{B C}-S_{A B C} \\ & -S_{A}-S_{B}+S_{A B} \\ & =S A B+S_{B C}-S_{A B C}-S_{B} \geq 0 \end{aligned} \end{split}\]

“Strong subadditivity of quantum information” - not easy to prove.