3.10 A first look at the density operator

The previous nine sections quietly assumed that the system has one definite state vector $\ket{\psi}$ .

This section handles the two situations where that is not enough — and in real experiments those two situations are practically the norm.

Two kinds of “not knowing”

The picture

Case one: classical ignorance

Silver atoms flying out of an oven with completely random spin directions. You know each atom has a definite spin state — you just don’t know which.

This is classical probability: 50% $\ket{\uparrow}$ , 50% $\ket{\downarrow}$ .

Note that this is not the superposition $\frac{1}{\sqrt2}(\ket{\uparrow}+\ket{\downarrow})$ — that is a definite state with spin along $+x$ , entirely different from “not knowing whether it is up or down”. How to tell them apart is spelled out below.

The mathematics

Case two: it has no state of its own

Take two particles in the entangled state

\ket{\Psi}=\frac{1}{\sqrt2}\big(\ket{\uparrow}_A\ket{\downarrow}_B-\ket{\downarrow}_A\ket{\uparrow}_B\big)\tag{3.10.1}

The whole has a definite state. But what state does particle A have on its own?

The answer: none. There exists no $\ket{\psi}_A$ such that $\ket{\Psi}=\ket{\psi}_A\otimes\ket{\phi}_B$ .

This is not a matter of ignorance — even with complete information about the whole, A still has no state vector of its own.

Both cases call for the same new tool.

The density operator

Pure state (you know it is $\ket{\psi}$ ):

\hat\rho=\ket{\psi}\bra{\psi}\tag{3.10.2}

This is the projection operator of section 3.2.

Mixed state (in $\ket{\psi_i}$ with probability $p_i$ ):

\hat\rho=\sum_i p_i\ket{\psi_i}\bra{\psi_i},\qquad p_i\ge0,\ \sum_ip_i=1\tag{3.10.3}

Verifying the trace formula ⟨A⟩ = Tr(ρÂ)basic~3 min

Take any orthonormal basis $\{\ket{n}\}$ :

\operatorname{Tr}(\hat\rho\hat A)=\sum_n\bra{n}\hat\rho\hat A\ket{n} =\sum_n\sum_ip_i\braket{n}{\psi_i}\bra{\psi_i}\hat A\ket{n}\tag{3.10.6}

Swap the two numbers (both are complex numbers, so they commute), then do the sum over $n$ using completeness $\sum_n\ket{n}\bra{n}=\hat 1$ :

=\sum_ip_i\bra{\psi_i}\hat A\Big(\sum_n\ket{n}\bra{n}\Big)\ket{\psi_i}=\sum_ip_i\bra{\psi_i}\hat A\ket{\psi_i}\tag{3.10.7}

which is exactly the required form. $\blacksquare$

Note that the trace is independent of the choice of basis, so this formula works in any representation.

The three properties of ρ

The picture

1. Hermitian: $\hat\rho^\dagger=\hat\rho$

Because each $\ket{\psi_i}\bra{\psi_i}$ is Hermitian and the $p_i$ are real.

2. Unit trace: $\operatorname{Tr}\hat\rho=1$

\operatorname{Tr}\hat\rho=\sum_ip_i\braket{\psi_i}{\psi_i}=\sum_ip_i=1\tag{3.10.8}

corresponding to “total probability 1”.

3. Positive semidefinite: $\bra{\phi}\hat\rho\ket{\phi}\ge0$

\bra{\phi}\hat\rho\ket{\phi}=\sum_ip_i|\braket{\phi}{\psi_i}|^2\ge0\tag{3.10.9}

corresponding to “probabilities are non-negative”.

The mathematics

The converse also holds

Any operator satisfying these three is the density operator of some physical state. So the three can serve directly as a definition:

A quantum state = a Hermitian, unit-trace, positive-semidefinite operator

This is more general than “a normalised state vector”, and closer to experiment — the most general thing you can prepare in a laboratory is a $\hat\rho$ .

Hermiticity lets you diagonalise:

\hat\rho=\sum_k\lambda_k\ket{k}\bra{k}\tag{3.10.10}

with $\lambda_k\ge0$ , $\sum_k\lambda_k=1$ . These eigenvalues are the classical probability distribution in the eigenbasis.

Purity: telling pure from mixed

The single number Tr(ρ²) is enough

\operatorname{Tr}(\hat\rho^2)=\sum_k\lambda_k^2\tag{3.10.11}

Pure state: a single $\lambda=1$ with the rest zero, so

\operatorname{Tr}(\hat\rho^2)=1\tag{3.10.12}

(Direct check: $\hat\rho^2=\ket{\psi}\braket{\psi}{\psi}\bra{\psi}=\hat\rho$ — a pure state’s $\hat\rho$ is a projection operator.)

Mixed state: $\sum_k\lambda_k=1$ but each is less than 1, and squaring makes them smaller, so

\operatorname{Tr}(\hat\rho^2)<1\tag{3.10.13}

Maximally mixed state (all $\lambda_k=1/d$ in a $d$ -dimensional space):

\operatorname{Tr}(\hat\rho^2)=\frac1d\tag{3.10.14}

For spin 1/2 ( $d=2$ ) that is $1/2$ — the minimum, corresponding to “knowing nothing at all”.

\frac1d\le\operatorname{Tr}(\hat\rho^2)\le1\tag{3.10.15}

This number is called the purity. On the Bloch sphere it has a direct geometric meaning: pure states sit on the surface, mixed states inside, and the maximally mixed state at the centre (section 9.1).

Superposition versus mixture: how experiment tells them apart

This is the most crucial point of the section — and the easiest to muddle when learning.

The picture

Superposition (a pure state)

\ket{+}=\frac{1}{\sqrt2}\big(\ket{\uparrow}+\ket{\downarrow}\big)\tag{3.10.16}

\hat\rho_\text{pure}=\frac12\begin{pmatrix}1&1\\1&1\end{pmatrix}\tag{3.10.17}

$\operatorname{Tr}\rho^2=1$ .

Mixture

“50% up, 50% down”

\hat\rho_\text{mix}=\frac12\begin{pmatrix}1&0\\0&1\end{pmatrix}\tag{3.10.18}

$\operatorname{Tr}\rho^2=1/2$ .

The mathematics

Measure along z: identical

Both give 50% up, 50% down. Measuring only z can never tell them apart.

Measure along x: they split at once

$\hat\rho_\text{pure}$ : it simply is $\ket{+_x}$ , so

P(+_x)=1\tag{3.10.19}

$\hat\rho_\text{mix}$ :

P(+_x)=\operatorname{Tr}(\hat\rho_\text{mix}\ket{+_x}\bra{+_x})=\frac12\tag{3.10.20}

The difference lives entirely in the off-diagonal elements. They are called the coherences, and they are where the capacity to interfere resides. What decoherence does is wipe them out (section 10.5).

One and the same ρ admits infinitely many 'preparation recipes'

The maximally mixed state can be written

\hat\rho=\frac12\ket{\uparrow}\bra{\uparrow}+\frac12\ket{\downarrow}\bra{\downarrow}\tag{3.10.21}

or just as well

\hat\rho=\frac12\ket{+_x}\bra{+_x}+\frac12\ket{-_x}\bra{-_x}\tag{3.10.22}

The two are the same operator (both equal $\frac12\hat 1$ ).

So the question “which pure states was this mixture made from?” has no unique answer — and no measurement can distinguish the recipes even in principle, because every prediction enters only through $\operatorname{Tr}(\hat\rho\hat A)$ , and $\hat\rho$ is the same.

This is a feature, not a bug. The security of quantum teleportation and quantum key distribution rests precisely on it: the reduced state an eavesdropper obtains is identical to the legitimate users’, and no information can be extracted from it (sections 9.5–9.6).

The reduced density operator

Back to case two from the opening.

Extracting a single particle's state from an entangled stateadvanced~6 min

For a composite system $\hat\rho_{AB}$ , trace out the degrees of freedom of B:

\hat\rho_A=\operatorname{Tr}_B\hat\rho_{AB}=\sum_j\bra{j}_B\hat\rho_{AB}\ket{j}_B\tag{3.10.23}

For the singlet $\ket{\Psi}=\frac{1}{\sqrt2}(\ket{\uparrow\downarrow}-\ket{\downarrow\uparrow})$ :

\hat\rho_{AB}=\frac12\big(\ket{\uparrow\downarrow}-\ket{\downarrow\uparrow}\big)\big(\bra{\uparrow\downarrow}-\bra{\downarrow\uparrow}\big)\tag{3.10.24}

Trace over B (taking $\ket{j}_B\in\{\ket{\uparrow}_B,\ket{\downarrow}_B\}$ ). The cross terms contain $\braket{\uparrow}{\downarrow}_B=0$ and all vanish, leaving

\hat\rho_A=\frac12\ket{\uparrow}\bra{\uparrow}+\frac12\ket{\downarrow}\bra{\downarrow}=\frac12\hat 1\tag{3.10.25}

On its own, A is maximally mixed, with $\operatorname{Tr}\rho_A^2=1/2$ .

This deserves a pause: the whole $\ket{\Psi}$ is pure ( $\operatorname{Tr}\rho_{AB}^2=1$ — we hold complete information about it), yet each of its parts is maximally mixed — we know nothing about them.

In the classical world this is impossible: know the whole and you know the parts. In the quantum world, information can reside entirely in the correlations, and in neither party alone.

This is the defining feature of entanglement. Section 9.3 turns “how entangled” into a number using the von Neumann entropy of $\hat\rho_A$ , $S=-\operatorname{Tr}(\rho\ln\rho)$ .

How ρ evolves: the von Neumann equation

Differentiate $\hat\rho=\sum_ip_i\ket{\psi_i}\bra{\psi_i}$ term by term with the Schrödinger equation:

\ii\hbar\frac{\dd\hat\rho}{\dd t}=[\hat H,\hat\rho]\tag{3.10.26}

Note it differs from the Heisenberg equation $\frac{\dd\hat A_H}{\dd t}=\frac{\ii}{\hbar}[\hat H,\hat A_H]$ by a minus sign — because $\hat\rho$ is a state, not an observable, and it evolves in the Schrödinger picture.

From $\hat\rho(t)=\hat U\hat\rho(0)\hat U^\dagger$ it follows that $\operatorname{Tr}(\rho^2)$ is invariant under unitary evolution: a pure state stays pure forever.

So how does decoherence turn pure states into mixed ones? Because decoherence is not the unitary evolution of an isolated system — it is “evolve unitarily together with the environment, then trace the environment out”. The tracing step is not unitary, and that is where the purity is lost. The evolution equation for open systems (the Lindblad equation) is in section 10.4.

Key formulas

Density operator
$\hat\rho=\sum_ip_i\ket{\psi_i}\bra{\psi_i}$
A pure state is the one-term special case
Expectation value
$\langle A\rangle=\operatorname{Tr}(\hat\rho\hat A)$
The trace is representation-independent — compute in any basis
Three properties
$\hat\rho^\dagger=\hat\rho,\quad\operatorname{Tr}\hat\rho=1,\quad\hat\rho\ge0$
The converse holds too — usable directly as a definition
Purity
$\frac1d\le\operatorname{Tr}(\hat\rho^2)\le1$
= 1 for pure states; = 1/d for the maximally mixed state
Reduced density operator
$\hat\rho_A=\operatorname{Tr}_B\hat\rho_{AB}$
A part of an entangled state is necessarily mixed
Von Neumann equation
$\ii\hbar\frac{\dd\hat\rho}{\dd t}=[\hat H,\hat\rho]$
Differs from the Heisenberg equation by a minus sign

Self-check4 questions

What comes next

Chapter 3’s toolkit is now complete. The final section settles the accounts: it redoes, in the new language, the derivations that consumed so many pages in chapter 2.

You will watch some proofs shrink from half a page to three lines — and others turn from “solve a partial differential equation” into “do some algebra”.

A first look at the density operator

Recommended first

After this section you should be able to

Two kinds of “not knowing”

The picture

The mathematics

The density operator

The three properties of ρ

The picture

The mathematics

Purity: telling pure from mixed

Superposition versus mixture: how experiment tells them apart

The picture

The mathematics

The reduced density operator

Key formulas

Self-check4 questions

What comes next

Two kinds of “not knowing”

◑The picture

∑The mathematics

The density operator

The three properties of ρ

◑The picture

∑The mathematics

Purity: telling pure from mixed

Superposition versus mixture: how experiment tells them apart

◑The picture

∑The mathematics

The reduced density operator

∎Key formulas

What comes next

The picture

The mathematics

The picture

The mathematics

The picture

The mathematics

Key formulas