Why isolated systems are not enough · Quantum Mechanics

Intuition

No real system is alone: an atom meets stray photons, a qubit its wiring, a nuclear spin the magnetic noise of its neighbours. The honest description takes the system and its environment together, in a pure state that evolves unitarily, and then looks at the system alone through the partial trace. Two things follow. The system’s state is in general mixed even though the whole is pure — and every mixed state arises this way: any density matrix is the reduced state of a pure state of the system and a partner. And the system’s own evolution is no longer unitary: its purity can fall and its entropy grow, in a way no Hamiltonian of the system alone reproduces.

A drop of ink in a glass of water: no molecule of ink is lost, but once spread through the water it cannot be gathered back by stirring. What a system gives to its environment is not destroyed either — only spread where no one can use it.

System and environment

Let the system start in $\ket{\psi}$ and the environment in $\ket{e_{0}}$ . The whole evolves unitarily; the system alone is its reduced state:

\begin{aligned}\ket{\Psi(t)}&=\hat{U}(t)\,\ket{\psi}\ket{e_{0}}\\ \hat{\rho}(t)&=\operatorname{tr}_{E}\ket{\Psi(t)}\bra{\Psi(t)}\end{aligned}

Properties

The whole, system with environment, evolves unitarily; the system is described by $\hat{\rho}=\operatorname{tr}_{E}$ of the whole.
Purification: $\hat{ρ} = \sum_{n} p_{n} ∣ n ⟩ ⟨$ is the reduced state of , with orthonormal of a partner — one for each nonzero .

Purification

\begin{aligned}&\hat{\rho}=\sum_{n}p_{n}\ket{n}\bra{n}\ \Longrightarrow\ \hat{\rho}=\operatorname{tr}_{E}\ket{\Psi}\bra{\Psi}\\ &\text{for}\ \ket{\Psi}=\sum_{n}\sqrt{p_{n}}\,\ket{n}\ket{e_{n}}\end{aligned}

Diagonalise the density matrix, pair each eigenvector with an orthonormal state of a partner, and weight each pair by the square root of its eigenvalue. The result is normalised, and tracing out the partner removes every cross term, because the partner’s states are orthogonal: what remains is the density matrix.

Proof steps

\hat{\rho}=\sum_{n}p_{n}\ket{n}\bra{n}

Diagonalise $\hat{\rho}$ by the spectral theorem.

\ket{\Psi}=\sum_{n}\sqrt{p_{n}}\,\ket{n}\ket{e_{n}},\qquad \braket{e_{m}|e_{n}}=\delta_{mn}

Pair each eigenvector with an orthonormal state of a partner.

\braket{\Psi|\Psi}=\sum_{m,n}\sqrt{p_{m}p_{n}}\braket{m|n}\braket{e_{m}|e_{n}}=\sum_{n}p_{n}=1

It is normalised.

\operatorname{tr}_{E}\ket{\Psi}\bra{\Psi}=\sum_{m,n}\sqrt{p_{m}p_{n}}\,\braket{e_{m}|e_{n}}\ket{n}\bra{m}=\hat{\rho}

Trace out the partner: only $m=n$ survives.

Applications

Practice

Never Alone

A system is open when it interacts with surroundings whose state we do not follow. Almost every real system is open to some degree.

Try it

Why must a real system usually be treated as open?