Decoherence

Intuition

Decoherence is what happens to a superposition when the environment keeps a record. Let a qubit in $a\ket{0}+b\ket{1}$ meet an environment that ends in $\ket{e_{0}}$ if the qubit is 0 and in $\ket{e_{1}}$ if it is 1. The populations stay $|a|^{2}$ and $|b|^{2}$ , but the coherence is multiplied by the overlap $\braket{e_{1}|e_{0}}$ . Once the environment tells 0 from 1, the overlap is near zero and the qubit behaves as a mixture: interference is gone. A large environment takes many small records, their overlaps multiply, and the coherence falls fast and never comes back. A macroscopic body has astronomically many such witnesses, which is why its superpositions are never seen. Decoherence does not, by itself, say which outcome occurs; that part of the measurement problem it leaves open.

A whisper in an empty room can still be taken back; in a crowded room each listener catches a fragment, and the fragments together can never be recalled. Decoherence spreads a which-way record among many listeners.

The size of a qubit’s coherence against time, when each of $N$ partners in the environment takes a record at its own rate. With $N=1$ the overlap $|\cos t|$ comes back to 1; with $N=8$ the product of the overlaps falls and stays near zero.

An interaction that records the qubit’s state in the environment without changing it:

\begin{aligned}(a\ket{0}+b\ket{1})\ket{e}&\to a\ket{0}\ket{e_{0}}+b\ket{1}\ket{e_{1}}\\ \rho_{01}&\to\rho_{01}\,\braket{e_{1}|e_{0}}\end{aligned}

Properties

The populations $|a|^{2}$ and $|b|^{2}$ are unchanged: the environment records which state, without moving it.
$|\braket{e_{1}|e_{0}}|\le1$ : a coherence can only shrink, and it vanishes when the records are orthogonal.

Coherence times overlap

\rho_{01}=ab^{*}\braket{e_{1}|e_{0}},\qquad \rho_{00}=|a|^{2},\ \ \rho_{11}=|b|^{2}

Let the interaction copy the qubit’s value into the environment. By linearity the superposition becomes an entangled state; tracing out the environment leaves the diagonal alone and multiplies each off-diagonal term by the overlap of the two records.

Proof steps

(a\ket{0}+b\ket{1})\ket{e}

A product of the qubit and its environment.

\hat{U}:\ \ket{0}\ket{e}\to\ket{0}\ket{e_{0}},\quad \ket{1}\ket{e}\to\ket{1}\ket{e_{1}}

The environment records the value without changing it.

\ket{\Psi}=a\ket{0}\ket{e_{0}}+b\ket{1}\ket{e_{1}}

Linearity.

\begin{aligned}\hat{\rho}&=|a|^{2}\ket{0}\bra{0}+|b|^{2}\ket{1}\bra{1}\\ &\quad+ab^{*}\braket{e_{1}|e_{0}}\ket{0}\bra{1}+\text{h.c.}\end{aligned}

Trace out the environment; h.c. is the Hermitian conjugate of the term before it.

\rho_{01}=ab^{*}\braket{e_{1}|e_{0}}

Only the coherence changes, by the overlap of the records.

Applications

Practice

Coherence Times Overlap

When the environment records the state of a qubit, its coherence is multiplied by the overlap of the two records.

\rho_{01}\to\rho_{01}\,\braket{e_{1}|e_{0}}

Try it

A qubit in $\tfrac{1}{\sqrt2}(\ket{0}+\ket{1})$ meets an environment whose records have $\braket{e_{1}|e_{0}}=0.4$ . What is $|\rho_{01}|$ afterwards?