The physical picture of decoherence

Last section’s Lindblad equation computed the exponential decay of the coherences down to the last detail, but one question got swept under the rug: the jump operator $\hat L$ was our choice. Pick $\hat\sigma_z$ and you erase the coherences along $x$ and $y$; pick something else and you erase something else. Who actually decides which basis the environment attacks?

Behind this question stands the oldest cold case in quantum mechanics.

The phenomenon first: where is the cat

The Schrödinger equation is linear: if an atom can be in a superposition of “decayed + not decayed”, then the cat coupled to it should be in a superposition of “dead + alive”. Yet no one has ever seen a superposed cat — indeed no one has ever seen so much as a dust grain in a superposition of “here + there”. Meanwhile, in the same laboratories, electrons, photons, even molecules tens of thousands of atomic masses heavy interfere perfectly cleanly.

Where is the dividing line? Why can the microscopic superpose and the macroscopic cannot? The old textbooks answered “measurement causes collapse”, but that merely renames the problem: there is no goggled observer standing next to the cat — who is measuring it?

The tools of this chapter are now enough for a physical answer: there is a measurer, and it never takes a day off — the environment itself.

The environment is a measurer that never shuts up

The key mechanism in one sentence: the state of the system gets written into the state of the environment. A dust grain at position $x_1$: the air molecules and photons bouncing off it fly away in certain directions. The grain at $x_2$: they fly off in other directions. Every scattered particle carries a record of “where the grain is” — the environment is photographing the system nonstop, never developing the film, and never pausing.

Coherence = the overlap of the environment's photographsbasic~8 minexpand

Step 1: a single scattering event. The system has two position states $\ket{x_1},\ket{x_2}$; the environment (one photon) starts in $\ket{E_0}$. The interaction writes the system’s state into the photon’s state (the system is heavy and barely recoils — this kind of “leave a record, change nothing” interaction is precisely the continuous version of chapter 9’s CNOT gate):

$$\ket{x_1}\ket{E_0}\to\ket{x_1}\ket{E_1},\qquad \ket{x_2}\ket{E_0}\to\ket{x_2}\ket{E_2}\tag{10.5.1}$$

By linearity, a superposition is carried into an entangled state:

$$\big(a\ket{x_1}+b\ket{x_2}\big)\ket{E_0} \;\to\; a\ket{x_1}\ket{E_1}+b\ket{x_2}\ket{E_2}\tag{10.5.2}$$

Step 2: look only at the system. We do not track the departed photon — trace out the environment (the routine of section 3.10; each cross term picks up an environment inner product):

$$\hat\rho_S= |a|^2\ket{x_1}\bra{x_1}+|b|^2\ket{x_2}\bra{x_2} +\Big[ab^*\braket{E_2}{E_1}\,\ket{x_1}\bra{x_2}+\text{h.c.}\Big]\tag{10.5.3}$$

The diagonal is untouched; the coherences have been multiplied by the factor $\braket{E_2}{E_1}$. The meaning of this inner product is plain: how hard the two “photographs” are to tell apart.

• $\braket{E_2}{E_1}=1$: the photon never saw where the grain was; the photographs are identical — the coherence is unscathed.
• $\braket{E_2}{E_1}=0$: the photographs are perfectly distinguishable; the environment “knows” the position — the coherences are wiped to zero, and what remains is exactly the diagonal-only classical mixture of section 10.1.

Step 3: scattering doesn’t happen just once. Every second, countless photons and molecules strike, each stamping its record independently. After $N$ scatterings,

$$\rho_{12}(N)=\rho_{12}(0)\prod_{n=1}^N\braket{E_2^{(n)}}{E_1^{(n)}}\tag{10.5.4}$$

Suppose each overlap is roughly the same number, with modulus $|\braket{E_2}{E_1}|=1-\epsilon$, and the scattering rate is $R$:

$$|\rho_{12}(t)|\approx|\rho_{12}(0)|\,(1-\epsilon)^{Rt} =|\rho_{12}(0)|\,\ee^{-t/\tau_{\text{dec}}}, \qquad \frac{1}{\tau_{\text{dec}}}\approx R\,\epsilon\tag{10.5.5}$$

The exponential decay is not an assumption — it is the inevitable result of multiplying many independent little labels together. Even if a single scattering sees almost nothing ($\epsilon\sim10^{-3}$), enough repetitions finish the coherence off regardless — and in the macroscopic world $R$ is unimaginably large.

For an object with a continuous position distribution, the same reasoning gives the standard form of scattering decoherence:

$$\rho(x,x';t)=\rho(x,x';0)\,\ee^{-\Lambda t\,(x-x')^2}\tag{10.5.6}$$

$\Lambda$ is called the localisation rate. The farther apart the superposition ($|x-x'|$ larger), the easier the photographs are to distinguish, and the faster it dies.

Who picks the pointer basis: the interaction decides

Now the opening question can be answered. The environment erases coherence between positions, rather than (say) between momenta, for a reason that lies not in the system but in the shape of the interaction: scattering, collisions, the Coulomb force — almost every coupling acts through position. What the environment can distinguish are the states on which the interaction takes different values.

The general rule: the states that (approximately) commute with the interaction Hamiltonian — and are therefore left almost undisturbed by the environment’s relentless photography — are the ones that can leave stable records. They are called the preferred basis or pointer states (the name comes from instrument pointers). Every other superposition entangles with the environment immediately and, as a state of the system’s own, vanishes forthwith. The jump operator $\hat L$ that “we chose” last section is, physically, chosen by exactly this rule — for macroscopic objects $\hat L$ is almost always a position-type operator, and the cat’s pointer states are “dead” and “alive”, not their superpositions.

The numbers are the point: absurdly fast

∎Key formulas

• Environmental labelling

  $$(a\ket{x_1}+b\ket{x_2})\ket{E_0}\to a\ket{x_1}\ket{E_1}+b\ket{x_2}\ket{E_2}$$

  The interaction writes the system state into the environment — the environment takes photographs

• The fate of the coherences

  $$\rho_{12}\to\rho_{12}\,\braket{E_2}{E_1}$$

  Overlap = how distinguishable the photographs are; orthogonal means coherence hits zero

• Exponential decay

  $$\rho(x,x^{\prime};t)=\rho(x,x^{\prime};0)\,\ee^{-\Lambda t(x-x^{\prime})^2}$$

  The inevitable product of many independent labels; the farther apart, the faster it dies

• Decoherence vs relaxation

  $$\frac{\tau_{\text{dec}}}{\tau_{\text{rel}}}\sim\Big(\frac{\lambda_{\text{th}}}{\Delta x}\Big)^2\sim10^{-41}\ (1\,\text{g},\ 1\,\text{cm})$$

  Macroscopic objects lose phase about 40 orders of magnitude faster than energy

?Self-check4 questions

1. 1.

   After one scattering event the system’s coherence is multiplied by ⟨E₂|E₁⟩. The physical meaning of this factor is:

   • The overlap of the environment’s two "photographs": the harder to distinguish, the more coherence survives
   • The overlap of the system’s two position states
   • The probability that the scattering occurred
   • The energy change of the environment

   Submit
2. 2.

   Suppose each scattering multiplies the coherence’s modulus by 0.999 (a single event sees almost nothing). How many scatterings does it take to press the coherence down to 1% of its initial value?

   scatterings10000% relative tolerance

   Submit
3. 3.

   Why is the pointer basis almost always (approximately) the position eigenstates?

   • Because scattering, collisions and the like couple through position, so position is what the environment distinguishes
   • Because the position operator is Hermitian
   • Because momentum is not conserved in quantum mechanics
   • Because human eyes can only see position

   Submit
4. 4.

   Which statements about decoherence and the measurement problem are correct? (Select all that apply.)

   Select all that apply

   • Decoherence explains why interference of macroscopic superpositions is unobservable
   • The diagonal mixture decoherence produces does not specify which outcome "actually happened"
   • Decoherence requires adding a new stochastic term beyond the Schrödinger equation
   • "Why does a single measurement have a unique outcome" remains an open interpretational question within the decoherence framework

   Submit

What comes next

Decoherence moved the coherence off the system into the system-environment correlations, and picked out the pointer states along the way. But the classical world has one stranger property still unexplained: objectivity. Ten people look at the same dust grain and see exactly the same position, and none of them disturbs it — yet quantum mechanics plainly says measurement disturbs the system, and unknown states cannot be cloned. What entitles everyone to “all know”?

The next section is this chapter’s outlook: quantum Darwinism — classicality = redundant backup copies of information in the environment.