Quantum Mechanics, Explained

Part III — The physics

What measurement predicts, and what it leaves open

An electron's spin is prepared pointing along $x$: an equal superposition of up and down along $z$. It passes a detector that registers up or down along $z$, and the detector clicks "up". What does Schrödinger's equation say happened? And why does nobody ever observe the effects of a detector being in a superposition of "clicked up" and "clicked down"?

Every essay since the formalism essay has used its rules for measurement: results are eigenvalues, probabilities are squared overlaps, and afterwards the system is in the state that matches the result. They have never failed. This last essay of the core sequence asks what those rules are, by treating the detector as a quantum system like any other. The calculation explains a great deal, in particular why the interference between a detector's possible readings cannot be seen in practice. It does not explain why one particular reading occurs, and the essay keeps those two questions apart.

Two kinds of change

Quantum mechanics, as used in the earlier essays, has two rules for how a state changes. Between measurements it follows Schrödinger's equation: smoothly, predictably, and linearly, so a superposition stays a superposition. At a measurement, the measurement rule applies: one result occurs at random, with the Born-rule probability, and the state is replaced by the one matching the result.

The second rule is not derived from the first. It is used, with complete success, to connect the theory to experiments. But a detector is made of atoms, and atoms obey Schrödinger's equation. So it is fair to ask: what does the first rule alone say when it is applied to the spin and the detector together?

A detector is a quantum system too

First, a piece of notation for two systems at once. If the spin is in state $|a\rangle$ and the detector in $|b\rangle$, the pair is in the product state $|a\rangle|b\rangle$, and the overlap of two product states is the product of the separate overlaps, $\langle a'|a\rangle\,\langle b'|b\rangle$, just as the probability of two independent events is the product of their probabilities.

Give the detector a starting state, $|\text{ready}\rangle$, and two possible final states, $|D_\uparrow\rangle$ ("the pointer reads up") and $|D_\downarrow\rangle$ ("reads down"). A good detector is one whose interaction with the spin does this:

$$\begin{aligned} |{\uparrow}\rangle\,|\text{ready}\rangle &\;\longrightarrow\; |{\uparrow}\rangle\,|D_\uparrow\rangle,\\[4pt] |{\downarrow}\rangle\,|\text{ready}\rangle &\;\longrightarrow\; |{\downarrow}\rangle\,|D_\downarrow\rangle . \end{aligned}$$

Now send in the spin prepared along $x$, $(|{\uparrow}\rangle + |{\downarrow}\rangle)/\sqrt2$. Schrödinger's equation is linear, so each part evolves as above and the results add:

$$\begin{aligned} &\tfrac{1}{\sqrt2}\bigl(|{\uparrow}\rangle + |{\downarrow}\rangle\bigr)|\text{ready}\rangle\\[4pt] &\;\longrightarrow\; \tfrac{1}{\sqrt2}\bigl(|{\uparrow}\rangle|D_\uparrow\rangle + |{\downarrow}\rangle|D_\downarrow\rangle\bigr). \end{aligned}$$

The final state is not a product of a spin state and a detector state. It cannot be factored: the spin has no pure state of its own, no single ket, and neither has the detector; only the pair does. (Each part can still be given a reduced description that yields the probabilities for measurements on it alone, but that description is a statistical mixture, not a ket.) Such a state is called entangled. It says that the spin and the pointer are perfectly correlated, with probability ½ for "up and reads up" and ½ for "down and reads down", exactly the Born-rule probabilities. What it does not do is pick one.

Where the interference goes

Before the detector, the spin was in a definite state, up along $x$, and a measurement along $x$ would have given "up" every time. That certainty came from the interference of the $\uparrow$ and $\downarrow$ parts. After the detector, measure the spin along $x$ again. The state "up along $x$" is $(|{\uparrow}\rangle + |{\downarrow}\rangle)/\sqrt2$, which overlaps each of $|{\uparrow}\rangle$ and $|{\downarrow}\rangle$ by $1/\sqrt2$. Taking that overlap on the spin part of the joint state leaves the detector in $\tfrac12\bigl(|D_\uparrow\rangle + |D_\downarrow\rangle\bigr)$, and the probability is the squared length of what is left. Writing $\gamma = \langle D_\downarrow|D_\uparrow\rangle$ for the overlap of the two detector states, taken real for simplicity, and using $\langle D_\uparrow|D_\uparrow\rangle = \langle D_\downarrow|D_\downarrow\rangle = 1$:

$$\begin{aligned}P(\text{up along }x) &= \tfrac14\bigl(1 + 1 + 2\gamma\bigr)\\[4pt] &= \tfrac12\bigl(1 + \gamma\bigr).\end{aligned}$$

This is the double-slit marker again, with the detector as the marker. If the two detector states are identical, $\gamma = 1$, the detector recorded nothing and the certainty survives. If they are orthogonal, $\gamma = 0$, the detector made a perfect record and the probability is ½: for any measurement on the spin alone, exactly what we would get if the spin were definitely up or definitely down, each half the time. The interference has not been destroyed. It is still in the joint state, but it now lives in the overlap of the detector's states, where no measurement on the spin alone can reach it.

Worked example: a detector made of many particles

A real detector does not have one pointer particle. Take a simple model of one with many: $N$ particles, each recording the spin independently and in the same way, so that each ends in one of two states with the same overlap $g$ between them. In this model the detector's overall states are products, so their overlap is a product too:

$$\gamma = g^N .$$

This is an illustration, not a formula for every detector: real record carriers are neither independent nor identical. What carries over is the trend. Each additional particle that picks up a record multiplies the overlap by a number no larger than one.

A weak record. Suppose each particle barely registers the spin, $g = 0.99$. After one particle, $\gamma = 0.99$ and $P(\text{up along }x) = 0.995$: the interference is nearly intact. After 100 particles, $\gamma = 0.99^{100} = 0.37$ and $P = 0.68$. After 1,000, $\gamma = 4\times10^{-5}$ and $P = 0.50002$.

A macroscopic record. A pointer that visibly moves involves something like $10^{20}$ atoms, each of which is somewhere slightly different depending on the reading. If each registered the difference even faintly and independently, as in the model, $\gamma$ would be a number below one raised to a power of about $10^{20}$: far too small for any feasible experiment to tell $P$ from ½.

Check: $N = 0$ gives $\gamma = 1$ and full interference, the no-detector case; and $\gamma$ can only shrink as more particles take part, never grow. A record, once spread over many particles, cannot be undone by anything we could do to the spin alone.

Interference surviving a detector made of more and more particles The probability of finding the spin up along x after a detector has acted, against the number of particles in the detector. It starts at 1 and falls towards one half. With a strong record per particle it reaches one half within a few particles; with a weak record of 0.99 per particle it falls more slowly but steadily. 0 10 20 30 40 50 60 0.5 0.75 1 number of detector particles N P(up along x) afterwards per-particle overlap 0.99 0.9 0.5 no record: full interference complete record: ½
The probability of finding the spin up along $x$ after a detector of $N$ particles has acted, for three strengths of record per particle. It starts at 1, full interference, and falls towards ½, the value for a spin that is definitely up or definitely down. The weaker the record per particle, the more particles it takes, but the fall is always exponential in $N$.

Decoherence, and what it does not settle

Nothing in this calculation needed a detector built on purpose. Any interaction that leaves a trace does the same. A dust grain one micrometre across, floating in air, is struck by around $4\times10^{16}$ gas molecules every second. If the grain were in a superposition of two positions further apart than the molecules' wavelength, about 0.03 nm, each collision would leave the molecule in nearly orthogonal states for the two positions. The overlap would vanish in a tiny fraction of a second. This process, the spreading of which-alternative information into the surroundings, is called decoherence. It is why objects exposed to ordinary, uncontrolled surroundings do not show interference between macroscopically different states, and in that sense why the everyday world looks classical. It is not an absolute barrier: in carefully isolated experiments, interference has been seen with molecules of up to about two thousand atoms, and in experiments with fullerene molecules it fades in a controlled way as gas is let into the apparatus.

Two questions need keeping apart here. Decoherence answers the first: why interference between the alternatives cannot be observed in measurements on the system, or on anything short of the whole environment that holds the records. For those measurements the alternatives behave like a list of possibilities with Born-rule probabilities. The second question is why one of them happens, and decoherence does not answer it. It does not choose among the alternatives. The joint state after the detector still contains both "up, and the pointer reads up" and "down, and the pointer reads down". Yet every run of the experiment shows one reading. How, or whether, one outcome becomes the outcome is the measurement problem. The measurement rule answers it by decree, and works. Whether that rule is a separate law of nature, a consequence of something not yet understood, or a statement about what a single observer can know, is not settled by the calculations in this essay or by any experiment so far.

Optional: the main interpretations, in brief

Each of these reproduces every prediction of the rules used in this book, except where noted.

Two questions

1. A partial record. A detector's two final states have overlap $\gamma = 0.6$. After it has acted, what is the probability of finding the spin up along $x$?

Worked answer

$P = \tfrac12(1 + 0.6) = 0.8$. The interference is partly there: more than the ½ of a perfect record, less than the certainty of no record.

The slip to avoid is treating a detector as either measuring or not. A detector whose two states overlap has made a partial record, and the interference is reduced in exact proportion to how much it recorded. That is the same continuum the double-slit essay found for an unreliable marker.

2. Prediction, observation, interpretation. Sort these three statements about the experiment in the opening. (a) "The probability of the detector reading up is ½." (b) "In this run, the detector read up." (c) "When the detector clicked, the wavefunction collapsed to spin up."

Worked answer

(a) is a prediction of quantum mechanics, the same in every interpretation, and it can be checked by repeating the run many times. (b) is an observation, a fact about one run, which no theory predicts in advance. (c) is an interpretation: one way of describing what connects (a) and (b). The many-worlds view would describe the same run without any collapse, and pilot-wave theory would say the electron's position followed a definite trajectory all along, which the magnet's field steered to one detector or the other.

The tempting wrong turn is to treat (c) as part of what experiments establish. Experiments establish (a) and (b). That the Born rule then works is not in dispute; what it means is.

Further reading

Where this leaves us

When the detector clicks, Schrödinger's equation says the spin and the detector have become entangled, perfectly correlated, with each reading carrying its Born-rule weight. The interference that made the spin's original state definite has moved into the overlap of the detector's states. For a detector coupled to ordinary surroundings, that overlap is driven so close to zero that no feasible measurement on the spin, or on the pointer, could reveal it. That is why nobody observes the interference a superposition of readings would show. It is a statement about what can be observed, not about which reading occurs. The equations do not say why this run showed "up". The prediction is a probability; the observation is one outcome; how the two are connected is where the interpretations part company.

That ends the core of this book. You have followed quantum mechanics from a hot poker to the measurement problem: quanta, amplitudes, states and their descriptions, wavefunctions, operators, uncertainty, the solvable systems, spin, identical particles, approximation, and measurement. The optional essays that follow take a different road. The electron in these essays moved slowly compared with light. Making quantum mechanics agree with relativity needs relativity first, and then leads to an equation that predicted spin and antimatter: the first optional essay starts with clocks and light.

Another road leads from the physics towards using quantum states: preparing, measuring, entangling and protecting them in order to compute and communicate. The companion book, Quantum Information, Explained, starts there. It assumes what this book has covered (amplitudes, spin ½, two-state systems and measurement), and its first chapter takes up the two-state system again under the name qubit. The entanglement with a detector and the decoherence described in this essay return there as working tools and as the noise that error correction has to undo.