The questions that can be asked together

Sharpness has to be paid for makes position and momentum sound like a special case of a general prohibition, and it is usually taught as though the prohibition were the rule. It is the exception. Almost every pair of quantities in quantum mechanics can be sharp at the same time, and there is an exact test for which pairs those are.

The test is one line of arithmetic. Two observables are represented by operators, operators can be multiplied, and multiplication in the wrong order need not give the same answer. The difference $AB - BA$ is the commutator, and a pair of observables has states with definite values for both exactly when their commutator vanishes.

What the commutator is a statement about #

The temptation is to read a non-zero commutator as a statement about instruments — that measuring one quantity jogs the other. That reading survives contact with a great many textbooks and it is not what the arithmetic says.

Nothing in the calculation above involves an apparatus. Two matrices were multiplied in both orders and subtracted. The result is a property of the pair of operators, fixed before any measurement is contemplated, and it decides something prior to measurement: whether a state with definite values for both quantities exists.

That is the useful statement, and it is stronger than the one about disturbance. If no such state exists then there is nothing for an ideal, gentle, non-disturbing measurement to find. The failure is not that the reading is spoilt in the taking; it is that there was no pair of numbers waiting to be read. The answer that was not there before is the same point arrived at from the other side.

The disturbance follows as a corollary. Since a state cannot have both values, a measurement that establishes one must leave the system in a state that lacks the other, and if the other had been established a moment before then that establishment is gone. Disturbance is what compatibility failing looks like in a laboratory, and it is a symptom rather than a cause.

The one that commutes with everything #

Three of the four entries in that grid are components of angular momentum and one is not. The total, $S^2$, commutes with all three, and for a spin-½ the reason is almost too easy: it is three-quarters of the identity matrix and the identity commutes with everything.

The easiness is misleading and the fact is general. For any angular momentum, of any size, $S^2$ commutes with each component even though the components do not commute with each other. That is why the states of an atom are labelled by a total angular momentum and one component, and never by two components — the pair of labels used everywhere in the order the shells fill is exactly the pair the commutators permit.

It is worth noticing how much of the structure of atomic physics that one fact fixes. A spectroscopic label like $^2P_{3/2}$ names a total spin, a total orbital angular momentum, and a total of the two — never a direction. The labels are not a convention chosen for tidiness; they are the largest set of simultaneously answerable questions there is, and the commutator grid is what says how large that set may be.

A shared set of answers #

The algebraic condition has a geometric face, and the geometric face is the one that explains what compatibility buys.

An observable sorts states into its eigenstates — the states on which it returns a definite value. Two observables that commute can be sorted by the same set: there is a basis of states each of which has a definite value for both. Two that do not commute have no such basis, and the grid above is that failure made visible. At any angle but zero, each state that answers the first question sharply is a superposition of states answering the second, and vice versa.

Every row and every column of that grid sums to one, at every angle, and the figure checks this rather than asserting it. That is not a decorative property. It says that whatever the first analyser found, the second finds something — the beam is not lost, it is redistributed. Compatibility is not about whether an answer comes back; it is about whether the answer that comes back was already determined.

And the transition between the two pictures is not gradual in the way the numbers suggest. The cells move continuously as the angle opens, but the kind of grid changes at once: it is a permutation or it is not. There is no partial sharing of an eigenbasis. Two observables either have a common set of answers or they have none in common at all, and the continuous-looking numbers are describing how badly the second case fails rather than how much of the first survives.

Repeatability is the operational test #

There is an operational version of all of this that needs no operators, and it is the one an experimenter would state: a measurement is compatible with another when inserting it changes nothing about the second’s outcome.

The three-analyser chain is the test. Measure, insert a second apparatus, measure the first quantity again. If the middle apparatus is asking a question already answered, the third reading reproduces the first with certainty. If it is asking an incompatible one, some of the first answer is destroyed, and the curve says how much.

The value at right angles is worth pausing on. It is a half, not zero. An apparatus perfectly incompatible with the first does not reverse the answer; it randomises it, and randomising leaves the original answer standing half the time. That is why a chain of analysers passes a beam that the first one had already excluded, which is the phenomenon the measurement that never touched it turns on, and it is also why the destruction cannot be undone by simply inserting a fourth apparatus to put things back.

The curve is $\cos^4(\theta/2) + \sin^4(\theta/2)$, which is the same $\cos^2(\theta/2)$ that governs a single spin measurement, applied twice. The commutator does not appear anywhere in it. That is the point of calling this the operational test: the algebra predicts the curve, and the curve can be measured by somebody who has never heard of a commutator.

The bound belongs to the state as much as to the pair #

The quantitative form of the incompatibility is Robertson’s relation: the product of the two spreads is at least half the size of the average commutator. It is the generalisation of the familiar $\hbar/2$, and its right-hand side is where all the trouble hides.

An average is taken in a state. So the bound is not a number belonging to the pair of observables — it is a number belonging to a pair of observables and a state, and it can vary from exact to vacuous as the state moves.

Both extremes are on that figure and neither is a pathology. On one circle of states the bound is achieved by every state on it, which is as tight as an inequality can be. On another, the average commutator vanishes identically while the two spreads are both large and their product is nowhere near zero. The relation is true on that circle and says nothing.

That is the practical warning. A textbook statement of the form “these two quantities obey $\Delta A\,\Delta B \geq c$” with $c$ a constant is only available when the commutator happens to be a constant — which is exactly the position and momentum case, where it is $i\hbar$ times the identity and the state drops out. Position and momentum are the special case, and the whole subject is taught from it.

For angular momentum, for a field and its conjugate, and for very nearly everything else, the bound moves with the state, and a relation whose right-hand side can be zero cannot be the reason a state has a spread. Something else has to say why the spread is there, and that something is the state itself.

What a complete set of commuting observables buys #

The positive use of all this is a piece of bookkeeping that quietly organises the whole of atomic and molecular physics.

Take an observable and measure it. The result narrows the state down to those with that eigenvalue, and if the eigenvalue is degenerate the narrowing is incomplete. Take a second observable that commutes with the first and measure that; the two labels together narrow further. Keep going until the labels specify the state uniquely. The collection is a complete set of commuting observables, and its size is the number of questions the system can answer at once.

For hydrogen that set has four members and their labels are the four quantum numbers. For a rigid rotor it is two. For a free particle it is the three components of momentum, which commute with one another — that is why a plane wave has a definite momentum vector and not merely a definite speed, and it is the fact behind everything has a wavelength.

The commutators decide the size of the set, and the size of the set decides the degeneracies, and the degeneracies decide which perturbation splits what. That chain runs from a piece of matrix arithmetic to the fine structure of a spectral line, and the line that is really two is one place it comes out.

There is a converse worth stating too. When two observables commute and one of them is the energy, the other is conserved — its distribution does not change as the state evolves. Every conservation law in quantum mechanics is of that form, which makes the commutator the quantum version of the argument what is conserved and why makes from symmetry.

Why a theory needs the failures at all #

A reasonable objection at this point is that a theory whose observables all commuted would be simpler and would still describe measurements, so the non-commuting pairs look like an avoidable complication.

They are not avoidable, and the reason is that some of the operators are doing a second job. A component of angular momentum is not only a quantity that can be measured; it is also the generator of rotations about its own axis, in the sense that acting with it repeatedly turns a state. Momentum generates translations, and energy generates the passage of time. Those roles are not additional facts bolted on — they are what makes the quantities conserved, by the argument a symmetry hands over.

Once an observable generates a motion, its commutator with another observable is fixed by what that motion does. Rotating about $z$ turns $S_x$ into $S_y$, so $[S_z, S_x]$ has to be proportional to $S_y$ and cannot be zero. The failure of the two components to be simultaneously definite is the same fact as the rotations about two different axes failing to commute — which is a statement about ordinary three-dimensional space that anyone can check with a book and two turns of the wrist.

So the incompatible pairs are exactly the pairs one of which moves the other, and the size of the commutator measures how much. A theory with no incompatible pairs would be a theory in which no observable generates any motion, which is a theory in which nothing happens. The prohibitions are the price of having dynamics at all, and reading them as a defect of the measurement process locates them in the wrong part of the theory entirely.

That also explains the one exception the last section noticed. Position and momentum have a commutator that is a constant because translation moves position by a fixed amount regardless of where the particle is, and it is the uniformity of space that makes the famous relation state-independent. Angular momentum’s commutators are not constants because a rotation moves a point by an amount that depends on where the point is.

Where the model stops #

The commutator condition is exact only for the full operators. Two quantities can commute to an excellent approximation and fail to commute exactly, and then the states are almost-but-not-quite shared. That is the ordinary situation in molecules, where the electronic and nuclear coordinates nearly separate, and “nearly” is where all the interesting spectroscopy lives.

A vanishing commutator does not mean the two quantities are independent. Compatible observables can be perfectly correlated: measuring one may determine the other completely, as it does for the two spins in a singlet, without any obstruction to measuring both. Compatibility is about whether both have values, not about whether they carry different information — the distinction that the correlation no instructions produce depends on.

Robertson’s bound is not the only one and is not the best one. Schrödinger’s refinement adds a term for the correlation between the two observables, and it is strictly stronger wherever that correlation is non-zero. The figures here draw Robertson’s because it is the one that is quoted; the slack they show is partly Robertson’s own weakness and not only the state’s doing.

Everything drawn here is a spin-½, which is the smallest interesting case. Two-dimensional operators have properties larger ones do not: every one of them is a linear combination of the identity and the three spin matrices, and the commutator of any two closes on the set immediately. Larger angular momenta keep the structure but lose the tidiness, and quantities with continuous spectra lose more than that — position and momentum have no normalisable eigenstates at all, so the phrase “the state with a definite position” is a limit rather than a state.

And the whole account assumes the observables are the ones written down. Which operators represent which laboratory quantities is an input to the theory rather than an output of it, and a proposed observable that nobody knows how to measure has a commutator with everything else whether or not it means anything. The arithmetic decides compatibility among a list; it does not supply the list.

What the pictures cannot show #

The grids are drawn with the magnitude of the commutator, and a magnitude discards the fact that the commutator is an operator. $[S_x, S_y]$ is not a number, it is $i\hbar S_z$, and which operator it turns out to be matters — that is the whole content of the angular-momentum algebra and the reason the three components close among themselves rather than generating something new. The figure checks that the size is right and cannot show which operator it belongs to.

The bound curves are drawn for pure states, which is where the relation is sharpest and where the geometry is a sphere. A mixed state sits inside the sphere, its spreads are larger, and its average commutator is smaller — so the slack grows in both directions at once, and the picture of a bound that is sometimes exact belongs to the surface only.

Where the ladder goes next #

The uncertainty ladder began with sharpness has to be paid for, which is Fourier analysis with $\hbar$ added to fix the units, and continued with the motion that cannot be stopped, where the relation acquires an energy cost. This rung replaces the single famous inequality with the condition behind it and finds that the condition is usually satisfied. The rungs after it: the time–energy relation, which is not of this form at all because time is not an observable; the way a commutator becomes a Poisson bracket in the limit where the old picture comes back; and the entropic uncertainty relations, which bound a quantity that does not vanish when the commutator does.

The habit worth carrying away is that a prohibition is only interesting alongside the permission it is carved out of. The uncertainty principle is a rule about a minority of pairs, and the majority — the pairs that commute, whose labels can all be written on the same state — is what makes quantum mechanics a description of anything definite at all.