A practical language for quantum states
An ammonia molecule is prepared with its nitrogen atom on one particular side of the molecule. Its energy is then measured. The result is always one of exactly two values, never anything in between, and which of the two comes up is a matter of chance. Where do the two values come from, and what sets the chances?
The earlier essays have given us states as arrows, amplitudes as overlaps between them, and wavefunctions as continuous lists of amplitudes. What is missing is a way to talk about measurements of anything, not just "which exit" or "where", in one consistent language. That language is usually called the formalism of quantum mechanics, and later essays refer back to this one as the formalism essay. This essay builds that language in four steps, each with its own notation: states, overlaps, measurable quantities, and averages. The ammonia molecule carries the whole calculation, and at the end we translate everything back to wavefunctions.
Step 1: a name for the state
In the essay on representations, a photon's polarisation was an arrow, and a basis gave that arrow a list of components. Paul Dirac's notation writes the arrow itself, independent of any basis, as $|\psi\rangle$, read "ket psi". The basis states get kets too, $|H\rangle$ and $|V\rangle$, and the photon polarised at 30° is
$$|\psi\rangle = 0.866\,|H\rangle + 0.500\,|V\rangle .$$That line is a superposition, in the sense of the wavefunction essay, now written explicitly: a state built from the basis states with complex coefficients, the amplitudes. The same ket written in the D/A basis would have different coefficients and be the same state. Nothing here is new except the brackets, which keep the state and its coordinates visibly apart.
Step 2: overlaps
The central number in the earlier essays was the amplitude for a system in one state to be found in another: the overlap of two arrows. Dirac writes it $\langle\varphi|\psi\rangle$, and since it is a bracket, the left half $\langle\varphi|$ is called a bra. Given the components of both states in any basis,
$$\langle\varphi|\psi\rangle = \varphi_1^*\,\psi_1 + \varphi_2^*\,\psi_2 ,$$multiplying matching components and adding, with the complex conjugate (marked $*$) on the left-hand state. For real components the conjugate changes nothing, and this is the overlap recipe from the representations essay. Three facts follow:
- The probability of finding $|\psi\rangle$ in $|\varphi\rangle$ is $|\langle\varphi|\psi\rangle|^2$.
- A state is normalised when $\langle\psi|\psi\rangle = 1$: it is certainly found in itself.
- Two states with $\langle\varphi|\psi\rangle = 0$ are orthogonal. A system in one is never found in the other, so the two can be told apart perfectly. The marker states in the double-slit essay destroyed the fringes completely exactly when they were orthogonal.
Step 3: a measurable quantity is a matrix
Consider an ideal measurement: one that returns a definite result from a list of possible values $a_1, a_2, \dots$, all different, and that disturbs the system no more than it must. For each result there is a state in which that result is certain: call it $|a_1\rangle$, $|a_2\rangle$, and so on. These states are orthogonal to one another, since a system certain to give $a_1$ never gives $a_2$, and together they form a basis. Everything about the measurement is contained in these results and states, and quantum mechanics packages them into a single matrix, called the operator for the quantity: the matrix whose eigenvectors are the states $|a_i\rangle$ and whose eigenvalues are the results $a_i$.
With that packaging, three rules describe such an ideal measurement:
- The only possible results are the eigenvalues of the operator.
- The probability of result $a_i$ in state $|\psi\rangle$ is $|\langle a_i|\psi\rangle|^2$.
- Immediately afterwards, the system is in the state $|a_i\rangle$ for the result that occurred. (This is the photon leaving a filter polarised along it.)
Real measurements can be less clean than this, and some quantities have several different states sharing the same result. The rules extend to those cases with more bookkeeping; the simple form is all the essays that follow need.
Operators for measurable quantities are always of a special kind, called Hermitian: each entry equals the complex conjugate of its mirror image across the diagonal. For real matrices that just means symmetric. Hermitian matrices are exactly the ones with real eigenvalues and orthogonal eigenvectors, which is what results and certainty-states have to be.
The ammonia molecule
An ammonia molecule, NH3, is a pyramid: three hydrogen atoms form a triangle, and the nitrogen sits either above or below its plane. Take these two arrangements as a basis, $|\text{up}\rangle$ and $|\text{down}\rangle$. By symmetry they have the same energy, call it $E_0$. If that were the whole story, the operator for the energy would be $E_0$ times the identity, and $|\text{up}\rangle$ would stay up forever.
But the nitrogen can get through the plane, even though classically it lacks the energy to: an effect called tunnelling, which the essay on confinement explains. That possibility appears as an amplitude connecting the two basis states, an off-diagonal entry in the energy matrix, which we write as $-C$:
$$\hat H = \begin{pmatrix} E_0 & -C \\ -C & E_0 \end{pmatrix}.$$The energy operator is traditionally written $\hat H$, after William Rowan Hamilton; the hat marks an operator. Its eigenvectors, checked by multiplying out, are the even and odd combinations
$$\begin{aligned} |\text{I}\rangle &= \tfrac{1}{\sqrt2}\bigl(|\text{up}\rangle + |\text{down}\rangle\bigr),\\[4pt] |\text{II}\rangle &= \tfrac{1}{\sqrt2}\bigl(|\text{up}\rangle - |\text{down}\rangle\bigr), \end{aligned}$$with energies $E_0 - C$ and $E_0 + C$. Those are the two values of the opening. Neither energy state has the nitrogen on one side; each is an equal superposition of both. The splitting $2C$ is measured directly: a molecule dropping from the upper state to the lower emits microwaves at 23.87 GHz, the frequency that ran the first maser in 1954, so $2C = h\nu = 9.87\times10^{-5}$ eV.
Step 4: averages
Repeat a measurement on many systems prepared in the same state and average the results. That average is the expectation value of the quantity, written $\langle A\rangle$. From the rules, each result $a_i$ turns up a fraction $|\langle a_i|\psi\rangle|^2$ of the time, so
$$\langle A\rangle = \sum_i a_i\,|\langle a_i|\psi\rangle|^2 = \langle\psi|\hat A|\psi\rangle .$$The second form, sandwiching the matrix between the state and its conjugate, gives the same number without finding any eigenvectors: multiply the list $|\psi\rangle$ by the matrix, then take the overlap with $|\psi\rangle$. It is an average over many repetitions. It need not be a value any single measurement can give.
Worked example: measuring ammonia's energy
Work in the up/down basis, and prepare a molecule in the state
$$|\psi\rangle = 0.866\,|\text{up}\rangle + 0.500\,|\text{down}\rangle ,$$mostly up. (It is the same pair of numbers as the photon at 30°.)
Possible results: the eigenvalues, $E_0 - C$ and $E_0 + C$. Nothing else.
Probabilities: overlaps with the eigenvectors, whose components are $(0.707, 0.707)$ and $(0.707, -0.707)$:
$$\begin{aligned} \langle\text{I}|\psi\rangle &= 0.707(0.866 + 0.500)\\[2pt] &= 0.966,\\[6pt] \langle\text{II}|\psi\rangle &= 0.707(0.866 - 0.500)\\[2pt] &= 0.259 . \end{aligned}$$Squaring: $P(E_0 - C) = 0.933$ and $P(E_0 + C) = 0.067$. They add to 1.
Expectation value, first way, from the probabilities:
$$\begin{aligned} \langle E\rangle &= 0.933\,(E_0 - C)\\[4pt] &\quad + 0.067\,(E_0 + C)\\[4pt] &= E_0 - 0.866\,C . \end{aligned}$$Second way, by matrix multiplication. First apply the matrix:
$$\begin{aligned} \hat H|\psi\rangle &= \begin{pmatrix} E_0 & -C \\ -C & E_0 \end{pmatrix}\begin{pmatrix}0.866\\0.500\end{pmatrix}\\[6pt] &= \begin{pmatrix} 0.866E_0 - 0.500C \\ 0.500E_0 - 0.866C\end{pmatrix}. \end{aligned}$$Then take the overlap with $|\psi\rangle$, multiplying matching components and adding:
$$\begin{aligned} \langle\psi|\hat H|\psi\rangle &= 0.866(0.866E_0 - 0.500C)\\[4pt] &+ 0.500(0.500E_0 - 0.866C)\\[4pt] &= E_0 - 0.866\,C . \end{aligned}$$The two routes agree. In real units, $C = 4.94\times10^{-5}$ eV, so the average lies $4.3\times10^{-5}$ eV below $E_0$.
Check: an eigenstate gives certainty. Prepare $|\text{I}\rangle$ instead. Its overlap with itself is 1 and with $|\text{II}\rangle$ is 0, so the result is $E_0 - C$ every time and the average equals it. A state is certain to give a value exactly when it is an eigenvector of the quantity being measured.
Check: average versus individual. Prepare $|\text{up}\rangle$. The overlaps are $0.707$ each, the probabilities ½ each, and $\langle E\rangle = E_0$: a value no single measurement ever returns.
Back to wavefunctions
Everything above was written for two-component lists, but none of it depended on the list being short. A wavefunction is the list of components of $|\psi\rangle$ in the position basis: $\psi(x) = \langle x|\psi\rangle$, the amplitude to be found at $x$. Sums over components become integrals, and each piece of the language carries over:
| Idea | Two-state list | Wavefunction |
|---|---|---|
| State | $(\psi_1, \psi_2)$ | $\psi(x)$ |
| Overlap $\langle\varphi|\psi\rangle$ | $\sum_i \varphi_i^*\psi_i$ | $\int \varphi^*(x)\,\psi(x)\,dx$ |
| Normalised | $|\psi_1|^2 + |\psi_2|^2 = 1$ | $\int|\psi|^2\,dx = 1$ |
| Operator | a 2×2 matrix | an instruction acting on $\psi(x)$ |
| Expectation value | $\sum_{ij}\psi_i^*C_{ij}\psi_j$ | $\int\psi^*\,\hat A\psi\,dx$ |
The operators become instructions. Position multiplies the wavefunction by $x$, so $\langle x\rangle = \int x\,|\psi|^2\,dx$, the average position. Momentum is the one found in the previous essay, $\hat p = -i\hbar\,d/dx$. Energy is the right-hand side of Schrödinger's equation, $\hat H = -(\hbar^2/2m)\,d^2/dx^2 + V(x)$. And the equation for a stationary state, $\hat H\varphi = E\varphi$, is now visibly an eigenvalue equation: the allowed energies of a bound particle are the eigenvalues of its energy operator, exactly as ammonia's two energies were the eigenvalues of its 2×2 matrix.
When two quantities cannot both be sharp
One more consequence decides a great deal. For the photon, the H/V question and the D/A question have no state in common that is certain for both: a photon definitely H gives D or A with probability ½ each, and vice versa. In matrix language, the two operators have no common eigenvectors, and a tell-tale sign is that the order of multiplying them matters. With results $+1$ and $-1$ for each question,
$$\begin{pmatrix}1&0\\0&-1\end{pmatrix}\begin{pmatrix}0&1\\1&0\end{pmatrix} = \begin{pmatrix}0&1\\-1&0\end{pmatrix},$$ $$\begin{pmatrix}0&1\\1&0\end{pmatrix}\begin{pmatrix}1&0\\0&-1\end{pmatrix} = \begin{pmatrix}0&-1\\1&0\end{pmatrix}.$$The difference $\hat A\hat B - \hat B\hat A$ is called the commutator, written $[\hat A, \hat B]$. When it is zero, the two quantities are compatible: they have a complete set of eigenstates in common, and every one of those states is sharp for both. When it is not zero, no such complete common set exists. Some particular states may still be sharp for both, but not all of either quantity's eigenstates can be; how much is ruled out depends on the pair. For the photon's two questions, no common state exists at all. Position and momentum are the most important pair, and the strictest: applied to any wavefunction, $\hat x\hat p - \hat p\hat x$ gives $i\hbar$ times the wavefunction, never zero, so no state is sharp in both. The next essay measures how unsharp they must be.
Optional: computing $[\hat x, \hat p] = i\hbar$
Apply both orders to an arbitrary wavefunction $\psi(x)$. Momentum first, then position: $\hat x\hat p\,\psi = x\,(-i\hbar\,\psi')$. Position first, then momentum: $\hat p\hat x\,\psi = -i\hbar\,(x\psi)' = -i\hbar\,(\psi + x\psi')$, by the product rule. Subtracting,
$$\begin{aligned} (\hat x\hat p - \hat p\hat x)\,\psi &= -i\hbar x\psi' + i\hbar\psi + i\hbar x\psi'\\[4pt] &= i\hbar\,\psi . \end{aligned}$$The $x\psi'$ terms cancel whatever $\psi$ is, leaving $i\hbar\psi$. The whole difference comes from the product rule: multiplying by $x$ and differentiating do not commute.
Optional: matrices and waves are one theory
The essay on representations told how Heisenberg's matrix mechanics and Schrödinger's wave mechanics turned out to be the same theory. The language of this essay shows why, and adds one piece. In Schrödinger's way of working, the state changes in time and the operators stay fixed. In Heisenberg's, the state stays fixed and the operators change. For a quantity with no explicit time dependence, Heisenberg's equation of motion is
$$\frac{d\hat A}{dt} = \frac{1}{i\hbar}\,[\hat A, \hat H].$$Every expectation value comes out identical either way; only the bookkeeping differs. Some older texts absorb the factor $1/i\hbar$ into the bracket's definition, which is why the equation is sometimes printed without it. Here the commutator is always the plain difference $\hat A\hat H - \hat H\hat A$.
Two consequences are worth seeing. A quantity that commutes with $\hat H$ has $d\hat A/dt = 0$: it is conserved. And for ammonia, the operator "up minus down" does not commute with $\hat H$, so a molecule started in $|\text{up}\rangle$ swings to $|\text{down}\rangle$ and back, completing a round trip every $1/\nu = 42$ picoseconds, at the maser frequency.
Two questions
1. Another preparation. An ammonia molecule is prepared in $|\psi\rangle = 0.6\,|\text{up}\rangle + 0.8\,|\text{down}\rangle$. What are the probabilities of the two energies, and what is $\langle E\rangle$?
Worked answer
First check it is normalised: $0.36 + 0.64 = 1$. The overlaps are $\langle\text{I}|\psi\rangle = 0.707(0.6 + 0.8) = 0.990$ and $\langle\text{II}|\psi\rangle = 0.707(0.6 - 0.8) = -0.141$. Squaring: $P(E_0 - C) = 0.98$ and $P(E_0 + C) = 0.02$. Then $\langle E\rangle = 0.98(E_0 - C) + 0.02(E_0 + C) = E_0 - 0.96\,C$.
The easy slips are to square the components $0.6$ and $0.8$ directly, which answers a different question (up or down?), or to forget that the negative overlap still gives a positive probability. Probabilities for a measurement always come from overlaps with that measurement's eigenstates.
2. The average that never happens. For a molecule prepared in $|\text{up}\rangle$, $\langle E\rangle = E_0$. Will an energy measurement on such a molecule ever give $E_0$? What does the expectation value tell you, and what does it not?
Worked answer
Never. The only possible results are the eigenvalues, $E_0 - C$ and $E_0 + C$, each with probability ½. The expectation value is the average over many repetitions, and here it falls exactly halfway between the only two values that occur.
It tells you where the results balance. It does not tell you what any one measurement gives, or how widely the results scatter. The tempting wrong turn is to read "expectation value" as "the value to expect". A dice roll has an average of 3.5, and no die shows 3.5.
Further reading
- Dirac, Principles of Quantum Mechanics (4th ed.), Ch. I §6 "Bra and Ket Vectors" and Ch. II §7–§10 (linear operators, eigenvalues and eigenvectors, observables): the language of this essay from its inventor.
- Feynman, Lectures on Physics, Vol. III, Ch. 8 "The Hamiltonian Matrix", §8-5 "The Hamiltonian matrix" and §8-6 "The ammonia molecule".
- Landau & Lifshitz, Quantum Mechanics (2nd ed.), §3 "Operators" and §11 "Matrices".
Where this leaves us
An ammonia molecule's energy comes out as one of two values because those are the eigenvalues of its energy operator, and the chances are the squared overlaps of the prepared state with the two eigenstates. The same four pieces, kets, overlaps, operators and expectation values, describe ideal measurements on any system, whether the state is a list of two numbers or a whole wavefunction.
The language has also exposed a limit. When two quantities' operators do not commute, they cannot share a complete set of sharp states, and for position and momentum no state at all is sharp in both. How sharp can an electron's position and momentum be at the same time, and what, precisely, does "sharp" mean? That is the uncertainty principle, and it is the subject of the next essay.