Physical Chemistry · Part 2 of 9

Operators, Eigenvalues & the Schrödinger Equation

Quantum Chemistry, Part 2 · 9 sections · about 22,204 words · CSIR-NET Chemical Sciences, GATE Chemistry & IIT-JAM

This is the grammar of the subject. Every quantum calculation you will ever do is an operator acting on a function, and almost every exam question reduces to recognising which operator, which function, and whether the two commute. This part builds that machinery slowly and then uses it to derive the uncertainty principle rather than quoting it. Two layers on every section: a slow, hand-held beginner path and a research-grade advanced/reference path.

The 9 sections in Part 2

1Operators, operator algebra and linearity Free below
2Hermitian operators: real eigenvalues and orthogonal eigenfunctions
3The eigenvalue equation, the correspondence rules and the Schrödinger equation
4Expectation values
5Superposition and the meaning of the expansion coefficients
6The measurement postulate, collapse, and the Ehrenfest theorem
7Commutators and their algebra
8[x, p] = iℏ and the angular-momentum commutators
9Compatible observables and the generalised uncertainty relation

Operators, operator algebra and linearity

Free extract

Section B.1 of Part 2, reproduced in full from the book — figures and all. No sign-in, no paywall on this section.

Beginner layer

An operator is an instruction that turns one function into another. That is the entire definition; everything else in this section is consequences.

Start with something familiar. The symbol d/dx is not a number and it is not a function. It is an instruction: hand it a function and it hands you back the derivative of that function. Hand it sin x and it returns cos x. Hand it x³ and it returns 3x². It does nothing on its own; it needs something to act on.

Quantum mechanics takes that idea and makes it the central object of the theory. The instruction ‘multiply by x’ is an operator in exactly the same sense: hand it sin x and it returns x sin x. So is ‘take the square root’, and so is ‘rotate the coordinate frame by 90° about z’. Some of these will turn out to represent measurable quantities and some will not, and B.1 and B.2 are about the test that separates them.

Operator: a rule Â which, applied to a function f, produces another function g, written Âf = g. The caret (the ‘hat’) distinguishes the operator Â from the number or observable A. An operator is meaningless until it is given something to act on: the symbol Â standing alone is an instruction waiting for an argument, exactly like the symbol √ standing alone.

An operator is an instruction: give it a function, it gives you back a function. That is the whole definition, and the three rows show the only three outcomes possible. In the first row d/dx turns a sine into a cosine — a genuinely different function, so sin 2πx is not an eigenfunction of d/dx. In the second row the operator is multiply by x, which is an operator even though nothing is differentiated; it distorts the Gaussian by weighting its right-hand side more heavily. In the third row the output is the input again, merely multiplied by −2: the shape is unchanged. That is the eigenvalue situation, and it is the only situation in which quantum mechanics gives a sharp, repeatable answer for a measurement.

Adding and multiplying operators

Operators can be combined, and the rules are almost the ones you expect — with one spectacular exception that the whole of B.7 is devoted to.

Sum. The sum of two operators is defined by what it does: (Â + B̂)f = Âf + B̂f. Apply each separately and add the results. So (d/dx + x×) acting on x² gives 2x + x³.

Product. The product of two operators means apply them one after the other, rightmost first: (ÂB̂)f = Â(B̂f). Read ÂB̂f from right to left, like a chain of instructions. B̂ goes first because it is nearest the function.

(Â + B̂)f = Âf + B̂f , (ÂB̂)f = Â(B̂f) — rightmost operator acts first

Powers. Â² means ÂÂ, i.e. apply Â twice. So (d/dx)² = d²/dx², which is why the kinetic-energy operator contains a second derivative: it is the momentum operator applied twice.

The identity and the inverse. The identity operator 1̂ leaves everything alone, 1̂f = f. An operator Â has an inverse Â⁻¹ if ÂÂ⁻¹ = Â⁻¹Â = 1̂. Many operators have no inverse: multiplication by x has no inverse as an operator on all functions, because dividing by x is undefined at x = 0.

And now the exception. For numbers, ab = ba always. For operators this is false in general, and the failure is not a technicality. Take Â = x× and B̂ = d/dx, and let both act on some function f:

x̂(d/dx)f = x f′ but (d/dx)(x̂f) = (d/dx)(xf) = f + x f′

The two results differ by f itself. The product rule of differentiation generated an extra term in one order and not in the other, so x̂(d/dx) and (d/dx)x̂ are different operators. Everything in B.7 to B.9 — the uncertainty principle included — grows out of that one observation.

Note: Always use a test function. When manipulating operators, never try to simplify them in the abstract. Write ÂB̂f with an explicit f, do the calculus, and strip f off only at the very end. Ninety per cent of the algebra errors in this topic come from students cancelling operators as though they were numbers. The identity above is invisible unless the f is written down.

Linear operators

Of all the operators one could invent, quantum mechanics uses only the linear ones. An operator is linear if it distributes over sums and lets constants slide through:

Â(c₁f₁ + c₂f₂) = c₁Âf₁ + c₂Âf₂ for all f₁, f₂ and all constants c₁, c₂

Differentiation is linear: the derivative of a sum is the sum of the derivatives, and constants come out. Multiplication by a fixed function is linear. Integration is linear. But ‘square the function’ is not, because (f + g)² ≠ f² + g²; ‘take the square root’ is not; ‘take the logarithm’ is not.

Linearity is not decoration; it is what makes superposition possible. An operator is linear if Â(c₁f₁ + c₂f₂) = c₁Âf₁ + c₂Âf₂ for all functions and all constants. On the left, d²/dx² is applied to f+g and then to f and g separately and the results added: the two curves coincide to within finite-difference rounding error. On the right the same test is run on the perfectly well-defined but non-linear operation ‘square the function’, and it fails by a wide margin, the gap being exactly the cross term 2fg. Postulate 2 demands linear operators precisely so that a superposition of states behaves like a superposition, which is the whole content of interference.

Why does quantum mechanics insist on linearity? Because of superposition. If ψ₁ and ψ₂ are possible states then so is c₁ψ₁ + c₂ψ₂, and interference — the two-slit pattern, the chemical bond, every spectroscopic selection rule — depends on being able to act on that combination term by term. A non-linear operator would scramble the terms together and destroy the whole structure. Linearity of the operators and linearity of the Schrödinger equation are the same statement seen from two sides.

Advanced / reference layer

The algebra properly: associativity, functions of operators, matrix representations, projectors, and where the analogy with matrices is exact.

Operators on a function space form an associative algebra over the complex numbers, and it is worth listing the axioms explicitly because every manipulation later is one of them:

Associativity of the product: (ÂB̂)Ĉ = Â(B̂Ĉ). Always true, and it is what allows brackets to be dropped.
Distributivity: Â(B̂ + Ĉ) = ÂB̂ + ÂĈ and (Â + B̂)Ĉ = ÂĈ + B̂Ĉ. Both hold, and both must be written out separately because the product is not commutative.
Scalars commute with everything: (cÂ) = (Âc) for a complex number c.
Commutativity of the product: FALSE in general. This is the single axiom of ordinary arithmetic that operators do not obey, and quantum mechanics is the physics of its failure.

Functions of operators. A function of an operator is defined by its power series. The one that matters most is the exponential:

e^Â ≡ 1̂ + Â + Â²/2! + Â³/3! + … = ∑_n=0^∞ Âⁿ/n!

This is not a decoration. The time-evolution operator of B.3 is Û(t) = e^−iĤt/ℏ; the translation operator is e^{−ia p̂/ℏ}, which is why momentum is called the generator of translations; the rotation operator is e^{−iφL̂_z/ℏ}, which is why angular momentum generates rotations and why the whole of Part 4 is really group theory in disguise. Note carefully that e^Âe^B̂ = e^Â+B̂ only if Â and B̂ commute; in general one needs the Baker–Campbell–Hausdorff expansion, whose first correction term is ½[Â,B̂].

e^Âe^B̂ = e^{Â+B̂+½[Â,B̂]+…} , equal to e^Â+B̂ only when [Â,B̂] = 0

Matrix representation. Choose a complete orthonormal set {ψ₁, ψ₂, …}. Any function is then a column of coefficients, and any linear operator is a matrix with elements

A_mn = ∫ ψ_m* Â ψ_n dτ ≡ ⟨m|Â|n⟩

Operator addition becomes matrix addition, operator multiplication becomes matrix multiplication (in the same order — and matrices famously do not commute either), and the eigenvalue equation becomes an ordinary matrix eigenvalue problem. This is not an analogy: it is Heisenberg's matrix mechanics, proved equivalent to Schrödinger's wave mechanics in 1926, and it is how every quantum-chemistry program on earth actually calculates. When Part 6 sets up a variational secular determinant and Part 7 diagonalises a Fock matrix, they are doing exactly this.

Hermiticity is a property you can check, and here it is checked. Each operator is written as a matrix in a real basis; the Hermitian test is whether the matrix equals its own conjugate transpose. x̂ passes exactly. Plain d/dx fails spectacularly — it is anti-Hermitian, M† = −M, and its eigenvalues come out purely imaginary, which is exactly why d/dx cannot represent an observable: no instrument returns 2.787 i. Multiplying by −iℏ converts the anti-Hermitian operator into a Hermitian one, and the eigenvalues become real; that factor of i in p̂ = −iℏ∂/∂x is therefore not cosmetic, it is what makes momentum measurable. The Hamiltonian is diagonal because the box eigenfunctions were used as the basis, and its diagonal entries reproduce n²π²/2 to three decimals.

Projection operators. A special and very useful class. Given a normalised function ψ_n, the operator P̂_n = |ψ_n⟩⟨ψ_n| picks out the component of any state along ψ_n:

P̂_nφ = ψ_n ∫ψ_n*φ dτ = c_nψ_n , P̂_n² = P̂_n

Projectors are idempotent — projecting twice is the same as projecting once, which is geometrically obvious — and they are Hermitian. The completeness of the eigenfunction set is the statement that the projectors sum to the identity, ∑_nP̂_n = 1̂, a relation used so constantly in perturbation theory that it has a nickname: inserting a resolution of the identity.

Dirac notation, in one paragraph. Write the state ψ_n as a ket |n⟩ and its complex conjugate as a bra ⟨n|. The overlap integral ∫ψ_m*ψ_ndτ becomes the bracket ⟨m|n⟩, the matrix element ∫ψ_m*Âψ_ndτ becomes ⟨m|Â|n⟩, and the expectation value becomes ⟨ψ|Â|ψ⟩. Nothing new is being said; the notation simply removes the integral signs and the dummy variables, and it makes the coordinate system invisible, which is why it is universal in the research literature. This book writes integrals in the beginner layers and brackets in the advanced ones, and you should be able to move between them without thinking.

Where the matrix analogy has to be handled with care. A finite matrix always has a complete set of eigenvectors if it is Hermitian, and its spectrum is a finite list. A differential operator on an infinite-dimensional function space may have a continuous spectrum (the free-particle Hamiltonian), a spectrum with both discrete and continuous parts (hydrogen: bound levels below zero, ionisation continuum above), or eigenfunctions that are not square-integrable (plane waves). Everything proved in B.2 is proved for the discrete, square-integrable case; the continuous case needs Dirac-delta normalisation and is handled properly in a functional-analysis course, not here. For the whole of CSIR-NET the discrete case is what is examined.

Evaluate (ÂB̂ − B̂Â)f for Â = x̂ and B̂ = d/dx, and say what the answer means Easy

Step 1 — put in a test function and do not remove it. Let f = f(x) be an arbitrary differentiable function. Everything that follows is calculus on f.

Step 2 — the first order. ÂB̂f = x̂(df/dx) = x f′. Nothing subtle happens: differentiate, then multiply by x.

Step 3 — the second order. B̂Âf = (d/dx)(xf). Now the product rule bites: (d/dx)(xf) = f + x f′.

Step 4 — subtract. (ÂB̂ − B̂Â)f = x f′ − (f + x f′) = −f.

Step 5 — strip the test function, but only now. Since this holds for every f, the operator identity is x̂(d/dx) − (d/dx)x̂ = −1̂. In words: the two orders differ by the identity operator.

Step 6 — what it means. Multiply through by −iℏ to turn d/dx into p̂_x and the result becomes [x̂, p̂_x] = iℏ, the most important equation in the subject. It is derived in full in B.8, but notice that it has just been obtained from nothing more than the product rule. The uncertainty principle is a consequence of elementary calculus plus the choice of operator for momentum.

Which of these operators are linear? (a) d³/dx³; (b) ‘multiply by x²’; (c) ‘take the complex conjugate’; (d) ‘integrate from 0 to x’; (e) ‘take the reciprocal’ Medium

The test. Â(c₁f₁ + c₂f₂) must equal c₁Âf₁ + c₂Âf₂ for all complex constants and all functions. One counter-example is enough to kill it.

(a) Linear. Differentiation of any order distributes over sums and passes constants through unchanged. All the operators of quantum mechanics that involve derivatives are linear for this reason.

(b) Linear. x²(c₁f₁ + c₂f₂) = c₁x²f₁ + c₂x²f₂. Multiplication by any fixed function is linear; this is why V̂ is linear whatever V is.

(c) NOT linear — and this one catches people. Complex conjugation distributes over the sum, but it also conjugates the constants: K̂(cf) = c*f* ≠ cK̂f unless c is real. An operator with this property is called antilinear. The time-reversal operator is antilinear, and that is a genuine and important fact, not a curiosity.

(d) Linear. Definite integration with a variable upper limit distributes over sums and passes constants out. Integral operators appear in scattering theory and in the exchange term of Part 7.

(e) NOT linear. 1/(f + g) is emphatically not 1/f + 1/g. Try f = g = 1: the left side is ½, the right side is 2.

The moral. Only (a), (b) and (d) could possibly represent observables. Linearity is a necessary condition; B.2 supplies the second necessary condition, Hermiticity, and together they are the content of postulate 2.

⚠ Common mistakes & exam traps

Cancelling operators as if they were numbers. ÂB̂ ≠ B̂Â in general, so you may not move an operator past another one, and you may not divide by an operator. Every time you are tempted, insert a test function and do the calculus instead.
Reading ÂB̂f left to right. The rightmost operator acts first. p̂x̂ψ means: multiply ψ by x, then differentiate. Getting this backwards reverses the sign of every commutator you will ever compute.
Thinking ‘multiply by x’ is not really an operator. It is, and it is the position operator. The fact that it does no calculus is irrelevant — it still takes a function and returns a different function.
Confusing the operator Â with the observable A or with its eigenvalue a. Â is an instruction, A is a physical quantity, a is a real number obtained by measuring. Sloppiness here makes the statement of postulate 3 incomprehensible.
Assuming every operator has an inverse. Most do not. In particular the projection operators are as far from invertible as possible: they destroy information by design.
Assuming e^Âe^B̂ = e^Â+B̂. True for numbers, false for operators unless they commute. This is the standard way to get a wrong answer in a Part-C question on time evolution or on the Trotter–Suzuki splitting used in simulation.

Easy

Evaluate (d²/dx² + 3x̂)(e^2x + sin x).

Show solution

Apply term by term — and you may do so precisely because both operators are linear.
d²/dx²(e^2x) = 4e^2x; d²/dx²(sin x) = −sin x.
3x̂(e^2x + sin x) = 3xe^2x + 3x sin x.
Total: 4e^2x − sin x + 3xe^2x + 3x sin x.
Note that the answer is not a constant times the original function, so (d²/dx² + 3x̂) has no eigenfunction here — which is exactly the situation for a real Hamiltonian with a non-trivial potential, and why Part 3 has to work so hard.

Med

Show that (ÂB̂)Ĉ = Â(B̂Ĉ) but that in general ÂB̂ ≠ B̂Â, using Â = x̂, B̂ = d/dx, Ĉ = x̂².

Show solution

Associativity. Both sides mean the same chain of instructions applied to f: do Ĉ, then B̂, then Â. There is no room for them to differ, because brackets only regroup the order in which the compositions are formed, not the order in which the operators act. Explicitly, (x̂(d/dx))x̂²f = x(d/dx)(x²f) = x(2xf + x²f′) = 2x²f + x³f′, and x̂((d/dx)x̂²)f gives the same.
Non-commutativity. x̂(d/dx)f = xf′ while (d/dx)x̂f = f + xf′. They differ by f, so ÂB̂ − B̂Â = −1̂.
Moral: the algebra of operators is associative but not commutative — exactly like matrix algebra, and for the same reason.

Hard

The operator K̂ takes the complex conjugate. Show that K̂² = 1̂ and explain why K̂ cannot represent an observable.

Show solution

K̂²f = K̂(f*) = (f*)* = f, so K̂² = 1̂ and K̂ is its own inverse.
It cannot represent an observable for a reason that has nothing to do with Hermiticity: it is not linear. K̂(cf) = c*f*, whereas linearity would demand cf*. An operator that conjugates the coefficients cannot preserve superpositions, so postulate 2 excludes it at the first hurdle.
K̂ is nevertheless important: the time-reversal operator in quantum mechanics is antiunitary and contains exactly this conjugation, which is the origin of Kramers degeneracy in odd-electron systems — a result you will meet again in EPR spectroscopy.

Hard

Evaluate (d/dx + x̂)(d/dx − x̂) as a single operator.

Show solution

Expand carefully, keeping the order, with a test function f:
(d/dx + x)(f′ − xf) = (f′ − xf)′ + x(f′ − xf)
= f″ − f − xf′ + xf′ − x²f = f″ − x²f − f.
So (d/dx + x̂)(d/dx − x̂) = d²/dx² − x² − 1̂. The naive answer d²/dx² − x², obtained by treating the operators as numbers, is wrong by exactly the identity operator — the same −1̂ that appeared in the worked example.
This factorisation is not an arbitrary exercise: with suitable factors of √2 these are the ladder operators of the harmonic oscillator, and the stray −1̂ is the origin of the zero-point energy ½ℏω. Part 3 does this properly.

Read the rest of Part 2

The remaining 8 sections of this part — Hermitian operators: real eigenvalues and orthogonal eigenfunctions, The eigenvalue equation, the correspondence rules and the Schrödinger equation, Expectation values, Superposition and… — and all nine parts of Quantum Chemistry are part of ChemVidya Full Access, along with the other books, 55 Study Notes and 6,000+ practice questions.

See plans Open in the app

Continue through Quantum Chemistry

← Part 1Foundations — Why Quantum Mechanics Exists All nine partsQuantum Chemistry Part 3 →Exactly Solvable Systems I — Boxes, Steps & Barriers

Related Physical Study Notes

Quantum Chemistry — BasicsPhysical · Quantum Chemistry The Phase Rule & Phase EquilibriaPhysical · Phase Equilibria Statistical ThermodynamicsPhysical · Statistical Thermodynamics Chemical KineticsPhysical · Chemical Kinetics