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Advanced Quantum Mechanics course sequence · Interactive lesson

Antiunitary operators and complex conjugation K

Time reversal is antiunitary, and an antiunitary operator breaks three habits Dirac notation builds: it conjugates every number it meets, it acts only to the right, and it swaps bra and ket in every matrix element. Learn the toolkit once, with K tied to a named basis, and Kramers, dipole moments and reciprocity become short proofs.

UAQM-01.02 · Lesson 2 of 9 in this unit. Position is not a completion score.
Starting model: Antiunitary operators and complex conjugation KPlane of mean position and momentum in oscillator units. Conjugation K_θ in the eigenbasis of x cos θ + p sin θ mirrors the state ψ, open circle at (1.5, 1.0), in the dashed line at θ = 0°, giving the filled circle at (1.50, -1.00). The faint circle is Kₓψ; the unitary e^(2iθN) turns it by 0° onto K_θ ψ.open: ψ, filled: Kθ ψ, faint: Kₓψ⟨x⟩⟨p⟩mirror line θ = 0°after Kθ:⟨x⟩ = 1.50⟨p⟩ = −1.00Kθ = e(2iθN) KₓKₓψ turned 0°[x′, p′] = −[x, p]
Start with a picture. Change the model in the lab below.
1

Build the model

Conjugate every number it meets, and bra and ket trade places in every matrix element.

The key idea

An antiunitary operator Θ maps kets one-to-one, complex-conjugates every number it passes, Θ(c|a⟩) = c*Θ|a⟩, and conjugates every overlap, ⟨Θa|Θb⟩ = ⟨a|b⟩*, so every transition probability is kept.

Example. Let K conjugate components in the Sz basis. K(|↑⟩ + i|↓⟩)/√2 = (|↑⟩ − i|↓⟩)/√2, so spin up along +y becomes spin up along −y, and ⟨Sy⟩ goes from +ħ/2 = +5.27 × 10⁻³⁵ J s to −5.27 × 10⁻³⁵ J s.

The complete physical model and assumptions

Of the two kinds of symmetry Wigner's theorem allows (topic 1.01, quoted), the antiunitary one, Θ(c|a⟩) = c*Θ|a⟩ with ⟨Θa|Θb⟩ = ⟨a|b⟩*, is the only way a symmetry can send i to −i, which time reversal will need. Its structure is simple once a basis is named: conjugating components in that basis is an antiunitary K, ΘK is then linear and unitary, so every antiunitary is Θ = UK, derived. The catch is that K is not one operator. Conjugating position amplitudes keeps |ψ(x)|² and sends |p⟩ to |−p⟩; conjugating momentum amplitudes sends |x⟩ to |−x⟩; rephasing the basis moves the momentum about which K reflects. Dirac notation needs the same care: Θ acts only to the right, an operator becomes the linear ΘAΘ⁻¹ with every number inside it conjugated, and ⟨a|A|b⟩ = ⟨Θb|ΘA†Θ⁻¹|Θa⟩, bra and ket exchanged, derived. With its ± for even and odd Hermitian A, that rule is what topics 1.07 to 1.09 run on. The model stops at physical operations: no Hamiltonian generates Θ, Θ cannot act on one particle of a pair alone, and no device applies it to an unknown state. The best physical approximation to conjugating an unknown qubit has fidelity 2/3, and single-photon realisations of the antiunitary universal NOT reached 0.618 ± 0.013 to 0.630 ± 0.008 (both quoted).

1Antilinear: real multiples pass, i is reversed

Wigner's theorem (topic 1.01, quoted) lets a symmetry keep every overlap ⟨a|b⟩, or keep its modulus and conjugate it. The second kind is antiunitary: Θ(c|a⟩ + d|b⟩) = c*Θ|a⟩ + d*Θ|b⟩ and ⟨Θa|Θb⟩ = ⟨a|b⟩*. Antilinear is not nonlinear. Sums and real multiples pass straight through and only i is reversed, so as operators ΘiΘ⁻¹ = −i and Θ(iħ)Θ⁻¹ = −iħ. Probabilities survive, since |⟨a|b⟩*|² = |⟨a|b⟩|². A one-dimensional space shows the whole structure: an antiunitary is c → e(iα)c*, a reflection of the complex plane in the line at angle α/2, which no rotation c → e(iβ)c can imitate. So no continuous family starting at the identity, and no Hamiltonian, can generate Θ, while a product of two antiunitaries is unitary (topic 1.01, quoted). Units throughout are SI with ħ = 1.0546 × 10⁻³⁴ J s kept explicit; nothing here is relativistic, so no metric signature enters.

2Every antiunitary is a unitary times K

Pick an orthonormal basis |n⟩, phases included, and let K conjugate components: K Σ cₙ|n⟩ = Σ cₙ*|n⟩. K is antiunitary, because ⟨Ka|Kb⟩ = Σ aₙbₙ* = ⟨a|b⟩*, and K² = 1. Given any antiunitary Θ, set U = ΘK. Then U(c|a⟩) = Θ(c*K|a⟩) = cΘK|a⟩, so U is linear, and ⟨Ua|Ub⟩ = ⟨Ka|Kb⟩* = ⟨a|b⟩, so U is unitary. Hence Θ = UK, derived: on a column of components Θ acts as c → U c*, conjugate first, then multiply. On a spin half, Θ|↑⟩ = |↓⟩ and Θ|↓⟩ = |↑⟩ define Θ = σₓK. Two particles share one K, conjugating product-basis components, and ΘA ⊗ ΘB = (UA ⊗ UB)K is well defined. K on particle A alone is not: c(|a⟩ ⊗ |b⟩) is both (c|a⟩) ⊗ |b⟩ and |a⟩ ⊗ (c|b⟩), and the two readings would conjugate c differently.

3K belongs to a basis, phases included

Conjugating components means something only once the basis is named. If |n′⟩ = W|n⟩, conjugation in the new basis is K′ = WKW⁻¹, and since KW⁻¹ = WᵀK, K′ = WWᵀK, derived, with Wᵀ the transpose in the old basis. So K′ = K exactly when W is real. For a spin half with |±y⟩ = (|↑⟩ ± i|↓⟩)/√2, WWᵀ = σz and Ky = σzKz. The two disagree about the same ket: Kz|+y⟩ = |−y⟩, while Ky|+y⟩ = |+y⟩. Kz keeps Sₓ and Sz and reverses Sy, the one imaginary Pauli matrix; Ky reverses Sₓ instead. Phases count too: rephasing |n⟩ → e(iφₙ)|n⟩ multiplies K by diag(e(2iφₙ)). Every such K squares to 1 and mirrors the Bloch sphere in one plane, so no basis gives a K that reverses all three spin components; topic 1.06 shows why spin's time reversal needs more than K.

4Conjugating position amplitudes reverses momentum

With ⟨x|p⟩ = e(ipx/ħ)/√(2πħ), Kₓ sends ψ(x) to ψ*(x), and conjugating ⟨x|p⟩ gives ⟨x|−p⟩, so Kₓ|p⟩ = |−p⟩, derived. Then Kₓ x Kₓ⁻¹ = x and Kₓ p Kₓ⁻¹ = −p, since −iħ d/dx loses the sign of its i. An electron packet with mean wavenumber 5.0 nm⁻¹, moving at ħk/mₑ = 5.79 × 10⁵ m s⁻¹, becomes one with the same |ψ(x)|² moving at −5.79 × 10⁵ m s⁻¹. Conjugating momentum amplitudes instead gives Kₚψ(x) = ψ*(−x), so Kₚ = ΠKₓ with Π the parity, a unitary. For an oscillator, with x and p in units of √(ħ/mω) and √(ħmω), conjugation in the rotated position kets e(iθN)|q⟩, eigenkets of x cos θ + p sin θ, is Kθ = e(2iθN)Kₓ, since the number operator N is real in the position basis: the plane of (⟨x⟩, ⟨p⟩) is mirrored in the line at angle θ. Check: Kₓ[x, p]Kₓ⁻¹ = −iħ = [x, −p], J s on both sides.

5Θ acts only to the right

For a linear operator ⟨a|A|b⟩ is the same whether A acts right or left. For an antilinear Θ it is not. The map |b⟩ → ⟨a|Θb⟩ is antilinear in |b⟩, so ⟨a|Θ is not a bra at all. The natural attempt to make one, using the antiunitary adjoint fixed by ⟨a|Θb⟩ = ⟨b|Θ†a⟩, with Θ† = Θ⁻¹, gives ⟨Θ†a|b⟩ = ⟨a|Θb⟩*: the complex conjugate of what was meant. Take K conjugating components in the basis |1⟩, |2⟩, with |a⟩ = (3|1⟩ + 4i|2⟩)/5 and |b⟩ = (|1⟩ + |2⟩)/√2. Then ⟨a|Kb⟩ = (3 − 4i)/(5√2) = 0.424 − 0.566i, while ⟨Ka|b⟩ = 0.424 + 0.566i: the same symbols, the opposite phase, with K² = 1 so K† = K. The convention from here on is that Θ acts on the ket to its right, written ⟨a|Θb⟩ or ⟨Θa|Θb⟩, and never as ⟨a|Θ|b⟩ with Θ free to move left.

6Transformed operators and the matrix-element rule

If states go to Θ|a⟩, operators go to A′ = ΘAΘ⁻¹, which is linear, since its two antilinear factors cancel; numbers inside A are conjugated, and ΘA†Θ⁻¹ = (ΘAΘ⁻¹)†. Now let |γ⟩ = A†|a⟩. Then ⟨a|A|b⟩ = ⟨b|γ⟩* = ⟨Θb|Θγ⟩ = ⟨Θb|ΘA†Θ⁻¹|Θa⟩, derived. For Hermitian A with ΘAΘ⁻¹ = ηA, η = ±1, this is ⟨a|A|b⟩ = η⟨Θb|A|Θa⟩: bra and ket exchange, and odd operators change sign. Check with Kz on a spin half: ⟨↑|Sy|↓⟩ = −iħ/2, Sy is odd, and −⟨↓|Sy|↑⟩ = −iħ/2. Two limits. At a = b, ⟨A⟩ in Θ|a⟩ is η⟨A⟩ in |a⟩, so an odd A has zero mean in any state Θ returns up to a phase. For a unitary U the same steps give ⟨Ua|UAU⁻¹|Ub⟩ with no exchange. The exchange is the fingerprint of antiunitarity, and Kramers, the dipole theorem and reciprocity (topics 1.07 to 1.09) apply this line.

2

Key equations

Read the units, assumptions and limits before substituting.

Antilinearity

Θ(c|a⟩ + d|b⟩) = c* Θ|a⟩ + d* Θ|b⟩

Definition. As an operator identity ΘcΘ⁻¹ = c*, so Θ(iħ)Θ⁻¹ = −iħ. Sums and real multiples pass straight through: antilinear means real-linear with i reversed.

|a⟩, |b⟩ any kets; c, d complex numbers, dimensionless; Θ carries no units, so Θ|a⟩ has the units of |a⟩

Antiunitarity

⟨Θa|Θb⟩ = ⟨a|b⟩*

Definition, one of the two options Wigner's theorem allows (topic 1.01, quoted). Norms survive, ‖Θa‖ = ‖a‖, while the phase of every overlap is reversed.

⟨a|b⟩ is dimensionless for normalised kets; its modulus, and every probability |⟨a|b⟩|², is unchanged

Every antiunitary is a unitary times K

Θ = U K, with K Σ cₙ|n⟩ = Σ cₙ* |n⟩ and U = ΘK

Derived: ΘK is a product of two antiunitaries, so unitary (topic 1.01). K depends on the basis: in |n′⟩ = W|n⟩ it becomes WWᵀK, equal to K only for real W.

|n⟩ an orthonormal basis, phases included; cₙ = ⟨n|a⟩ dimensionless; K² = 1; U linear and unitary

Conjugation in the position basis

Kₓ|p⟩ = |−p⟩

Derived by conjugating ⟨x|p⟩. So Kₓ x Kₓ⁻¹ = x and Kₓ p Kₓ⁻¹ = −p: |ψ(x)|² is kept and every momentum reversed. In the momentum basis the roles swap.

Kₓ sends ψ(x) to ψ*(x) in the basis with ⟨x|p⟩ = e(ipx/ħ)/√(2πħ); p in kg m s⁻¹; ħ = 1.0546 × 10⁻³⁴ J s

The transformed operator

A′ = Θ A Θ⁻¹

Derived: two antilinear factors make a linear one. Then ⟨Θa|A′|Θb⟩ = ⟨a|A|b⟩*, not ⟨a|A|b⟩: a real mean survives, an off-diagonal phase does not.

A any linear operator, in its own units; A′ is linear; numbers inside A are conjugated; (ΘAΘ⁻¹)† = ΘA†Θ⁻¹

The matrix-element rule

⟨a|A|b⟩ = ⟨Θb|ΘA†Θ⁻¹|Θa⟩

Derived. When a Hermitian A has ΘAΘ⁻¹ = ηA, η = ±1, it shortens to ⟨a|A|b⟩ = η⟨Θb|A|Θa⟩, with no complex conjugate left anywhere.

Θ acts only to the right; A any linear operator, so both sides carry the units of A, J s for a spin component
3

Interactive lab

Predict first. Explore the model. Compare the result.

Explore the model

1.5 √(ħ/mω)
−2 to 2 √(ħ/mω); steps of 0.1.
1.0 √(ħmω)
−2 to 2 √(ħmω); steps of 0.1.
0 °
0 to 180 °; steps of 5.

Try this. Start at θ = 0, where Kθ is Kₓ: the filled circle drops to (⟨x⟩, −⟨p⟩), same position, momentum reversed. Sweep to 90° for Kₚ, which reverses ⟨x⟩ instead. At every angle Kₓψ and Kθ ψ differ by a turn of 2θ, a unitary, while ψ itself is only ever mirrored.

Live results

⟨x⟩ AFTER Kθ
1.50 √(ħ/mω)
⟨p⟩ AFTER Kθ
-1.00 √(ħmω)
KEPT x cos θ + p sin θ
1.50
TURN FROM Kₓψ TO Kθ ψ
0 °

Results follow this lesson’s model and assumptions. Displayed values are rounded.

Interactive Advanced Quantum Mechanics modelPlane of mean position and momentum in oscillator units. Conjugation K_θ in the eigenbasis of x cos θ + p sin θ mirrors the state ψ, open circle at (1.5, 1.0), in the dashed line at θ = 0°, giving the filled circle at (1.50, -1.00). The faint circle is Kₓψ; the unitary e^(2iθN) turns it by 0° onto K_θ ψ.open: ψ, filled: Kθ ψ, faint: Kₓψ⟨x⟩⟨p⟩mirror line θ = 0°after Kθ:⟨x⟩ = 1.50⟨p⟩ = −1.00Kθ = e(2iθN) KₓKₓψ turned 0°[x′, p′] = −[x, p]
Plane of mean position and momentum in oscillator units. Conjugation Kθ in the eigenbasis of x cos θ + p sin θ mirrors the state ψ, open circle at (1.5, 1.0), in the dashed line at θ = 0°, giving the filled circle at (1.50, −1.00). The faint circle is Kₓψ; the unitary e(2iθN) turns it by 0° onto Kθ ψ.

Predict → change one variable → check

Plan a change before running it. Every other input stays at the saved baseline.

Saved baseline. Mean position ⟨x⟩ of ψ: 1.5 √(ħ/mω); Mean momentum ⟨p⟩ of ψ: 1.0 √(ħmω); Angle θ of the basis that K conjugates: 0 °

Baseline 1.5; choose −2 to 2, in steps of 0.1.
4. Compared with the saved baseline, this result will…

This experiment is kept only while this page is open. It does not change saved course progress.

4

Catch the common trap

The tempting error is to treat Θ like a unitary: let it act on the bra as well, and write ⟨Θa|ΘAΘ⁻¹|Θb⟩ = ⟨a|A|b⟩ because ⟨Ua|UAU⁻¹|Ub⟩ = ⟨a|A|b⟩. For an antiunitary that sandwich is the complex conjugate, ⟨a|A|b⟩* = ⟨b|A†|a⟩, and that conjugate is exactly why the correct rule exchanges bra and ket. Expectation values hide the slip, since they are real; off-diagonal elements expose it. In the Sz basis ⟨↑|Sy|↓⟩ = −iħ/2 = −5.27i × 10⁻³⁵ J s, and K Sy K⁻¹ = −Sy. The unitary habit predicts ⟨K↑|(−Sy)|K↓⟩ = −iħ/2, but the matrices give +iħ/2. Kramers' theorem and reciprocity are built from the phases of off-diagonal elements like this one, so a dropped conjugate gives a wrong theorem, not a small numerical error.

5

Practice & worked examples

3 authored examples. Make an attempt, then reveal the reasoning one step at a time.

EasyExample 1

On a spin half, Θ is defined by Θ|↑⟩ = |↓⟩ and Θ|↓⟩ = |↑⟩, extended antilinearly. Write Θ as UK with K conjugating Sz-basis components, find Θ|a⟩ for |a⟩ = (|↑⟩ + i|↓⟩)/√2, test ⟨Θa|Θb⟩ = ⟨a|b⟩* with |b⟩ = (|↑⟩ + |↓⟩)/√2, and find which spin components Θ reverses.

Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.

0 of 5 steps revealed.

MediumExample 2

Kₓ conjugates amplitudes in the standard position basis, where ⟨x|p⟩ = e(ipx/ħ)/√(2πħ). A colleague conjugates instead in the rephased basis |x⟩′ = e(ik₁x)|x⟩, with k₁ = 3.0 nm⁻¹. Express K′ through Kₓ, find K′pK′⁻¹, and give the mean velocity of an electron packet with mean wavenumber k₀ = 5.0 nm⁻¹ before and after each operation.

Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.

0 of 6 steps revealed.

HardExample 3

Two electron packets of width σ = 1.0 nm are ψⱼ(x) = (2πσ²)(−1/4) exp[−(x − xⱼ)²/(4σ²) + ikⱼx], with x₁ = 0, k₁ = 1.0 nm⁻¹ and x₂ = 1.0 nm, k₂ = 2.0 nm⁻¹. Compute ⟨ψ₁|p|ψ₂⟩. Then predict from the matrix-element rule ⟨Kₓψ₂|p|Kₓψ₁⟩ and ⟨Kₚψ₂|p|Kₚψ₁⟩, where Kₚ conjugates momentum-space amplitudes, and verify both.

Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.

0 of 6 steps revealed.

6

Quick check

Select an answer and explain it before checking.

Kₓ is complex conjugation in the position basis, so Kₓ x Kₓ⁻¹ = x and Kₓ p Kₓ⁻¹ = −p. Two atomic states have ⟨φ|Lz|ψ⟩ = (0.6 − 0.8i)ħ. What is ⟨Kₓψ|Lz|Kₓφ⟩?
7

Review & continue

An antiunitary operator Θ maps kets one-to-one, complex-conjugates every number it passes, Θ(c|a⟩) = c*Θ|a⟩, and conjugates every overlap, ⟨Θa|Θb⟩ = ⟨a|b⟩*, so every transition probability is kept.

  • Explain the relationship without looking at the equation.
  • Check a sign, a unit and a model assumption in your working.
  • Return to the lab or worked example if a step is still unclear.

Extra challenging questions

Connect several ideas, explore supplied data, and compare your reasoning with a worked solution.