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Advanced Quantum Mechanics course sequence · Interactive lesson

Permanent electric dipole moments violate both P and T

Every search for an electron's electric dipole moment is a search for broken time reversal, and this lesson proves why. Inside an isolated multiplet a moment can only point along J, and P and T each forbid that on their own. Then come the cases that look like counterexamples, hydrogen's n = 2 and ammonia, and why they are not.

UAQM-01.08 · Lesson 8 of 9 in this unit. Position is not a completion score.
Starting model: Permanent electric dipole moments violate both P and TStark map of two opposite-parity levels, split by Δ/h = 1058 MHz and joined by a dipole matrix element w = 1.73 e a₀; by default hydrogen's 2s₁/₂ and 2p₁/₂. At E = 100 V cm⁻¹ the lower level carries ⟨d_z⟩ = 0.67 e a₀; it saturates above Δ/2w = 239 V cm⁻¹. Dashed: ±wE, the Δ = 0 case.energy/h (MHz)dashed: Δ = 0, lines ±wElower level: ⟨dz⟩ = 0.67 e a₀evenodd0applied field E (V cm⁻¹)400
Start with a picture. Change the model in the lab below.
1

Build the model

A dipole locked to the angular momentum is odd under parity and under time reversal.

The key idea

A permanent electric dipole moment is a nonzero ⟨d⟩ in a stationary state of an isolated system with no applied field, seen as an energy shift −⟨d⟩⋅E that stays linear in E however weak the field is made.

Example. Hydrogen's 2s₁/₂ level in 10 V cm⁻¹ moves 0.46 MHz, quadratically in E, because its dipole is induced. An electron moment at the HfF⁺ bound, 4.1 × 10⁻³⁰ e cm, would move a level 4.1 × 10⁻²⁵ eV in 100 kV cm⁻¹, linearly.

The complete physical model and assumptions

Take a rotation-invariant system and a level of angular momentum j that is degenerate with nothing but its own 2j + 1 values of m. Inside that multiplet Quantum Mechanics II's projection theorem leaves the dipole operator d = Σ qₐrₐ no freedom: every matrix element of d is one number c times the matching element of J, so a moment can only point along the angular momentum. The two discrete symmetries then act on that single number. Parity reverses d and leaves J alone, so a parity-invariant Hamiltonian demands c = −c; time reversal leaves d alone and reverses J, so a T-invariant one demands c = −c as well. Either symmetry alone kills the moment, so a nonzero c needs both to fail, and because the weak interaction already breaks P, a measured electron or neutron moment is a test of T. The proof rests on three assumptions and stops when one is removed: opposite-parity levels that are exactly degenerate (hydrogen's 2s and 2p), opposite-parity levels so close that a laboratory field mixes them (ammonia's 23.79 GHz inversion pair, every polar molecule's parity doublets), or surroundings that break rotation invariance (an ion at a crystal site with no inversion centre). Each gives a real, measurable dipole, and none of them violates P or T.

1Permanent means linear in a vanishing field

Put a system in a weak uniform field and expand a level's energy W in the field E: W = W₀ − ⟨d⟩⋅E − ½αE² + … . A moment is permanent when the linear coefficient survives as E → 0, that is, when a stationary state of the isolated system has ⟨d⟩ ≠ 0 with no field applied; an induced moment, αE, vanishes with the field. The lesson works in SI units, quoting moments in the e cm the experiments use (1 e cm = 1.602 × 10⁻²¹ C m); nothing here is relativistic, so the course's metric (+, −, −, −) never enters. For scale, e a₀ = 8.48 × 10⁻³⁰ C m = 5.29 × 10⁻⁹ e cm, and the best electron bound, 4.1 × 10⁻³⁰ e cm, is twenty-one orders of magnitude below it.

2Inside one multiplet, d has nowhere to point but J

Let the level be a single multiplet |α j m⟩, m = −j … j, degenerate with no other state. Quantum Mechanics II's projection theorem, quoted, says every vector operator restricted to it is a multiple of J: ⟨α j m′|d|α j m⟩ = c⟨j m′|J|j m⟩ with c = ⟨J⋅d⟩/(ħ²j(j + 1)). J⋅d is Hermitian, so c is real, and the stretched state m = j carries ⟨dz⟩ = cjħ, the number called the moment. For j = 0 there is nothing to point along and ⟨d⟩ = 0 by rotation invariance alone, the trivial limit. The magnetic moment obeys the same theorem, μ = γJ, and is allowed: μ and J are both axial and both reverse under T, so γ is even under both symmetries. The dipole is polar and T-even, the opposite of J on both counts.

3Parity: a polar vector cannot be locked to an axial one

Suppose [P, H] = 0. P also commutes with rotations, so it maps the isolated multiplet onto itself and commutes with Jz and with the ladder operators; each |α j m⟩ is then a parity eigenstate with one common eigenvalue η = ±1. Since P d P⁻¹ = −d and P J P⁻¹ = +J, conjugating d = cJ by P gives −d = cJ, so c = −c and c = 0, derived. Directly: ⟨d⟩ = ⟨ψ|P⁻¹(−d)P|ψ⟩ = −η²⟨d⟩. Parity alone is a weak protection, though. The weak interaction breaks it, measured since 1957, and atomic parity violation has been seen in caesium. Yet a moment needs more than broken P, because time reversal forbids it independently.

4Time reversal reaches the same verdict antiunitarily

Suppose [Θ, H] = 0. Then Θ maps the multiplet onto itself, and because Θ Jz Θ⁻¹ = −Jz, it sends |α j m⟩ to a phase times |α j −m⟩ (topic 1.06). For an antiunitary Θ, ⟨Θψ|A|Θψ⟩ = ⟨ψ|Θ⁻¹AΘ|ψ⟩*, and dz is Hermitian and T-even, so ⟨dz⟩ in the state −m equals ⟨dz⟩ in m: T makes the moment even in m. The projection theorem makes it cmħ, odd in m. At m = j ≠ 0 both hold only if c = 0, derived. The phase cancels, so Θ² = ±1 never enters. With either symmetry enough, a nonzero moment needs P and T both broken. By the CPT theorem, quoted here, it then breaks CP too, which is why these searches probe new sources of CP violation.

5Hydrogen, ammonia and polar molecules break the assumptions

In the Coulomb model hydrogen's 2s and 2p are degenerate, so n = 2 is not one j multiplet. A field mixes them into (|2s⟩ ± |2p₀⟩)/√2, shifted by ±3 e a₀ E (Quantum Mechanics II): linear, yet those states are not parity eigenstates and carry ⟨L⟩ = 0, so their dipole is not tied to J, neither proof applies, and H stays P- and T-invariant. The Lamb shift, 1057.8 MHz (quoted), restores isolation: with w = √3 e a₀ the 2s₁/₂ shift stays quadratic below Δ/2w ≈ 239 V cm⁻¹. Quantum Mechanics I's ammonia is the same two-level problem, Δ/h = 23.79 GHz and w = 1.47 D = 0.578 e a₀, so Δ/2w ≈ 16 kV cm⁻¹. A polar molecule's 1.85 D, water's say, is a body-frame number: a free molecule's stationary states have definite parity, and a lab field merely mixes close opposite-parity rotational levels. An ion at a crystal site with no inversion centre is the third escape, with rotation invariance itself gone.

6What the experiments bound, and how small it is

For an electron H = −dₑ σ⋅Eeff, but a bound non-relativistic electron feels zero average field, since ⟨F⟩ = 0 in a stationary state: Schiff's theorem, quoted. The signal is relativistic and grows steeply with nuclear charge, so experiments use heavy polar molecules, fully polarised by fields of order 100 V cm⁻¹ because their parity doublets are so close. The electron then feels a calculated Eeff of about 78 GV cm⁻¹ in ThO and 23 GV cm⁻¹ in HfF⁺ (quoted). The measured bounds on |dₑ|, at 90% confidence, are 1.1 × 10⁻²⁹ e cm from ThO and 4.1 × 10⁻³⁰ e cm from HfF⁺ (quoted). At the HfF⁺ bound, dₑEeff = 9.4 × 10⁻²⁰ eV, and the splitting 2dₑEeff is h × 4.6 × 10⁻⁵ Hz. The neutron's bound, 1.8 × 10⁻²⁶ e cm (quoted), keeps any charge separation along its spin below 2 × 10⁻²⁶ cm, two parts in 10¹³ of its own size. Every measurement so far is consistent with zero.

2

Key equations

Read the units, assumptions and limits before substituting.

The multiplet's only vector is J

⟨α j m′|d|α j m⟩ = c ⟨j m′|J|j m⟩, with c = ⟨J⋅d⟩/(ħ² j(j + 1))

Quoted from Quantum Mechanics II's projection theorem. It turns the whole question into one real number c; for j = 0 it gives ⟨d⟩ = 0 from rotation invariance alone.

d = Σ qₐrₐ in C m, J in J s, c in C m J⁻¹ s⁻¹; one j multiplet, degenerate with no other level

Parity forces c = 0

P d P⁻¹ = −d and P J P⁻¹ = +J, so −cJ = cJ

Derived: conjugate d = cJ by P. A polar vector cannot be locked to an axial one if H is parity-invariant.

P unitary; every state of the isolated multiplet is a parity eigenstate, because P commutes with H and with J

Time reversal forces c = 0

Θ d Θ⁻¹ = +d and Θ|j m⟩ ∝ |j −m⟩ give c m ħ = −c m ħ

Derived: T makes ⟨dz⟩ even in m, rotations make it c m ħ, odd in m, and only c = 0 is both. Θ² never enters, so integer and half-integer j fall alike.

Θ antiunitary, with ⟨Θψ|dz|Θψ⟩ = ⟨ψ|dz|ψ⟩ for Hermitian, T-even dz; take m = j ≠ 0

The moment's Hamiltonian

HEDM = −(d/jħ) J⋅E

Derived from d = cJ. J⋅E is odd under P and under T, and it shifts level m by −dEm/j: linear in E, reversing with E, the signature every search looks for.

d = c j ħ, the moment of the m = j state, in C m (quoted in e cm; 1 e cm = 1.602 × 10⁻²¹ C m); E in V m⁻¹; spin ½ gives −dₑ σ⋅E

Two opposite-parity levels in a field

W± = ±√((Δ/2)² + (wE)²)

Derived by diagonalising the 2 × 2 Hamiltonian. Δ → 0 gives the linear ±wE of hydrogen's degenerate n = 2; wE ≫ Δ/2 gives a linear shift with no broken symmetry.

W± the level energies in J, from the doublet's centre; Δ the zero-field splitting in J; w = |⟨+|dz|−⟩| in C m; E in V m⁻¹

Weak fields: the moment is induced

⟨dz⟩ = αE, with α = 2w²/Δ, for wE ≪ Δ/2

Derived from ⟨dz⟩ = −∂W₋/∂E. However large α is, ⟨dz⟩ vanishes as E → 0: a big polarisability is not a permanent moment.

α in C m² V⁻¹, for the lower level, the upper one having −α; hydrogen's pair, with w = √3 e a₀ and Δ/h = 1057.8 MHz, gives α = 6.15 × 10⁻³⁴ C m² V⁻¹, so 2s₁/₂, the upper level, has ⟨dz⟩ = −αE
3

Interactive lab

Predict first. Explore the model. Compare the result.

Explore the model

1058 MHz
0 to 2000 MHz; steps of 2.
1.73 e a₀
0.5 to 3 e a₀; steps of 0.01.
100 V cm⁻¹
0 to 400 V cm⁻¹; steps of 5.

Try this. At hydrogen's 1058 MHz, drag E up from zero: the levels start flat, so ⟨dz⟩ starts at zero, and only past about 239 V cm⁻¹ do they turn linear. Now drag Δ to 0: the curves fall onto the dashed lines, a linear Stark effect from zero field with no symmetry broken.

Live results

⟨dz⟩, LOWER LEVEL
0.67 e a₀
FRACTION OF w
39 %
STARK SHIFT OF EACH LEVEL
44.4 MHz
SATURATION FIELD Δ/2w
239 V cm⁻¹

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

Interactive Advanced Quantum Mechanics modelStark map of two opposite-parity levels, split by Δ/h = 1058 MHz and joined by a dipole matrix element w = 1.73 e a₀; by default hydrogen's 2s₁/₂ and 2p₁/₂. At E = 100 V cm⁻¹ the lower level carries ⟨d_z⟩ = 0.67 e a₀; it saturates above Δ/2w = 239 V cm⁻¹. Dashed: ±wE, the Δ = 0 case.energy/h (MHz)dashed: Δ = 0, lines ±wElower level: ⟨dz⟩ = 0.67 e a₀evenodd0applied field E (V cm⁻¹)400
Stark map of two opposite-parity levels, split by Δ/h = 1058 MHz and joined by a dipole matrix element w = 1.73 e a₀; by default hydrogen's 2s₁/₂ and 2p₁/₂. At E = 100 V cm⁻¹ the lower level carries ⟨dz⟩ = 0.67 e a₀; it saturates above Δ/2w = 239 V cm⁻¹. Dashed: ±wE, the Δ = 0 case.

Predict → change one variable → check

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

Saved baseline. Opposite-parity splitting Δ/h: 1058 MHz; Dipole matrix element w: 1.73 e a₀; Applied field E: 100 V cm⁻¹

Baseline 1058; choose 0 to 2000, in steps of 2.
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 dipole operator is even under time reversal, so a permanent moment breaks only parity.” The error is treating the moment as d alone. A moment of an isolated multiplet is the relation d = cJ, a fixed link between the dipole and the angular momentum, and the measurable is the sign and size of c. Time reversal leaves d alone and reverses J, so the time-reversed system obeys d = −cJ, a different law unless c = 0. The energy that reveals the moment, −(d/jħ) J⋅E, couples a T-even vector to a T-odd one and is odd under T exactly as σ⋅E is. Compare the magnetic moment μ = γJ: μ and J both reverse under T and neither changes under P, so neither symmetry constrains γ, and spin magnetic moments are commonplace.

5

Practice & worked examples

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

EasyExample 1

Show that H = −dₑ σ⋅E is odd under P and under T. Then, for dₑ = 4.1 × 10⁻³⁰ e cm, the HfF⁺ bound, find the splitting 2dₑE between spin along and against E, as an energy and a frequency, in a 100 kV cm⁻¹ laboratory field and in HfF⁺'s effective field of about 23 GV cm⁻¹ (quoted).

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

Hydrogen's 2s₁/₂ lies 1057.8 MHz above 2p₁/₂ (the Lamb shift, quoted), and for mⱼ = ½ the two are joined by |⟨2s₁/₂|dz|2p₁/₂⟩| = √3 e a₀. In a two-level model, find the Stark shift and induced dipole of 2s₁/₂ in 10 V cm⁻¹, the field Δ/2w at which the response turns linear, and the 2s lifetime the field causes, given the 2p lifetime of 1.60 ns (quoted).

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.

HardExample 3

The ¹⁹⁹Hg ground level is a nuclear-spin doublet, F = I = ½, isolated from every other level. Let dA be its ⟨dz⟩ in the m = +½ state. In a weak field B along z, E = 10.0 kV cm⁻¹ is applied parallel, then antiparallel, to B. Find the change Δν in the spin-precession frequency, evaluate it at dA = 7.4 × 10⁻³⁰ e cm (the ¹⁹⁹Hg bound, quoted), and explain why an E-split doublet at B = 0 would prove T violation.

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.

Ammonia's two lowest inversion levels, 23.79 GHz apart, show a Stark shift that is quadratic in weak fields and becomes linear in strong ones, with a slope set by the molecule's 1.47 D dipole. What does the linear regime say about P and T?
7

Review & continue

A permanent electric dipole moment is a nonzero ⟨d⟩ in a stationary state of an isolated system with no applied field, seen as an energy shift −⟨d⟩⋅E that stays linear in E however weak the field is made.

  • 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.