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 Incoming links: Translation of a scalar creation operator

Past exam of the mathematics course of the University of Cambridge / 2005 / iii / Paper 49 / 1 / Solution Created 2026-10-03 Updated 2026-10-14

Use the Minkowski metric

η = diag (1, - 1, - 1, - 1)

and the free real scalar field Lagrangian density

L = \frac{1}{2} \partial_{μ} ϕ \partial^{μ} ϕ - \frac{1}{2} m^{2} ϕ^{2}

. Its conjugate canonical momentum is

π = \dot{ϕ}

. Spatial translation symmetry and Noether's theorem give the canonical stress-energy tensor

T^{μν} = \partial^{μ} ϕ \partial^{ν} ϕ - η^{μν} L, \partial_{μ} T^{μν} = (□ ϕ + m^{2} ϕ) \partial^{ν} ϕ = 0.

Vanishing boundary flux then makes the four-momentum of a free real scalar field constant. In particular the contravariant spatial momentum is

P = - \int π \nabla ϕ d^{3} x .

The minus sign is from

\partial^{i} = - \partial_{i}

; it is essential for the following mode calculation.

In the quantized theory, insert the mode expansions into this integral, using a Hermitian symmetric ordering or normal ordering. Write

\int_{p} = \int d^{3} p / (2 π)^{3}

. The spatial integral is

(2 π)^{3} δ^{(3)} (p \pm k)

, leaving

P = \frac{1}{2} \int_{p} p (a_{p} a_{p}^{†} + a_{p}^{†} a_{p} - a_{- p} a_{p} - a_{- p}^{†} a_{p}^{†}) .

Each last term integrates to zero: replace

p

by

- p

and commute the two annihilation operators or the two creation operators. The mixed terms reorder by the canonical commutation relation. Their vacuum contribution is proportional to the integral of the odd vector

p

and vanishes with an inversion-symmetric regulator; normal ordering removes it directly. Thus the normal-ordered free scalar four-momentum has spatial part

P = \int_{p} p a_{p}^{†} a_{p} .

These manipulations can be made first with box modes and a symmetric momentum cutoff, then used on finite-particle wave packets.

The commutator derivation identity gives

[P_{i}, a_{q}^{†}] = \int_{p} p_{i} a_{p}^{†} [a_{p}, a_{q}^{†}] = q_{i} a_{q}^{†}, [P_{i}, a_{q}] = - q_{i} a_{q} .

For

U (y) = e^{- i P \cdot y}

, the commutator expansion for exponential conjugation therefore gives the translation of a scalar creation operator

U (y) a_{q}^{†} U (y)^{†} = e^{- i q \cdot y} a_{q}^{†} .

The vacuum has zero momentum, so

U ∣0 ⟩ = ∣0 ⟩

. Hence the one-particle momentum eigenstate obeys

U ∣ q ⟩ = e^{- i q \cdot y} ∣ q ⟩

: translation preserves its momentum and multiplies it by the associated phase.

The annihilation operator receives the opposite phase. Substituting both phases into the field expansion gives

U (y) ϕ (x) U (y)^{†} = ϕ (x + y) .

This is the specified conjugation convention for the spatial translation operator. On states, the active translation moves a wave packet by

y

, giving a position wavefunction

ψ (x - y)

. Conjugating the field by

U^{†}

instead would give the opposite argument shift. Infinitesimally the displayed result is

[P_{i}, ϕ] = i \partial_{i} ϕ

.

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