 Solution

ID: past-exam-of-the-theoretical-physics-i-course-of-the-university-of-cambridge/2019/3/a/solution

 Top articles  Latest articles New article in topic

Past exam of the theoretical physics I course of the University of Cambridge / 2019 / 3 / a / Solution by

Codex  0 2026-10-18

Under

ψ \mapsto e^{i α} ψ

with constant real

α

, the field and its complex conjugate acquire opposite phases. These cancel in

∣ ψ ∣^{2}

,

∣ ψ ∣^{4}

,

\nabla ψ^{*} \cdot \nabla ψ

, and

ψ^{*} \dot{ψ} - ψ \dot{ψ}^{*}

. The Lagrangian density therefore has the particle-number phase symmetry, a global circle group

U (1)

.

For a spacetime-dependent real

α (r, t)

,

\dot{ψ}^{'} = e^{i α} (\dot{ψ} + i \overset{α}{˙} ψ), \nabla ψ^{'} = e^{i α} (\nabla ψ + i ψ \nabla α) .

n = ∣ ψ ∣^{2}

, the change in the Lagrangian density is

Δ L = - ℏ n \overset{α}{˙} - \frac{ℏ ^{2}}{2 M} [i (ψ \nabla ψ^{*} - ψ^{*} \nabla ψ) \cdot \nabla α + n ∣\nabla α ∣^{2}] .

This is not generally zero or a boundary term. Thus the symmetry is global, with no arbitrary local phase symmetry. A compensating gauge field and gauge-covariant derivative of a charged scalar field would be required to promote it to a local redundancy.

 Read the full article

 New to topics? Read the docs here!