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In particle physics, one checks the invariance of the Lagrangian under global transformations of the fields $\psi\rightarrow e^{i\alpha}\psi$ and local transformations $\psi\rightarrow e^{i\alpha(x)}\psi$.

The latest was the motivation for introducing covariant derivative and the photon in the case of $U(1)$ transformation.

But at the basis, what is the motivation to have invariance of a given defined Lagrangian under a transformation of phase $e^{i\alpha}$ of the fields?

Qmechanic
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2 Answers2

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In particle physics, what is the motivation to have invariance of a defined lagrangian under a transformation of phase $e^{iα}$ of the fields?

Based on your comments on the other answers, I think you might just be looking for the straightforward answer: Charge conservation.

A derivation of Noether's theorem starts by considering a classical field transformation $$ \phi(x) \to \phi(x) + \eta\Delta \phi(x)\;, $$ where $\eta$ is "small," and where the transformation only changes the the Lagrange density by a total divergence: $$ \mathcal{L}\to\mathcal{L}+\eta\partial_\mu K^\mu\;. $$

Such a transformation is called a "symmetry," since the equations of motion are not changed.

Given the above type of symmetry transformation we have a conserved current: $$ j^\mu = \frac{\partial \mathcal{L}}{\partial \partial_\mu\phi}\Delta \phi(x) - K^\mu\tag{1}\;. $$


If you consider the specific transformation $$ \phi(x)\to e^{i\alpha}\phi(x)\;, $$ $$ \phi^*(x)\to e^{-i\alpha}\phi^*(x)\;, $$ for a complex scalar field $\phi(x)$, and the specific Lagrange density $$ \mathcal{L} = |\partial_\mu \phi|^2 - V(|\phi|^2)\;, $$ then $K^\mu=0$ and the conserved current of Eq. (1) above is $$ j^\mu = i\left(\phi\partial^\mu \phi^* - \phi^*\partial^\mu\phi\right) $$ and the conserved charge is $$ Q=\int d^3x j^0(x)\;,\tag{2} $$ which is often interpreted as the electric change associated with the complex $\phi$ field. (E.g., in the context of "poor man's QED.")

The charge of Eq. (2) is conserved because $$ \frac{dQ}{dt} = \int d^3 x \frac{dj^0}{dt} = \int d^3 x\vec \nabla \cdot \vec j = \oint d^2\vec S\cdot \vec j = 0\;, $$ since we assume the current is localized somewhere (and so is zero far away at "infinity").

hft
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A counter example: the Majorana mass term for right-handed neutrino $$ m\psi^\dagger\psi_c \rightarrow me^{-2i\alpha}\psi^\dagger\psi_c $$ breaks the phase symmetry under $\psi\rightarrow e^{i\alpha}\psi$, where the charge conjugation $\psi_c$ transforms as $\psi_c\rightarrow e^{-i\alpha}\psi_c$.

Even though Majorana mass has not been confirmed yet via "neutrinoless double beta decay", the very contemplation of this possibility means that there is nothing sacrosanct about the invariance of lagrangian under a phase transformation $\psi\rightarrow e^{i\alpha}\psi$.

There are plenty other counter examples in condensed matter physics. For example, the BCS theory of superconductivity involves terms like $$ \psi_{\uparrow}^\dagger\psi_{\downarrow}^\dagger $$ which breaks the phase symmetry as well.

MadMax
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