I'm sorry for the triviality of my questions.

Why is $\bar{\psi} = e^{i \theta}\bar{\psi}$, where $\theta$ is a real number, used as the global gauge transformation? Why $e^{i \theta}$; what's the physical significance or benefit?

Why is $\bar{\psi} = e^{i \theta(x)} \psi$ the local gauge transformation? What does $\theta$ being a function of $x$ instead of a real number change such that it no longer applies globally?

The Dirac Lagrangian is globally but not locally invariant. How can this be? Is local not a subset of global as far as invariance is concerned?

  • 4
    $\begingroup$ Global in this case means that at any space(time) point the transformation acts in the same way, i.e. it is independent w.r.t. the space coordinate. A local transformation, instead, may change from one space-time point to another (and thus it is more difficult to be satisfied). $\endgroup$
    – yuggib
    Commented May 2, 2015 at 7:31
  • 1
    $\begingroup$ Be aware that the idea of a "gauge transformation" is far bigger than just the specific case of the Dirac Lagrangian and the specific case of multiplying by a phase. $\endgroup$
    – ACuriousMind
    Commented May 2, 2015 at 11:30

2 Answers 2


Multiplying by $e^{i\theta}$ is a rotation of $\theta$ in the complex plane. Physically it changes the phase of a plane wave by an angle $\theta$. This is a global symmetry because we arbitrarily choose a reference point for measuring the phase of plane waves. If we change the phase of all plane waves by an equal amount then this is equivalent to just moving our reference point. The usual analogy is to add some distance $d$ to all mountain heights on Earth. This just moves our sea level reference by a distance $d$ and doesn't actually change the mountains.

A local gauge transformation is not a subset of a global gauge transformation. In this respect the name is a bit misleading. Sticking with our analogy of mountain heights, a local gauge transformation would be to add a different distance $d(x)$ to each mountain height, where $x$ is some function of the position of the mountain. Obviously mountain heights on Earth are not invariant under this local transformation, because some heights would change more than others.

In the context of quantum mechanics, observables are calculated using some equation. For our system to be invariant under a local gauge transformation means the equations that describe observables must give the same results when we apply some local gauge transformation. This severely restricts the form those equations can have.


In fact, global gauge transformations are a subset of local gauge transformation: changing the same amount everywhere is a special case (ie, more restricting) of changing the phase of each point independently.

In the Dirac Lagrangian $$\mathcal{L} = \bar{\psi}(i\gamma^\mu\partial_\mu - m)\psi$$ you have to derive $\psi$. If you make a global transformation $\psi \to e^{-i\theta}\psi$ $\bar{\psi} \to \bar{\psi} e^{i\theta}$, $$\mathcal{L} \to \bar{\psi} e^{i\theta}(i\gamma^\mu\partial_\mu - m)e^{-i\theta}\psi$$ then $e^{-i\theta}$ is a constant, gets out the derivative and cancels with the term from $\bar{\psi}$, so the lagrangian is invariant. But if you make a local transformation $\psi \to e^{-i\theta(x)}\psi$ $\bar{\psi} \to \bar{\psi} e^{i\theta(x)}$, you have to derive both the spinor and the phase change using the Leibniz rule $$\mathcal{L} \to \bar{\psi} e^{i\theta(x)}(i\gamma^\mu\partial_\mu - m)e^{-i\theta(x)}\psi = \bar{\psi} (i\gamma^\mu\partial_\mu - m)\psi + \bar{\psi}\gamma^\mu (\partial_\mu\theta) \psi$$ If $\partial_\mu \theta \neq 0$ (that is, the transformation depends on the coordinates and it isn't global), the extra term spoils gauge invariance (unless you add a gauge field).


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