I've been reading Lancaster & Blundell, and in Chapter 14 they focus on the Lagrangian $$ \mathcal{L}=(\partial^\mu\psi)^\dagger(\partial_\mu\psi) - m^2\psi^\dagger\psi. $$ To impose invariance to the transformation $\psi\rightarrow\psi\exp(i\alpha(x))$, where $\alpha(x)$ is a coordinate-dependent phase, they replace the derivatives in $\mathcal{L}$ with covariant derivatives $$ D_\mu = \partial_\mu + iqA_\mu. $$ Invariance then follows if we also admit the transformation $$ A_\mu\rightarrow A_\mu-\frac{1}{q}\partial_\mu\alpha(x). $$

Now, my question is a simple one: why are we 'allowed' to change the Lagrangian seemingly arbitrarily? I see how this change leads to the invariance of $\mathcal{L}$ with respect to the transformation $\psi\rightarrow\psi\exp(i\alpha(x))$, but surely in doing so we change the dynamics of the field $\psi$? Expansion of the `new' Lagrangian would seem to suggest that the EL equations do indeed result in different dynamics.

Many thanks for your help.

  • 11
    $\begingroup$ I always didn't like how textbooks explained this... it's not that you're "allowed" to do this or that. Your first Lagrangian represents one theory, adding a gauge field represents a completely different theory. The motivation is that the change is relatively simple, not that the change does nothing. $\endgroup$
    – knzhou
    Commented Oct 19, 2020 at 7:32

2 Answers 2


This is indeed true and is what is called the gauge principle. It tells us that if we make a global symmetry local, we need to add a corresponding gauge field such that the total Lagrangian still remains invariant under this local gauge transformation. This is a new dynamical field which has its own equations of motion and can couple to the fermion leading to interactions.

In this case the original Lagrangian is invariant under $U(1)$ as $\psi \to \psi e^{i \alpha}$, note that also $\partial_\mu \psi \to \partial_\mu \psi e^{i \alpha}$. We say that these fields transform in the fundamental representation of $U(1)$.

Now after making our transformation local: $\alpha \equiv \alpha(x)$ it's easy to see that $\partial_\mu \psi \not\to \partial_\mu \psi e^{i \alpha(x)}$
To account for this as we still want our field to transform in the fundamental representation we have to introduce a gauge field $A_\mu(x)$ and a covariant derivative $\mathcal{D}_\mu$ such that $\mathcal{D}_\mu \psi \to \mathcal{D}_\mu \psi e^{i\alpha(x)}$. This last transformation dictates how $A_\mu(x)$ should transform.

  • 1
    $\begingroup$ I'm not familiar with the physics convention, when you say $\mathcal D_\mu \psi e^{i\alpha(x)}$, do you mean $(\mathcal D_\mu \psi) e^{i\alpha(x)}$ or $\mathcal D_\mu (\psi e^{i\alpha(x)})$? $\endgroup$
    – Keshav
    Commented Nov 3, 2020 at 20:06
  • $\begingroup$ The second one, $\mathcal{D}_\mu$ acting on everything on the right if there are no brackets. $\endgroup$
    – Guliano
    Commented Nov 3, 2020 at 20:22

As was mentioned in some of the comments, the Lagrangians $$ \mathcal{L}=(\partial^\mu\psi)^\dagger(\partial_\mu\psi)-m^2\psi^\dagger\psi $$ and $$ \mathcal{L}=(D^\mu\psi)^\dagger(D_\mu\psi)-m^2\psi^\dagger\psi+\frac{1}{4}F_{\mu\nu}F^{\mu\nu} $$ represent distinct theories each with their own properties.

The usual way to motivate the transition from the "ungauged" theory and the "gauged" one is to note that if we want invariance under the transformation $\psi\rightarrow e^{i\alpha}\psi$ for $\alpha=\alpha(x)$ an arbitrary real function, then taking a Lagrangian that's already invariant in the special case where $\alpha$ is a constant and replacing all the derivatives of $\psi$ by covariant derivatives $D_\mu$, would be good enough to construct a Lagrangian which is also invariant under the local transformations.

There is another way of looking at things, however, which may feel a little less ad hoc. Though this viewpoint can be described in terms of this example of $\psi$ fields, it's slightly more natural to start with the example of a vector field.

So, suppose that $V^a$ are the components of some vector field - note that these are only the components. The vector field itself, meaning the abstract object which is invariant under coordinate changes is $V=V^a\boldsymbol{e}_a$ where the $\boldsymbol{e}_a$ form a basis of vectors at each point in space (technically called frame fields). For example, in two dimensions, we could take $\boldsymbol{e}_0=\boldsymbol{\hat r}$ and $\boldsymbol{e}_1=\boldsymbol{\hat \theta}$.

Now the key assumption is that the physics of our system should not depend on the basis vectors we choose to represent our vector fields in - that is, if we changed to Cartesian unit vectors instead of polar unit vectors, the components $V^a$ would certainly need to change, but the object $V=V^a\boldsymbol{e}_a$ should not.

Since any change in the basis vectors $\boldsymbol{e}_a$ will be a (linear) map from a linear space to itself, these can be represented by matrices $U^a_b$ so under a change in basis we would have $\boldsymbol{e}^\prime_a=U^b_a\boldsymbol{e}_b$. If we are truly to be independent of the basis vectors, we will be able to perform such a transformation point by point, these basis change matrices may have arbitrary dependence on the spacetime point, $U^a_b=U^a_b(x)$. In order for $V$ to be independent of these changes, the components must transform by the inverse of $U$, $V^{\prime\,a}=U^{-1\, a}_b V^b$.

Finally now, we want to build our Lagrangian out of $V$ and its derivatives. So long as our manifold has a metric, we can build arbitrarily high derivatives out of the differential $d$ and the Hodge dual $*$. If we compute the differential of $V$ in terms of the components, we would find $$ dV=(dV^a)\boldsymbol{e}_b+V^a(d\boldsymbol{e}_b). $$ The differential of the components is simple because these are all $0$-forms (scalars), and so $d V^a=\partial_\nu V^adx^\nu$. For the differential of the basis vectors, we can first note that the result must

a) be a 1-form

b) be some combination of unit vectors again.

These two statements together imply the differential must take the generic form $$ d\boldsymbol{e}_a=(A_\mu)_a^b\boldsymbol{e}_bdx^\mu $$ where $A_{\mu\,b}^a$ is some unknown function, suggestively named. Putting this result back into the calculation of $dV$, we find $$ dV=\partial_\mu V^a\boldsymbol{e}_adx^\mu+V^aA_{\mu\,a}^b\boldsymbol{e}_bdx^\mu. $$ Collecting the differentials, unit vectors, and components together, this becomes $$ dV=\boldsymbol{e}_adx^\mu(\delta^a_b\partial_\mu+A_{\mu\,b}^a)V^b=\boldsymbol{e}_adx^\mu(D_\mu)^a_bV^b. $$ In the last line we have identified the covariant derivative $D$. This differs slightly from the covariant derivative in the question by overall scalings of $A$ (the $iq$) which could have been absorbed into our definition of $A$.

This expression also differs slightly from what's in the question by the additional indices $a$ and $b$ floating around. In the case of the complex scalar field, we are not dealing with a vector, but instead some object $\tilde \psi=\psi z$ where now $z$ is some complex number with $|z|=1$. This now plays the role our $\boldsymbol{e}$'s played before (but has no indices).

Since $z$ must have modulus 1, we can only transform to a new $z$ by $z^\prime=e^{iq\alpha}z$ where $\alpha=\alpha(x)$ in the same way the change of basis matrix $U$ was allowed to vary point to point (and $q$ has been put in for convenience). Since there are no indices on this $z$, our calculation of the differential would yield $$ d\tilde \psi=dx^\mu zD_\mu\psi=dx^\mu z(\partial_\mu+iqA_\mu)\psi. $$

As a fun side note, observe that if in the example of a vector we renamed $A$ to $\Gamma$ and called the gauge potential a Christoffel symbol instead, we would immediately reproduce the covariant derivative from general relativity.

  • $\begingroup$ You have given an idea of how we can geometrically approach gauge theory. However, this is not an answer to the question. $\endgroup$
    – NDewolf
    Commented Nov 3, 2020 at 20:30
  • $\begingroup$ @NDewolf The question asked had two parts, first whether this Lagrangian is the same, and second why we should be 'allowed' to make this change. I started by answering that the Lagrangian are indeed different which answers the first question. The second part of the question is why such a change is reasonable. My answer is constructed to address this from a geometrical perspective. $\endgroup$ Commented Nov 3, 2020 at 23:00
  • $\begingroup$ The question was specifically related to how this is "allowed" with respect to the dynamics. Your answer is just a geometric way of phrasing invariance. $\endgroup$
    – NDewolf
    Commented Nov 4, 2020 at 7:57

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