I'm trying to understand the connection between Noether charges and symmetry generators a little better. In Schwartz QFT book, chapter 28.2, he states that the Noether charge $Q$ generates the symmetry, i.e. is identical with the generator of the corresponding symmetry group. His derivation of this goes as follows: Consider the Noether charge

\begin{equation} Q= \int d^3x J_0(x) = \int d^3 x \sum_m \frac{\delta L}{\delta \dot \phi_m} \frac{\delta \phi_m}{\delta \alpha} \end{equation}

which is in QFT an operator and using the canonical commutation relation $$[ \phi_m(x) ,\pi_n(y)]=i \delta(x-y)\delta_{mn},$$ with $\pi_m=\frac{\delta L}{\delta \dot \phi_m}$ we can derive

\begin{equation} [Q, \phi_n(y)] = - i \frac{\delta\phi_n(y)}{\delta \alpha}. \end{equation}

From this he concludes that we can now see that "$Q$ generates the symmetry transformation".

Can anyone help me understand this point, or knows any other explanation for why we are able to write for a symmetry transformation $e^{iQ}$, with $Q$ the Noether charge (Which is of course equivalent to the statement, that Q is the generator of the symmetry group)?

To elaborate a bit on what I'm trying to understand: Given a symmetry of the Lagrangian, say translation invariance, which is generated, in the infinite dimensional representation (field representation) by differential operators $\partial_\mu$. Using Noethers theorem we can derive a conserved current and a quantity conserved in time, the Noether charge. This quantity is given in terms of fields/ the field. Why are we allowed to identitfy the generator of the symmetry with this Noether charge?

Any ideas would be much appreciated

  • $\begingroup$ Hi Jakob, I found an additional term $\int d^{3} \mathbf{x} \sum_{m} \pi_m(\mathbf{x})[\frac{\delta\phi_m(\mathbf{x})}{\delta\alpha},\phi_n(\mathbf{y})]$ while evaluating $[Q, \phi_n(\mathbf{y})]$. How would you argue that this is zero? Thank you very much for your help. $\endgroup$ Commented Feb 19, 2018 at 15:14
  • $\begingroup$ see also Conserved charges and generators. $\endgroup$ Commented May 11, 2018 at 19:25

2 Answers 2


Consider an element $g$ of the symmetry group. Say $g$ is represented by a unitary operator on the Hilbertspace $$ T_g = \exp(tX) $$ with generator $X$ and some parameter $t$. It acts on an operator $\phi(y)$ by conjugation $$ (g\cdot\phi)(y) = T_g^{-1}\phi(y) T_g = e^{-tX}\phi(y) e^{tX} = \big[ 1 + t[X,\cdot]+\mathcal{O}(t^2)\big]\phi(y)$$ On the other hand the variation of $\phi$ is defined as the first order contribution under the group action, e.g $$ g\cdot\phi = \phi + \frac{\delta \phi}{\delta t}t+\mathcal{O}(t^2) $$ Since in physics we like generators to be hermitian, rather than anti-hermitian one sends $X\mapsto iX$ and establishes $$ [X,\phi] = -i\frac{\delta \phi}{\delta t} $$

Also, this answer and links therein ought to help you further.

  • $\begingroup$ Thank you! On little/big thing I don't understand: Why does the group element act on an operator $\phi(y)$, by conjugation? $\endgroup$
    – jak
    Commented Sep 27, 2014 at 12:51
  • $\begingroup$ My first guess would be that this is because we look at $\phi$ in the adjoint representation. Ths would mean that $\phi$ lives in the tangent space above the identity, i.e. $T_e$, which is the space the generators live in (=the Lie algebra). The natural product of this space is the commutator. If we consider the adjoint representation of the group, we map each element to a linear operator on $T_e$. The action of each group element is then given by the commutator. $g \circ \phi = [g,\phi] \approx i [X, \phi]$, which is close to, but unfortunately not exactly what you wrote $\endgroup$
    – jak
    Commented Sep 27, 2014 at 13:02
  • $\begingroup$ I confused a few points... There are of course two maps, one for the group and one for the Lie algebra, both onto the space of Linear Operators on $T_e$: $\mathrm{Ad}_g(X) = gX g^{-1}, \qquad \mathrm{ad}_X(Y) = [X,Y] $. Therefore, my question is solely, why the field $\phi$ lives in the Lie algebra, i.e. $\phi \in T_e$. Then its clear, because the only possible homomorphism for the group onto its own Lie algebra is given by $Ad_g$ as defined above, and the action of $g$ on $\phi$ is given by conjugation. $\endgroup$
    – jak
    Commented Sep 27, 2014 at 13:34
  • 1
    $\begingroup$ @JakobH Sorry for getting back to you just now. The field $\phi$ is a linear operator on Fockspace. As usual the group reps. act on states as $T_g\vert \varphi \rangle$ and on operators as a similarity trafo by conjugation $T_g^{-1} \phi T_g$. I don't think there's more to it. Although, gauge fields transform under the adjoint representation. My knowledge in this matter is unfortunatly sketchy, though i'm planning to catch up :) $\endgroup$
    – Nephente
    Commented Sep 27, 2014 at 20:56
  • 1
    $\begingroup$ Thanks a lot for your help! I was wondering if it is possible to derive from \begin{equation} Q= \int d^3x J_0(x) = \int d^3 x \sum_m \frac{\delta L}{\delta \dot \phi_m} \frac{\delta \phi_m}{\delta \alpha} \end{equation} using $[X,\phi] = -i\frac{\delta \phi}{\delta t}$ \begin{equation} \rightarrow Q= i \int d^3 x \sum_m \frac{\delta L}{\delta \dot \phi_m} [X,\phi_m] \end{equation} that $Q=X$ without using the canonical commutator. Anyway, your answer and comment answer my original question perfectly $\endgroup$
    – jak
    Commented Oct 5, 2014 at 9:54

I would like to make an addition to Nephente's answer, because you asked this in your comment, and I also think this is also somewhat part of the full picture here.

Why does the group element act on an operator $\phi$, by conjugation?

This is by no means a mathematically strict answer, but still can be made one.
Consider our $\phi$ acts on a state $|\varphi\rangle$. $$ |\psi\rangle = \phi|\varphi\rangle. $$ Let's state, that our symmetry operator is represented by the following, which is the same operation on every state.
$$ |\varphi'\rangle = T_g^{-1}|\varphi\rangle,\\ |\psi'\rangle = T_g^{-1}|\psi\rangle. $$ From this, we can deduce (by inserting $\mathbb{1} = T_g\ T_g^{-1}$), that $$ |\psi'\rangle = \underbrace{T_g^{-1} \phi\ T_g\ }_{A}\underbrace{T_g^{-1} |\varphi\rangle}_{|\varphi'\rangle}. $$ We can see from this, that $A$ is how we expect the transformed $\phi$ to behave for $|\varphi'\rangle$, $|\psi'\rangle$. Because this is true for $\forall|\varphi\rangle$, $\forall g$ for a given $\phi$, we can conclude, that $$ \phi' = T_g^{-1}\phi\ T_g. $$

(Note, that I think when Nephante wrote, that $T_g$ is how the symmetry operator is represented on the Hilbert space, he really meant it is $T_g^{-1}$, because he then later states, that operators transform by $T_g^{-1}\phi\ T_g$.)


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