# Connection between conserved charge and the generator of a symmetry

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

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

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

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

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

• 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. – Ziruo Zhang Feb 19 '18 at 15:14
• see also Conserved charges and generators. – AccidentalFourierTransform May 11 '18 at 19:25

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 $$[X,\phi] = -i\frac{\delta \phi}{\delta t}$$

• Thank you! On little/big thing I don't understand: Why does the group element act on an operator $\phi(y)$, by conjugation? – jak Sep 27 '14 at 12:51
• 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 – jak Sep 27 '14 at 13:02
• 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. – jak Sep 27 '14 at 13:34
• @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 :) – Nephente Sep 27 '14 at 20:56
• @JakobH Do you need further clarification? – Nephente Oct 1 '14 at 14:15

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 $$T_g^{-1}\phi\ T_g$$.)