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For the Klein Gordon field, the conserved charge for translation in space is given by: $$\vec{P}=\frac{1}{2}\int d^{3}k \, \vec{k}\{a^{\dagger}_{k}a_{k}+a_{k}a^{\dagger}_{k}\}$$

If we were to find the generators for a space translation, we would find that, $$P_{j}=i\partial_{j},$$ where $j=1,2,3$.

If we act both of the above operators on the field $\phi$, the result matches! My question is whether both of these, the generators and the conserved charges of a symmetry are always the same thing? What would be a simple way to see this connection?

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    $\begingroup$ Related: physics.stackexchange.com/q/24596/2451 & physics.stackexchange.com/q/69271/2451 $\endgroup$ – Qmechanic Sep 20 '17 at 15:36
  • $\begingroup$ Just working in plain QM, suppose we have a continuous symmetry $U$ parametrized by $\theta$, so $[U(\theta), H] = 0$. Now, $U(\theta) = e^{i \theta A}$ for Hermitian $A$ (so $A$ is the generator) and taking $\theta \to 0$ gives $[A, H] = 0$ (so $A$ is the conserved quantity). That's all! $\endgroup$ – knzhou Sep 25 '17 at 10:36
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    $\begingroup$ Just to be sure: are you asking about classical mechanics, quantum mechanics, or both? $\endgroup$ – AccidentalFourierTransform Sep 26 '17 at 10:47
  • $\begingroup$ Sorry, I have been travelling for the past few days. I intended to ask about QFT, but feel free to bring out the differences between that and classical mechanics in the present context. $\endgroup$ – Arnab Barman Ray Sep 29 '17 at 11:55
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    $\begingroup$ @ArnabBarmanRay I tried to address both cases in my answer below; in fact, I tried to state the result in as much generality as possible. I hope it is clear enough. Cheers! $\endgroup$ – AccidentalFourierTransform Sep 29 '17 at 20:10
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OP is wondering whether the conserved charge associated to a continuous symmetry always generates the symmetry itself. We can say, in full generality, that the answer is

Yes.

Let us see how this works.

Classical mechanics.

We use a notation adapted to classical field theory rather than point-particle mechanics, but the former includes the latter as a special sub-case so we are losing no generality.

Consider a classical system which may or may not include gauge fields and/or Grassmann odd variables. For simplicity, we consider a flat space-time. Assume the system is invariant under the infinitesimal transformation $\phi\to\phi+\delta\phi$. According to Noether's theorem, there is a current $j^\mu$ $$ j^\mu\sim \frac{\partial\mathcal L}{\partial\dot\phi_{,\mu}}\delta\phi $$ which is conserved on-shell, $$ \partial_\mu j^\mu\overset{\mathrm{OS}}=0 $$

This in turns implies that the associated Noether charge $Q$ $$ Q\overset{\mathrm{def}}=\int_{\mathbb R^{d-1}} j^{0}\,\mathrm d\boldsymbol x $$ is conserved, $$ \dot Q\overset{\mathrm{OS}}=0 $$

In Ref.1 it is proved that the charge $Q$ generates the transformation $\delta\phi$,

$$ \delta\phi=(Q,\phi) $$

where $(\cdot,\cdot)$ is the DeWitt-Peierls bracket. This is precisely our claim. The reader will find the proof of the theorem in the quoted reference, as well as a nice discussion about the significance of the result.

Furthermore, a similar statement holds when space-time is curved, but this requires the existence of a suitable Killing field (cf. this PSE post).

Moreover, for standard canonical systems, Ref.1 also proves that $(\cdot,\cdot)$ agrees with the Poisson bracket $\{\cdot,\cdot\}$.

Quantum mechanics.

This is in fact a corollary of the previous case. Ref.1 proves that, up to the usual ordering ambiguities inherent to the quantisation procedure, the DeWitt-Peierls bracket of two fundamental fields agrees with the commutator $[\cdot,\cdot]$ of the corresponding operators.

If we assume that the classical conservation law $\partial_\mu j^\mu\equiv 0$ is not violated by the regulator (i.e., if the symmetry is not anomalous), then we automatically obtain the quantum analogue of our previous result, to wit

$$ \delta\phi=-i[Q,\phi] $$

as required.

References

  1. Bryce DeWitt, The Global Approach to Quantum Field Theory.
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Any operator that commutes with the Hamiltonian (and does not have any explicit time dependence) is conserved in time, as can be trivially seen from the operator's Heisenberg equation of motion. Of course, if an operator $A$ commutes with the Hamiltonian, then $f(A)$ does as well for any analytic function $f$, so we trivially get an infinite-dimensional space of (non-algebraically-independent) conserved quantities.

If the symmetry is continuous, so that we can index the unitary symmetry transformations $U(\theta)$ by a continuous parameter $\theta$, than the symmetry generator $T := i \frac{dU}{d\theta}|_{\theta = 0}$ is a conserved quantity, and is generally the canonical representative of the equivalence class of (algebraically dependent) conserved quantities that we identify as "the" conserved quantity corresponding to the symmetry. But we could just as well have chosen $U$ itself instead. (For many-body systems and in field theory, $T$ is typically more natural than $U$ to work with because it can be represented as a spatial sum/integral over local terms.)

If the symmetry is discrete, than we cannot define a generator, and we must work with the unitary operator $U$ itself, which is still a valid conserved quantity (e.g., the parity of the field, so that we don't have to worry about scalars time-evolving into pseudoscalars or vice versa).

So any generator of a continuous symmetry is a conserved quantity corresponding to that symmetry, but not every conserved quantity corresponding to a symmetry is the generator of a continuous symmetry.

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My question is whether both of these, the generators and the conserved charges of a symmetry are always the same thing?

Yes, they are. To see that let's consider a generic symmerty of the action, that has a conserved current $j^\mu$ due to the Noether's theorem. In that case we can define a conserved current as $$Q(t)=\int d^3x J^0 ~.$$

Since the fields (i.e. $\phi$) are operators, at the first order they transform under the symmetry as $$\phi\rightarrow\phi'=e^{iT}\phi e^{-iT}\simeq(1+iT)\phi(1-iT)=\phi+i[T,\phi]~,$$ where $T$ is the generator of the transformation in the field rappresentation. So we have $$\phi'-\phi=\delta\phi=i[T,\phi]~.$$

If we show that $-i\delta\phi=[Q,\phi]$, than the game is over. Let's do this in the two cases of spacetime translations and internal symmetries.

Spacetime translations: in the case of spacetime translations the Noether's theorem give us 4 conserved currents, as $$\partial_\mu T^{\mu}_\nu=0$$ and so, from the expression of $T^{\mu}_\nu$, we have $$Q_\nu(t)=\int d^3 xT^{0}_\nu=\int d^3 x(\pi\partial_\nu \phi-\mathcal L g^{0}_{\nu})$$ form which (using canonical commutation rules) $$[Q_\nu(t),\phi(y,t)]=\int d^3x[\pi(x,t)\partial_\nu \phi(x,t)-\mathcal L g^{0}_{\nu},\phi(y,t)]=-i\partial_\nu\phi(y,t)$$ that is equal to $-i\delta\phi$ (for a given translation of $a^\mu$) $$\phi'=\phi+a^\mu\partial_\mu\phi$$

Internal simmetries:: in the case of internal symmetry we have that $$j^\mu=\frac{\partial \mathcal L}{\partial \partial^\mu\phi}\delta\phi$$ and so the conserved current is $$Q(t)=\int d^3x j^0=\int d^3 \pi(x,t)\delta\phi(x,t)~.$$ We can see that this case is simplier since $$[Q(t),\phi(y,t)]=\int d^3x[\pi(x,t),\phi(y,t)]\delta\phi(x,t)=-i\delta\phi(y,t)~.$$

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