I want to obtain some clarification on the concept of Noether charge. Given conserved current $J^\mu$ e.g. in free scalar field theory in $(n+1)$ dimensional Minkowski spacetime $M$, i.e. $\partial_\mu J^\mu=0$, we can define Noether charge \begin{equation} Q := \int_{M} d^{n}x \, J^0\,. \tag{1} \end{equation} The standard argument for conserved charge is simply the total derivative \begin{equation} \frac{dQ}{dt} = \int_M d^nx \partial_0 J^0 = \int_M d^nx \left(-\partial_i J^i\right) = \int_{\partial M} dS_\mu J^\mu = 0\,.\tag{2} \end{equation} Here we have used the "differentiation under integral sign", set $c=1$ so that $x^0 \equiv t$ and using Gauss' theorem to convert volume integral to surface integral with directed area $dS_\mu$, and finally the integral over boundary $\partial M$ vanishes as required by field theory where $J^\mu $ vanishes at the boundary.

On the other hand, from equation of motion we know that $Q$ is a constant of motion if we have \begin{equation} \frac{\partial Q}{\partial t} = \{Q,H\}\tag{3} \end{equation} where $H$ is the Hamiltonian of the free theory (in quantum theory the Poisson bracket becomes commutator up to some complex number $i\hbar$). What this tells us is that conserved charge can have explicit time dependence, since the Poisson bracket needs not vanish.

While this seems to boil down to distinction between partial and total derivative, I am not so sure about the physics anymore:

  1. does this mean that when someone says Noether charge is time-independent, it is misleading because time dependence can be explicit but the "advective" part can counterbalance the explicit time dependence?

  2. Is there a cleaner, better way to think of "conservation" with respect to $t$ in this context?

  3. Is "constant of motion" strictly referring to quantities whose Poisson bracket with the Hamiltonian vanishes, or the ones whose total derivative with respect to time vanishes? Note that the latter implies the former but not the other way round.


2 Answers 2


Hamilton's equation for a quantity $Q$ reads

$$ \frac{dQ}{dt}~=~\{Q,H\}+\frac{\partial Q}{\partial t} .$$

Therefore a constant of motion $Q$ satisfies

$$ \frac{dQ}{dt}~=~0 \qquad \Leftrightarrow\qquad \frac{\partial Q}{\partial t}~=~\{H,Q\}.$$ In other words, there is no contradiction. There just seems to be a sign mistake (or a non-standard sign convention for the Poisson bracket) in OP's last equation (3).


I'm not entirely sure of your question, so I will attempt to clear up any confusion with a definition. By definition, we say that a current is conserved if and only if its 4-divergence is zero. So, what that means is

$$ \partial _\mu j^\mu =0$$

If you write out the derivatives this means that

$$ \frac{\partial j^0}{\partial t} - \nabla \cdot \mathbf{j} = 0$$

This is just the usual continuity equation from electrodynamics. That is, since $j^0 \equiv \rho(x,t)$ we have the usual conservation of charge statement that

$$\frac{\partial \rho}{\partial t} = \nabla \cdot \mathbf{j} $$

This statement is completely general, and you can see that $\frac{\partial j^0}{\partial t} =0$ if and only if you have a "divergence free" current density.


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