In classical field theory, due to Noether's theorem, corresponding to every continuous symmetry there is a conserved current/charge. However, to arrive at this conclusion one has to assume that the Euler-Lagrange EoM holds, $$\partial_\mu\frac{\partial\mathcal{L}}{\partial \left(\partial_\mu\phi\right)}-\frac{\partial\mathcal{L}}{\partial\phi}=0.$$ Now, when we promote the fields to the status of operators to build quantum fields, we impose the following commutation relation on the fields, $$\left[\hat{\phi}(\mathbf{x},t),\hat{\pi}(\mathbf{y},t)\right]=i\delta^{(3)}(\mathbf{x}-\mathbf{y}).$$ This means that that the operator $\hat{\phi}(x)$ do not satisfy any Euler-Lagrange type EoMs. Then what happens to the conserved quantities/currents/charges in a quantum theory? How can the Noether theorem apply to quantum fields, if there's no EoM for the field operators?

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    $\begingroup$ Possible duplicate: Is there a formulation of Noether’s theorem for the path integral formalism? $\endgroup$
    – Qmechanic
    Commented Nov 7, 2020 at 4:39
  • $\begingroup$ @Qmechanic The question sounds similar but I want to know what happens in the canonical quantization formulation. I am not much familiar with path integrals and I have no idea what Ward-Takahashi identities are! $\endgroup$ Commented Nov 7, 2020 at 5:30
  • $\begingroup$ @annav With this search keyword I mostly find QFT notes with mostly the same approach where Noerther's theorem is discussed in Classical Field Theory and never it is mentioned clearly when the quantum fields are discussed. $\endgroup$ Commented Nov 7, 2020 at 5:59
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    $\begingroup$ @FaberBosch To clarify: when you say that the field operators in QFT don't have EoMs, do you mean that we can't write them in Euler-Lagrange form? Whether that's true or not, the field operators in QFT do obey EoMs (in the Heisenberg picture, which is the picture we always use in classical field theory), even though they don't commute. We use the EoMs to determine whether a given quantity is conserved. We don't need to use a lagrangian for this, so whether the lagrangian formalism works for non-commuting operators may or may not be relevant, depending on exactly what you're asking. $\endgroup$ Commented Nov 7, 2020 at 16:37
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    $\begingroup$ @FaberBosch the operators $\phi(x)$ and $\pi(x)$ are already predetermined. You're probably confusing them with the expectation values and/or eigenvalues. $\endgroup$ Commented Nov 7, 2020 at 17:14

2 Answers 2


Let me make some comments on this topic as it's an important point that often gets glossed over till late into most standard treatments of QFT. There are a couple major places where symmetries come in, and I will try and describe them here.

First, let's recall that in a classical theory, Noether's theorem has two important implications. The first is the existence of conserved charges which can be useful in solving the equations of motion and so on. But her theorem also tells us that these charges generate the transformation they're associated with via the Poisson bracket. That is, suppose we have a continuous transformation $T_\alpha$ parametrized by $\alpha$ which acts on our fields $\phi$ by $$ \phi^\prime = T_\alpha[\phi]. $$ If this transformation is a symmetry, then Noether's theorem tells us there will be an associated charge $Q$ which is consrved: $$ \frac{d Q}{d t}=\{Q,H\}=0 $$ and which generates the symmetry. That is, for any function $F$ on phase space, $$ \frac{d F(T_\alpha[\phi])}{d\alpha}\biggr|_{\alpha=0}=\{F,Q\}. $$ Meaning if we take the derivative of $F$ along the flow generated by the transformation $T$, this is equivalent to computing the Poisson bracket of $F$ and $Q$.

The reason I belabor this point is because it carries over to the quantized version of the theory: the charges $Q$ which are conserved $[H,Q]=0$ and are associated to a symmetry generate that symmetry via the commutator, up to factors of $i$ and $\hbar$.

For example, we know the angular momentum operator is the generator of rotations. The linear momentum operator of the generator of translations, and so on.

So Noether's theorem provides a link between transformations on our Hilbert space and symmetries.

Next, we know that for any collection of pair-wise commuting operators, we can simultaneously diagonalize them all. This is very useful for organizing our Hilbert space. For example, when considering the hydrogen atom, we use the fact that the Hamiltonian commutes with $L_z$ and $L^2$ to write a basis of states in the form $|E_n,\ell,m\rangle$. If you've only ever seen Griffith's presentation of the hydrogen atom problem, I highly recommend looking in Sakurai's book. There operator algebras (which are the clear way to understand the role of symmetry) are stressed as opposed to tedious PDE problems.

Finally, there are Ward identities, which unfortunately are typically not discussed until late into a standard treatment of QFT. Essentially these are relations between inner products controlled by symmetry. That is, suppose we have some collection of operators $\mathcal{O}_1,\ldots,\mathcal{O}_n$ and wish to compute $$ \langle 0|\mathcal{O}_1\ldots\mathcal{O}_n|0\rangle. $$ Symmetries will tell us something about how this expectation value relates to other expectations. The easiest way to derive the identities comes from what's known as a field redefinition in a path integral approach, but these can also be obtained from an operator point of view.

Schematically, these identities state that if $\frac{d}{d\alpha}$ is the derivative with respect to a continuous symmetry (in the same sense as in the classical case above), then we must have the identity $$ \frac{d}{d\alpha}\langle 0|\mathcal{O}_1\ldots\mathcal{O}_n|0\rangle=-i\sum_{k=1}^n\langle0|\mathcal{O}_1\ldots T[\mathcal{O}_k]\ldots\mathcal{O}_n|0\rangle. $$ Again, this is only schematic to give some idea of how it goes.

The power of the Ward identities is that the hold non-perturbatively and as a result are one of the few statements we can really confidently make about a theory without relying upon some perturbative expansion. Whether classical or quantum, that is always the power of Noether's theorem: it tells us things we otherwise would never be able to compute.

In any case, there is some very nice information about these things out there, unfortunately I have tended to find it very scattered about.


The simplest possible Lorentz-invariant equation of motion for a field is $\Box \phi = 0$. The classical solutions are plane waves. For example a solution is $\phi(x) = a_p(t) e^{i \vec p \cdot \vec x}$, which is the equation of motion of a harmonic oscillator.

A general solution is
$\phi(x, t) = \int \frac{d^3 p}{(2 \pi)^3} (a_p e^{-i p x} + a_p^\dagger e^{i p x})$
where $a_p$ and $a_p^\dagger$ are respectively the annihilation and creation operators, and $[a_k, a_p^\dagger] = (2 \pi)^3 \delta^3 (\vec p - \vec k)$ are the equal time commutation relations.

The operator canonically conjugate to $\phi(x)$ at $t = 0$ is $\pi(\vec x) = \partial_t \phi(x) |_{t = 0}$. If you compute the commutator, you get
$[\phi(\vec x), \pi(\vec y)] = i \delta^3 (\vec x - \vec y)$

All what above is consistent with the Euler-Lagrange equations of motion, hence the Noether theorem holds for the quantized fields as well.


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