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My background about this argument:

a) Let's consider a classical field theory, where $\mathcal{L}(\phi(x),\partial_{\mu}\phi_(x))$ is the lagrangian density. A symmetry is a transformation $\phi(x) \rightarrow\phi'(x')=f(\phi(x))$ of the fields that leaves the lagrangian density $\mathcal{L}$ invariant (more general definitions are possible, but I hope this one is sufficient). Noether's theorem establishes that continous symmetry groups correspond to conserved currents and charges.

b) In the corresponding quantum field theory, fields are promoted to fields operators acting on the Hilbert space of states. In this case, symmetries are represented by probability conserving operators. According to Wigner's theorem, these are unitary operators (or antiunitary) on the Hilbert space of states. Conserved quantities should instead arise as operators that commute with the Hamiltonian. The latter is "imported" in the quantum field theory just by considering directly the classical expression, where field operators appear instead of classical fields.

Example:

Dirac free lagrangian density is: $$\mathcal{L}=\overline{\psi}(i\gamma^\mu\partial_\mu-m)\psi$$

Parity is a (discrete) symmetry of this lagrangian at a classical level, since ($'$ denotes parity-transformed quantities): $$\mathcal{L'}=\overline{\psi'}(i\gamma^\mu\partial_\mu'-m)\psi'=\overline{\gamma^0\psi}(i\gamma^\mu\partial_\mu'-m)\gamma^0\psi=\psi^+(i\gamma^\mu\partial_\mu'-m)\gamma^0\psi=\psi^+\gamma^0(i\gamma_\mu\partial^\mu-m)\psi=\mathcal{L}.$$

Questions:

a) Does the classical symmetry imply the existence of an operator that in some sense "represents" parity symmetry at a quantum level?

b) If yes, what is the connection between the operator and the classical parity transformation?

c) Does a conserved quantity at a quantum level (operator that commute with hamiltonian) arise in this case as a consequence of the existence of the parity symmetry?

d) Is it possible to generalize the considerations about the special simple case of parity?

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    $\begingroup$ This post (v4) seems to be a quite broad question. $\endgroup$
    – Qmechanic
    Apr 5, 2022 at 16:18

1 Answer 1

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Often, a classical symmetry will become a symmetry of the quantum theory, but not always. A symmetry of the classical theory which fails to be a symmetry of the quantum theory is called an anomaly.

One way to express the idea of a conserved current for a symmetry of a quantum level is the Ward identity (see, eg, Eq 6.45 of the chapter "Symmetries in QFT" in David Skinner's Quantum Field Theory II course) \begin{equation} \partial_\mu \big\langle j^\mu(x) \prod_{i=1}^N O(x_i)\big\rangle = -\sum_{n=1}^N \delta(x-x_i)\big\langle \delta O(x_i) \prod_{j\neq i} O(x_j)\big\rangle \end{equation} where $j^\mu(x)$ is the Noether current for the symmetry (which is an operator in QFT), $O(x_i)$ is a local operator defined at the point $x_i$, and $\delta O(x_i)$ is the transformation of $O(x_i)$ under the symmetry. The terms on the right hand side are called contact terms, and are quantum corrections to the classical Noether's theorem.

When the symmetry is anomalous, there is an additional term in the Ward identities due to the anomaly \begin{equation} \partial_\mu \big\langle j^\mu(x) \prod_{i=1}^N O(x_i)\big\rangle = -\sum_{n=1}^N \delta(x-x_i)\big\langle \delta O(x_i) \prod_{j\neq i} O(x_j)\big\rangle + \big\langle F(x) \prod_{i=1}^N O(x_i) \big\rangle \end{equation} where the last term represents the failure of the current to be conserved due to the anomaly.

Note that when we take the $O(x_i)$ to simply be $1$, we get the special case \begin{equation} \partial_\mu \big\langle j^\mu(x) \big\rangle = F(x) \end{equation} where $F=0$ if the symmetry is not anomalous.

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  • $\begingroup$ Thanks for your answer. My question anyway focuses on the case in which the symmetry is in fact not anomalous and since I don't know the path integral formalism, I was not able to understand the reference. I will try to modify the question and include an example to clarify the point. $\endgroup$ Apr 5, 2022 at 13:54
  • $\begingroup$ @Antonio19932806 You don't need path integrals to at least understand the statement: if the symmetry is not anomalous, then the classical symmetry implies an analog of Noether's theorem in quantum field theory called the Ward Identities. The Ward Identities say that the divergence of the expectation value of the Noether current for that symmetry (which is an operator in QFT) times a product of local operators is zero, up to contact terms, which represent quantum corrections to Noether's theorem. $\endgroup$
    – Andrew
    Apr 5, 2022 at 13:57
  • $\begingroup$ In fact I understand this statement, but unfortunately this only partially solves the main point of my question. I apologize with you for phrasing the question in such a way was not so clear. $\endgroup$ Apr 5, 2022 at 15:08
  • $\begingroup$ @Antonio19932806 Perhaps what you are asking is more along the lines of this statement: "In classical mechanics, the Noether charge $Q$ for a symmetry generates infinitesimal transformations via the Poisson bracket, $\{X, Q\} = \delta X$ (where $X$ is some observable and $\delta X$ is the infinitesimal transformation of that observable under the symmetry). In quantum mechanics / quantum field theory, this becomes the operator statement $[X, Q] = \delta Q$, where $[A, B]$ is the commutator of $A$ and $B$." $\endgroup$
    – Andrew
    Apr 5, 2022 at 15:33
  • $\begingroup$ You can use the commutator (which gives the infintesimal transformation) to build up to finite transformations, using the Baker-Campbell-Hausdorff formula. $\endgroup$
    – Andrew
    Apr 5, 2022 at 15:48

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