2
$\begingroup$

If $A$ is a classical symmetry, it is possible that after quantization $A$ is no longer a symmetry. One of the ways to see this in the operator formulation of quantum mechanics is the following.

Let $H$ and $A$ be a self-adjoint operators on the Hilbert space $\mathcal{H}$. We consider $H$ as quantum hamiltonian and $A$ as its symmetry: $A$ and $H$ commute, meaning that operator-valued spectral measures commute.

If $\psi \in \mathcal{H}$ is a state, then $$ \frac{d}{dt} \langle A(t)\rangle= \frac{d}{dt}\langle \psi(t), A\psi(t)\rangle =\langle -i H \psi(t), A\psi(t)\rangle+ \langle\psi(t), A (-iH)\psi(t)\rangle\\ =i(\langle H \psi(t), A\psi(t)\rangle - \langle\psi(t), A H\psi(t)\rangle). \quad (\ast) $$ Quantum Hamiltonian $H$ is densely defined on $\mathcal{H}$ with domain $D(H)$ and if $A$ does not preserved $D(H)$ the first term $\langle H \psi(t), A\psi(t)\rangle$ is not equal to $\langle\psi(t), HA\psi(t)\rangle$ and $\frac{d}{dt}\langle A(t)\rangle \neq 0$. In other words even if $H=H^{\dagger}$ on $D(H)$ it is not necessarily so on $A(\mathcal{H})=\text{range}(A)$.

My question is how to derive formula $\ast$ using path integral formulation? Somehow the fact that Feynman measure is not invariant under symmetry $A$ has to play a role in such computation, but I don't see how to do it. If I start with $$ \int \mathcal Dp(s) \mathcal Dq(s) A(p(t),q(t))e^{iS(p(s),q(s))}, $$ with some boundary conditions at $s=0$ and $s=T$ and $0<t<T$, take derivative $\frac{d}{dt}$ I don't see how I can get anything equivalent to $H-H^{\dagger}$ on $A(\mathcal H)$.

$\endgroup$
  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$ – ACuriousMind Nov 22 '16 at 17:34
1
$\begingroup$

Typically, genuine anomalies are restricted to QFT (we need Hilbert's infinite hotel). However, there are some examples in QM. The best one (I think) is the $1/r^2$ potential in 3 dimensions. This is experimentally realized in bound states of three bosons, and has been studied experimentally. If the 2-body subsystem has a bound with zero binding energy, then the 3-body Schroedinger equation has a $1/r^2$ potential in hyperspherical coordinates.

The $1/r^2$ has a classical scale symmetry, which is broken to a discrete scale symmetry by the anomaly, see for example here. This is seen as a geometric series of three body bound states.

The problem is usually studied in QM or by summing Feynman diagrams, but there are attempts to discuss the anomaly directly using the path integral, see, for example, here.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.