We know that classically, if we have some theory $\mathcal{L}$ such that the action $\int d^4 x \mathcal{L}$ is invariant under time translation, then we can use Noether's theorem to find that (the spatial integral over)

$$\sum\limits_i \partial_t \phi_i \frac{\partial \mathcal{L}}{\partial (\partial_t \phi_i)} - \mathcal{L}$$

is classically a conserved quantity, and we happen to call it "energy".

This is a classical field theory result. But how do we go about showing a similar conservation law in QFT?

  • $\begingroup$ Noether's theorem. A Lagrangian is a Lagrangian, whether it's classical or quantum mechanical. There are so-called anomalies, which arise in path integrals because the $d[\phi]$ in the path integral have a determinant which can be conserved as well. $\endgroup$ – webb Oct 30 '14 at 18:25
  • $\begingroup$ @webb part of the derivation of Noether's theorem requires the assumption that the Euler-Lagrange equation is satisfied. Maybe somebody can correct me if I'm wrong, but I believe this implies that any conservation law derived from Noether's theorem is strictly classical. $\endgroup$ – Bridgeburners Oct 30 '14 at 18:33
  • $\begingroup$ Dirac equation, Schrodinger equation, etc. are all derived from minimizing an action that has conserved currents associated with symmetries in the action. For example, the invariance of the Schrodinger equation Lagrangian under unitary transformations of $\psi$ implies conservation of the norm. $\endgroup$ – webb Oct 30 '14 at 18:39
  • 1
    $\begingroup$ The precise quantum analogue to Noether's theorem are the Ward-Takahashi identities. $\endgroup$ – ACuriousMind Oct 30 '14 at 20:11
  • $\begingroup$ Bridgeburners, well, not exactly. Quantum operator equations are known to look like classical ones, so in most cases the operator 'conservation' is present. $\endgroup$ – Prof. Legolasov Oct 30 '14 at 20:12

In canonical quantization one constructs the Hamiltonian formalism. Energy conservation is therefore manifest (since Hamiltonian is time-independent and commutes with itself).

Quantum-mechanically, the Hamiltonian of the system can be expressed via particle creation-annihilation operators. So, the total energy of the field is also the total energy of all particles and is quantum-mechanically conserved.

You can get a feel of this conservation by computing the evolution of the wave-functional by expanding it into a sum of energy eigenstates multiplied by exponentials:

$$ \Psi = \sum_i e^{-i E_i t / \hbar} \cdot \Psi_i. $$

Note that $E_i$ do not depend on $t$, so energy is in fact conserved (this applies to all QM with time-independent Hamiltonian and not just to QFT).


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.