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Let us say I have a classical field theory, with a field $\phi(\vec r,t)$ which satisfies the relevant Euler-Lagrange equation for the Lagrangian $\mathscr{L}$. Is the general procedure (i.e. one that will work for any $\mathscr{L}$) for going from this classical field to the quantum field operators in QFT?

I know that one can look at commutators, but is this really enough to uniquely specify the field operators?

Quantum spaghettification
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1 Answers1

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The Euler-Lagrange equation has a set of solutions. In classical physics, we consider only solutions which are c-number valued (or Grassmann-number valued if you include "classical" fermion fields); to quantise, we use solutions whose value is a linear operator from a Hilbert space to a Hilbert space (usually the same one). The trick is to ensure they satisfy canonical commutation relations, CCRs, (or canonical anticommutation relations, CARs, for fields satisfying Fermi-Bose stats) analogous to canonical relations. Just as Poisson brackets satisfy $\{q_i,\,p_j\}=\delta_{ij},\,\{q_i,\,q_j\}=\{p_i,\,p_j\}=0$ in discrete classical mechanics, and the first of these generalises in classical field theory to $\{\phi(t,\,\mathbf{x}),\,\pi(t,\,\mathbf{y})\}=\delta(\mathbf{x}-\mathbf{y})$, for a QFT we take $[\hat{\phi}(t,\,\mathbf{x}),\,\hat{\pi}(t,\,\mathbf{y})]_\pm=\text{i}\hbar\delta(\mathbf{x}-\mathbf{y})$ where we use a commutator or anticommutator accordingly (and of course also the $\hat{\phi}$-$\hat{\phi}$ and $\hat{\pi}$-$\hat{\pi}$ (anti)commutators vanish).

Linear ELEs lead to an especially simple result. Suppose a real classical $\phi$ satisfies a linear ELE with solutions $\phi=\sum_\sigma(\phi_\sigma a_\sigma +\text{c.c.})$ with complex coefficients $a_\sigma$ and complex functions $\phi_\sigma$; then the quantised counterpart is the Hermitian field $\hat{\phi}=\sum_\sigma(\phi_\sigma \hat{a}_\sigma +\text{h.c.})$. (You can obtain a similar expression for $\hat{\pi}$ from this.) The spacetime-constant coefficients are promoted to operators, the $\hat{a}_\sigma$ being annihilation operators so $\hat{\phi}| 0\rangle=\sum_\sigma\phi_\sigma^\ast\hat{a}_\sigma^\dagger| 0\rangle$ is a linear combination of $1$-particle states, with $| 0\rangle$ the vacuum ket.

Quantum spaghettification
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J.G.
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  • Hi, thanks for your answer. Please could you go into a little more detail of how we use the CCRs and CARs to find the field operators? i.e. how do we know that $\hat{\phi}=\sum_\sigma(\phi_\sigma \hat{a}_\sigma +\text{h.c.})$ is the correct field operator? – Quantum spaghettification Jul 31 '17 at 12:38
  • @Quantumspaghettification That's the most general Hermitian solution of a linear ELE. We don't deduce it from canonical relations, but we can use this solution format to restate those relations equivalently as (anti)commutator relations on the ladder operators. – J.G. Jul 31 '17 at 14:02
  • Sorry to be a pain, but how do we know that it is the most general? i.e. why does swapping the complex coefficients for creation and annihilation operators make the most general Hermitian field? – Quantum spaghettification Jul 31 '17 at 14:38
  • @Quantumspaghettification Because the solution basis is still the same set of functions; we're just changing the "scalars" used as coefficients to span a vector space. – J.G. Jul 31 '17 at 14:46
  • Oh, ok that makes sense - apart from one thing; the coefficients $a_\sigma$ are arbitrary constants whilst the operators $\hat a_\sigma$ are not (correct me if I am wrong). This seems like we are going from a general expression to a specific expression - how is the generality kept? – Quantum spaghettification Jul 31 '17 at 15:04
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    @Quantumspaghettification The only thing that's specific is their (anti)commutators, deduced from canonical relations. You can use the formula for $\hat{\phi}$ to obtain a Hamiltonian in which the role of these operators in particle number is obvious; you then conclude all $\hat{a}$s annihilate the vacuum ket. – J.G. Jul 31 '17 at 15:42