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In classical field theory, when you have a free real scalar field $\phi$ with Lagrangian (density): $$ L = \frac{1}{2} \, \eta^{\mu \nu} \, \partial_{\mu} \phi \,\partial_{\nu} \phi - \frac{1}{2} m^2 \phi^2,$$ where $\eta^{\mu \nu}$ is the Minkowski metric with signature $(+, -, -, -)$, the corresponding Euler-Lagrange equations are the Klein-Gordon equations: $$ \eta^{\mu \nu} \partial_{\mu} \partial_{\nu} \, \phi + m^2 \phi = 0. $$

Moreover, it turns out that after quantization, in QFT, the quantized field $\hat{\phi}(x)$ in the Heisenberg picture also obeys the Klein-Gordon equations.

This vaguely reminds me of the Cayley-Hamilton theorem, which states that if $A$ is a complex $n \times n$ matrix, then $p(A) = 0$, where $p(x)$ is the characteristic polynomial of $A$. Indeed, we know that any eigenvalue $\lambda$ of $A$ satisfies $p(\lambda) = 0$, and the CH theorem tells us that when you promote a generic eigenvalue $\lambda$ to the "operator" $A$, then $A$ satisfies the same equation, namely $p(A) = 0$.

I realize it is a "formal" question for most physicists, but I wonder if what I am thinking of is essentially true.

Malkoun
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This is the ultimate "soft" question, so I can't possibly give you a "hard" answer, but, offhand, no.

Indeed, in your vision, the C-H theorem amounts to a functor "promoting" a characteristic polynomial with n roots $\lambda_n$ to the same polynomial of a diagonal (direct sum) operator with these roots along its diagonal; and then scrambling the operator to a generic hermitian one (you seem to think of strictly diagonalizable operators, the all but trivial case of the theorem, which is fine for your purposes).

However, quantization is not quite a map between quantum operators and their eigenvalues, despite the coherent state vision you might conceivably be basing this on. That is to say, the complex coefficients $\alpha(p), \alpha^*(p)$ you are quantizing to $a(p), a^\dagger(p)$ are in no way eigenvalues of your quantum result--and the linear K-G equation is not readily interpretable as a characteristic polynomial of the $\alpha(p), \alpha^*(p)$s... Quantization here is just arraying a slew of decoupled classical momentum (normal) modes with arbitrary coefficients to the same modes with quantum coefficients, and the similarity is predicated on the identical component equations.

In a sense, this is the trivial part encoding Lorentz invariance. The real fun starts when you project the nontrivial implications of the quantized system, and how features unthinkable for the classical precursor start emerging.

But, as indicated, absent further specific hard questions, it is hard to comment on you vision.

Cosmas Zachos
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  • Thank you Cosmas Zachos for your answer. By the way, what do you mean by the distinction between "soft" and "hard" questions? I am not a physicist and I am kind of curious (learning more about physicists' jargon). By the way, I am not necessarily assuming that the matrix is diagonalizable. It may have some bigger Jordan blocks, but I think I understand your point. Quantization is not really analogous to promoting an eigenvalue to a matrix, is what I think you are saying. However, I am not yet fully convinced that I am wrong either. I will think a bit more. – Malkoun Jan 15 '22 at 01:55
  • Ok, your answer makes sense. There may be some theorem which generalizes the physics fact mentioned in my post, but chances are it is not quite the Cayley-Hamilton theorem. Roughly speaking, in the Hamiltonian formalism, Poisson brackets become commutators (for bosons) after quantization, so the evolution equation for observables, with Poisson brackets, becomes the operator evolution equation, with commutators. I guess the physics fact I have mentioned basically boils down to that, right? – Malkoun Jan 15 '22 at 02:57
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    Yes, PBs go to commutators among other things, but there is a whole field, deformation quantization, fleshing it out; it is not a functor. Real fields are Hermitean, so diagonalizable. Soft is common informal evocation of point of view subjectivity, in contrast to yes or no problems… But closing doors is an impossible task… – Cosmas Zachos Jan 15 '22 at 03:23