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If the time evolution function $K(A,B;t)$ is such that $$\psi(A;t) = \int K(A,B;t)\psi(B;0)dB.$$ We know that this function has all the information about how systems evolve. So from this we should be able to determine the Hamiltonian or the Lagrangian.

Then the Hamiltonian operator can be defined in terms of this as $$i\hat{H}\delta(A,B) = \dot{K}(A,B;0).$$ Here I put $\hbar=1$. I think this is the best we can do. To determine the Hamiltonian in terms of conjugate momenta would require anti-commuting variables.

This follows from $$\psi(t) = e^{-i t \hat{H}}\psi(0).$$

I was trying to find the Lagrangian $L[A]$ in terms of the $K(A,B;t)$

The best I could come up with is the horrible looking limit:

$$L[A]=L(A,\dot{A}) = \lim_{u\rightarrow 0} \left(\frac{\dot{K}(A+ u \dot{A} , A ; u )}{K(A+ u \dot{A} , A ; u )} - \frac{\dot{K}(0, 0 ; u )}{K(0 , 0 ; u )} \right)$$

which probably only works for zero dimensional systems. Is there a known simpler expression for the Lagrangian in terms of the evolution function (also sometimes called the propagator in non-relativistic cases).

The question is, is there a better formula than this? (Assuming this formula is correct)

Edit: Deleted unnecessary information.

Qmechanic
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  • As I said, I do disagree with your expressions. You should then recast your question as: "given all matrix elements of the hamiltonian, how do you find the corresponding Lagrangian"? It is absolutely crucial to avoid any and all path integral tangents, which would only reinforce misconceptions. – Cosmas Zachos Jan 09 '23 at 17:59
  • OK, I deleted unnecessary information. But I disagree, I am not asking about matrix elements of a Hamiltonian. I am asking for a formula directly from the propagator to the Lagrangian - simpler than the limit I made. –  Jan 09 '23 at 18:19
  • The limit you made is misguided, as ̂ eigenvalues A,B are not time-dependent. The classical limit of the Hamiltonian leads to the (classical!) Lagrangian, normally, cf. – Cosmas Zachos Jan 09 '23 at 18:33
  • No, what I'm doing is just inserting time dependent fields $A(t)$ into the function K. No, we must use the quantum mechanics Hamiltonian, the classical Hamiltonian does not preserve the anti-commuting quantities. –  Jan 09 '23 at 21:15
  • Fields? You are assuming a wavefunctional Schrödinger equation? Quantum Hamiltonian-Lagrangian function, instead of plain Lagrangian? Anticommuting quantities? You must probably specify all this in your question – Cosmas Zachos Jan 09 '23 at 21:28

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I'm not sure I understand your expressions. Let's nondimensionalize to $\hbar=1$ and consider your eigenvalues A,B of $\hat x$, which, of course, do not depend on the time. Functional integration is overkill here.

The conventional definition of the propagator is but $$ K(A,B;t)\equiv \langle A| \hat U(t)|B\rangle= \delta(A-B) -it \langle A| \hat H|B\rangle + O(t^2), $$ so that the coordinate matrix elements of the Hamiltonian are just $$ \langle A| \hat H|B\rangle =i\lim_{u\to 0}\frac{K(A,B;u)-\delta(A-B)}{u}=i\dot{K}(A,B;0), \qquad \leadsto \\ \hat H = \int\!\!dAdB~~|A\rangle \langle A|\hat H|B\rangle \langle B|~, $$ from which you may transition to the Lagrangian, if needs be, in the standard manner.

For the free particle, you should be able to reproduce $$ \langle A| \hat H|B\rangle = -\frac{\partial_A^2}{2m} ~\delta(A-B)~. $$

Cosmas Zachos
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  • Yes, you have reproduced the Hamiltonian in terms of K which I already did just rewritten it in terms of bra-ket notation. I'm looking for the Lagrangian in terms of K. There is a lot of complicated equations hidden in the simple phrase "you may transition to the Lagrangian, if needs be, in the standard manner". This gets us no closer to writing a formula for L in terms of K. Sorry. –  Jan 09 '23 at 17:33
  • I assume you appreciate how the classical limit of the operator hamiltonian determined through the matrix elements above can be instantly Legendre-transformed to the classical Lagrangian, no? E.g., $H=p^2/2m$ above... – Cosmas Zachos Jan 09 '23 at 19:36
  • Taking the classical limit of a Hamiltonian might lose information if we are not using anti-commuting variables? But if this is the case then I would take this as an answer if you could express it as an equation. –  Jan 09 '23 at 21:17
  • Lose information? What kind of nonclassical object Lagrangian do you have in mind? This is not NRQM? – Cosmas Zachos Jan 09 '23 at 21:30
  • A classical hamiltonian obviously has less information that a quantum mechanical hamiltonian since since we are removing the ordering. No-one said nonclassical Lagrangian. –  Jan 09 '23 at 22:39
  • Emphasize it in the question. In conventional NRQM, including path integrals, only classical Lagrangians are featured. – Cosmas Zachos Jan 10 '23 at 00:17