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The Time-dependent Schrödinger equation is given by $$i\hbar \frac{d}{dt}|\Psi(t)\rangle=\hat{H}|\Psi(t)\rangle $$

From Classical Mechanics, we know that $$\mathcal{L}=\dot{q}p-H$$ which should change in quantum mechanics as

$$\hat{\mathcal{L}}=\frac{1}{2}(\hat{\dot{q}}\hat{p}+\hat{p}\hat{\dot{q}})-\hat{H}$$

The question is, Is is right to use this relation in replacing $\hat{H}$ from Schrödinger equation so that $$i\hbar\frac{d}{dt}|\Psi(t)\rangle=\left[\frac{1}{2}(\hat{\dot{q}}\hat{p}+\hat{p}\hat{\dot{q}})-\hat{\mathcal{L}}\right]|\Psi(t)\rangle$$

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Certainly there's no reason why you couldn't define the operator $\hat L$ such that this is true. But in general, doing so won't buy you any of the power that comes from a Lagrangian formalism in classical mechanics. For example, you will not have a action minimization principle would this allow you to implement a Lagrangian formulation for Noether's theorem.

Another way to make use of the Lagrangian again in quantum mechanics is via the path integral. There (under some assumptions) the classical Lagrangian can be used to great effect, included results which tie to a sort of action minimization principle (if you're interested in effective field theories for example) and Noether's theorem (which results then in Ward identities).

Since it was asked about in the comments, let me give a (very) brief description of the path integral and how it relates to the Lagrangian. For more details, any book on quantum field theory will have (to varying levels of detail) what I am about to describe. If you want to see the path integral in the context of quantum mechanics, I would suggest the section on it in J.J. Sakurai's book.

I will first note two things. The idea is somewhat different that what is described in the question and there is never a need for a Lagrangian "operator." I will mention that in volume 1 of Weinberg's quantum field theory book he does introduce such an operator, but only as a stepping stone to other things, and that book, while exceptionally detailed is not for the faint of heart. I will also mention that the path integral (and thereby the introduction of Lagrangian and Lagrangian-like things into quantum mechanics) forms the essential machinery of modern theoretical physics from all of particle physics and the standard model to large swaths of modern condensed matter theory and other subfields as well (but those are the ones I know use it most). What I describe below barely scratches the surface of what this machinery can do.

The idea goes as follows. Suppose we want to compute the transition amplitude from a state $|x_i\rangle$ at say $t=0$ to the state $|x_f\rangle$ at a later time $t$. Then the amplitude we need to compute is $$ \langle x_f|e^{-iHt}|x_i\rangle. $$ But instead of moving the state forward in time by an amount $t$ using the time evolution operator $e^{-iHt}$, we could equivalently evolve by an amount $t/N$ consecutively $N$ times. That is, we could write $$ \langle x_f|\prod_{n=1}^Ne^{-iHt/N}|x_i\rangle. $$ Between each of these factors of the time evolution operator, I will insert the identity twice, once as a complete set of position states and once as a complete set of momentum states: $$ 1=\int dpdx|p\rangle\langle p|x\rangle\langle x|. $$ Doing so, we find that every factor of the time evolution operator is sandwiched between a momentum and position eigenstate while the extra $\langle p|x\rangle$ factors become things like $e^{ipx}$. The important thing about the sandwiching of the time evolution operators is that now (up to some care about operator orderings), we can replace all the momentum and position operators in $H$ by the eigenvalues, which are now just numbers rather than operators.

If we keep careful track of the position and momentum eigenvalues now up in the exponentials, we find that in the limit $N\rightarrow\infty$, the thing we obtain after all these insertions can be made to look roughly like $$ e^{i\int(p\dot x-H)dt} $$ and so the transition amplitude we were looking to calculate looks like $$ \int\mathcal{D}x\mathcal{D}pe^{i\int(p\dot x-H)dt} $$ where the integrals $\mathcal{D}$ are taken over all possible values of $x$ and $p$ at each and every time between our initial and final times, hence the name path integral. In many common examples we can identify this thing in the exponent as not only the integral of the Lagrangian, but in fact the classical action.

Now, there are many details, tricky points, and caveats that go with this idea of the path integral, but this is the essential idea of it.

Thinking about it more, there is a book by Anthony Zee called Quantum Field Theory in a Nutshell which works out the derivation of the path integral within the first 10-20 pages with a reasonable amount of detail. Though of course to get all the caveats one would likely need to refer to Weinberg's book, but that's certainly not the place to go for a first pass at these ideas.

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  • $\begingroup$ So nothing can stop me from defining lagrangian operators in such a way, But this not gurrenty whether or not it's good $\endgroup$ – Young Kindaichi Nov 12 '20 at 7:27
  • $\begingroup$ Can I derive a path integral formulasim from here? $\endgroup$ – Young Kindaichi Nov 12 '20 at 7:28
  • $\begingroup$ @YoungKindaichi I updated the answer to include more information about the path integral. I do not include all detail because that would probably be too much for this post, but I have included a couple references where you can fill in varying levels of detail for yourself. $\endgroup$ – Richard Myers Nov 12 '20 at 8:03
  • $\begingroup$ Thanks, So can I proceed by introducing lagrangian at first and then see if it reproduce the path integral. Are there any other input I will need? $\endgroup$ – Young Kindaichi Nov 12 '20 at 10:14
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    $\begingroup$ A quick question: isn't the operator $\hat{\dot{q}}$ not well defined in quantum mechanics? Also, for reference, Dr. Shankar's excellent book on quantum mechanics also covers path integrals in some detail, and more advanced statistical mechanics texts should as well. $\endgroup$ – John Dumancic Nov 12 '20 at 19:36

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