# Lagrangian Quantization

1. Jun 26, 2006

### eljose

Let be the Lagrangian of a particle:

$$L(q,\dot q,t)$$ my question is if we can get its quantizied version in the form:

$$p\dot q-L(q,\dot q,t)| \Psi>=E_{n}|\Psi >$$

of course we know how to quantizy the Momentum operator the question is..¿how do we quantizy the "celerity" $$\dot q$$ operator acting over an state?..thanks.

2. Jun 26, 2006

### masudr

You can't write the Hamiltonian like that and still use the $\dot{q}$. Just as in classical Hamiltonian mechanics, you have to eliminate $\dot{q}$ in favour of $p = \partial L / \partial \dot{q},$ and so you shouldn't have any "generalized velocity" (which is what you have called celerity) quantities appearing in the expression.

Just to be picky as well, I would ask that you put brackets around your operator acting on $|\psi\rangle$ on the left-hand-side.

Alternatively, you might want to eliminate the $p$ that you have put on the left-hand-side with the relation between $p$ and $\dot{q}$ given by $p = \partial L / \partial \dot{q}.$ In this case, I guess the operator $\dot{q}$ would be given by $\hat{\dot{q}} =d \hat{x}/dt.$

Last edited: Jun 26, 2006
3. Jun 26, 2006

Staff Emeritus
I was just reading the perspective on Julian Schwinger in the History of Physics section of the arxiv. It mentioned his papers on the quantum action principle in fully relativistic form. Apparently he was concerned to have a way to define relativistic QM without starting with a classical theory, quantizing it, and then throwing it away.

Has anybody looked at these papers? Can anybody comment on the methodology?

4. Jun 27, 2006

### eljose

The problem with Hamiltonian Quantization is that in several theories you have $$H\Psi= 0$$ so there wouldn,t be any time evolution of the system..

"self-adjoint" could you give the references to the arxiv papers about Julian Schwinger?..i would like to take a look at them.

5. Jun 27, 2006

### masudr

Well if H = 0, then by definition

$$\sum_{i=1}^N p_i\dot{q}_i - L(\vec{q}, \dot{\vec{q}}, t) = 0$$

so you will get the same problem, or have I missed something?

6. Jun 27, 2006