# Method for deduce Schrödinger Equation

1. Nov 28, 2014

### Garrulo

Is there any method for deduce Schrödinger equation from quantization of action??

2. Nov 28, 2014

### vanhees71

The closest way in this direction is the Feynman path-integral approach, from which you can derive the Schrödinger equation. Of course, action is not quantized to begin with :-).

3. Nov 28, 2014

### Garrulo

But then, when is quantized the action?

4. Nov 28, 2014

### vanhees71

Where did you get the idea from that action is quantized? That's an idea that is outdated for nearly 89.5 years now! :-)

5. Nov 28, 2014

### Garrulo

But then, how is deduced the Schrödinger Equation??. All the phyiscs theories are the result of a minimal or maximal quantity of a magnitud. k for thermodynamics, c for special relativity

6. Nov 28, 2014

### vanhees71

I don't understand this ideas. There is no way to derive quantum theory from simpler principles. Today, it's a fundamental theory. It's also not possible to derive Newtonian classica mechanics or classical electromagnetic theory from something simpler than the fundamental laws (Newton's space-time model and the dynamical equations of motion, whose principle form can be deduced from space-time symmetries but specific force laws like Newton's gravitational force; or Einstein-Minkowski space of special relativity, from which principle dynamical field equations can be derived, but the specific form of the Maxwell equations as the Abelian gauge theory of a massless vector field must be deduced from experience).

7. Nov 28, 2014

### Staff: Mentor

You can see Leonard Susskind derive the Schrödinger equation in his 2012 course on Quantum Mechanics, lecture 4. It is on YouTube here

8. Nov 28, 2014

### dextercioby

In a different formulation/axiomatization of Quantum Mechanics, the $\frac{d \psi(t)}{dt} = \frac{1}{i\hbar} H\psi(t)$ is no longer an axiom, but a theorem. It's just a restatement of the known fact that the Hamiltonian is a generator of a strongly continuous unitary representation of a one-parameter subgroup of the symmetry group of the theory: the Galilei group in non-specially relativistic physics and the Poincare group in specially relativistic physics.

While what I wrote in the above paragraph is just fancy mathematics, there's a physical digression on this issue: the nice discussion in the 2nd Chapter of Prof. Sakurai's book on QM. The only derivation of the SE I can remember without resorting to harmonic analysis and functional analysis.

Last edited: Nov 28, 2014
9. Nov 28, 2014

### Garrulo

But in 2.1.24 and 2.1.25 he assumes the linearity of the equation, ignoring terms of O(dx^2) (Sorry, I don´t manage Latex in this forum)

10. Nov 28, 2014

### Garrulo

But he assumes in chapter E=\hbar⋅ω an exact relation. In the moment that quadratic terms there are, minors but there are (maybe, not I am assuming, simply not discharging, the linearity breaks Sorry, I don´t know use Latex in this forum

11. Nov 28, 2014

### Staff: Mentor

12. Nov 30, 2014

### dextercioby

$$E=\hbar\omega$$ comes from the DeBroglie pilot wave assumption for material particles (other than photons). It's assumed exact, no linearity assumption implied.

13. Nov 30, 2014

### Staff: Mentor

Its true basis is symmetry - see Chapter 3 - Ballentine - QM - A Modern Development.

Thanks
Bill