# Schrodinger equation from commuation relations

1. Mar 4, 2009

### erkokite

I was wondering- is it possible to derive an equation of motion for example, the Schrodinger equation from the uncertainty principle (in commutator form)?

i.e. Is it possible to derive the Schrodinger equation from the following:

$$\left[\hat{x},\hat{p}\right]=ih$$

I gave it a shot, but of course, all I came out with is ih=ih (which of course is correct), and is the derivation of the uncertainty principle. However, I was actually wondering if I could use the commutation relation as an equation of motion for a wave function. Is this possible?

Thanks.

2. Mar 4, 2009

### AEM

You might find chapter 16, sections 19 through 24 (pages 378-383) of Bohm's Quantum Theory interesting. He doesn't do what you ask, but he shows some connections between Schrodinger's equation, the Hamiltonian formulation mechanics and Heisenberg's representation of quantum mechanics that may be relevant to your thinking.

3. Mar 5, 2009

### confinement

That commutator together with the notion that observables are represented with operators is enough to get you the time-independent shrodinger equation H psi = E psi. To get the time-dependent wave equation we also need the commutator [E,t] = i hbar, but then t is not really an operator so no books contain this relation, but it would follow from the classical poisson bracket for those variables along with Dirac's prescription for first quantization.

4. Mar 5, 2009

### Peeter

Bohm's chapter 9 has something pretty close. He shows how the commutator of the Hamiltonian (which at that point is an operator to be determined designated H) can be found in the expression for the expected value of an operator. He then uses the expected values and correspondence with classical particle equations (Newton's laws) to come up with Schrodinger's equation (ie: coming up with the equations and Erhenfest's theorem in one shot).

5. Mar 22, 2009

### dextercioby

You can't simply derive

$$\frac{d\psi (t)}{dt}=\frac{1}{i\hbar}\hat{H} \psi (t)$$ from

$$[\hat{x},\hat{p}]_{-}=i\hbar \hat{1}$$

but you can derive both starting with the commutations relations of the Galilei algebra.