# Schrödinger equation

1. Dec 13, 2012

### Mr-T

Does the Schrödinger equation completely neglect the uncertainty principle? If so, wouldn't this imply that our intensity distribution has its own probability distribution?

2. Dec 14, 2012

### tom.stoer

The Schrödinger equation predicts the wave function with certainty; but from this wave function the uncertainties of observables can be derived exactly

3. Dec 14, 2012

### Staff: Mentor

The momentum-space wave function $\Phi(p,t)$ is basically the Fourier transform of the position-space wave function $\Psi(x,t)$. The uncertainty principle comes from the properties of Fourier transforms. Any pair of functions that are related by Fourier transforms has a similar uncertainty principle.

4. Dec 14, 2012

### Mr-T

I understand what both of you are saying and I appreciate the replies.

In the Schrödinger equation we input values for energy/mass assuming we know with 100% certainty what these values for energy/mass are. Due to the input of these values is where my question holds its regards.

5. Dec 14, 2012

### tom.stoer

No

The input is a wave function, the output is a wave function at a later time. This predicts with certainty that a system will be in a state A' at time t' > t provided that it was in state A at time t; A is specified by a wave function or a state vector |A>.

In case of the time-indep. SE the input is not energy, the input is nothing! The outputs are a) the allowed energy eigenvalues and b) the corresponding eigenfunctions. The SE does not tell you in which state the system is, in only tells you what the allowed state are

6. Dec 14, 2012

### Mr-T

If you are not inputting any information into the T-I SE then how do you know what particle it is talking about?!

7. Dec 14, 2012

### Jorriss

Do you mean you specify a potential, then solve the SE equation for a given potential? Or you plug in the values of the eigenvalues?

8. Dec 14, 2012

### dextercioby

The remark by Tom is an overstatement, an exaggeration. The input is the specific form of the Hamiltonian in terms of fundamental observables such as position, momentum, spin.

9. Dec 14, 2012

### Mr-T

If all direct observables have some uncertainty, won't this mess up our intensity distribution even more than the fouriers already do?

10. Dec 14, 2012

### Staff: Mentor

OK, I think I see where you were going with your original question...

In the time-dependent Schrodinger equation $H\Psi=E\Psi$ the Hamiltonian is written as if all of its inputs were exactly known. For example, if we're dealing with two charged particles, there will be a $\frac{1}{r1-r2}$ term somewhere in it, where r1 and r2 are the positions of the two particles. You should read that as saying not that the two particles are at those exact positions, but rather that if they were in those positions that would be the exact distance between them. The uncertainty principle doesn't stop us from talking about how things would be if we knew exactly where a particle was, it just forbids us from knowing exactly where it is.

Once I have the Hamiltonian written down, I solve Schrodinger's equation; and as tom.stoer said in #2, the uncertainty principle is inherent in the ψ that comes out.

11. Dec 14, 2012

### Mr-T

Ahh yes, talking in this fashion resolves my concerns.

Thank you nug

12. Dec 15, 2012