# B Why exp(ikx-iwt) and not exp(ikx+iwt)?

1. Aug 12, 2016

### Isaac0427

The time independent free particle is described by the equation $e^{i(kx-\omega t)}$. Why do we use this equation instead of $e^{i(kx+\omega t)}$. I know the answer is "because that's how nature works," but I want to know why we think nature works that way. Where did we get these equations from? Why was $e^{i(kx-\omega t)}$ postulated and not $e^{i(kx+\omega t)}$?

Thanks!

2. Aug 12, 2016

### nashed

The time exponent part is just the solution to a time independent Hamiltonian inserted into Schrodinger's equation, there really isn't much more to it.... now if you want to ask how come Schrodinger's equation describes nature then you're out out of luck pal.

3. Aug 12, 2016

### robphy

4. Aug 12, 2016

### Staff: Mentor

Both forms are solutions to Schrödinger's equation. See for yourself by substituting each one into the SE.

5. Aug 12, 2016

### Isaac0427

But with the exp(ikx+iwt), the energy would be negative. This is because the energy operator is iħ∂t instead of -iħ∂t as it is in the momentum operator (of course in the momentum operator the derivative is with respect to position). I've always thought the energy operator was defined that way because of the sign on iwt. So I'm getting that this is just a convention so that the function evolves with time in a certain way.

6. Aug 12, 2016

### Staff: Mentor

You're right, the solution with $+i \omega t = +iEt/\hbar$ isn't valid. Somehow I was thinking of these versions: $e^{-i(\omega t - kx)}$ and $e^{-i(\omega t + kx)}$ which puts the ± on the kx.

As to why it has to be $-i \omega t = -iEt/\hbar$, have you seen how to solve the free-particle SE by using separation of variables, i.e. $\Psi(x,t) = \psi(x)f(t)$?