Sol'n of Schrodinger's Eq: A e^{kx-wt} & A Sin(kx-wt)

[zodas]

The general solution of Schrödinger's equation is givrn by --------
\Psi= A e^{kx-wt}.
And this satisfies the equation .

But the general solution of 3-D sinosidal wave is given by
Psi= A Sin(kx-wt)
And this also satisfies the Schrödinger's equation.

Schrödinger is credited to find the solution as complex phase factor (to signify matter waves) .

Now the question is what is the need of depicting matter waves as complex phase factor ?

 

[LostConjugate]

I think it has to do with keeping the total energy linear while taking a second derivative. For example: (d2/dt2)(exp[iEt]) = -E^2exp[iEt]. Where as i(d/dt)(exp[iEt]) = -E(exp[iEt]).

In a complex wave equation (i) acts as a derivative because it changes the phase by the same amount (90 degrees) while preserving linear total energy in the solution.

 

[bapowell]

The general solution of Schrödinger's equation is givrn by --------
\Psi= A e^{kx-wt}.
And this satisfies the equation .

You're missing an "i" in that argument of the exponential.

But the general solution of 3-D sinosidal wave is given by
Psi= A Sin(kx-wt)
And this also satisfies the Schrödinger's equation.

No it doesn't. Not without some funky potential.

 

[LostConjugate]

No it doesn't. Not without some funky potential.

I understand that it does.

http://quantummechanics.ucsd.edu/ph130a/130_notes/node13.html

 

[bapowell]

I understand that it does.

http://quantummechanics.ucsd.edu/ph130a/130_notes/node13.html

No. That's the time independent solution. zodas is claiming that the full time dependent wave function \Psi(x,t) = A sin(kx-wt) satisfies the SE. Just work it out...it's not difficult.

 

[SpectraCat]

I understand that it does.

http://quantummechanics.ucsd.edu/ph130a/130_notes/node13.html

Where is the time dependence of the wavefunction? It is not covered in that link. The point is that for an eigenstate of a quantum system, the time-dependent phase is always complex. So it is the "omega-t" term in the sine function mentioned by the OP that makes it not a solution of the TDSE. Try plugging that sine function into the TDSE and see what you get ... you will find that, as bapowell said, it requires a "funky" potential.

 

[LostConjugate]

Got it.

 

[zodas]

Thanks guys !