Quick QM: Why Does Schrodinger Eq Being 1st Order Matter?

[Thrice]

The schrodinger eq is 1st order in t. Why does that matter to the probability interpretation?

Should be an easy question, but I can't seem to get it.

 

[quasar987]

Here's a suggestion you might want to investigate. When solving the SE by the method of separation of variables, we find that the time dependant part of the solution is \exp{iEt/\hbar}, and the position dependant part satisfies the time-independant SE. Denote \psi(x) the solution to the time independant SE for a given potential. Then the general solution to the SE is \Psi(x,t)=\psi(x)e^{iEt/\hbar}, and according to the Born interpretation, \Psi \Psi^* is a probability density function for the position of the particle. But \Psi \Psi^* = \psi\psi^*. I.e. the probability density is is time independant!

So the question is, would the time dependant part of the \Psi still be such that the probability is time independant if the t "dependance" of the SE was not of first order?

 

[Daverz]

Interestingly, Schrödinger originally started with a relativistic equation, but didn't know what to do with the negative probability densities that resulted, so came up with the final non-relativistic equation instead.

 

[Thrice]

Here's a suggestion you might want to investigate. When solving the SE by the method of separation of variables, we find that the time dependant part of the solution is \exp{iEt/\hbar}, and the position dependant part satisfies the time-independant SE. Denote \psi(x) the solution to the time independant SE for a given potential. Then the general solution to the SE is \Psi(x,t)=\psi(x)e^{iEt/\hbar}, and according to the Born interpretation, \Psi \Psi^* is a probability density function for the position of the particle. But \Psi \Psi^* = \psi\psi^*. I.e. the probability density is is time independant!

So the question is, would the time dependant part of the \Psi still be such that the probability is time independant if the t "dependance" of the SE was not of first order?

Ok I understand what you said there (I think), but I can't make the jump to what would happen if it wasn't 1st order in t. I know how to derive the equation from dealing with wave packets, I know why it's 1st order & how to get the time independent SE eigenvalue equation.

Call it a lack of imagination. I can't see how it could be different. I know the difference between the SE & the familiar wave equation, but I can't see how to get probabilities out of the latter.

 

[quasar987]

It should have to do with what Daverz said. I.e. that a different order in t will give us negative probabilities... but how could that be? Whatever the solution \Psi(x,t) to a modified SE, the complex conjugate of \Psi is still it's norm squared, which is still positive no matter what.

 

[Thrice]

Bump. I was going through old posts & needed some closure here. Thanks to whoever moved this.