Wave Function Doubt and Derivation

[sarvesh0303]

Homework Statement

I was reading up on the Wave Function used in the Schrödinger Wave Equation. However one source said that
ψ(x,t)=e^(-i/hbar*(px-Et))
Another source had this
ψ(x,t)=e^(i/hbar*(px-Et))

Which one of these is true and could someone give a derivation for the correct wavefunction?

Homework Equations

The Attempt at a Solution

 

[Simon Bridge]

You can see for yourself by substituting into the 1DSE.
In fact - that is an important exercise: do that and get back to us.

 

[sarvesh0303]

I know I can! But I want to try deriving the 1DSE by using this value. So I want to know another way of deriving it. I read that it can be derived by taking
ψ(x,t)=A(cos(2∏x/λ-2∏ηt)+isin(2∏x/λ-2∏ηt))

where η=frequency
and then the euler formula was used.

My doubt with this derivation is that would every wave (its wavefunction) always be of the form mentioned above.
If so then since isin(2∏x/λ-2∏ηt) is a complex term, then wouldn't it imply that every wave must have a complex component?
This part really confuses me!

 

[Simon Bridge]

I know I can! But I want to try deriving the 1DSE by using this value.

The 1DSE is not really something you derive is it? You will benefit by substituting both forms of the wavefunction you were asking about into the Schrödinger equation to see which one is the "real" solution. Have you tried this yet? If not - do it. If you have, please report what you found.

So I want to know another way of deriving it.

"it" what? The wavefunction for a free particle or the 1DSE? You have mentioned trying to derive both now.

My doubt with this derivation is that would every wave (its wavefunction) always be of the form mentioned above.
If so then since isin(2∏x/λ-2∏ηt) is a complex term, then wouldn't it imply that every wave must have a complex component?
This part really confuses me!

In general, wave-functions are complex - this is correct. Why would this confuse you? It is why you have to premultiply by the complex conjugate before integrating.