Why Is the Wave Function Often Expressed as Psi = Cos(kr - wt) + i Sin(kr - wt)?

[AL-Hassan Naser]

why is psi = cos (k r - w t) + i sin ( k r - w t) = e^ [ i ( k r - w t)]?
my question precisely is why not:
1. psi = sin (k r - w t) + i cos ( k r - w t) ?
2. psi = sin (k r - w t) + i sin ( k r - w t) ?
3. psi = cos (k r - w t) + i cos ( k r - w t) ?

why not any of these three? is there a reason it is only the first form?

this means that there always has to be reflection between the real and imaginary parts of the wave function and i don't know why??Also,

when solving the Schrodinger's equation we assume that
PSI (r,t) = psi (r) X phi (t)

Why this form in particular not another form including addition or subtraction wouldn't this be possible and why?

thanks

 

[moriheru]

You should clarify what you are talking about. Are you talking about harmonic oscillators(seeminlgy not but just for example), steppotencials(aswell not I know) or any other wavefunction?

 

[AL-Hassan Naser]

I am talking about the wave function form used in Schrodinger's equation

 

[AL-Hassan Naser]

Also I am sorry but my background in Chemical Engineering so I didn't study quantum at all before so please give me any simple detail that can help me

thanks

 

[moriheru]

K is the wavenumber and one has simply reformed the basic exponetntial with the wavenumber k. I don't know if that helps but it is a start.

 

[bhobba]

I am talking about the wave function form used in Schrodinger's equation

The precise form has to do with the solution of the free particle Schroedinger equation. Substitute into it to see which are possible solutions:
http://www.physics.ox.ac.uk/Users/smithb/website/coursenotes/qi/QILectureNotes3.pdf

There are also key existence theorems in Partial differential equations that if you find solutions in one form, other forms don't exist. Best to consult a book on PDE's if that interests you.

You next question probably is why the Schroedinger equation and complex numbers. That's a deep issue. The use of imaginary numbers is an almost magical part of QM.

Check out the following:
http://arxiv.org/abs/1204.0653

Also get a hold of a copy of Ballentine - Quantum Mechanics - A Modern Development:
https://www.amazon.com/dp/9810241054/?tag=pfamazon01-20

Give the first three chapters a read.

But the deepest reason of all is the requirement of continuous transformations between so called pure states:
http://www.scottaaronson.com/democritus/lec9.html
http://arxiv.org/pdf/quantph/0101012.pdf

Thanks
Bill

 

[Simon Bridge]

why is psi = cos (k r - w t) + i sin ( k r - w t) = e^ [ i ( k r - w t)]?
my question precisely is why not:
1. psi = sin (k r - w t) + i cos ( k r - w t) ?
2. psi = sin (k r - w t) + i sin ( k r - w t) ?
3. psi = cos (k r - w t) + i cos ( k r - w t) ?

... looks like you are asking, if the first one is a solution (to the time dependent Schrodinger equation) then why are the others not solutions? Is this correct?

I'd answer with a question: why would you expect the others to be solutions?
Maybe they are. Did you substitute the others into the equation and see?

Note: the Scrodinger equation is usually set up for a particular situation - the specifics of the situation can rule out some wave-functions that may otherwise be valid solutions. The first line you wrote is not the only possible solution. Context is everything.

when solving the Schrodinger's equation we assume that
PSI (r,t) = psi (r) X phi (t)

Why this form in particular not another form including addition or subtraction wouldn't this be possible and why?

... it is not assumed, it is a property of the solution you first wrote down: $$\Psi(\vec r, t) = e^{i(\vec k\cdot \vec r - \omega t)} = e^{i\vec k\cdot \vec r} e^{- i\omega t} = \psi(\vec r)\chi(t)$$ It's a bit like asking why the equation for a line is assumed to be: ax+by=c ... why not some other form including multiplications and divisions?