B The deduction of Schrodinger equation, I'm stuck

[Joao]

Hi everyone! Please, I'm trying to understand the Schrodinger equation, and I've understood it this far, which which is a miracle, hehehe:
(Laplacian)(psi) plus ((2phi)/h)^2.2m (E-V)(psi)

I know that hbar = h/(2phi)

But how that turns into

(Laplacian)(psi)+2m/(hbar)^2.(E-V)(psi)

My math isn't good enough... =( can someone please point me how it happened? =)

Thanks a lot!

 

[Joao]

Or, in other words, how this:
∇^2Ψ+{[2π/h]^2}.2m(E-V)ψ

Assuming

ħ=h/2π

Becomes

∇^2ψ+{2m/ħ^2}(E-V)Ψ

Sorry for the bad English! =)

Thanks!

 

[PeterDonis]

I'm trying to understand the Schrodinger equation

It might help if you would give a reference for where you are getting this from. I'm not sure what you are writing down is a correct expression for the Schrodinger Equation.

Also, please use the PF LaTeX feature, it makes your equations much easier to read. See here for help:

https://www.physicsforums.com/help/latexhelp/

 

[Joao]

It might help if you would give a reference for where you are getting this from. I'm not sure what you are writing down is a correct expression for the Schrodinger Equation.

Also, please use the PF LaTeX feature, it makes your equations much easier to read. See here for help:

https://www.physicsforums.com/help/latexhelp/

Thanks a lot! Sorry for the ugly equations!

I'm based on this YouTube video:



So, I'm starting with the Helmholtz equation:

$$∇^2Ψ+(\frac {2π} {υ})^2Ψ$$

And my goal is to the time independent Schrodinger equation

$- \frac {ħ^2} {2m} ∇^2 Ψ +VΨ = EΨ$

Miraculously, I 've kind of understood until here:

$∇^2 + (\frac {2π} {h})^2 . 2m (E-V)Ψ$

And I know that

$ħ= \frac {h} {2π}$

But I don't understand how that turned into this:

$∇^2Ψ + \frac {2m} {ħ^2} (E-V)Ψ$

Please... Can someone please tell me what happened? I don't understand it... =(

Thanks and sorry again for my confusion on the first and second posts!

 

[Demystifier]

You must be missing something very very trivial, so trivial that it is very difficult to say what it is.

 

[Joao]

Thanks everyone! I just woke and now I understood!

$ħ = \frac {h} {2π}$
That is the same as h = ħ2π

Just substitute h for ħ2π in the first equation, cancel 2π, ^2 and multiply the 1 with 2m...

Sorry to bother. Thanks all, funny how a night of sleep changes one's perspective.