"Derivation" of the Schrödinger Equation

Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
4 replies · 2K views
BearY
Messages
53
Reaction score
8
When reading a textbook I came across some reasoning about Schrödinger Equation.
It started with the wave function $$\nabla^2\psi=k^2\psi$$
I am a bit lost at this point. Where does the right side of the equation come from? What should I review to fix that part of my knowledge?
 
Physics news on Phys.org
Try to find out this equation is for Schroedinger Wave equation for what? If you also give us what is written prior to this and beyond this we may be able to help you better.
 
Let'sthink said:
Try to find out this equation is for Schroedinger Wave equation for what? If you also give us what is written prior to this and beyond this we may be able to help you better.
The reasoning says this is the spatial part of the wave equation, which to my understanding, is $$\nabla^2\psi$$.
After this, we have $$k=\frac{2\pi mv}{h}$$
and $$mv^2 = 2(E−V)$$
and thus $$\nabla^2\psi = \frac{8\pi^2m}{h^2}(E-V)\psi$$
and rearrange to get the $$E_k + V = E$$

To my understanding, this reasoning just to show that Schrödinger Equation is indeed a wave function that was talked about earlier in the book.
 
BearY said:
The reasoning says this is the spatial part of the wave equation, which to my understanding, is $$\nabla^2\psi$$.
After this, we have $$k=\frac{2\pi mv}{h}$$
and $$mv^2 = 2(E−V)$$
and thus $$\nabla^2\psi = \frac{8\pi^2m}{h^2}(E-V)\psi$$
and rearrange to get the $$E_k + V = E$$

To my understanding, this reasoning just to show that Schrödinger Equation is indeed a wave function that was talked about earlier in the book.
Read in between the lines and try to answer my question this equation is wave equation for what?
Also make a comparison with the differential equation for a classical plane wave.