Quantum vs. Classical Mechanic graphing

[terp.asessed]

Hey, I am curious if there's a correspondence between Classical and Quantum Mechanics graphs in terms of Potential (or kinetic) Energy as a function of x, aside from equations?

 

[ShayanJ]

In classical mechanics, particles cause potentials and particles are also affected by potentials. So there some kind of a consistency between them because a particle which is affected by another particle's potential field, can itself have a similar potential field. There are only some cases where you can have potentials which aren't related to some kind of a usual particles configuration, like a uniform electric field which can be caused only by charges at infinity.
But in QM, things are different. The potential used in the Schrödinger's equation is the same as the potential used in classical mechanics and so there is no difference in that sense. But if you want to know the electric potential field of an electron, then things get different. You should use the modulus squared of the electrons' wave function(times -e) as the charge density and find the electric potential but that depends on the wave function's form and so things are very different from classical mechanics here.So different that there can be no comparison in the way you want! Also the consistency I mentioned in the case of classical mechanics isn't present here and I think that's one important reason that pushed physicists for formulating Quantum theories for fields.
In QFT things are even more different!

 

[terp.asessed]

Thanks for the info! Btw, if you don't mind, could you pls expand on:

The potential used in the Schrödinger's equation is the same as the potential used in classical mechanics and so there is no difference in that sense.

I thought that it only applied in the case where energy level (n) in the QM is very large to the point the wave behaves more like Classical than Quantum Mechanics?

 

[ShayanJ]

I think its better to explain in using an example. Consider a particle in the potential V(x)=\frac{1}{2} k x^2. As you can see, there is nothing here that tells us we want to do it classically or quantum mechanically. That's exactly what I mean. The procedures, equations, interpretations and solutions are different, but the potential is the same!
You use V(x)=\frac{1}{2} k x^2 for solving the classical problem and when you want to solve the quantum mechanical problem, you use the same thing and you do nothing to make it quantum mechanical!

 

[terp.asessed]

Ok, thanks!