Proving Real & Imaginary Parts of Complex Wavefunction

[reilly]

Complex numbers and variables have been part of physics, engineering, chemistry, biology for a long time. Why? The best way is to study the math of complex variables and see for yourself. Complex variables provide a depth of analysis hard to achieve with real variables alone. Things like contour integration, conformal mapping and 2-D potential theory are basic in the physics toolbox. You develop intuition about complex variable things by working with them-- over time by continuing to work with them, you will begin to view their use as second nature. So, why use vector spaces, or groups, or other mathematical approaches? They help get the job done. .

What's the job? The job is what physicists say it is -- sometimes physicists are more abstract and mathematical than mathematicians , or were during the heyday of axiomatic field theory. Sometime's its strictly back of the envelope, like Fermi's computation of the TNT equivalent of the first A Bomb at Almogordo -- or his famous orals question: how far can a bird fly?

Physicists do not form a monolithic community, albeit there are certain common threads among the subcultures or schools or groups... The point is, approaches to math can be all over the map, approaches to research and methodology can be all over the map -- ultimately, at least for the professional, it's a matter of style, convenience and practicality.

And, with regard to rigor, it took quite a few years for the mathematicians to catch up with Dirac and realize the delta function is cool. Intuition is a powerful tool, just as is rigorous logic and math. The style issue: when and how do you blend the use of these tools?

Complex variables expand the language and tool set of physics--they are here to stay.

Regards,
Reilly Atkinson

 

[caribou]

I`m not saying the result is wrong or questionable. It's very plausible if you physically interpret this answer.
I find the derivation quite horrid. Things are done I was told that weren't allowed, like treating dm/dt as a fraction. It would be much more elegant to set up a differential equation and solve it. This doesn't even have to be done in a physics class, but in a lecture on DE's.

I`m sure I can think of more examples, but something like this make me go: -< (Is this kosher?)

Yeah, I also like to try and do things right as well, like not using a "dt" as a "delta t" for example.

Infinitesimals are another one, with the "infinitesimals" of scientists and engineers actually being ficticious concepts in the branch of mathematics they use.

 

[Hurkyl]

That's not necessarily true -- there are nonstandard models of analysis that provide honest to goodness infinitessimals.

(But they do take come care to use properly. e.g., a ratio of infinitessimals is only infinitessimally close to the derivative, not exactly equal)

 

[caribou]

Yeah but as I say, they're not using that branch of mathematics.

 

[selfAdjoint]

Yeah but as I say, they're not using that branch of mathematics.

It's a model of what they're doing. They could represent their equations and integrals in terms of nonstandard analysis if they chose to.

 

[Sterj]

A question, If a particle has a wave function like this f(x,t)=e^i(kx-wt) ,is then the probability to find the particle = |f(x,t)|^2?

 

[dextercioby]

No,that's the probability density...It's 1,which means that the wave-function,non square integrable Lebesgue,does not describe a physical state of a quantum system...

Daniel.

 

[Sterj]

that means, if f(x,t) want's to be a wave function |f(x,t)| must be equal to 1. And then the probability density is the integral(ff*dV)

 

[dextercioby]

What?The probability density that the quantum system be found at the moment 't' in the point \vec{r} is \mathcal{P}=|\Psi(\vec{r},t)|^{2} and that's that...

Daniel.

 

[Sterj]

ahh, sure, it has to be so.
A particle can be expressed as a wave, but why the hell it can also be expressed as an oscillator, I mean an oscillator isn't really the same thing as a wave. And why is the ground state energy of a harmonical oscillator the 0-point energy of a particle?
thanks

 

[Raparicio]

Outer product

We have a complex vector space equipped with an inner product, complete with respect to the norm defined by the inner product (i.e. the Hilbert space). Elements of it are vectors. And since it is a complex space, each vector can be multplied by a scalar complex number.
Masud.

Has HIlbert Space an outer product (grassman)? Could it have this?

 

[dextercioby]

Nope,it doesn't.A Hilbert space is what it is.An inner product Banach space.

Daniel.