Why do we associate the energy eigenstates with the "wavefunction" and "position"?by jeebs Tags: associate, eigenstates, energy, position, wavefunction 

#1
Apr1511, 09:44 AM

P: 326

This is something that has bothered me for a long time but that I never got around to asking. I suspect I'll feel like an idiot for not knowing this but anyway...
One of the first things I was introduced to in quantum mechanics courses is the timeindependent "schrodinger" equation, or in other words, the "energy eigenvalue equation", [tex] H\Psi = E\Psi [/tex], where [tex]\Psi[/tex] are the eigenfunctions of H. We were also introduced to the idea that you could take this object [tex] \Psi [/tex], square it, and that would give you the probability of finding your particle at a given position. [tex]\Psi[/tex] gets called the "wavefunction" but I notice that the use of that word seems to dry up when you start getting into doing your quantum mechanical calculations using Dirac notation. I have never understood where this dual use of the energy eigenstate [tex]\Psi[/tex] came from, and it seems like it should be an important thing to know. I mean, we have our set of observables like momentum, energy, etc. and their corresponding operators, eigenstates and eigenvalues  why is it the energy eigenstates in particular that we take as our "wavefunction"? We could find different eigenstates for any operator that does not commute with H, right? Why isn't our "wavefunction" the eigenstate of some other operator? This can't have been an arbitrary choice? 



#2
Apr1511, 12:20 PM

P: 788

A quantum mechanical system can have a state that is not an eigenstate of H. But if you have some Hermitian operator you can express any state as a linear combination of eigenstates of that operator. So for instance if we learn everything there is to learn about eigenstates of H, then we can form any state whatsoever as a linear combination of eigenstates of H, and so we know all there is to know. You are right in pointing out that H is not special in this way, but we want to know what the eigenstates of H are because these are the energy levels of the system, which are important.
I think at this level you should make a distinction between the state and the wavefunction. The state is some vector in Hilbert space, written like [tex] \psi \rangle[/tex]. The wave function (in position space) is a function [tex]\psi(x)[/tex] which is shorthand for what in Dirac notation is written [tex]\langle x  \psi \rangle[/tex] where [tex] x \rangle[/tex] is a position eigenstate. This then is a complex number whose squared magnitude gives the probability density to find the particle at x. Again the position basis is not unique, you can define a momentum space wave function [tex]\psi(p) = \langle p  \psi \rangle[/tex] whose squared magnitude gives the probability density for a momentum measurement to give the value p. 



#3
Apr1511, 12:24 PM

Mentor
P: 11,225





#4
Apr1511, 12:36 PM

Sci Advisor
P: 1,867

Why do we associate the energy eigenstates with the "wavefunction" and "position"?
In the early days of QM, there was Schrödinger, who worked with the wave function. "wave mechanics/Wellenmechanik") Then there was Heisenberg, who worked with matrices. Dirac tied it all together with his operator formalism. Schrödinger was originally concerned with electrons in atoms. His original interpretation of the wave function's square was as the electron density around the atom. So that's basically where it came from.
So the question: "What's so special about the energy/position basis?" is a good one. There isn't anything special about them. You can of course form any number of hermitian operators with their own corresponding basis. It's basically just a historic artifact of the original problem they were tackling. The whole 'waveparticle duality' thing is also an artifact of that. That said, the position/momentum basis is still usually the most convenient one if you're trying to solve the wave function for electrons in atoms. But from the modern viewpoint, there's indeed nothing special about it. 



#5
Apr1511, 02:21 PM

Sci Advisor
PF Gold
P: 1,942

However, in as far as chemists are interested in the charge distirbution, they need to convert whatever they get at the end into a charge density, and this is a positiondependent field. The latter is not far from Schroedinger's view  see the entry ''How do atoms and molecules look like?'' in Chapter A6 of my theoretical physics FAQ at http://www.mat.univie.ac.at/~neum/ph...faq.html#atoms 



#6
Apr1511, 03:13 PM

Sci Advisor
P: 1,867

The mathematical functions used to approximate the spatial part of the spinorbitals are arbitrary and have no real physical significance. Gaussians are used out of purely computational considerations. The fact that they're a eigenbasis for the harmonicoscillator Hamiltonian is useful if you want to illustrate that they form a complete set, and therefore are suitable to use as basis functions, but that's about as far as that extends here. There are methods in general use (and many more that have been experimented with) that make use of other basis functions, such as ordinary exponential functions and even splines. Not to mention that in DFT methods, the basis functions don't represent the wave function of any real thing at all. Since there are methods that do use actual gaussian wave packets (e.g. this) to model electron transfer in complexes, I think you're liable to cause a lot of confusion here. 



#7
Apr1511, 03:19 PM

P: 326

Don't get me wrong, I'm all too familiar with treating particles as waves when it's useful to, but when I think of an electron bound to a nucleus, a wave is not the word I would use... unless the electron orbitals do go through oscillating changes in shape for things more complicated than the H ground state? 



#8
Apr1511, 04:05 PM

Sci Advisor
P: 1,867

The hydrogen atom has states which are spherical harmonics (apart from the radial exponential function), which are in fact just a 3dimensional standing wave when confined to a sphericallysymmetric 'box'. Although I believe it's called the 'wave function' more out of analogy to the classical dispersion relation than to the classical wave equation. While the hydrogen wavefunction corresponds directly to the electronic density, the wave function of any manyparticle system doesn't really exist in 'real' space, but a multidimensional 'configuration space', because it's a function of all coordinates of all electrons. E.g. helium would be Ψ(r_{1}, r_{2}) with the coordinates of both electrons, its square giving the probability of finding electron 1 at point r_{1} simultaneously with electron 2 at point r_{2}. So to get the actual electron density (the probability of finding any electron at some point in space), you have to integrate the square of the wave function over all of space for all coordinates but one. (Or in other terms, you have the 1particle density, the twoparticle density and so on) The 'probability cloud' description can be a bit misleading, since it's easy to forget that the electrons are still moving (in some, nonclassical sense), even if the resulting particle density does not change with time. Otherwise you might ask why we'd need quantum mechanics to begin with, since classical electrostatics doesn't actually have any problems modeling a static chargedensity cloud. 


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