# Why 3n-dimensional function?

1. Sep 5, 2008

### Marty

One of the key points against the physical reality of the electron wave function is the argument that the function does not exist in ordinary 3-dimensional space, but rather in 3n-dimensional phase space where n is the number of electrons. I wonder why this needs to be so. Can someone comment on why the Schroedinger function does not work if we attempt to build up multiple-electron situations by superposition of wave functions in ordinary space?

2. Sep 5, 2008

### Maaneli

If you are working with a physical situation where the wavefunctions of the N particles have little or no joint support in configuration space, then you could certainly construct a 3N-dimensional wavefunction for N-particles that factorizes into N 3-dimensional wavefunctions. For example,

psi(x1,x2) = psi(x1)$$\otimes$$psi(x2).

Moreover, in the factorizable case, you can even use the nonlinear Hartree-Fock representation for the 3-dimensional wavefunctions interacting via their self-fields.

However if there is a physical situation where there is considerable overlap of N particles in configuration space, then the wavefunction is not factorizable and this is called an entangled state.

3. Sep 5, 2008

### Marty

I wonder if you could give an example of such an entangled state. I am thinking perhaps of any atom beyond Helium in the periodic table. Could you show why there is a problem with attempting to describe such states with simple 3-dimensional wave functions?

4. Sep 5, 2008

### Marty

...where I mean to interpret the square of the wave function as the charge density.

5. Sep 5, 2008

### Maaneli

The most famous example of an entangled state is the singlet-state. And I think you're right about entangled states in any atom beyond Helium (should be examples in Sakurai's or Schiff's textbooks). The reason there is a problem with attempting to describe such states with 3-D wavefunctions is therefore as I said: those wavefunctions share a common support in configuration space and therefore ar not factorizable. A slight caveat is that you could in fact construct a wavefunction with the center of mass coordinates, in which case it is factorizable after all, even for entangled states (see Holland's text "The Quantum Theory of Motion"). To your follow-up comment, I need you to clarify why you are asking that, as it seems like a separate issue to me.

6. Sep 5, 2008

### Marty

I can think of two singlet states: the H2 molecule and the helium atom. I think these are fully described by saying that both ground state wave functions, spin up and spin down, are fully occupied. So I don't need any 6-dimensional functions in these instances. Would you say I'm right?

I have to say I find your reason excessively formal. I am looking for a more physical reason. For example, in the case of Lithium, if I take the electrons to be occupying the two s1 orbitals (up and down) plus a third electron in a 2p orbital....do I get the wrong charge distribution when I add the wave functions and then take the square?

That would be an example of what I would call a physical reason.

I can't comment on your caveat but I hope I've explained what I'm trying to ask.

7. Sep 6, 2008

### atyy

Even Newtonian mechanics does not exist in 3-D space. To fully describe a system, we have to know each particle's position (x,y,z) and momentum (px,py,pz). So 1 particle exists in a 6-D space. N particles exist in a 6N-D space. Anyway, I guess from your argument, it's a good thing we no longer believe in Newtonian mechanics! Can you honestly say point particles are intuitive? I did like absolute space though (and still do).

8. Sep 7, 2008

### Marty

I'm still hoping someone can give a physical example of the practical difference between 3-dimensional vs 3n-dimensional wave functions. Why for example don't we describe the electron cloud around a Lithium atom as the superposition of two s orbitals and one p orbital?

9. Sep 7, 2008

### Haelfix

Already in classical (wave) mechanics, you can have >3 dimensional configurations space. For instance a classical gauge theory.

The real difference between quantum mechanics and classical mechanics is the interpretation of the observables and operators that underlie the dynamics. After quantization, we are dealing with probabilities (so the square of the wavefunction).

You can formulate classical mechanics in a way that also deals with probabilities, however then you don't have superposition.

So no matter what, there is something extra.

10. Sep 8, 2008

### atyy

Density functional theory does this, and is often computationally efficient, but I don't know the extent of its validty:
http://www.physics.ohio-state.edu/~aulbur/dft.html
http://www.unc.edu/~shubin/dft.html