Wavefunction of a 2-particle system.

[carbon9]

Hi everybody,

I'm studying Hartree-Fock theory. It is written that, in Hartree approach, for the two electrons in positions r1 and r2 (these are vectors, of course), the joint wavefunction of this 2-particle system is given as

Psi(r1, r2)=Psi1(r1) * Psi2(r2)

I'm confused a lot at this point. I have a basic knowledge of quantum mechanics and as much as I know, the wavefunction of a particle is a function of spatial coordinates (r), as in infinite square well, the wavefunction is given as

Psi(x)=sqrt(2/L) * sin(n*pi*x/L)

here, wavefunction is a function of x. It can be plotted as a function of position x and the squared modulus of wavefunction is also a function of x which gives the probability that we find it there.

Let us turn back to the 2-particle case. The wavefunction of the 2-particle system is Psi(r1, r2) that seems to be meaningless to me since there is only one coordinate system and the wavefunction had to be given something like

Psi(r)=somefunction( Psi1(r), Psi2(r)) (All Psi's are a function of independent variable r)

but it is given as

Psi(r1, r2)=Psi1(r1) * Psi2(r2)

r1 and r2 being the coordinates of particle-1 and particle-2.

Could anybody please give an explanation of this situation? We do not know r1 and r2 exactly in quantum mechanics. But how can we write the wavefunction of 2-particles as a function of r1 and r2?

 

[Hurkyl]

Could anybody please give an explanation of this situation?

What you have observed is an algebraic statement of what makes quantum mechanics different from classical mechanics.

 

[shouvikdatta8]

Dear carbon9,

I don't know anything about the Hartree-Fock theory. But, I've studied QM in my undergrad course.

First, consider what a wavefunction means. Its a function which gives info about the state of the system. For, single particle position-space wavefunctions, (like the ones you were talking about here) their modulus squared gives the probabililty of getting the particle between x and x+dx (or in 3D its r and r+dr.) Similarly, the modulus squared of 2-particle wavefunctions must give the probability of finding one particle in the range r1 & r1+dr1 AND the other particle in the range r2 & r2+dr2. In statistical lingo, AND = multiplication. So, here you just multiply 2 probabilities to get a combined probability.

I hope the above explanation will answer your question. The mathematical operation is consistent with the physical reasoning.

Regards,

Shouvik

 

[carbon9]

Dear Hurkyl and shouvikdatta8, thanks for your answers.

I think I've now understood something. I now think something like the picture attached to this message. Is this picture correct?

Psi(r1, r2, ..., rn) gives the combined "AND"ed probability.

Regards

 

[Dr Transport]

Dear Hurkyl and shouvikdatta8, thanks for your answers.

I think I've now understood something. I now think something like the picture attached to this message. Is this picture correct?

Psi(r1, r2, ..., rn) gives the combined "AND"ed probability.

Regards

Now you're getting there...

 

[shouvikdatta8]

Hi carbon9,

I don't see the reason why you need to consider a 3n-dimensional system. 3 coordinates are just sufficient to represent position. Also, in Quantum Mechanics you can't just pin-point positions like r1,r2.. etc. You'll have to consider an infinitesimal volume about r1,r2.. r_n, (in accordance to uncertainty requirements) so that the 'i'th particle lies between r_i and r_i + dr_i.

Regards,

Shouvik.

 

[carbon9]

Hi Shouvik,

Since I'm studying n-particles, the wavefunction is said to be of 3n-dimension? Is this wrong?

Regards

 

[Dr Transport]

Hi Shouvik,

Since I'm studying n-particles, the wavefunction is said to be of 3n-dimension? Is this wrong?

Regards

No, but to get your head around this concept, try thinking about 2 particles in 1-d then move forward.

 

[atyy]

Because of the problem you mentioned, for two (or more) identical particles, the wavefunction must be symmetrized/anti-symmetrized according to whether the particles are bosons are fermions. Bosons is obvious for me - you can interchange positions without changing the wavefunction. Fermions are weird, but not so weird if I remember that the wavefunction is only the probability amplitude, not the probability. Apparently there's a third sort of particle called the anyon in 2D...

 

[shouvikdatta8]

Observe, \Psi(r1,r2) ... since r1 and r2 are vectors, they have 3 components each and hence the wavefunction is 6-dimensional for 2 particles. However, the position space as you had considered in your diagram must be of 3-dimensions.

 

[carbon9]

Thanks for replies.

I think my problem lies in the misunderstanding the dimensions of functions. OK, wavefunction of 2-particles is 6-dimensional. But now consider a system of 2-particles in the real space with the following wavefunctions:

1. Psi1(r1)
2. Psi2(r2)

Then, as much as I know, the squared modulus is the probability of finding a particle at coordinate r1+dr1 and r2+dr2 for particle 1 and particle 2, respectively. OK?

But when it comes to think particle-1 AND particle-2, I still get confused. The wavefunction of 2-particle SYSTEM is

Psi(r1, r2)=Psi(x1, y1, z1, x2, y2, z2)=Psi1(x1, y1, z1)*Psi2(x2, y2, z2) is this right? I think so.

Theeen, let us come to the crucial point. What is the probability of finding particle-1 at r1+dr1 AND particle 2 at r2+dr2?

Probability density function(x1, y1, z1, x2, y2, z2)=|Psi1(x1, y1, z1)*Psi2(x2, y2, z2)|^2 ? If this is right?

Vital question is: How can we plot the wavefunction of 2-particle system?
Another: How can plot the probability density function of 2-particle system in 3-D? Since, if my expression of probability density function is correct, then it is also 6-dimensional and we can not plot it?

 

[atyy]

The wavefunction of 2-particle SYSTEM is

Psi(r1, r2)=Psi(x1, y1, z1, x2, y2, z2)=Psi1(x1, y1, z1)*Psi2(x2, y2, z2) is this right? I think so.

Theeen, let us come to the crucial point. What is the probability of finding particle-1 at r1+dr1 AND particle 2 at r2+dr2?

Since the particles are identical, the question doesn't make sense, as you expected. The wave function has to be symmetrized for bosons, or anti-symmetrized for fermions.
http://farside.ph.utexas.edu/teaching/qmech/lectures/node59.html
http://galileo.phys.virginia.edu/classes/752.mf1i.spring03/IdenticalParticlesRevisited.htm

 

[shouvikdatta8]

You needn't worry about symmetry of wavefunctions for the time being. You CANNOT plot wavefunctions of 2-particle system in 3D because it's a function of six variables (or coordinates.) You can just imagine. . .
You are right in the other cases...

 

[reilly]

Think about throwing two baseballs from x=y=z=0. Throw many times in any direction you want, then you will have an event space, of ball1 at r1, and ball 2 at r2. As in(let z=0)


Ball1 Ball2
Throw

1 (5,2,0) (3,7,0)

2 (2,2,0) (5,8,0)

3 (4,9,0) (6,7,0)

and so on. There are two types of probabilities here; one for single events for ball1 and ball2, and for joint events.(This is standard probability theory.)

That is, there are probability: Prob(r1=(x1,y1,0) and Prob(r2=(x2,y2,0). Here these two single event probabilities describe independent events, and each deals with the coordinates of a single ball. These coordinates refer to the same space but to different balls; they refer to measured variables -- events as they are often called in probability theory.

Then there is the joint probability, Prob(r1=(x1,y1,0) and Prob(r2=(x2,y2,0), simultaneously, which is given here by

Prob{ r1=(x1,y1,0) x Prob(r2=(x2,y2,0)|},

a function of 6 variables, with values in the same space.

Now suppose that the tosses are not independent; after a throw of ball1, the thrower must throw ball2 in a direction + or - 30 degrees from the first throw. The product form no longer holds, which is usually the case in QM. Then, if we let the probabilities = wave function squared, the probability that ball1 is at s1 and ball2 is at s2 is given by

|Psi( r1 = s1; r2=s2|^^2.

Note this approach is constantly used in scattering theory, for both single particle detection, say the electron only in electron-proton scattering, and coincidence detection in which both particles are detected.

If you use some fancy computer graphics, like 3-d plots then you can plot multiparticle wave functions or probabilities --similar plots are often found in books dealing with special functions, with pre-computer-age plots in the form of contour maps. By hand, with a few crayons or colored pencils, you ought to be able to do your own plots -- try
psi= Sin(|r1+r2|), for example -- sorry about the lack of normalization. -- where r1 and r2 are 3- d vectors.
Regards,
Reilly Atkinson

 

[carbon9]

Dear atyy, shouvik and reilly,

Thanks for your answers. The links given by atyy were quite useful in understanding symmetric and antisymmetric wavefunctions and Pauli exclusion principle.

Also, I think I've understood that shouvik said, that we can not plot a 6-D function in 3D space. OK.

I thank to reilly also, especially the review of probabilty theory was a must for QM as he pointed. But, at reilly's post, I could not find a way for plotting psi= Sin(|r1+r2|) when r1 and r2 are vectors in the 3D-space. For example, I tried to plot in in MATLAB but I could not unfortunately. If there's an example plot on Internet, I'd like to look at it so, I can I think understand better.

Regards,

 

[adolescent00]

You're missing the point of the Hartree Fock approximation. You were correct in thinking that, in a real system, the position of an electron is affected by the positions of all the other electrons (coloumbs law). The result is a series of overlap intergrals, usually called Haa or Sab, produced, from the hamiltonian operating on the wave fuction, when calculating the expectation value of the energy. The hartree fock model simplfies this by using the following approximation; when calculating the energy of a system, an electron is a fixed particle and all othe electrons move independently of this fixed electron. Hence you can use separation of variables, and write the wavefunction as Psi(r1,r2)=psi(r1)*psi(r2).

 

[carbon9]

Thanks adolescent00.