Pauli Exclusion Principle

Surely many will know or at least have heard the principle of exclusion of Pauli. Yes, that which says that “two identical particles can not simultaneously occupy the same state” or, more precisely, “two identical fermions can not occupy simultaneously the same quantum state”. Something that some have said that two particles can not be in the same place and others say they can not have the same wave function.

Now, why?. Although it may sound complicated is the logical sequence of solving Schrödinger’s equations (what a difficult name) for two identical particles.

I’m going to assume that some quantum mechanics you know, or at least the basics and that you do not work by hearing. That you know that every particle has a wave equation that gives us the probability of finding it in an area of ​​space and that thanks to this wave equation we know its energy and, more important, its wave function. And that the energy calculation, from Schrödinger’s equation is based on potential energy.

Now imagine the problem of the infinite well where we consider an area of ​​potential 0 (potential energy) with two walls (at x = 0 and x = L) where the potential energy is infinite and, whatever we get there, there he remains for much energy.

This imaginary problem is the basis for learning to calculate and solve the wave function through a differential equation drawn from the linear application which is the Schrödinger equation (if all this sounds complicated, read it slowly).

If in the problem of the infinite well we put two particles and add that these do not interact with each other, we will have Schrödinger say:

$latex -\frac{h^{2}}{2m}(\frac{\partial^2 \Psi (x_{1},x_{2})}{\partial x_{1}^2} + \frac{\partial^2 \Psi (x_{1},x_{2})}{\partial x_{2}^2})+U\Psi (x_{1},x_{2})=E\Psi (x_{1},x_{2})$

Where we have the coordinates of the two particles that have the same mass.

As the particles do not interact with each other (what a luck), we can say that:

$latex U=U_{1}(x_{1})+U_{2}(x_{2})$

Something that helps us solve the Schrödinger equation using the boundary conditions where U=0 when x=0 or x=L where, we go at the edges of the well.

By pulling the skein, we obtain that:

$latex \Psi_{nm}=\Psi_{n}(x_{1}\Psi_{m}(x_{2}$

That is, the wave function is (because it is probability) the multiplication of the probabilities of each wave function of each particle separately. That is, if the particle n (the first) is at energy level 1 and the particle m (the second) is at energy level 2, the wave function would be.

$latex \Psi _{12}=Asen\frac{\Pi x_{1}}{L}sen\frac{2\Pi x_{2}}{L}$

But, if instead of the wave function, we use probability (to normalize) where, we know that:

$latex P=\int \Psi\overline{\Psi}dx=\int \Psi^{2}dx$

And because we can not distinguish particle 1 and 2 because they are identical we will have to:

$latex \Psi^{2}(x_{1}, x_{2})=\Psi^{2}(x_{2}, x_{1})$

Wow! so, what can we do now. So simple, we construct some new functions that we will call:

$latex \Psi_{S}=A(\Psi_n(x_{1})\Psi_m(x_{2}+\Psi_n(x_{2})\Psi_m(x_{1})$
$latex \Psi_{A}=A(\Psi_n(x_{1})\Psi_m(x_{2}-\Psi_n(x_{2})\Psi_m(x_{1})$

Where the difference of the sign will tell us if the functions are symmetric or not since if it is antisymmetric the wave function is 0 and if not zero. Resolved!.

Thus, if the two particles are identical, the wave function is antisymmetric, and since the quantum number of the particles can not be the same, it is zero.

And all this roll that comes ?. Very simple, considering the last formula with the last conclusion, two particles can not have the same quantum numbers, since in that case their wave is zero. And if it is null it can not be given. This is Pauli’s exclusion principle mathematically told and interpreted.

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