- #1

LeonhardEuler

Gold Member

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## Main Question or Discussion Point

I've been thinking about how the requirement of anti-symmetry of the wavefunction is introduced in multi-electron problems and I am left puzzled over some aspects of it.

Various types of symmetry come automatically in classical physics. If you are studying water flowing in a cylindrical pipe, you know, if you think about it, that the velocity will not depend on [tex]\theta[/tex]. If you don't realize this, it's just an inconvenience and you will get the same unique solution when you solve the Navier-Stokes equations.

Apparently, this is not how it works in QM. I say this because antisymmetry of the wavefunction is introduced as an independent postulate from the Schrödinger equation in all the books I've read. Treating the Schrödinger equation as a boundary value problem like any other, it seems the solution(s) must be either symmetric or antisymmetric under interchange of electrons due to their indistinguishably. My understanding is that we must restrict ourselves to looking at the antisymmetric ones.

Now for what I find confusing. The variational principle is often stated by saying that the expectation of the energy for any wavefunction will be greater than or equal to the expectation for the ground state, with equality holding only for the ground state itself. This makes no reference to antisymmetry. Either the lowest energy wavefunction is antisymmetric without imposing any restriction, making the restriction unnecessary, or it is not antisymmetric, meaning either the variational principle or antisymmetry of the wavefunction is false.

I looked at a proof of the variational principle in a book I have, and it depends on expanding the arbitrary wavefunction as a linear combination of eigenfunctions of the Hamiltonian. Perhaps the way out of this is that the books I have just have an imprecise statement of the variational principle and it should be amended to say the "expectation of the energy for any

(e.g. f(x,y) = sin(x) - sin(y), break the 2 sin's into Taylor expansions, but when defining [tex]\phi_n[/tex]'s, mix and match different order terms, like

[tex]\phi_1 = x[/tex], [tex]\phi_2 = -\frac{x^3}{6} - y[/tex] [tex]\phi_3 = \frac{x^5}{120}+\frac{y^3}{6}[/tex] etc.)

Another bothersome thing is that the idea of the distinction between fermions and bosons seems a little artificial and unable to handle generality. By this I mean, consider the following scenario: It is discovered that there are actually two types of electrons that vary just extremely minutely in their mass or charge. If you have a helium atom with two types of electrons, you no longer expect perfect antisymmetry, but you expect the result to be extremely similar to the identical electron case because the situations are so similar. How do you actually express the way a fermion is different from a boson in this case, since there will now be neither symmetry nor antisymmetry? Is the difference really how they behave in cases of exact symmetry, or is there something more fundamental, or more physical to explain the difference?

Various types of symmetry come automatically in classical physics. If you are studying water flowing in a cylindrical pipe, you know, if you think about it, that the velocity will not depend on [tex]\theta[/tex]. If you don't realize this, it's just an inconvenience and you will get the same unique solution when you solve the Navier-Stokes equations.

Apparently, this is not how it works in QM. I say this because antisymmetry of the wavefunction is introduced as an independent postulate from the Schrödinger equation in all the books I've read. Treating the Schrödinger equation as a boundary value problem like any other, it seems the solution(s) must be either symmetric or antisymmetric under interchange of electrons due to their indistinguishably. My understanding is that we must restrict ourselves to looking at the antisymmetric ones.

Now for what I find confusing. The variational principle is often stated by saying that the expectation of the energy for any wavefunction will be greater than or equal to the expectation for the ground state, with equality holding only for the ground state itself. This makes no reference to antisymmetry. Either the lowest energy wavefunction is antisymmetric without imposing any restriction, making the restriction unnecessary, or it is not antisymmetric, meaning either the variational principle or antisymmetry of the wavefunction is false.

I looked at a proof of the variational principle in a book I have, and it depends on expanding the arbitrary wavefunction as a linear combination of eigenfunctions of the Hamiltonian. Perhaps the way out of this is that the books I have just have an imprecise statement of the variational principle and it should be amended to say the "expectation of the energy for any

*antisymmetric*wavefunction is greater etc...". But this is still a little unsatisfactory to me, because how do we know any antisymmetric function can be expanded in just the*antisymmetric*eigenfunctions of the Hamiltonian? Certainly antisymmetric functions can have expansions containing terms that are not themselves antisymmetric, as I can think of examples.(e.g. f(x,y) = sin(x) - sin(y), break the 2 sin's into Taylor expansions, but when defining [tex]\phi_n[/tex]'s, mix and match different order terms, like

[tex]\phi_1 = x[/tex], [tex]\phi_2 = -\frac{x^3}{6} - y[/tex] [tex]\phi_3 = \frac{x^5}{120}+\frac{y^3}{6}[/tex] etc.)

Another bothersome thing is that the idea of the distinction between fermions and bosons seems a little artificial and unable to handle generality. By this I mean, consider the following scenario: It is discovered that there are actually two types of electrons that vary just extremely minutely in their mass or charge. If you have a helium atom with two types of electrons, you no longer expect perfect antisymmetry, but you expect the result to be extremely similar to the identical electron case because the situations are so similar. How do you actually express the way a fermion is different from a boson in this case, since there will now be neither symmetry nor antisymmetry? Is the difference really how they behave in cases of exact symmetry, or is there something more fundamental, or more physical to explain the difference?