Computational Chemistry 4.6 - Anti-Symmetry Principle

Short lecture on the anti-symmetry principle for electronic wavefunctions.
One of the primary problems with approximating an N-electron wavefunction as a Hartree product of one-electron spin orbitals does not comply with the antisymmetry principle: wavefunctions must switch signs under the exchange of any pair of electrons. The antisymmetry principle fixes the problem that the molecular Hamiltonian does not contain any spin dependence, and all spin-dependence can be embedded within anitsymmetry requirements. Antisymmetry can be achieved by using a linear combination of all possible permutations of electrons and spin orbitals with a sign swap for every index pair exchange. Additionally these permutations must be normalized with the number of terms.
Notes Slide: i.imgur.com/Xj...
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[@user-to3fx2do4d]

Hello,sir. Generally, the anti-symmetric wave function is not the eigenfunction of the system. Example. Please suppose the system composed of two electrons. Electron 1 is in a hydrogen atom. Electron 2 is in a helium ion He+. rA, rB:the position of each nucleus. ❘rA-rB❘>>1. H1=(p1^2/2m)-(e^2/4πε0❘r1-rA❘), H2=(p2^2/2m)-(2e^2/4πε0❘r2-rB❘), H1φA(r1)=EAφA(r1), H2φB(r2)=EBφB(r2). Hartree product φA(r1)φB(r2) is the eigenfunction of operator H1+H2, and lead us the energy of the system EA+EB. But, another product φA(r2)φB(r1) is not the eigenfunction of operator H1+H2. (Please calculate it. It's very easy.) Therefore, the anti-symmetric wave function ψ=φA(r1)φB(r2)-φA(r2)φB(r1) is not the eigenfunction of the system.

[@orion4049]

You said the antisymmetry principle is a postulate, but that's not true though? What I know is that the permuted wavefunction only changes by a factor c, and it's easy to prove that c can only be +1 or -1 (and it is -1 in this case as electrons are fermions)

[@TMPChem]

The fact that electrons are fermions is an inbuilt assumption in your conclusion, and is equivalent to the first statement. It is indeed straightforward to demonstrate that c^2 = 1. It's impossible to say whether c is +1 or -1 without an assumption. That assumption is the antisymmetry principle.

[@theorangebox8464]

Why does it need to be antisymmetric? I.e., why isn't psi(1,2)=psi(2,1)?

[@starship9629]

Pauli exclusion principle

[@seitdrs2467]

Hi, the yellow equation left at 2:46 you mention that 1 and 2 are just 'dummy', I did not understand that part. Could you please expand on that.

[@TMPChem]

When I say "dummy indices", it means that 1 and 2 are arbitrary labels, and I could have just as easy switched which one is 1 and which one is 2. Also, the choice is independent for each integral, meaning I'm allowed to exchange 1 and 2 for some integrals, but not others as I please. This can be used to demonstrate equality of some terms which initially don't look equal until the dummy index exchange occurs.

[@youngtaepark2166]

@@TMPChem I am also very confused about that part. What I understand is that in the yellow equation, 1 and 2 are dummy labels and in below sky blue equations, 1 and 2 are not dummy labels. Did I understand this correctly? If this is correct, I would appreciate it if you could explain why.

[@solsticetwo3476]

박영태 - In my opinion, the mistake is in the second equality of the yellow equation. I mean, phi1(2).phi2(1)=phi1(1).phi2(2), is wrong. If it were true, then, the blue equations would be both zero. So, since it is not true, then, [phi(1,2)=phi1(1).phi2(2) ] =/= [phi1(2).phi2(1) = phi(2,1)]. But the difference is not because of the sign (one is not the negative of the other, which is the anti symmetry requirement). While, defining the composite wavefunction as a linear combination, they are now different and opposite each other: phi(1,2)= -phi(2,1). The coordinates (electron identity) cannot be dummy variables in this context (you need to preserve r1, r2, etc). That is why is needed a composite function that considers the "distinguishability" and uses it for the desired outcome. The Slater determinant fulfill that requirement. The result makes the electrons indistinguishable in the sense that each of them has to be evaluated (permuted) across all the wavefunctions.

[@gooblepls3985]

@@solsticetwo3476 I agree 100% on this.

[@Irrazzo]

For Psi(1,2,3) on the right, if one is adding a (-1) prefactor for each pairwise permutation, shouldn't the sign sequence be: + - - + + -, instead of + - + - + -? Since the sequential pairwise permutation count is 0,1,1,2,2,3. Great video series, thank you very much!

[@TMPChem]

Hi Irrazzo. You are correct. The signs of the middle two terms should be switched. Each term gets a positive sign if its permutation was achieved with an even number of pairwise swaps and vice versa for odd.

[@theultimatereductionist7592]

You incorrectly wrote -Psi1(2)*Psi2(3)*Psi3(2) when you should have written -Psi1(2)*Psi2(3)*Psi3(1). You wrote "2" for the argument of Psi3 when you meant "1".

[@rimjhimshukla365]

How do we get the intuition that there would be N! terms for a system of N particles?

[@gooblepls3985]

This is just due to N! describing how many permutations of a (finite, discrete) set of objects exist: If you have N objects that can be arranged in some order, you first have N choices for which object is put in the first position. After making this choice final, you have (N-1) choices for which object is put in the second position, and so on, up until there is only one object left, where there only exists one option (putting it in the final position). So overall you get N*(N-1)*(N-2)*...*1 = N! Since we have to consider all possible orders of N electrons, we get exactly [the number of permutations for N electrons] terms.

[@apburner1]

First.