Question

18. Every electron in the universe is the same as any other electron. It is a symmetry of physics, that if you interchange any pair of electrons, the Schrödinger equation is invariant. Consider one pair of electrons which we call electron-1 and electron-2. Define the interchange operator P12 to simple interchange the two electrons. H4142 = EU142 HP120142 = EP120102 a) Use the above equations to show that the P12 operator commutes with the Hamiltonian. b) Use that fact that if we operate twice with P12 we get back to the same state, to show what the eigenvalues of the P12 operator are. c) The overall state of two electrons must be antisymmetric under interchange. Usually, we can factorize the spin state of two electrons from the spatial state: y = Uspin Uspace. Since the overall state of two electrons has to be antisymmetric under interchange, we can have a symmetric spin state with an antisymmetric space state, or vice versa. Lets delve into the spin states of two electrons. Its very common to use arrows to designate whether the spin of an electron is up or down since these are the only two states of an electron and to use the first arrow for electron-1 and the second for electron-2. So, for example, the state says that electron-1 has spin up and electron two has spin down. Since there are only 2 spin states for any electron, there are just 4 states of two electrons: 11 th, 1, and H. Of these four states, which are symmetric under interchange and which are antisymmetric. d) From the two states that are neither symmetric or antisymmetric, make one linear combination that is antisymmetric and one that is symmetric. Multiply by a constant to normalize these states to one pair of electrons. e) Write the properly-normalized, symmetric-under-interchange spin state(s) of two electrons? f) What is the properly-normalized, antisymmetric-under-interchange spin state of two electrons?

Answer 1

HY, Y₂ = EY, Y₂ . HP2 Y Y ₂ = E Piz Y, Yz. where Piz is interchange opesatous, [H, PJ Tila. = = = (using the H Piz Y, Y₂ - Piz B HY Z E Piz Y, Yz - Pia (E Y Y₂) Epiz Y Y Z - EP 12 Y Y . relations]

So lie. b). his = Piatt [ H, Pia] T K = 0. H and Piz commutes, and Now, HAF = E Tz ls. Pia (4 th 2) = € Piatt. . the electrons) So, Piz ( 4, 42) = Y Y, (iner interchanging & Piz [ Piz (Y, G)] = Piz ( P2 %) = Piz? (4,142) = Ki Yaa * Piz? C4, 43) = 1 M Pan So, I is eigen value of P . so, Piz has eigen values T = ± 1. Now, to the states using all orientations of spins are 117), Itty 1+"), Itt), : P12 191) = 109). (no change in state) so, 17th state is symmetric This is not an, eigenstate equation. So, let is. not an eigenstate

This is also not an eigenstake equation. in Pielst) - 1st. (same state ofter interchange) So, this is symmetric state Hits d. lots, tas using this two states and taking lineas combinations - * l Ispin) = 1 +7 + 14t) Pia (Ispin) = Piz [ Itty + HAY] - Piz Itty + Piz (+7) = 1HT) +17+). F So, Ispin) = 174) + Hity is a synneksie stake. If a be the noumalization constant then a Ispin) = [ 1767 +147) < Tspin / Espin) = [ {1+1 + (+11] [[Ass that 1 - 02.[ <74174) + (+11473] as, 1947 and 14t) are orthogonal so (70147) - <+1171) 0.

. Espir à 1/2 [ 174 + 16A)], Sinibaly, taking a sign het a Yspin - 14) - 1417. Piz tspin= Piz [lity - 1+1)] = Pizlo) - Piz 147). = 17) - 17+). | -[174) - 1+7)] so, tspen is antisymmetrie. using nosmalization - - tspin= / [ loty - 147)] es properly nosnalized symmetric under interchange spins states are D 171), la 16), and Yspin = [ Its)+HA)] f) Propedy normalized antisymmetric spin functiona Espin- / [ 1843-1+1)]