/sci/ - Science & Math

>>11517521
>masses for the second and first generations are produced respectively by one-loop and two-loop radiative corrections
Wtf why?

>>11466294
Remember that [math]a^\dagger[/math] is a sum of both [math]\hat{x}[/math] and [math]\hat{p}[/math], and the Fourier kernel [math]\exp(-ipx) = \langle p,x\rangle[/math] is a braket of the eigenstates of both. In other words it is an eigenstate of [math]a^\dagger[/math].

>>11409478
Yes.
>PDEs on [math]H^1[/math]
>unitary [math]*[/math]-reps of [math]C^*[/math]-algebras
>covariant reps of group algebras [math]L^1(G)[/math]
>irreps of affine Lie's
>

>>11404059
"Charge" refers to labels of the irreps of a certain compact Lie group. For [math]U(1)[/math] the label is [math]\mathbb{Z}[/math], which refers to an elementary charge. In this sense spin is a certain charge.
>Do opposite spins experience a binding force
Yes but only because of the term [math]S_z^2[/math] is in the Hamiltonian, not because spin is a charge. "Charge" and "force" are fundamentally different notions.

>>11382371
Ok so you're just looking at a single atom. Again, the reason you can even talk about angular momentum eigenstates is because [math]{\bf L}^2[/math] is a symmetry, i.e. [math]{\bf L}^2[/math] commutes with the Hamiltonian. This gives rise to selection rules that completely forbid any mixing between the electrons in each shell: think of it as the fact that the eigenspace of [math]H[/math] decomposes into a direct sum of Verma's [math]\bigoplus_{l,|m|\leq l} V_{l,m}[/math] when you diagonalize the Casimirs [math]{\bf L}^2, L_z[/math] of [math]\mathfrak{su}(2)[/math] along with [math]H[/math], and the spectral theorem for [math]H[/math] says that these eigenstates completely cover the entirety of your Hilbert space. It makes no sense in the first place to mix them, not to mention the fact that paired-electron states explicitly breaks the charge [math]U(1)[/math] gauge symmetry. This is why you need SOC or BCS terms to explicitly break these symmetries before you can mix them.
Problem is Mercury probably does have these kinds of terms (Hg is actually superconducting) in the tight-binding Hamiltonian. People typically assume a SC term, consistent with the leftover symmetries, in the mean-field/Hartree-Fock Hamiltonian whose order parameter [math]\Delta[/math] is determined perturbatively.

>>11281467
>>11281600
Lmao yes please do this