→ chungweitw:這只是在說SO(3)會有這些結果. 07/18 01:13
→ chungweitw:我覺得不知道這些, 對於學習並沒有影響. 07/18 01:14
程度很差 不太了解這段文字的意義(不是翻譯問題)
不知道能不能請高手解釋 感謝
◎To put the separation method of solving PDEs in perspective,let us view it
as a consequence of a symmetry of the PDE.Take the stationary Schrodinger
equation HΨ=EΨ as an example with a potential V(r) depending only on the
radial distance r.Then this PDE is invariant under rotaions that comprise the
group SO(3).Its diagonal generator is the orbital angular momentum operator
2
Lz=-i(∂/∂φ),and its quadratic(Casimir) invariant is L .Since both commute
with H,we end up with three separate eigenvalue equations:
2
HΨ=EΨ, L Ψ=l(l+1)Ψ, LzΨ=mΨ.
2 2 2 2
Upon replacing Lz in L by its eigenvalue m ,the L PDE becomes Legendre's ODE
,and similarly HΨ=EΨ becomes the radial ODE of the separation method in
2
spherical polar coordinates upon substituting the eigenvalue l(l+1) for L .
◎For cylindrical coordinates the PDE is invariant under rotations about the
z-axis only,which form a subgroup of SO(3),This invariance yields the generator
Lz=-i(∂/∂φ) and separate azimuthal ODE LzΨ=mΨ as before.Invariance under
translations along the z-axis with the generator -i(∂/∂z) gives the separate
ODE in the z-variable provided the boundary conditions obey the same symmetry.
The potential V=V(ρ) or V=V(z) depends on one variable,as a rule.
◎In general,there are n mutually commuting generators H with eigenvalues m
i i
of the (classical) Lie group G of rank n and the corresponding Casimir
invariants C with eigenvalues c ,which yield the separate ODEs
i i
H Ψ=m Ψ C Ψ=c Ψ
i i i i
in addition to the radial ODE HΨ=EΨ.
--
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◆ From: 218.173.167.219
※ 編輯: kuromu 來自: 218.173.167.219 (07/17 14:04)