［Phy撚圍爐區］數撚爆人PO又唔開（2）
清水灣哂銀時
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以下內容我自己諗，有錯請指正
1.) Feynman propagator 只係 momentum space 入面既time-ordered operator，即係Fourier transform個<0| \phi(x,0) \phi(x,t) |0>，用time-ordered operator因為我地計緊個amplitude <\phi (t=-\infty) | \phi (t=\inf)>，就好似用Hamiltonian計緊time-evolution咁。
2.) Feynman integral divergence有兩種，IR同UV，兩樣野好唔同。IR divergences通常因為低能量既時候wavefunctions可以form bound states、或者有soft-colinear emission，比較難搞。
3.) UV divergences因為極高能量既時候而家個theory唔make sense，應該有其他renormalizable theory (例如string theory 將個integration phase space cut剩個fundamental domain所以啲amplitude係finite)，而宜家個theory只係個UV theory既low energy effective theory。個low energy Lagrangian既divergent operators個coefficients應該用effective field theory既角度睇，詳情可以睇Wilson's Effective Action。
大致上係︰啲infinite coefficients唔係無限，只係啲operators with finite coefficients係零。
(好似係) 
1.) Feynman propagator 只係 momentum space 入面既time-ordered operator，即係Fourier transform個<0| \phi(x,0) \phi(x,t) |0>，用time-ordered operator因為我地計緊個amplitude <\phi (t=-\infty) | \phi (t=\inf)>，就好似用Hamiltonian計緊time-evolution咁。
2.) Feynman integral divergence有兩種，IR同UV，兩樣野好唔同。IR divergences通常因為低能量既時候wavefunctions可以form bound states、或者有soft-colinear emission，比較難搞。
3.) UV divergences因為極高能量既時候而家個theory唔make sense，應該有其他renormalizable theory (例如string theory 將個integration phase space cut剩個fundamental domain所以啲amplitude係finite)，而宜家個theory只係個UV theory既low energy effective theory。個low energy Lagrangian既divergent operators個coefficients應該用effective field theory既角度睇，詳情可以睇Wilson's Effective Action。
大致上係︰啲infinite coefficients唔係無限，只係啲operators with finite coefficients係零。
(好似係) 數學上嚟講 SO(3) 係一個 group 而L^2(R^3)係一個vector space
當你有一個group 同一個 vector space 你就可以討論 representation of a group in a vector space
Representation 係講緊對於每一個R in the group (而家嘅情況中個group 係SO(3)) 你都會有一個linear operator
P(R): L^2(R^3) -> L^2(R^3)
使得:
如果你有R, R' in SO(3) 而你想計下作用P(R')再作用P(R)會得出咩, 咁你可以先將R 同R' 乘埋(佢地可以相乘因為SO(3) 係一個group)再計P(RR')
上面用數學表達嘅話即係:
P(R) P(R') = P(RR')
當個vector space 有埋hermitian form, 咁unitary 係指啲linear operators P(R) 會preserve 埋呢個Hermitian form
正如有人覆過 其實P(R) 基本上就係做緊R^3 上旋轉 所以會係unitary 都幾natural
當你有一個group 同一個 vector space 你就可以討論 representation of a group in a vector space
Representation 係講緊對於每一個R in the group (而家嘅情況中個group 係SO(3)) 你都會有一個linear operator
P(R): L^2(R^3) -> L^2(R^3)
使得:
如果你有R, R' in SO(3) 而你想計下作用P(R')再作用P(R)會得出咩, 咁你可以先將R 同R' 乘埋(佢地可以相乘因為SO(3) 係一個group)再計P(RR')
上面用數學表達嘅話即係:
P(R) P(R') = P(RR')
當個vector space 有埋hermitian form, 咁unitary 係指啲linear operators P(R) 會preserve 埋呢個Hermitian form
正如有人覆過 其實P(R) 基本上就係做緊R^3 上旋轉 所以會係unitary 都幾natural
有冇ching留意呢個新既Feynman path integral note? https://arxiv.org/pdf/2201.03593.pdf 
I'm a mathematician, not a physicist, so here is a purely algebraic perspective: recall that a tangent vector is just an element v_p in T_p(M) and a cotangent vector is just an element p_v of the cotangent space T_p(M)*, which is by definition, just the set of homomorphisms from T_p(M) to R, i.e.e T_p(M)* = Hom(T_p(M), R), now since the momentum p is a homomorphism which takes a tangent vector and output a scalar, so it's a covector. But of course this doesn't give you any physical intuition of why momentum should be a cotangent vector.
