x*****d
发帖数: 427 | 1 正在读Weinberg场论的第一章，繁杂无比（呵呵，可能是因为自己
反应不过来），有大侠能简单的解释一下粒子作为群表示的观点究竟
是怎么一回事吗？ | S*********g
发帖数: 5298 | 2 Under a Poincare transformation, an electron is still an electron. It
will not
be transformed into a spin-1 particel. This is the meaning of the
invariant.
Your second question is also related to this invariant.To answer it,
let's talk about the spinor representaion of the Poincare group, which
can be used to describe an electron.
For an electron, you can use a spinor (A1, A2, A3, A4) to describe an
electron as you learned in the Dirac equation in Quantum Mechanics.
Under the Poincare transforma
【在 x*****d 的大作中提到】
: 正在读Weinberg场论的第一章，繁杂无比（呵呵，可能是因为自己
: 反应不过来），有大侠能简单的解释一下粒子作为群表示的观点究竟
: 是怎么一回事吗？
| s********k
发帖数: 107 | 3 I think it is the tranformation should satisfy U_{deg} * U = 1
and Linear space is not so complicated. Just in QM, remeber Lx, Ly, Lz matrix
reprensation in L=1/2, 1, 2, .... So, different Sz composes a linear space
for rotation.
【在 x*****d 的大作中提到】
: 正在读Weinberg场论的第一章，繁杂无比（呵呵，可能是因为自己
: 反应不过来），有大侠能简单的解释一下粒子作为群表示的观点究竟
: 是怎么一回事吗？
| y***u
发帖数: 25 | 4 Come on, Lorentz group is not a unitary group!
What Superstring said in principal is correct. But we should include more
groups
to classify the particles thoughly (theoretically, we can identify each kind
of particles by using all symmetries). Lorentz invariance must be hold by all
particles.
Lorenz group has spinor representations (spin half integer partiles) as well
as tensor
ones(spin integer particles). There are two Casimier operators in Poincare's
algebra, one is P^2=m^2, another one is
【在 s********k 的大作中提到】
: I think it is the tranformation should satisfy U_{deg} * U = 1
: and Linear space is not so complicated. Just in QM, remeber Lx, Ly, Lz matrix
: reprensation in L=1/2, 1, 2, .... So, different Sz composes a linear space
: for rotation.
| y***u
发帖数: 25 | 5 Unitary group is SU(N), but Lorentz group is SO(3,1). You can check if Lorentz
group is unitary or not by a particular representation. Say, in the spinor
representation, the generator is {\Sigma}^{u v}=i/2
{\gamma^{u},{\gamma^{v}}+x^u P^v-x^v P^v. We know \gamma^u{\dagger}=\gamma_u
(the metric is +1,-1,-1,-1). By using the anticommutation of gamma matrices,
you can see as u,v=i,j (spatial component) the generators are hermitian which
means the corresponding subgroup is unitary. But as u,v=0,i, t
【在 s********k 的大作中提到】
: I think it is the tranformation should satisfy U_{deg} * U = 1
: and Linear space is not so complicated. Just in QM, remeber Lx, Ly, Lz matrix
: reprensation in L=1/2, 1, 2, .... So, different Sz composes a linear space
: for rotation.
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