- #1
hgandh
- 26
- 2
From Chapter 5.9 Weinberg's QFT Vol 1, massless fields are defined as:
\psi_l(x)=(2\pi)^{-3/2}\int d^{3}p\sum_{\sigma}[k a(p,\sigma)u_l(p,\sigma)e^{ipx}+\lambda a^{c\dagger}(p,\sigma)v_l(p,\sigma)e^{-ipx}]
With coefficients defined by the conditions:
u_{\bar{l}}(p,\sigma) =\sqrt{|k|/p^0} \sum_{l}D_{\bar{l}l}(L(p))u_l(k,\sigma)
v_{\bar{l}}(p,\sigma) =\sqrt{|k|/p^0} \sum_{l}D_{\bar{l}l}(L(p))v_l(k,\sigma)
u_{\bar{l}}(p,\sigma) exp(i\sigma \theta(k,W) =\sqrt{|k|/p^0} \sum_{l}D_{\bar{l}l}(W)u_l(k,\sigma)
v_{\bar{l}}(p,\sigma) exp(-i\sigma \theta(k,W) =\sqrt{|k|/p^0} \sum_{l}D_{\bar{l}l}(W)v_l(k,\sigma)
Where D_{\bar{l}l}(L(p)) is a general, irreducible representation of the homogenous Lorentz group restricted to standard boosts, L(p) that take the standard momentum k = (0,0,k) into arbitrary momentum p and D_{\bar{l}l}(W) is the Lorentz representation restricted to the little group for massless particles. Now Weinberg says that the equations for v are just the complex conjugates of the equations for u so that we can adjust the constants k and \lambda so that
v_l(p,\sigma)=u_l(p,\sigma)^*
However, taking the complex conjugates of the equations of u:
u_{\bar{l}}(p,\sigma)^* =\sqrt{|k|/p^0} \sum_{l}D_{\bar{l}l}(L(p))^*u_l(k,\sigma)^*
u_{\bar{l}}(p,\sigma)^* exp(-i\sigma \theta(k,W) =\sqrt{|k|/p^0} \sum_{l}D_{\bar{l}l}(W)^*u_l(k,\sigma)^*
This is where I get stuck. The above will be true if D_{\bar{l}l}(L(p))^*=D_{\bar{l}l}(L(p)) and D_{\bar{l}l}(W)^*=D_{\bar{l}l}(W). However, this does seem to necessarily be true. Is there another way to prove Weinberg's claim?
\psi_l(x)=(2\pi)^{-3/2}\int d^{3}p\sum_{\sigma}[k a(p,\sigma)u_l(p,\sigma)e^{ipx}+\lambda a^{c\dagger}(p,\sigma)v_l(p,\sigma)e^{-ipx}]
With coefficients defined by the conditions:
u_{\bar{l}}(p,\sigma) =\sqrt{|k|/p^0} \sum_{l}D_{\bar{l}l}(L(p))u_l(k,\sigma)
v_{\bar{l}}(p,\sigma) =\sqrt{|k|/p^0} \sum_{l}D_{\bar{l}l}(L(p))v_l(k,\sigma)
u_{\bar{l}}(p,\sigma) exp(i\sigma \theta(k,W) =\sqrt{|k|/p^0} \sum_{l}D_{\bar{l}l}(W)u_l(k,\sigma)
v_{\bar{l}}(p,\sigma) exp(-i\sigma \theta(k,W) =\sqrt{|k|/p^0} \sum_{l}D_{\bar{l}l}(W)v_l(k,\sigma)
Where D_{\bar{l}l}(L(p)) is a general, irreducible representation of the homogenous Lorentz group restricted to standard boosts, L(p) that take the standard momentum k = (0,0,k) into arbitrary momentum p and D_{\bar{l}l}(W) is the Lorentz representation restricted to the little group for massless particles. Now Weinberg says that the equations for v are just the complex conjugates of the equations for u so that we can adjust the constants k and \lambda so that
v_l(p,\sigma)=u_l(p,\sigma)^*
However, taking the complex conjugates of the equations of u:
u_{\bar{l}}(p,\sigma)^* =\sqrt{|k|/p^0} \sum_{l}D_{\bar{l}l}(L(p))^*u_l(k,\sigma)^*
u_{\bar{l}}(p,\sigma)^* exp(-i\sigma \theta(k,W) =\sqrt{|k|/p^0} \sum_{l}D_{\bar{l}l}(W)^*u_l(k,\sigma)^*
This is where I get stuck. The above will be true if D_{\bar{l}l}(L(p))^*=D_{\bar{l}l}(L(p)) and D_{\bar{l}l}(W)^*=D_{\bar{l}l}(W). However, this does seem to necessarily be true. Is there another way to prove Weinberg's claim?
