- #1

- 23

- 2

## Main Question or Discussion Point

From Chapter 5.9 Weinberg's QFT Vol 1, massless fields are defined as:

[tex] \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}][/tex]

With coefficients defined by the conditions:

[tex]u_{\bar{l}}(p,\sigma) =\sqrt{|k|/p^0} \sum_{l}D_{\bar{l}l}(L(p))u_l(k,\sigma)[/tex]

[tex]v_{\bar{l}}(p,\sigma) =\sqrt{|k|/p^0} \sum_{l}D_{\bar{l}l}(L(p))v_l(k,\sigma)[/tex]

[tex]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)[/tex]

[tex]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)[/tex]

Where [itex]D_{\bar{l}l}(L(p))[/itex] is a general, irreducible representation of the homogenous Lorentz group restricted to standard boosts, [itex]L(p)[/itex] that take the standard momentum [itex]k = (0,0,k)[/itex] into arbitrary momentum [itex]p[/itex] and [itex]D_{\bar{l}l}(W)[/itex] is the Lorentz representation restricted to the little group for massless particles. Now Weinberg says that the equations for [itex]v[/itex] are just the complex conjugates of the equations for [itex]u[/itex] so that we can adjust the constants [itex]k[/itex] and [itex]\lambda[/itex] so that

[tex]v_l(p,\sigma)=u_l(p,\sigma)^*[/tex]

However, taking the complex conjugates of the equations of [itex]u[/itex]:

[tex]u_{\bar{l}}(p,\sigma)^* =\sqrt{|k|/p^0} \sum_{l}D_{\bar{l}l}(L(p))^*u_l(k,\sigma)^*[/tex]

[tex]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)^*[/tex]

This is where I get stuck. The above will be true if [itex]D_{\bar{l}l}(L(p))^*=D_{\bar{l}l}(L(p))[/itex] and [itex]D_{\bar{l}l}(W)^*=D_{\bar{l}l}(W)[/itex]. However, this does seem to necessarily be true. Is there another way to prove Weinberg's claim?

[tex] \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}][/tex]

With coefficients defined by the conditions:

[tex]u_{\bar{l}}(p,\sigma) =\sqrt{|k|/p^0} \sum_{l}D_{\bar{l}l}(L(p))u_l(k,\sigma)[/tex]

[tex]v_{\bar{l}}(p,\sigma) =\sqrt{|k|/p^0} \sum_{l}D_{\bar{l}l}(L(p))v_l(k,\sigma)[/tex]

[tex]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)[/tex]

[tex]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)[/tex]

Where [itex]D_{\bar{l}l}(L(p))[/itex] is a general, irreducible representation of the homogenous Lorentz group restricted to standard boosts, [itex]L(p)[/itex] that take the standard momentum [itex]k = (0,0,k)[/itex] into arbitrary momentum [itex]p[/itex] and [itex]D_{\bar{l}l}(W)[/itex] is the Lorentz representation restricted to the little group for massless particles. Now Weinberg says that the equations for [itex]v[/itex] are just the complex conjugates of the equations for [itex]u[/itex] so that we can adjust the constants [itex]k[/itex] and [itex]\lambda[/itex] so that

[tex]v_l(p,\sigma)=u_l(p,\sigma)^*[/tex]

However, taking the complex conjugates of the equations of [itex]u[/itex]:

[tex]u_{\bar{l}}(p,\sigma)^* =\sqrt{|k|/p^0} \sum_{l}D_{\bar{l}l}(L(p))^*u_l(k,\sigma)^*[/tex]

[tex]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)^*[/tex]

This is where I get stuck. The above will be true if [itex]D_{\bar{l}l}(L(p))^*=D_{\bar{l}l}(L(p))[/itex] and [itex]D_{\bar{l}l}(W)^*=D_{\bar{l}l}(W)[/itex]. However, this does seem to necessarily be true. Is there another way to prove Weinberg's claim?