Weinberg: The Quantum Theory of Fields I, Eq. 2.5.14

  • Thread starter Living_Dog
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  • #1
Living_Dog
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First off I am a physics instructor at a local school and am trying to learn QFT from books. So this is not homework for a class at all.

I am trying to re-derive Eq. 2.5.14 from the above equation to it and it is in fact that "above" equation which is my hurdle. It is on pg. 66 for those of you with the book. Here is some previous information (for those of you without it) which I have struggled with the passed few days (actually weeks but I only get to spend a few hours each week due to other activities):

page 63:

momentum operator: [tex] P^\mu \Psi_{p,\sigma} = p^\mu \Psi_{p,\sigma} [/tex] (2.5.1)

unitary operator of a boost on a single particle state: [tex] U(\Lambda)\Psi_{p,\sigma} = \Sigma_{\sigma '} C_{\sigma'\sigma}(\Lambda,p) \Psi_{\Lambda p,\sigma} [/tex] (2.5.3)

"In general, it may be possible by using suitable linear combinations of the [tex]\Psi_{p,\sigma}[/tex] to choose the [tex]\sigma[/tex] labels in such a way that the matrix [tex]C_{\sigma'\sigma}(\Lambda,p)[/tex] is block-diagonal; ..."

page 63-64:

"Our task now is to work out the structure of the coefficients [tex]C_{\sigma ', \sigma}(\Lambda,p)[/tex] in irreducible representations of the inhomogeneous Lorentz group."

He defines a standard 4-momentum, [tex]k^\mu[/tex], which is invariant under a subgroup, [tex]W[/tex], of the homogeneous Lorentz group, [tex]L[/tex]. So [tex]W \subset L[/tex], and is used to boost [tex]k^\mu[/tex] to an arbitrary 4-momentum [tex]p^\mu[/tex] via:

[tex]p^\mu = {L^{\mu}}_{\nu}(p) k^\nu[/tex].​

Note: [tex]L(p) \in W[/tex]

single particle state of momentum 'p' defined:

[tex]\Psi_{p,\sigma} \equiv N(p) U(L(p))\Psi_{k,\sigma}[/tex].​

Therefore:

[tex]U(\Lambda) \Psi_{p,\sigma} = N(p) U(\Lambda) U(L(p)) \Psi_{k,\sigma}[/tex].​

Using the multiplication rule for these unitary operators: U(A)*U(B) = U(A*B), then:

[tex]U(\Lambda) \Psi_{p,\sigma} = N(p) U(\Lambda L(p)) \Psi_{k,\sigma}[/tex].​

Inserting a factor of [tex]1 = U(L(\Lambda p)L^{-1}(\Lambda p))[/tex] then the above becomes, after applying the multiplication rule:

[tex]U(\Lambda) \Psi_{p,\sigma} = N(p) U(L(\Lambda p)L^{-1}(\Lambda p)) U(\Lambda L(p)) \Psi_{k,\sigma} = N(p) U(L(\Lambda p)) U(L^{-1}(\Lambda p))\Lambda L(p)) \Psi_{k,\sigma}[/tex].​

The second unitary factor takes [tex]k[/tex] to [tex]p[/tex] then to [tex]\Lambda p[/tex] and then back to [tex]k[/tex]. This forms the little group, [tex]W[/tex], of the homogeneous Lorentz group - it keeps [tex]k[/tex] the same. Namely, [tex]k^\mu = {W^{\mu}}_{\nu} k^\nu[/tex] and

[tex]W \equiv L^{-1}(\Lambda p)\Lambda L(p)[/tex]. (2.5.10)​

Note: I think that the standard 4-momentum is just another way of saying the 4-momentum in the particle's in its rest frame.

For any such [tex]W[/tex]:

[tex]U(W)\Psi_{k,\sigma} = \Sigma_{\sigma'} D_{\sigma'\sigma}(W)\Psi_{k,\sigma'}[/tex] (2.5.8)​

The irreducible matrices [tex]D_{\sigma'\sigma}(W)[/tex] satisfy the following relationship:

[tex]D_{\sigma'\sigma}(\overline{W}W) = \Sigma_{\sigma''}D_{\sigma'\sigma''}(\overline{W})D_{\sigma''\sigma}(W)[/tex] (2.5.9)​

Note: we may chose the states with standard 4-momentum to be orthonormal, i.e.:

[tex]<\Psi_{k',\sigma'}|\Psi_{k,\sigma}> = \delta^3([/tex]k'-k[tex])\delta_{\sigma'\sigma}[/tex]​

The next step, to determine the normalization factor, [tex]N(p)[/tex], Weinberg starts from the bra-ket of two distinct single particle states, namely:

[tex]<\Psi_{p',\sigma'}|\Psi_{p,\sigma}> = N(p)<U^{-1}(L(p))\Psi_{p',\sigma'}|\Psi_{k,\sigma}>[/tex]​

and obtains:

[tex] = N(p)N^*(p'){D^*}_{\sigma\sigma'}(W(L^{-1}(p),p'))\delta^3([/tex]k'-k[tex])[/tex].​

This is the equation which I cannot derive. How does one go from the next to last to this last equation???


-joe
____________________________________
 
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Answers and Replies

  • #2
meopemuk
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[tex]<\Psi_{p',\sigma'}|\Psi_{p,\sigma}> = N(p)<U^{-1}(L(p))\Psi_{p',\sigma'}|\Psi_{k,\sigma}>[/tex]​

and obtains:

[tex] = N(p)N^*(p'){D^*}_{\sigma\sigma'}(W(L^{-1}(p),p'))\delta^3([/tex]k'-k[tex])[/tex].​

This is the equation which I cannot derive. How does one go from the next to last to this last equation???


It is useful to note that the product [itex]<\Psi_{p',\sigma'}|\Psi_{p,\sigma}>[/itex] is non-zero only if [itex]p=p'[/itex]. So, I will replace [itex]p' \to p[/itex] and [itex]k' \to k[/itex] where convenient. Then


[tex]<\Psi_{p',\sigma'}|\Psi_{p,\sigma}> = N(p)<\Psi_{p',\sigma'}|U(L(p))\Psi_{k,\sigma}> = N(p)<U^{-1}(L(p))\Psi_{p',\sigma'}|\Psi_{k,\sigma}> = N(p)<U^{-1}(L(p))\Psi_{p,\sigma'}|\Psi_{k,\sigma}> [/tex]

[tex]= N(p)N^*(p) < U^{-1}(L(p))U(L(p))\Psi_{k',\sigma'}|\Psi_{k,\sigma}> = N(p)N^*(p) < \Psi_{k',\sigma'}|\Psi_{k,\sigma}> = |N(p)|^2 \delta_{\sigma, \sigma'} \delta(\mathbf{k-k'}) [/tex]


Eugene
 
  • #3
Living_Dog
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0
It is useful to note that the product [itex]<\Psi_{p',\sigma'}|\Psi_{p,\sigma}>[/itex] is non-zero only if [itex]p=p'[/itex]. So, I will replace [itex]p' \to p[/itex] and [itex]k' \to k[/itex] where convenient. Then


[tex]<\Psi_{p',\sigma'}|\Psi_{p,\sigma}> = N(p)<\Psi_{p',\sigma'}|U(L(p))\Psi_{k,\sigma}> = N(p)<U^{-1}(L(p))\Psi_{p',\sigma'}|\Psi_{k,\sigma}> = N(p)<U^{-1}(L(p))\Psi_{p,\sigma'}|\Psi_{k,\sigma}> [/tex]

[tex]= N(p)N^*(p) < U^{-1}(L(p))U(L(p))\Psi_{k',\sigma'}|\Psi_{k,\sigma}> = N(p)N^*(p) < \Psi_{k',\sigma'}|\Psi_{k,\sigma}> = |N(p)|^2 \delta_{\sigma, \sigma'} \delta(\mathbf{k-k'}) [/tex]


Eugene

Holy mackeral!!! I didn't see the forest for the trees! I kept seeing:

[tex]U^{-1}(L(p))U(L(p))[/tex]​

as

[tex]U^{-1}(L(p))U(L(p'))[/tex].​

because he applies [tex]p = p'[/tex], after this result, at the end. So I was stuck thinking that it is only applied at the end. I never thought to apply it ahead of that. (such a dope! - I am...)

Thanks ever so much, now I can make some more progress.
 

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