Question about Weinberg Book QFT1 (5.1.13)

In summary: I was confused because the extra factor made me perplexed. I wasn't sure if it was included in Weinberg's derivation or not. I was able to derive it from the unnumbered equation immediately prior to it, but I was still confused because the argument of a in that equation doesn't match the argument of u in the equation that includes the extra factor.
  • #1
PRB147
127
0
TL;DR Summary
I cannot derive Weinberg book QFT volume 1, (5.1.13), please help.
According to (5.1.6)
$$U_0(\Lambda,a)\psi_\ell^+(x)U^{-1}_0(\Lambda,a)=\sum\limits_{\ell \bar{\ell}}D_{ \ell \bar{\ell} }(\Lambda^{-1})\psi^+_{\bar{\ell}}(\Lambda x+a).$$ (5.1.6)
According to definition 5.1.4:
$$\psi^+_{\bar{\ell}}(\Lambda x+a)=\sum\limits_{\sigma n}\int d^3{\bf p
} u_\ell(\Lambda x+a;{\bf{p}},\sigma,n)a({\bf{p}},\sigma,n)$$
If we change the integral variable $${\bf p}\rightarrow \Lambda {\bf p}$$ and using Lorentz invariant $$d^3{\bf p}={p_0}{\frac{d^3(\Lambda {\bf p})}{(\Lambda p)^0}}$$, then,
$$\psi^+_{\bar{\ell}}(\Lambda x+a)=\sum\limits_{\sigma n}\int d^3(\Lambda {\bf p}) \left(\frac{p_0}{(\Lambda
p)^0}\right)u_\ell(\Lambda x+a;\Lambda{\bf{p}},\sigma,n)a(\Lambda{\bf{p}},\sigma,n)$$
If the above relation is correct, then I cannot derive equation (5.1.13).
Because of the extra factor below $$\left(\frac{p_0}{(\Lambda
p)^0}\right)$$.
It is this factor that made me perplexed, this extra factor make my derivation be different from (5.1.13), my result is $$\sqrt{\left(\frac{(\Lambda
p)^0}{p_0}\right)}$$ instead of $$\sqrt{\left(\frac{p_0}{(\Lambda
p)^0}\right)} \textrm{in book (5.1.13).}$$
 
Last edited:
Physics news on Phys.org
  • #2
Try to read the latex guide
Edit: looks better now, thanks
 
Last edited:
  • Like
Likes PRB147
  • #3
As far as I can tell this extra factor is correct and included in Weinberg‘s derivation (in my edition it is mentioned at the top of p. 194, and is then included in the unnumbered equations before (5.1.13)). Do you agree with those? You have an extra

##\sqrt{(\Lambda p)^0/p^0}##

from (5.1.11), so overall you have ##\sqrt{p^0/(\Lambda p)^0}##.

Is your actual problem maybe deriving (5.1.13) from the unnumbered ones immediately prior to it (which include your extra factor)?
 
  • Like
Likes PRB147 and malawi_glenn
  • #4
Dr.AbeNikIanEdL said:
As far as I can tell this extra factor is correct and included in Weinberg‘s derivation (in my edition it is mentioned at the top of p. 194, and is then included in the unnumbered equations before (5.1.13)). Do you agree with those? You have an extra

##\sqrt{(\Lambda p)^0/p^0}##

from (5.1.11), so overall you have ##\sqrt{p^0/(\Lambda p)^0}##.

Is your actual problem maybe deriving (5.1.13) from the unnumbered ones immediately prior to it (which include your extra factor)?
Thank you very much for your quick reply, as you know the unnumbered equation in page 194 closely below the sentence "it is necessary and sufficient that" is obtained from 5.1.6 and 5.1.11. the left hand side of this unnumbered equation can be derived from 5.1.6 and above here, while the right hand side from 5.1.11. However, in both sides, there exists the same integral ##\int d^3\bf{p}##, so, both sides need to change to ##\int d^3(\Lambda {\bf p})##.
So, both sides produce an extra factor ##p^0/(\Lambda p)^0##, leading to an inconsistency withthe book. Alternatively speaking, the factor on the right side should be ##\sqrt{(\Lambda p)^0/p_0}##, which is the inverse of the factor in 5.1.13.
 
Last edited:
  • #5
Ah, no for the left hand side there is not change of variables, just a renaming. You can write equation (5.1.4) as

## \Psi^+_l(x) = \sum_{\sigma,n} \int d^3(\Lambda p) u_l(x;\mathbb{p}_\Lambda,\sigma,n)a(\mathbb{p}_\Lambda,\sigma,n) ##

by just renaming the integration variable. Note also the arguments of ##u_l## and ##a## changed to ##\mathbb{p}_\Lambda##, and indeed that is also the argument of ##u## on the left hand side in the unnumbered equation. If you do the transformation, the arguments would not change and nothing like (5.1.13) would follow since the argument of ##a## for example would also not match.
 
  • Like
Likes PRB147, vanhees71 and malawi_glenn
  • #6
Thank you! Yes, you are right.
 

Similar threads

  • High Energy, Nuclear, Particle Physics
Replies
5
Views
1K
  • High Energy, Nuclear, Particle Physics
Replies
1
Views
1K
  • High Energy, Nuclear, Particle Physics
Replies
1
Views
1K
Replies
2
Views
995
Replies
5
Views
2K
Replies
1
Views
587
  • Atomic and Condensed Matter
Replies
4
Views
2K
  • High Energy, Nuclear, Particle Physics
Replies
4
Views
1K
  • High Energy, Nuclear, Particle Physics
Replies
1
Views
2K
  • Quantum Physics
Replies
9
Views
852
Back
Top