Problem Related to Photons with Mass

  • #1
diffidus
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1
Homework Statement
I am trying to understand example 13.6 given in the book 'Quantum Field Theory for the Gifted Amateur' by Lancaster and Blundell. The problem is: Consider a massive photon in the rest frame so that P^μ=([m,0,0,0). If we boost the particle in the z direction to P^μ=([E,0,0,p) can we calculate the product of the polarisation vectors:
Relevant Equations
I have included the relevant equations below. I tried multiplying the matrices together but could not get the correct answer. Is there anybody out there that can help?
Before boost we have
1705446058588.png

Then using the Lorentz boost:
1705446402498.png

I want to calculate:

1705446519033.png

I tried multiplying the matrices together but I never get the stated answer which should be:

1705446624839.png
 
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  • #2
I get the stated answer except the minus signs. Just using the result of the previous example:
1705452372876.png
 
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  • #3
Hill said:
I get the stated answer except the minus signs. Just using the result of the previous example:
View attachment 338712
Hill

So I think you got your result by using the previous example values for ε as you have said. Then, I assume you carried out the product sum with with λ =1,3. Thanks for this answer, it gets me closer but then there is still the annoying minus signs, which makes me think there must be another subtlety and that worries me.
 
  • #4
diffidus said:
the annoying minus signs
I think they are a typo.
 
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  • #5
Hill said:
I think they are a typo.
That would be great, but I am not certain since in the book that I refer to, it also gives the product involving εε in terms of a projection operator:
1705510713921.png

and that
1705510999703.png

When I use this method it gives the matrix including the minus signs. So I am still not confident. Anyhow, thanks for your effort, it is really appreciated.
 
  • #6
diffidus said:
When I use this method it gives the matrix including the minus signs.
Could you show how it gives the minus signs? I don't see it.
 
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  • #7
Hill said:
Could you show how it gives the minus signs? I don't see it.
Apologies - I looked back at my calculation and it appears that I managed to get a minus sign in there. I have redone the calculation and I now get positive values and so I am now completely on the same page as you. In the matrix quoted in the book - it must be a typo. Thanks very much for your help in solving a frustrating problem.
 
  • #8
Hill said:
Could you show how it gives the minus signs? I don't see it.
I feel a bit like a bad penny now but I have looked over my calculation and I can now see where the minus signs come from.
The problem is given in terms of a boost in the z-direction:
1705605199925.png

From this we get:

1705605311588.png

Which can be calculated independently using:

1705605377518.png

Using this second method the minus signs in the A matrix arise since in the projection formula the momentum vectors are covariant and so should be:
1705605539257.png

So for example:

1705605617413.png

Similarly for A30.
The bad news is this still leaves me with the original problem! The good news, however, is that I think I have spotted the issue.

In the original problem equation 13.29 gives the boost matrix as (multiplied by 1/m):

1705605850554.png

However, I think there is a typo here as it should be:

1705605929161.png


When this boost matrix is used and followed through the rest of the derivation everything is resolved - phew!

Once again thanks for your help with this it has helped me a lot as it got me to think about my calculations and, naively, I never considered typos.
 

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  • #9
diffidus said:
I feel a bit like a bad penny now but I have looked over my calculation and I can now see where the minus signs come from.
The problem is given in terms of a boost in the z-direction:
View attachment 338787
From this we get:

View attachment 338788
Which can be calculated independently using:

View attachment 338789
Using this second method the minus signs in the A matrix arise since in the projection formula the momentum vectors are covariant and so should be:
View attachment 338790
So for example:

View attachment 338791
Similarly for A30.
The bad news is this still leaves me with the original problem! The good news, however, is that I think I have spotted the issue.

In the original problem equation 13.29 gives the boost matrix as (multiplied by 1/m):

View attachment 338794
However, I think there is a typo here as it should be:

View attachment 338795

When this boost matrix is used and followed through the rest of the derivation everything is resolved - phew!

Once again thanks for your help with this it has helped me a lot as it got me to think about my calculations and, naively, I never considered typos.
Thank you for clearing this point. I think now that there is no typo at all, and the resolution for the minus sign in the original problem is the same: the polarization vectors in 13.30 are contravariant, but they are covariant in 13.40. This gives us the minus signs in the answer, doesn't it?
 
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  • #10
A photon doesnt have a mass. it's massless.
 
  • #11
billtodd said:
A photon doesnt have a mass. it's massless.
Yes. However,
1705922995375.png

(Lancaster, Blundell, Quantum Field Theory for the Gifted Amateur)
 
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  • #12
I remember this Lagrangian called proca lagrangian density. otherwise I cannot recall.
 
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