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Doubt with differential forms (YM)

  • Thread starter Dox
  • Start date
  • #1
Dox
26
1
Hi everyone,

Homework Statement



I've been studying a paper in which there is a connection given by,

[tex]A = f(r)\sigma_1 dx+g(r)\sigma_2 dy,[/tex]​

where [tex]\sigma[/tex]'s are half the Pauli matrices. I need to calculate the field strength,

[tex]F = dA +[A,A].[/tex]​

Homework Equations



[tex]A = f(r)\sigma_1 dx+g(r)\sigma_2 dy,[/tex]

[tex]F = dA +[A,A][/tex]​


The Attempt at a Solution



I have computed it, but a factor is given me problems. I would say,

[tex]dA = f' \sigma_1 dr\wedge dx + g'\sigma_2 dr\wedge dy[/tex]​

and

[tex][A,A] = 2 f g \sigma_3 dx\wedge dy,[/tex]​

with a factor 2 coming from the fact that there are two contributions... like a binomial.

Is it OK or there is a half factor hidden in the definition of [tex][A,A][/tex]?

Thank you so much.

DOX​
 

Answers and Replies

  • #2
fzero
Science Advisor
Homework Helper
Gold Member
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289
If the [tex]\sigma_i[/tex] are equal to 1/2 the Pauli matrices, then their commutation relation is

[tex][\sigma_i,\sigma_j] = i\epsilon_{ijk}\sigma_k.[/tex]

If you recheck your calculation, you'll find that there's no factor of 2,


[tex]
[A,A] = f g \sigma_3 dx\wedge dy.
[/tex]
 
  • #3
Dox
26
1
If the [tex]\sigma_i[/tex] are equal to 1/2 the Pauli matrices, then their commutation relation is

[tex][\sigma_i,\sigma_j] = i\epsilon_{ijk}\sigma_k.[/tex]

If you recheck your calculation, you'll find that there's no factor of 2,


[tex]
[A,A] = f g \sigma_3 dx\wedge dy.
[/tex]
Thank you fzero. I've found out that people use to write a commutator for this factor but it is just a wedge product... that's why I was getting a different factor of 2.

THX.
 

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