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Pauli matrices

  • Thread starter smallgirl
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  • #1
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1. Consider the 2x2 matrix [tex]\sigma^{\mu}=(1,\sigma_{i})[/tex] where [tex]\sigma^{\mu}=(1,\sigma)
[/tex] where 1 is the identity matrix and [tex]\sigma_{i}[/tex] the pauli matrices. Show with a direct calcuation that [tex]detX=x^{\mu}x_{\mu}[/tex]








3. I'm not sure how to attempt this at all...
 

Answers and Replies

  • #2
I think you're missing a part out, where we define how the big X is related to the little x's
As a hint, how would you normally go about finding the determinant of a 2x2 matrix?
 
  • #3
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Well usually ab - cd assuming a, b are elements in the first row..

However not sure what to do next...
 
  • #4
vanhees71
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Write down [itex]X[/itex] explicitly and then calculate its determinant by just doing the algebraic manipulations. It's nothing deep in this calculation, but its result is very important for the description of spin-1/2-particles in relativistic quantum theory!
 
  • #5
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Well I don't know what x is therefore I don't know how to write X, and I'm not sure how sigma would change to x
 
  • #6
George Jones
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Isn't [itex]X[/itex] given as [itex]X = x^\mu \sigma_\mu[/itex]?
 
  • #7
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Yes...
 
  • #8
George Jones
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For an [itex]n \times n[/itex] matrix [itex]A[/itex] and a number [itex]a[/itex], what is [itex]det \left(aA\right)[/itex] in terms of [itex]detA[/itex]?
 
  • #9
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[tex]det(cA)=c^{n}(detA)
[/tex]
 
  • #10
George Jones
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[tex]det(cA)=c^{n}(detA)
[/tex]
So [itex]det \left(x^\mu \sigma_\mu\right) = [/itex] ?
 
  • #11
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Well,you need to write it as the product of four vectors,use the explicit form of σ matrices,and calculate the determinant of resulting 2×2 matrix.Got it.you will find it's use in showing the relation SO(3,1)=SL(2,C)/Z2.
 
  • #12
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Well from George's previous post, I had realised the conclusion beforehand, however I'm not sure how that will result in the answer I need.. but I guess if I follow andrien it should fall out?
 
  • #13
George Jones
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Each [itex]x^\mu[/itex] is a number and each [itex]\sigma_\mu[/itex] is a number. As an intermediate step, apply
[tex]det(cA)=c^{n}(detA)
[/tex]
to [itex]det \left(x^\mu \sigma_\mu\right)[/itex]. What do you get?
 
  • #14
vanhees71
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No! [itex]\sigma_{\mu}[/itex] are complex [itex]2 \times 2[/itex] matrices, namely
[tex]\sigma_0=\mathbb{1}[/tex] and [itex]\sigma_{j}[/itex] ([itex]j \in \{1,2,3 \}[/itex]) the three Pauli matrices. This means that in standard representation of the Pauli matrices with [itex]\sigma_3[/itex] diagonalized you have
[tex]X=x^{\mu} \sigma_{\mu}=x^0 \mathbb{1} + \vec{x} \cdot \vec{\sigma}=\begin{pmatrix}
x^0+x^3 & x^1-\mathrm{i} x^2 \\
x^1 + \mathrm{i} x^2 & x^0-x^3
\end{pmatrix}.
[/tex]
Now evaluate its determinant!
 
  • #15
George Jones
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each [itex]\sigma_\mu[/itex] is a number.
No!
Oops! The above was a slip of my fingers while typing. Obviously, each [itex]\sigma_\mu[/itex] is a matrix, and this what I meant to type.
 
  • #16
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sweeet! I see it now, I had the error of not including x^0 when writing out X which was why I couldn't solve the problem. Thank you!!
 
  • #17
George Jones
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Now that you see it using the explicit matrix given by vahees71, I will write out the method at which I hinted.

[tex]det\left( x^\mu \sigma_\mu \right) = \left( x^\mu \right)^2 det \sigma_\mu = \left( x^0 \right)^2 - \left( x^1 \right)^2 - \left( x^2 \right)^2 - \left( x^3 \right)^2 ,[/tex]

since [itex]det\sigma_0 = 1[/itex] and [itex]det\sigma_i = -1[/itex] for each [itex]i[/itex].

I think it is important to see the calculation done both ways, i.e., summing before taking the determinant and summing after taking determinants.
 
  • #18
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yeah I noticed that the det for sigma(i) was -1, but I didn't know what the determinant for sigma(0) was, so I guess I wasn't able to put everything together...as I wasn't sure how to include sigma(0)
 
  • #19
George Jones
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But [itex]\sigma_0[/itex] is just the identity matrix ...
 

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