Pauli matrices

1. Consider the 2x2 matrix $$\sigma^{\mu}=(1,\sigma_{i})$$ where $$\sigma^{\mu}=(1,\sigma)$$ where 1 is the identity matrix and $$\sigma_{i}$$ the pauli matrices. Show with a direct calcuation that $$detX=x^{\mu}x_{\mu}$$

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

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?

Well usually ab - cd assuming a, b are elements in the first row..

However not sure what to do next...

vanhees71
Gold Member
Write down $X$ 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!

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

George Jones
Staff Emeritus
Gold Member
Isn't $X$ given as $X = x^\mu \sigma_\mu$?

Yes...

George Jones
Staff Emeritus
Gold Member
For an $n \times n$ matrix $A$ and a number $a$, what is $det \left(aA\right)$ in terms of $detA$?

$$det(cA)=c^{n}(detA)$$

George Jones
Staff Emeritus
Gold Member
$$det(cA)=c^{n}(detA)$$

So $det \left(x^\mu \sigma_\mu\right) =$ ?

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.

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?

George Jones
Staff Emeritus
Gold Member
Each $x^\mu$ is a number and each $\sigma_\mu$ is a number. As an intermediate step, apply
$$det(cA)=c^{n}(detA)$$

to $det \left(x^\mu \sigma_\mu\right)$. What do you get?

vanhees71
Gold Member
No! $\sigma_{\mu}$ are complex $2 \times 2$ matrices, namely
$$\sigma_0=\mathbb{1}$$ and $\sigma_{j}$ ($j \in \{1,2,3 \}$) the three Pauli matrices. This means that in standard representation of the Pauli matrices with $\sigma_3$ diagonalized you have
$$X=x^{\mu} \sigma_{\mu}=x^0 \mathbb{1} + \vec{x} \cdot \vec{\sigma}=\begin{pmatrix} x^0+x^3 & x^1-\mathrm{i} x^2 \\ x^1 + \mathrm{i} x^2 & x^0-x^3 \end{pmatrix}.$$
Now evaluate its determinant!

George Jones
Staff Emeritus
Gold Member
each $\sigma_\mu$ is a number.

No!

Oops! The above was a slip of my fingers while typing. Obviously, each $\sigma_\mu$ is a matrix, and this what I meant to type.

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!!

George Jones
Staff Emeritus
Gold Member
Now that you see it using the explicit matrix given by vahees71, I will write out the method at which I hinted.

$$det\left( x^\mu \sigma_\mu \right) = \left( x^\mu \right)^2 det \sigma_\mu = \left( x^0 \right)^2 - \left( x^1 \right)^2 - \left( x^2 \right)^2 - \left( x^3 \right)^2 ,$$

since $det\sigma_0 = 1$ and $det\sigma_i = -1$ for each $i$.

I think it is important to see the calculation done both ways, i.e., summing before taking the determinant and summing after taking determinants.

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)

George Jones
Staff Emeritus
But $\sigma_0$ is just the identity matrix ...