Invariance of the determinant under spin rotations

[ShayanJ]

Homework Statement

Show that the determinant of a $2 \times 2$ matrix $\vec\sigma \cdot \vec a$ is invariant under $\vec \sigma\cdot \vec a \rightarrow \vec \sigma\cdot \vec a' \equiv \exp(\frac{i\vec \sigma \cdot \hat n \phi}{2})\vec \sigma\cdot \vec a \exp(\frac{-i\vec \sigma \cdot \hat n \phi}{2})$.

Homework Equations

$\sigma_1=\left( \begin{array}{cc} 0 \ \ \ \ \ \ \ \ 1 \\ 1 \ \ \ \ \ \ \ \ 0 \end{array} \right)$
$\sigma_2=\left( \begin{array}{cc} 0 \ \ \ \ -i \\ i \ \ \ \ \ \ 0 \end{array} \right)$
$\sigma_3=\left( \begin{array}{cc} 1 \ \ \ \ \ \ \ \ 0 \\ 0 \ \ \ \ -1 \end{array} \right)$

$\exp(\frac{\pm i\vec \sigma \cdot \hat n \phi}{2})=\cos{(\frac \phi 2)}\pm i(\vec\sigma \cdot \hat n)\sin{(\frac \phi 2)}$

The Attempt at a Solution

I tried to do it by matrix multiplication but the calculations turned out to be so much tedious that I thought there should be a better way. If this is the only way, just tell me. Otherwise just point to the right direction. Any other hint is welcome too.
Thanks

 

[fzero]

You should be able to use the commutation relation for the Pauli matrices $[\sigma_i , \sigma_j ] = 2 i \epsilon_{ijk} \sigma_k$ (sum over $k$ assumed) rather than actually write out explicit matrix products.

 

[ShayanJ]

I used the properties of Pauli matrices and got the following:

$\cos^2 \frac \phi 2 \ (\vec \sigma \cdot \vec a)-i\sin \phi \left[(\vec a \times \hat n)\cdot \vec \sigma \right]+\sin^2\frac \phi 2 \left[(\vec \sigma \cdot \hat n)(\vec a \cdot \hat n )+i \vec a \cdot \vec \sigma- i(\hat n \cdot \vec \sigma)(\vec a \cdot \hat n)\right]$

But I don't know how to proceed!

 

[fzero]

I used the properties of Pauli matrices and got the following:

$\cos^2 \frac \phi 2 \ (\vec \sigma \cdot \vec a)-i\sin \phi \left[(\vec a \times \hat n)\cdot \vec \sigma \right]+\sin^2\frac \phi 2 \left[(\vec \sigma \cdot \hat n)(\vec a \cdot \hat n )+i \vec a \cdot \vec \sigma- i(\hat n \cdot \vec \sigma)(\vec a \cdot \hat n)\right]$

But I don't know how to proceed!

I think that in the $\sin^2(\phi/2)$ term, the 1st and 3rd terms should have the same coefficient (to give an overall $2(\hat n \cdot \vec \sigma)(\vec a \cdot \hat n)$) and there shouldn't be an $i$ in the 2nd term. If you want to recheck the term, the identity $\sigma_i \sigma_j = \delta_{ij} I + i \epsilon_ijk}\sigma_k$ should be useful (though you might already be aware of it).

So you will have an expression $\vec{A} \cdot \vec{\sigma}$ and you should show that $\det ( \vec{A} \cdot \vec{\sigma}) = - \vec{A}\cdot \vec{A}$, by explicit muliplication or otherwise. Then apply this to your particular expression.

 

[ShayanJ]

I corrected some mistakes and I got $\vec A \cdot \vec \sigma$ where $\vec A=\cos^2 \frac \phi 2 \vec a +\sin \phi (\vec a \times \hat n)+\sin^2 \frac \phi 2 \left[2(\vec a \cdot \hat n)\hat n-\vec a \right]$.
Now my problem is, $\vec A \cdot \vec A$ is not a straightforward expression and I can't write it as a function of only $\vec a \cdot \vec a$.

 

[strangerep]

Can't you just use general properties of determinants?
I.e., $\det(AB) = \det A \; \det B$ and $\det A^{-1} = 1/\det A$, etc ?

Am I missing something here??

 

[ShayanJ]

thanks strangerep. The determinants of those exponentials are 1 and so things are very simple.
I have this bad habit of always trying the hardest way first!

 

[strangerep]

The determinants of those exponentials are 1 [...]

Actually, you don't even need to know their value...

 

[ShayanJ]

Actually, you don't even need to know their value...

I don't understand!

 

[strangerep]

Well, I already noted these properties of determinants:
$\det(AB) = \det A \; \det B$ and $\det\left(A^{-1}\right) = 1/\det A$, so...
$$ \det\left(ABA^{-1}\right) ~=~ ... ~?$$

[Edit: you answered my post before I'd finished. Work out the above...]

 

[ShayanJ]

Well, I already noted these properties of determinants:
$\det(AB) = \det A \; \det B$ and $\det\left(A^{-1}\right) = 1/\det A$,

Yes, but those are useful in proving the invariance only when you know that the determinant of those exponentials are one. But you said I don't need to know the value of the determinants. This is what I don't understand.

 

[strangerep]

Re-read my post #10...

 

[ShayanJ]

I get it now, thanks.