Invariance of the determinant under spin rotations

In summary: You are saying that the determinant of those exponentials is 1, which is why you don't need to know their value in order to apply the invariance principle.
  • #1
ShayanJ
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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
 
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  • #2
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.
 
  • #3
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!
 
  • #4
Shyan said:
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.
 
  • #5
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 ##.
 
  • #6
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??
 
  • #7
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!
 
  • #8
Shyan said:
The determinants of those exponentials are 1 [...]
Actually, you don't even need to know their value... :oldwink:
 
  • #9
strangerep said:
Actually, you don't even need to know their value... :oldwink:
I don't understand!
 
  • #10
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...]
 
  • #11
strangerep said:
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.
 
  • #12
Re-read my post #10...
 
  • #13
I get it now, thanks.
 

FAQ: Invariance of the determinant under spin rotations

1. What is the invariance of the determinant under spin rotations?

The invariance of the determinant under spin rotations is a fundamental concept in quantum mechanics that states the determinant of a quantum state remains unchanged when the state is subjected to a spin rotation. This means that the probability amplitudes for different spin states remain the same despite the rotation.

2. Why is the invariance of the determinant under spin rotations important?

This concept is important because it allows us to easily calculate the probabilities of different spin states without having to take into account the orientation of the coordinate system. It also provides a mathematical basis for the conservation of angular momentum in quantum systems.

3. How does the invariance of the determinant under spin rotations relate to the Pauli exclusion principle?

The Pauli exclusion principle states that no two identical fermions can occupy the same quantum state simultaneously. This principle is closely related to the invariance of the determinant under spin rotations, as it ensures that the probability amplitudes for different spin states remain unchanged, thus preventing two fermions from occupying the same state.

4. Can the invariance of the determinant under spin rotations be observed experimentally?

Yes, the invariance of the determinant under spin rotations can be observed experimentally through various techniques such as spin-resolved photoemission spectroscopy and neutron scattering. These experiments have confirmed the relationship between spin rotations and the conservation of angular momentum in quantum systems.

5. Are there any exceptions to the invariance of the determinant under spin rotations?

While the invariance of the determinant under spin rotations holds true in most cases, there are some exceptions. One example is in the presence of a magnetic field, which can cause a change in the spin state and therefore affect the determinant. However, these exceptions are typically accounted for in the mathematical equations used to describe quantum systems.

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