What is the purpose and meaning of Pauli/sigma matrices

1. Jan 10, 2012

mraptor

What is the purpose and meaning of Pauli/sigma matrices ?

What distinguish them from other Operators ?

thanx

2. Jan 11, 2012

muppet

The Pauli matrices describe how spin-1/2 particles transform under rotations.

To give a concrete example of why they're needed, suppose you've got an electron that's in a "spin up" eigenstate about the z-axis, which we might write as (1,0) in vector notation. (First component is the coefficient of a spin-up eigenstate, second that of the spin-down, and their moduli squared must sum to one). Now rotate your co-ordinate system by 90 degrees about the y-axis, so that the x-axis is rotated into the z-axis. Clearly, your electron won't be in a z-spin eigenstate any more, becase the z-axis has changed. In fact, as you've just changed what you're calling the direction about which it has a definite spin, it will be in an x-spin eigenstate. So how do you describe this transformation mathematically- what operator turns a spin-z eigenstate into a spin-x one, and how would you work out the result if we'd picked a general rotation where the answer was less obvious? If you were dealing with some vector in space, like the momentum vector, you'd know exactly how to rotate it, but you're dealing with a two-component vector in the subset of Hilbert space concerned with "spin up" and "spin down".

The general answer is that, for spin-1/2 particles, you implement rotations on such two-component spin vectors (a,b) with the matrix $e^{i\frac{1}{2}\vec{\theta}\cdot\vec{\sigma}}$, where $\vec{\theta}$ is a 3-vector pointing along the axis of rotation, whose modulus is the angle of rotation, and $\vec{\sigma}$ is the vector of Pauli matrices, and the exponential of a matrix is defined by the power series. For really small (infinitesmal) angles, we can truncate this power series to just the leading order term, so that the Pauli matrices implement the infinitesmal rotation; we say they are the generators of the rotation.

Analagous constructions exist for particles of any spin; the general rule is that for a particle of spin j, you need a set of 3 (2j+1)x(2j+1) matrices that satisfy the same commutation relations as the Pauli matrices. Note that for j=1, 2j+1=3; the ordinary rotations of 3d space do the trick.

If you want to know more, any decent book on QM should do, but Landau+ Lifschitz has a really good treatment of spinors. If you really want to get into the maths to understand it properly, you need to learn about the idea of representation theory- I'd recommend "Groups, Representations and Physics" by Jones for explaining precisely what a representation actually is, but the most popular treatment, especially for Lie groups like the rotation group, is "Lie Algebras in particle physics" by Georgi.

Hope that helps.

3. Jan 11, 2012

mraptor

Thanx alot , I understood at least the first part ;)
Let me try to explain it with my words..
So in a sense i measure electron/s and find his spin.. but for whatever reason i have to let say rotate the system f.e. to take the second measurement or prepare the system for the something else..
I can use those matrices/operators to keep the correct answer.
i.e. keep the things in sync...

4. Jan 13, 2012

eaglelake

Every observable in quantum mechanics is represented by a Hermitian operator. Spin ½ is the observable represented by the Pauli spin matrices, one for each component. If we do an experiment where we measure the z-component, say, of the spin, then the eigenvalues of the spin operator (the Pauli spin matrix $$\sigma _z$$) are the possible results of the spin measurement. And its eigenvectors become the basis vectors in spin space; all spin ½ state vectors are written as the superposition of the eigenvectors.

In short, solving the eigenvalue equation for a Pauli spin matrix allows us to determine the possible results for a spin measurement and the probability distribution of those results. We do this for any quantum observable, but the calculations for spin ½ are the simplest to make since spin ½ space is only two-dimensional.

Best wishes

5. Jan 13, 2012

cygnet1

I am new to this field, which is far from intuitive for me! I want to make sure I understand what you're saying.

I think I understand that the quantum state vector depends on your choice of coordinate system. Suppose you're given a state vector |ψz> with respect to the "standard" (x, y, z) system. (I'm not sure whether this is left or right-handed.) We can decompose this vector as a linear combination of the eigenvectors (ez1, ez2) of the σz Pauli matrix:

$|\psi_z \rangle = a_{z1} |e_{z1}\rangle + a_{z2} |e_{z2}\rangle = a_{z1} \left (\begin{array}{c}1\\0\end{array} \right) + a_{z2} \left (\begin{array}{c}0\\1\end{array} \right)$

The spin-up probability when measured in the z-direction is then |az1|2, and the spin-down probability is |az1|2. Correct?

We also can decompose the same state into components of the σx eigenvectors

$|\psi_z \rangle = a_{x1} |e_{x1}\rangle + a_{x2} |e_{x2}\rangle = \frac{1}{\sqrt 2}a_{x1} \left (\begin{array}{c}1\\1\end{array} \right) + \frac{1}{\sqrt 2}a_{x2} \left (\begin{array}{c}1\\-1\end{array} \right)$,

to obtain the probability |ax1|2 of a spin-up measurment in the x-direction. If we denote the Ux matrix to be

$U_x = \frac{1}{\sqrt 2} \left ( \begin{array}{cc} 1 & 1 \\ 1 & -1 \end{array} \right )$,

then the "rotated" quantum state |ψx> becomes

$|\psi_x \rangle = \left ( \begin{array}{c} a_{x1} \\ a_{x2} \end{array} \right ) = U_x^\dagger | \psi_z \rangle = e^{i \pi \sigma_y/4} |\psi_z \rangle.$

Am I on the right track?