Pauli matrices

1. Oct 5, 2008

Lojzek

I have some questions about Pauli matrices:

1. How do we calculate them? Which assumptions are needed?
Are the assumptions related to properties of orbital angular momentum in any way?

2. How do we prove that the Pauli matrices (the operators of spin angular momentum) are the generators of the group of rotations (on a particle with spin 1/2)?

Last edited: Oct 5, 2008
2. Oct 5, 2008

Rettaw

You don't really calculate them, maybe you mean why did Pauli construct them?

As I recall it they where constructed to have the properties of the angular momentum algebra, that is (up to constants I can never recall)
$$[\sigma_i , \sigma_j] = \epsilon_{i,j,k}\sigma_k$$
where the epsilon thins is the levi-cevita tensor (among other names).

As I'm unsure of what people mean when they talk about the generators of something I can't help you much with that, but it will probably ammount to showing that the pauli matrices obey the relation I wrote above.

3. Oct 5, 2008

Ben Niehoff

Calculate

$$A = e^{\alpha \sigma_1 + \beta \sigma_2 + \gamma \sigma_3}$$

and show that A is a member of SU(2).

Hint: You will save yourself some tedious algebra if you use the properties of the matrix exponential, combined with the fact that the Pauli matrices are traceless and Hermitian.

4. Oct 6, 2008

PhilDSP

Another important way to view Pauli matrices is that they are representations of infinitesimal rotation. So they are generators of both rotation groups SU(2) and SO(3).

If you generate a matrix which rotates a 3 dimensional object in each dimension, then take the limit of its effects approaching zero as it operates on each dimension, then reduce that matrix so that the effect on each dimension is separated to its own matrix, you will produce the Pauli matrices.

A good book with that derivation is "Spinors in Physics" by Jean Hladik.

5. Oct 6, 2008

samalkhaiat

I will be as clear as I possibly can, assuming that you are asking
1) How can we arrive at the Pauli matrices?
2) What role do they play in the transformation law of spinors?

I) Math

Any 2x2 Special Unitary matrix (under which a 2-component spinor transforms) can be written as

$$\exp (i A)$$

where A is traceless hermitian 2x2 matrix.
Thus, we can write

$$A = \left( \begin{array}{rr} \theta_{3}/2 & a^{*} \\ a & - \theta_{3}/2 \\ \end{array} \right)$$

(notice that $Tr(A) = 0, \ \mbox{and} \ A^{\dagger} = A$ )

Now, if we write

$$a = \frac{1}{2} \theta_{1} + \frac{i}{2}\theta_{2}$$

$$a^{*} = \frac{1}{2}\theta_{1} - \frac{i}{2}\theta_{2}$$

the matrix A becomes

$$A = \frac{\theta_{1}}{2}\left( \begin{array}{rr} 0 & 1 \\ 1 & 0 \\ \end{array}\right) + \frac{\theta_{2}}{2} \left( \begin{array}{rr} 0 & -i \\ i & 0 \\ \end{array}\right) + \frac{\theta_{3}}{2} \left( \begin{array}{rr} 1 & 0 \\ 0 & -1 \\ \end{array}\right)$$

or

$$A = \frac{1}{2}\theta_{i}\sigma_{i}$$

Now you can show that the matrices $\sigma_{i}/2$ generate the Lie algebra of “rotation”

$$\left[ \sigma_{i}/2 , \sigma_{j}/2 \right] = i \epsilon_{ijk}\sigma_{k}/2$$

6. Oct 6, 2008

samalkhaiat

II) Physics

Since the spin of a one particle state

$$S_{i} = \int \ d^{3}x \Psi^{\dagger} \sigma_{i} \Psi \ \ \ \ (1)$$

is a 3-vector, it has to obey the same transformation law of coordinates

$$\bar{x}_{i} = x_{i} + \omega_{ij}x_{j}\ \ \ (2)$$

under infinitesimal rotation about the three axes;

$$\omega_{23} = \theta_{1}, \ \omega_{31} = \theta_{2}, \ \omega_{12} = \theta_{3}$$

are the infinitesimal angles about $x_{1}, x_{2}, \ \mbox{and}\ x_{3}$

Thus under infinitesimal rotation, we have

$$\bar{S}_{i} = S_{i} + \omega_{ij}S_{j} \ \ \ (3)$$

It should be noted from eq(1) that the transformation properties of the spin vector rest entirely upon the 2-component wavefunction (spinor)

$$\Psi = \left( \begin{array}{c} a \\ b \\ \end{array} \right)$$

$$\Psi^{\dagger} = \left( a^{*} \ b^{*} \right)$$

So, eq(3) has to be obtained from the transformations

$$\bar{\Psi} = ( 1 + B ) \Psi, \ \mbox{and} \ \bar{\Psi}^{\dagger} = \Psi^{\dagger}( 1 + B^{\dagger}) \ \ \ (4)$$

where B is (unknown) infinitesimal 2x2 matrix. To determined B, we use the fact that rotations do not change the probability density

$$\Psi^{\dagger}\Psi = |a|^{2} + |b|^{2}$$

Thus

$$\bar{\Psi}^{\dagger}\bar{\Psi} = \Psi^{\dagger} ( 1 + B^{\dagger})( 1 + B) \Psi = \Psi^{\dagger}\Psi$$

only if

$$B^{\dagger} = - B \ \ \ \ (5)$$

This means that coordinate transformations (rotations) induce a unitary transformations on wavefunctions (vectors) in the physical Hilbert space.

So,

$$\bar{S}_{i} = \int \ d^{3}x \Psi^{\dagger} ( 1 - B) \sigma_{i} ( 1 + B) \Psi$$

or

$$\bar{S}_{i} = S_{i} + \int \ d^{3}x \Psi^{\dagger}[ \sigma_{i} , B ] \Psi \ \ (6)$$

Comparing this expression with eq(3), we find

$$\left[ \sigma_{i}, B \right] = \omega_{ij}\sigma_{j} \ \ \ (7)$$

There is a cute method for solving these (three) commutator equations, but I will only give you the answer here. Indeed, you can check that

$$B = \frac{i}{2}\left( \omega_{23}\sigma_{1} + \omega_{31}\sigma_{2} + \omega_{12}\sigma_{3}\right) = \frac{i}{2}\theta_{i}\sigma_{i}$$

unambiguously solve the system eq(7).

Regards

sam

7. Oct 8, 2008

Lojzek

Thanks for the replies. I tried to put the information into a correct sequence to
avoid the possibility of a circular proof or "something true follows from this sentence"
type proof:

1. We assume that the rotation operator is linear. The operator can be represented by 2x2 matrix since the spin space is 2 dimensional.

2. The rotation operator must be unitary (so that scalar product is invariant to rotations).

3. The determinant of rotation matrix must be +-1. To preserve volume? I understand
this must be true for rotation of the space part of wavefunction, but I am not sure about spin part.
I think that a matrix with det=+-1 acting on the spin part of wavefunction would preserve
dc(up)*dc(down) volume (where psi=c(up)*|up>+c(down)*|down>). Is this preservation of volume necessary?

4. Now we know that R is a member of SU(2). It is possible to prove that any element of SU(2) can be expressed as exp(i*omega*sigma/2), where sigma is a vector of Pauli matrices and omega is a vector of real numbers.
So sigma are the generators of the group of rotations.

5. We can check that the commutators between components of spin respect the same algebra as orbital angular momentum (however I don't know whether we should get the same algebra despite a different dimension of the space).

Can someone check if this is the correct path?

8. Oct 8, 2008

regards

sam