michael879 said:
The pauli matrices, when used in the spin operator, transform like a 3-vector
.
No, they DO NOT transform much in the same way that ordinary numbers do not transform. This seems to become a common misunderstanding now days. So, let me explain to you how the spin vector transforms under an SO(3) rotation. The spin vector of one particle state is given by
S_{ i } = \int d^{ 3 } x \ \psi^{ \dagger } ( x ) \sigma_{ i } \psi ( x ) \ , \ \ (1)
where \psi ( x ) is the spinor “wave function”. Being a 3-vector, it has to obey the same (infinitesimal) transformation law of the coordinates:
\bar{ x }_{ i } = x_{ i } + \omega_{ i j } x_{ j } \ ,
where
\omega_{ 1 2 } = \theta_{ 3 } , \ \omega_{ 2 3 } = \theta_{ 1 } , \ \omega_{ 3 1 } = \theta_{ 2 } ,
are the infinitesimal rotation angles about the coordinates z, x and y. So, the spin vector must transform according to
\bar{ S }_{ i } = S_{ i } + \omega_{ i j } S_{ j } \ . \ \ \ (2)
Now comes the important point: under space rotation, the transformation of the spin vector rests entirely upon the spinors (NOT THE SIGMA’S). That is
\bar{ S }_{ i } = \int d^{ 3 } x \bar{ \psi }^{ \dagger } \sigma_{ i } \bar{ \psi } \ . \ \ (3)
So, eq(2) has to be achieved by the infinitesimal transformations
\bar{ \psi } = ( 1 + \eta ) \psi , \ \ \ \bar{ \psi }^{ \dagger } = \psi^{ \dagger } ( 1 + \eta^{ \dagger } ) \ . \ \ (4)
To determine this transformation law (i.e. to find \eta), first we observe that
\bar{ \psi }^{ \dagger } \bar{ \psi } = \psi^{ \dagger } ( 1 + \eta^{ \dagger } ) ( 1 + \eta ) \psi \equiv \psi^{ \dagger } \psi \ ,
is a scalar only if
\eta^{ \dagger } = - \eta \ . \ \ \ (5)
Putting (5) and (4) in (3), we find
\bar{ S }_{ i } = S_{ i } + \int d^{ 3 } x \ \psi^{ \dagger } ( \sigma_{ i } \eta - \eta \sigma_{ i } ) \psi \ . \ \ (6)
Now, comparing (6) with (2) leads to
[ \sigma_{ i } , \eta ] = \omega_{ i j } \sigma_{ j } \ .
This system of equations can be solved unambiguously by
\eta = \frac{ i }{ 2 } ( \omega_{ 1 2 } \sigma_{ 3 } + \omega_{ 2 3 } \sigma_{ 1 } + \omega_{ 3 1 } \sigma_{ 2 } ) = \frac{ i }{ 2 } \vec{ \theta } \cdot \vec{ \sigma } \ .
We recognise (the exponential of) this (I hope) as the correct SO(3) = SU(2) / Z_{ 2 } unitary transformation of spinors. So, you must keep this in mind, coordinate transformations do not change the
matrices which
generate the unitary transformations in Hilbert space. So, when you make coordinate transformations, Dirac and Pauli matrices must be treated like ordinary numbers.
Maybe you should read a thread before you comment on it...
Yeah thanks, maybe next time :)
My question here is really simple, and still hasn't been answered.. Let me just try to make it as specific as possible:
I have a massive system in a state with some angular momentum, let's say \hbar. If I were to measure its total angular momentum (assuming this is possible, I don't really see how one would), would I always measure \sqrt{2}\hbar? From the responses in this thread the answer is yes, which led to the question: why? In classical relativistic mechanics the total angular momentum of any system is not Lorentz invariant, and can be altered by boosting the system. How can total angular momentum go from being the magnitude of a 3-vector, to being a Lorentz scalar in the transition from classical to quantum mechanics?? Surely all classical systems can be described as very large quantum systems, so shouldn't all angular momenta be lorentz scalars??
The magnitude of spin angular momentum IS the eigen-value of the INVARIANT Casmir operator. In the non-relativistic (SO(3)) case, this SO(3)-INVARIANT is given by s ( s + 1 ), and in the (massive) relativistic (SO(3 , 1)) case, this SO(1,3)-INVARIANT is given by - m^{ 2 } s ( s + 1 ).
