# The connection between classical and quantum spins

I have two questions, which are connected with each other.

The first question.

In a classical relativistic (SRT) case for one particle can be defined (in a reason of "antisymmetric" nature of angular momentum) 4-tensor $$L_{\alpha \beta} = x_{\alpha }p_{\beta} - x_{\beta }p_{\alpha } = (\mathbf L, \mathbf G),$$ where $\mathbf L = [\mathbf r \times \mathbf p ], \quad \mathbf G = \frac{E}{c}\mathbf r - ct\mathbf p$ are angular momentum and center of energy vectors. It is called angular momentum tensor.

In a case of many particles it's convenient to introduce Pauli-Lubanski 4-vector of spin: $$S_{\nu} = \frac{1}{2}\varepsilon_{\nu \alpha \beta \gamma }L^{\alpha \beta}U^{\gamma},$$ where $U^{\gamma}$ is summary 4-speed. The physical meaning of this vector is representation of own angular momentum of particles (without summand connected with motion of system as one body):

$$S_{0} = (\mathbf L \cdot \mathbf U), \quad \mathbf S = \mathbf L U_{0} - [\mathbf G \times \mathbf U] = \frac{E}{M}\left( \mathbf L - [\mathbf R_{E} \times \mathbf P]\right).$$

By the way, for one particle in classical case $$\mathbf S = \mathbf L - [\mathbf R_{E}\times \mathbf P ] = [\mathbf R_{E}\times \mathbf P ] - [\mathbf R_{E}\times \mathbf P ] = 0 .$$ For many particles in general case $\mathbf L$ include orbital angular momentum, so $\mathbf L - [\mathbf R_{E}\times \mathbf P ] \neq 0$.

There, if we introduce an operator representation, it will be easy to show, that $$[\hat {L}_{x}, \hat {L}_{y}] = [\hat {S}_{x}, \hat {S}_{y}] = i\hat {L}_{z},$$ where equality of first and second commutators is possible in the rest frame (add: i.e., for a case $\hat {\mathbf P}| \Psi_{\mathbf P = 0} \rangle = 0|\Psi_{\mathbf P = 0} \rangle$). That, after simple transformations, leads us to the discrete spectra of eigenvalues of $\hat {L}_{z}$ (or $\hat {s}_{z}$ in a case of the rest frame): $$m = s, s - 1, ..., -s,$$

where s is the "full momentum" (or spin in the rest frame).

So, the question: can we talk about classical analogue of spin in a case of quantum mechanics, when we analyze the quantum system (not one particle)? In a classical case the analogue of spin is an own angular momentum determined above.

The second question.

In a field theory it is well known that angular-spin tensor, which is a current of a Lorentz-symmetry of Lagrangian (Noether formalism). Does it connect with a spin of the quantum field in terms of first question? If the answer is positive, there is another question: has Belifante procedure an analogue of a quantum case?

• Not sure I quite understand the question - are you asking for a connection between quantum and classical spin like in the Einstein de Haas effect? – twistor59 Jul 14 '13 at 17:13
• I'm asking about the interpretation of the spin as a purely quantum phenomenon, which disappears in a classical limit. But in that case I can't introduce operator of the spin using method stated above. – user8817 Jul 14 '13 at 20:07
• I feel like your statement about the "equality of first an second commutators is possible in the rest frame" is strange; what is $\hat{S}_x$? It would appear that you are just showing that that the algebra of angular momentum and spin is the same. – levitopher Jul 14 '13 at 22:27
• As I wrote, 4-vector of spin (i.e., it's 3-vector part) representates an own angular momentum relative to an arbitrary inertial system: $$\mathbf S = \mathbf L U_{0} - [\mathbf G \times \mathbf U] = \frac{E}{M}\left( \mathbf L - [\mathbf R_{E} \times \mathbf P]\right),$$ where $\mathbf R_{E}$ is the radius-vector of the center of energy. – user8817 Jul 14 '13 at 23:45
• In operator's representation $$\hat {\mathbf S } = \hat {\mathbf L}\hat {U}_{0} - [\hat {\mathbf G}\times \hat {\mathbf U} ] = \frac{E}{M}\left( \hat {L} - [\hat {\mathbf R}_{E} \times \hat {\mathbf P}]\right).$$ In a case, when eigenvalue of $\hat {\mathbf P}$ is equal to zero (i.e., at rest frame), the components of \hat {\mathbf S } is equal to full angular momentum $\hat {\mathbf L }$. – user8817 Jul 14 '13 at 23:48