mma said:
The generator functions hence the physical quantities of the elementary particles in these representations will be just as we expect.
I feel that this should be explained.
First of all, what does the strange expression dq \wedge dp \, mean?
This is nothing else than a brief notation for \sum_k{dq_k \wedge dp_k} \,. So, the matrix of \omega \, in the \frac{\partial}{\partial q_1},\frac{\partial}{\partial q_2},\frac{\partial}{\partial q_3}, \frac{\partial}{\partial p_1},\frac{\partial}{\partial p_2},\frac{\partial}{\partial p_3} \, basis is
\begin{pmatrix}<br />
0 & I \\<br />
-I & 0 \\<br />
\end{pmatrix}, where I \, is the 3x3 identity matrix and 0 \, is the zero matrix.
The symplectic representation A \, of a one-parameter subgroup \exp(ta)\, of the group G on the symplectic manifold (M, \omega) \, is a symplectic flow on M. A simplectic flow on M is a \Bbb{R} \times M \mapsto M : (t,x) \mapsto A_{exp(ta)}(x) function that keeps the symplectic form.
The function f \, is the
generator function of this flow by definition, if the vector field X =j^{-1}_\omega(df) is the infinitesimal generator of the flow, i.e. X is the Killing vector field belonging to a \in \mathfrak{g} in the given representation. It means that in a x \in M point X(x) = \frac{d}{dt}\left.{\!\!\frac{}{}}\right|_{t=0}A_{\exp(ta)}(x)
The equation X = j^{-1}_\omega(df) means that for any vector field Y \,, df(Y) = \omega(X,Y)\,. In our case
j^{-1}_\omega(df) = (d_pf, -d_qf) and j^{-1}_{\omega\sigma}(df) = (\frac{\partial f}{\partial z}, -\frac{\partial f}{\partial \phi}) where d_qf = (\frac{\partial f}{\partial q_1},\frac{\partial f}{\partial q_2},<br />
\frac{\partial f}{\partial q_3}) and d_pf = (\frac{\partial f}{\partial p_1},\frac{\partial f}{\partial p_2},<br />
\frac{\partial f}{\partial p_3}).
The canonical physical quantities are the generator functions of the one-parameter flows. Namely, the generator function of a one-parameter flow of translation, rotation and time-shift is called momentum, angular momentum and energy, respectively.
Let's see if this definition in our case really yields the usual physical quantities plus the less usual spin.
Energy
For calculating the generator function belonging to energy, let's take a one-parameter flow of time-shift. The flow is now A_t(q,p,s) = (q + \frac{p}{\mu}t, p,s) \,.
The generator function H of this flow of this flow must satisfy
j^{-1}_\omega(dH) = (d_pH,-d_q H) = \frac{d}{dt}\left.{\!\!\frac{}{}}\right|_{t=0}\mathrm{proj}_1(A_t(q,p,s)) = (\frac{p}{\mu},0)
and
j^{-1}_{\omega \sigma}(dH) =(\frac{\partial H}{\partial z}, -\frac{\partial H}{\partial \phi}) = \frac{d}{dt}\left.{\!\!\frac{}{}}\right|_{t=0}\mathrm{proj}_2(A_t(q,p,s)) = (0, 0)
That is,
\frac{\partial H}{\partial p_1} = \frac{p_1}{\mu}, \frac{\partial H}{\partial p_2} = \frac{p_2}{\mu}, \frac{\partial H}{\partial p_3} = \frac{p_3}{\mu},
\frac{\partial H}{\partial q_1} = 0,\frac{\partial H}{\partial q_2} = 0, \frac{\partial H}{\partial q_3} = 0,
\frac{\partial H}{\partial \phi} = 0 and \frac{\partial H}{\partial z} = 0.
The solution of these equation is H = \frac {p^2}{2\mu} + const
This is the energy of the free particle.
Momentum
In order to calculate momentum, let's take a one-parameter flow of translation along the k-th coordinate line. For the sake of definiteness, take k=1. The flow is now A_t(q,p,s) = (q + (t,0,0), p,s) \,.
According the definitions above, the generator function P_1 of this flow of this flow must satisfy
j^{-1}_\omega(dP_1) = (d_ pP_1,-d_q P_1) = \frac{d}{dt}\left.{\!\!\frac{}{}}\right|_{t=0}\mathrm{proj}_1(A_t(q,p,s)) = ((1,0,0),(0,0,0))
and
j^{-1}_{\omega \sigma}(dP_1) =(\frac{\partial P_1}{\partial z}, -\frac{\partial P_1}{\partial \phi}) = \frac{d}{dt}\left.{\!\!\frac{}{}}\right|_{t=0}\mathrm{proj}_2(A_t(q,p,s)) = (0, 0)
That is,
\frac{\partial P_1}{\partial p_1} = 1,\frac{\partial P_1}{\partial p_2} = 0, \frac{\partial P_1}{\partial p_3} = 0,
d_q P_1 = 0 \,,
\frac{\partial P_1}{\partial \phi} = 0 and \frac{\partial P_1}{\partial z} = 0.
The solution of these equation is P_1 = p_1 + const \,. Performing these calculations for arbitrary k =1,2,3 instead of k=1 we get P_k = p_k + const \,. This is the k-th component of the momentum.
Angular momentum
Now take the one-parameter flow of the rotations to see the angular momentum. Let's take one-parameter flow of rotations around the k-th axis. Now our flow is
A_t(q,p,s) = (R_k(t)q , R_k(t)p , R_k(t)s) \,
where R_k(t) \, the matrix of the rotation by t around the k-th axis. Now the generator function J_k \, of this flow of this flow must satisfy
j^{-1}_\omega(dJ_k) = (d_p J_k, -d_q J_k) = \frac{d}{dt}\left.{\!\!\frac{}{}}\right|_{t=0}\mathrm{proj}_1(A_t(q,p,s)) = (\dot{R}_k(0)q , \dot {R}_k(0)p) and
j^{-1}_{\omega \sigma}(dJ_k) =(\frac{\partial J_k}{\partial z}, -\frac{\partial J_k}{\partial \phi}) = \frac{d}{dt}\left.{\!\!\frac{}{}}\right|_{t=0}\mathrm{proj}_2(A_t(q,p,s)) = \dot{R}_1 s
For the calculations let's take again k = 1.
The matrix of the rotation about the 1st axis is
R_1(t) = \begin{pmatrix}<br />
1 & 0 & 0 \\<br />
0 & \cos t &-sin t\\<br />
0 & \sin t & cos t\\<br />
\end{pmatrix}, the derivative of it at t=0 is
\dot{R}_1(0) = \begin{pmatrix}<br />
0 & 0 & 0 \\<br />
0 & 0 & -1\\<br />
0 & 1 & 0\\<br />
\end{pmatrix}
Applying this matrix we get j_\omega^{-1}(dJ_k) = (d_pJ_1, -d_q J_1) = ((0, -q_3, q_2)), (0, -p_3, p_2) \,. That is,
\frac{\partial J_1}{\partial p_1} = 0, \frac{\partial J_1}{\partial p_2} = -q_3, \frac{\partial J_1}{\partial p_3} = q_2 \,,
\frac{\partial J_1}{\partial q_1} = 0, \frac{\partial J_1}{\partial q_2} = -p_3, \frac{\partial J_1}{\partial q_3} = p_2 \,.
The solution for J_1 \, is: J_1 = -q_2p_2 + q_2p_3 + S_1(s)\,.
If we calculate for k=2 and 3 too, we get J_k = \varepsilon_{klm}q_lp_m + S_k(s)\,, i. e.
J = q \times p + S(s) \,, where S(s) will be calculated in the following.
Spin
Now continue to calculate the generator function of the one-parameter flow of the rotations by calculating S(s) \, from the condition
j^{-1}_{\omega\sigma}(dJ_k) = (\frac{\partial J_k}{\partial z}, -\frac{\partial J_k}{\partial \phi}) = \frac{d}{dt}\left.{\!\!\frac{}{}}\right|_{t=0}\mathrm{proj}_2(A_t(q,p,s)) = \dot{R}_k(0)s
We know already that J_k(p,q,s) = L_k(p,q) + S_k(s) \,, where L_k = (q \times p)_k.
Substituting J_k \, in our equation with L_k(p,q) + S_k(s) \, we get the condition
(\frac{\partial S_k}{\partial z}, -\frac{\partial S_k}{\partial \phi}) = \dot{R}_k(0)s
Here s \, is a point on the \Bbb{S}^2_\sigma \, sphere.
We will use the following charts on the \Bbb{S}^2_\sigma \,.
1. We embed \Bbb{S}^2_\sigma \, into \Bbb{R}^3 \, and use the s_1, s_2 \, and s_3 \, coordinates of \Bbb{R}^3 \, having s_1^2 + s_1^2 + s_1^2 = \sigma^2.
2. We leave the (0,0,\sigma) \, and (0,0,-\sigma) \, points from the sphere and also the half circle connecting them and passing through (-\sigma, 0,0) \,. We coordinatize this slashed sphere by the
\phi(s) := \mathrm{sign}(s_2)\mathrm{arccos}\frac{s_1}{\sqrt{s_1^2 + s_2^2}},
z(s) := s_3 \,
functions. We can coordinatize similarly the open set made by leaving from the sphere the half circle connecting the points (0, \sigma,0) \, and (0, -\sigma, 0) \, passing through (\sigma, 0,0) \,. These two open sets cover the sphere.
In this local coordinate chart R_3(t)(\phi, z) = (\phi + t, z) \,, hence \dot{R}_3(0)(\phi, z) = (1 , 0), so our condition for the generator function S_3 \, is
(\frac{\partial S_3}{\partial z}, -\frac{\partial S_3}{\partial \phi}) = (1 , 0)
The solution is: S_3 = z = s_3 + const \,. The other two components yields similar result, so we can state
S_k = s_k + const \, is the generator function of the rotation of the sphere around the k-th axis.
Putting together this with our previous result for the angular momentum, we can say that the total generator function of such rotation on the manifold ((\Bbb{R}^3 \times \Bbb{R}^3) \times \Bbb{S}^2_\sigma, \omega \times \omega_\sigma) is:
J_k = (q \times p)_k + s_k \,.
This is the k-th component of the angular momentum including spin. In the case of spin length \sigma = 0 \,, this is the usual classical, non-relativistic expression for the angular momentum.
Voilá
