# Symplectic mechanics - elementary particles

1. Nov 30, 2008

### nughret

In Souriau's book of symplectic mechanics he describes an elementary dynamical system on which the Poincare group is dynamic and acts transitively. He then describes a massive particle with spin where the spin is some positive number. When we consider this particle in the presence of an external electromagnetic field then this spin has similar properties to that considered in QM, such as the elementary particle has a magnetic moment parallel to the spin. I asked my lecturer if it was then the case that spin is a property of relativity rather than QM but he has not read Souria's book but said that he had heard that Souriau had his own intepretation of spin. I was then wondering if anyone knew of the intepretation of this intrinsic spin described by Souriau, if the system is quantized will this correspond to spin in QM?

2. Dec 1, 2008

### mma

Spin isn't a relativistic nor a quantummechanical feature. Souriau also introduce spin in the norelativistic case, see p.187 of his book.

3. Dec 2, 2008

### nughret

Do you mean that you can "insert" an internal structure that then transforms correctly to represent the spin? It seems to me though that when we define an elementary symplectic structure that transforms according to the fact that the Poincare group is dynamic that the notation of spin emerges naturally. It doesn't for example if we define the angular momentum vector as l = r x p.

Yes here though he takes the limit |v|<< c of the relativistic case to get the structure. It may be the case that spin also comes out naturally in a system on which the Galilei group is dynamic and acts transitively. Souriau doesn't go into the detail of this though so I am going to start working through to see what happens.
If you have studied Souriau's book then I will have some other questions which you could help with.

4. Dec 2, 2008

### Haelfix

I don't see how you can derive spin without working from a special relativity framework. You can surely add it adhoc to a nonrelativistic system, but actually deriving it from first principles.. I have never seen that done before.

5. Dec 3, 2008

### mma

Unfortunately, I don't have a deep insight in this field, even I know Souirau's book only in very scrahtchy, but I have read in a lecure note the fllowing theorem.

Let $$\Bbb{S}^2_\sigma$$ denote the sphere with radius $$\sigma > 0$$ and $$\omega_\sigma = d\phi \wedge dz$$.

The equivalence classes of the transitive, differentiable, faithful semi-symplectic representations of the Galilean group are characterised by a $$\mu > 0$$ and a $$\sigma \geq 0$$ real numbers. The $$A^{(\mu,\sigma})$$ representative of the equivalence class belonging to $$(\mu, \sigma)$$ on $$((\Bbb{R}^3 \times \Bbb{R}^3) \times \Bbb{S}^2_\sigma, \omega \times \omega_\sigma)$$ can give on the elements $$(0,\xi,v,R)$$ and $$(\tau ,0,0, I)$$ as

$$A^{(\mu,\sigma)}_{(0,\xi,v,R)}(q,p,s) = (Rq + \xi, Rp - \mu v, Rs)$$,

$$A^{(\mu,\sigma)}_{(\tau,0,0,I)}(q,p,s) = (q + \frac{p}{\mu}\tau, p, s)$$

($$(q,p,s) \in ((\Bbb{R}^3 \times \Bbb{R}^3) \times \Bbb{S}^2_\sigma$$)

So we get all reansitive, differentiable, semi-symplectic representation to which the commutator cocycle belong to the $$\mu \neq 0$$ weak cohomology class.

The geneator functions hence the physical quantities of the elementary paticles in these representations will be just as we expect.

Last edited: Dec 3, 2008
6. Dec 3, 2008

### mma

Of course $$\mu$$ is the mass, $$\sigma$$ is the spin of the elementary particle. The evolution space is $$(\Bbb{R}^3 \times \Bbb{R}^3) \times \Bbb{S}^2_\sigma$$, $$\omega = dq \wedge dp$$

7. Dec 3, 2008

### shoehorn

I haven't seen precisely this done but it's almost trivial to "derive" spin in a non-relativistic framework by looking at the behaviour of a point particle using a graded Berezin algebra.

Highly unusual, admittedly, but possible nonetheless.

8. Dec 4, 2008

### mma

Could you explain this shortly, please?

9. Dec 7, 2008

### mma

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
$$\begin{pmatrix} 0 & I \\ -I & 0 \\ \end{pmatrix}$$, 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 $$d_qf = (\frac{\partial f}{\partial q_1},\frac{\partial f}{\partial q_2}, \frac{\partial f}{\partial q_3})$$ and $$d_pf = (\frac{\partial f}{\partial p_1},\frac{\partial f}{\partial p_2}, \frac{\partial f}{\partial p_3})$$.

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

$$R_1(t) = \begin{pmatrix} 1 & 0 & 0 \\ 0 & \cos t &-sin t\\ 0 & \sin t & cos t\\ \end{pmatrix}$$, the derivative of it at t=0 is
$$\dot{R}_1(0) = \begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & -1\\ 0 & 1 & 0\\ \end{pmatrix}$$

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

Last edited: Dec 7, 2008
10. Dec 8, 2008

### shoehorn

Not really. You could, however, take a look at chapter 11 of Peter Freund's SUSY book to see how this is done.

11. Dec 8, 2008

### mma

Thank you for the reference!

12. Dec 12, 2008

### nughret

Thanks for the in depth post mma very helpful. So is it the case that if we have any group with a subgroup that acts like SO(3) on our manifold we can form a concept of spin that acts as we would expect?