Editing Canonical transformation (section)

== Infinitesimal canonical transformation ==
Consider the canonical transformation that depends on a continuous parameter <math>\alpha </math>, as follows:

<math display="block">\begin{align}
& Q(q,p,t;\alpha) \quad \quad \quad & Q(q,p,t;0)=q \\  
& P(q,p,t;\alpha) \quad \quad \text{with} \quad & P(q,p,t;0)=p    \\   
\end{align}  </math>

For infinitesimal values of <math>\alpha </math>, the corresponding transformations are called as ''infinitesimal canonical transformations'' which are also known as differential canonical transformations.

=== Explicit construction ===
Consider the following generating function:

<math display="block">G_2(q,P,t)= qP + \alpha G(q,P,t)  </math>

Since for <math>\alpha=0 </math>, <math>G_2 = qP </math> has the resulting canonical transformation, <math>Q = q </math> and <math>P = p </math>, this type of generating function can be used for infinitesimal canonical transformation by restricting <math>\alpha </math> to an infinitesimal value. 

From the conditions of generators of second type:

<math display="block">\begin{align}
{p} &= \frac{\partial G_{2}}{\partial {q}} = P + \alpha  \frac{\partial G}{\partial {q}} (q,P,t) \\
{Q} &= \frac{\partial G_{2}}{\partial {P}} = q + \alpha  \frac{\partial G}{\partial {P}}  (q,P,t) \\
\end{align}</math>


Since <math>P = P(q,p,t;\alpha) </math>, changing the variables of the function <math>G </math> to <math>G(q,p,t) </math> and neglecting terms of higher order of <math>\alpha </math>, gives:<ref>{{Harvnb|Johns|2005|p=452-454}}</ref>

<math display="block">\begin{align}
{p} &= P + \alpha  \frac{\partial G}{\partial {q}} (q,p,t) \\
{Q} &= q + \alpha  \frac{\partial G}{\partial p}  (q,p,t) \\
\end{align}</math>

Infinitesimal canonical transformations can also be derived using the matrix form of the symplectic condition.<ref name=":1">{{Cite web |last=Hergert |first=Heiko |date=December 10, 2021 |title=PHY422/820: Classical Mechanics |url=https://people.nscl.msu.edu/~hergert/phy820/material/pdfs/w14.pdf |url-status=live |archive-url=https://web.archive.org/web/20231222161338/https://people.nscl.msu.edu/~hergert/phy820/material/pdfs/w14.pdf |archive-date=December 22, 2023 |access-date=December 22, 2023}}</ref> The function <math>G(q,p,t) </math> is very significant in infinitesimal canonical transformations and is referred to as the generator of infinitesimal canonical transformation.

=== Active and passive transformations ===
{{See also|Active and passive transformation}}
In the active view of transformations, the coordinate system is changed without the physical system changing, whereas in the passive view of transformation, the coordinate system is retained and the physical system is said to undergo transformations.

==== Active view of transformation ====
Thus, using the relations from infinitesimal canonical transformations, the change in the system states under active view of the canonical transformation is said to be:

<math display="block">\begin{align}
& \delta q = \alpha \frac{\partial G}{\partial p}  (q,p,t)   \quad \text{and} \quad \delta p = - \alpha  \frac{\partial G}{\partial q}  (q,p,t) , \\  
\end{align}  </math>


or as <math>\delta \eta = \alpha J \nabla_\eta G </math> in matrix form.


For any function <math>u(\eta)   </math>, it changes under active view of the transformation according to:

<math display="block">\delta u = u(\eta +\delta \eta)-u(\eta) = (\nabla_\eta u)^T\delta\eta=\alpha (\nabla_\eta u)^T J (\nabla_\eta G) = \alpha \{ u,G \}   . </math>

==== Passive view of transformation ====
Considering the change of Hamiltonians in the [[Active and passive transformation|passive view]], i.e., for a fixed point,<math display="block">K(Q=q_0,P=p_0,t) - H(q=q_0,p=p_0,t) = \left(H(q_0',p_0',t) + \frac{\partial G_{2}}{\partial t}\right) - H(q_0,p_0,t) = - \delta H +\alpha \frac{\partial G}{\partial t} = \alpha\left(\{ G,H\}+\frac{\partial G}{\partial t} \right)=\alpha\frac{dG}{dt}    </math>

where <math display="inline">(q=q_0',p=p_0') </math> are mapped to the point, <math display="inline">(Q=q_0,P=p_0) </math> by the infinitesimal canonical transformation, and similar change of variables for <math>G(q,P,t) </math> to <math>G(q,p,t) </math> is considered up-to first order of <math>\alpha </math>. Hence, if the Hamiltonian is invariant for infinitesimal canonical transformations, its generator is a constant of motion.

=== Generators of dynamical symmetry transformations ===
Consider the transformation where the change of coordinates also depends on the generalized velocities.

<math display="block">\begin{align}
q^r\to q^r+\delta q^r\\
\delta q^r=\epsilon\phi^r(q,\dot{q},t)\\
\end{align}</math>

If the above is a dynamical symmetry, then the lagrangian changes by:

<math display="block">\delta L=\epsilon\frac d {dt}F(q,\dot q,t)</math>

and the new Lagrangian is said to be dynamically equivalent to the old Lagrangian as it ensures the resultant equations of motion being the same. The change in generalized velocity and momentum term can be derived as:

<math display="block">\begin{align}
p=\frac{\partial L}{\partial \dot q}, \quad& \dot q=\frac {dq}{dt}\\
\delta p_r=\frac{\partial^2L}{\partial q^s\partial\dot q^r}\delta q^s+\frac{\partial^2 L}{\partial \dot q^s\partial \dot q^r}\delta \dot q^s,\quad&\delta \dot q^r=\epsilon \frac{\partial \phi^r}{\partial q^s} \dot q^s+\epsilon \frac{\partial \phi^r}{\partial \dot q^s}\ddot q^s+\epsilon\frac{\partial\phi^r}{\partial t}
\\
\end{align}</math>

==== Generator of transformation ====
Using the change in Lagrangian property of a dynamical symmetry:

<math display="block">\frac d{dt}F=\frac{\partial F}{\partial q^r}\dot q^r+\frac{\partial F}{\partial \dot q^r}\ddot q^r+\frac{\partial F}{\partial t}=\frac{\delta L}{\epsilon}=\left(\frac{\partial L}{\partial q^r}\phi^r+\frac{\partial L}{\partial \dot q^r}\frac{\partial \phi^r}{\partial t}\right)+p_s\frac{\partial \phi^s}{\partial q^r}\dot q^r+p_s\frac{\partial \phi^s}{\partial \dot q^r}\ddot q^r</math>

Since the <math>\ddot q</math> terms appear only once in either side, it's coefficients must be equal for this to be true, giving the relation: <math display="inline">p_s\frac{\partial \phi^s}{\partial \dot q^r}=\frac{\partial F}{\partial \dot q^r} </math> using which, it can be shown that 

<math display="block"> \{q^r,\epsilon (p_s\phi^s-F)\}=\delta q^r,\quad \{p_r,\epsilon(p_s\phi^s-F)\}=\delta p_r+\epsilon\left(\frac{\partial L}{\partial q^s}-\frac{d}{dt}\frac{\partial L}{\partial \dot q^s}\right)\frac{\partial \phi^s}{\partial \dot q^r}</math>

Hence, the term <math>p\phi-F</math> generates the canonical dynamical symmetry transformation if either the Euler Lagrange relation gives zero, or if <math>\frac{\partial \phi_s}{\partial \dot q^r}=0\,\forall s,r</math> which is a infinitesimal point transformation. Note that in the point transformation condition, the quantity generates the transformation regardless of if the Euler Lagrange equations are satisfied and since they do not depend on the dynamics of the problem are said to be a purely kinematic relation.<ref>{{Cite journal |last=Mallesh |first=K. S. |last2=Chaturvedi |first2=Subhash |last3=Balakrishnan |first3=V. |last4=Simon |first4=R. |last5=Mukunda |first5=N. |date=2011-02-01 |title=Symmetries and conservation laws in classical and quantum mechanics |url=https://link.springer.com/article/10.1007/s12045-011-0020-5 |journal=Resonance |language=en |volume=16 |issue=2 |pages=129–151 |doi=10.1007/s12045-011-0020-5 |issn=0973-712X|url-access=subscription }}</ref>

{| class="toccolours collapsible collapsed" width="80%" style="text-align:left"
!Proof
|-
|
Firstly, the change in momentum can be expressed in a more useful form as follows:<math display="block">\delta p_r=\frac{\partial^2L}{\partial q^s\partial\dot q^r}\delta q^s+\frac{\partial^2 L}{\partial \dot q^s\partial \dot q^r}\delta \dot q^s=\frac{\partial}{\partial \dot q^r}\left(\frac{\partial L}{\partial q^s}\delta q^s+\frac{\partial L}{\partial \dot q^s}\delta \dot q^s\right)-\frac{\partial L}{\partial  q^s} \frac{\partial}{\partial \dot q^r}(\delta q^s)-\frac{\partial L}{\partial \dot q^s} \frac{\partial}{\partial \dot q^r}(\delta\dot  q^s)=\frac{\partial}{\partial \dot q^r}(\delta L)-p_s\frac{\partial}{\partial \dot q^r}(\delta \dot q^s)-\frac{\partial L}{\partial q^s}\frac{\partial}{\partial \dot q^r}(\delta q^s)</math>

Simplifying the required poisson brackets,

<math> \begin{align}
\{q^r,\epsilon (p_s\phi^s-F)\}=\epsilon \left(\phi_r+\frac{\partial \dot q^m}{\partial p_r}\cancelto{=0}{\left(p_s\frac{\partial \phi^s}{\partial \dot q^m}-\frac{\partial F}{\partial \dot q^m}\right)}\right)&=\delta q^r\\
\{p_r,\epsilon(p_s\phi^s-F)\}=\epsilon\left(-p_s\frac{\partial \phi^s}{\partial q^r}+\frac{\partial F}{\partial q^r}+\cancelto{=0}{\left(\frac{\partial F}{\partial \dot q^m}-p_s\frac{\partial \phi^s}{\partial \dot q^m}\right)}\left(\frac{\partial \dot q^m}{\partial q^r}\right)_{q,p,t}\right) &=\epsilon\left(-p_s\frac{\partial \phi^s}{\partial q^r}+\frac{\partial F}{\partial q^r}\right)\\
\end{align} </math>


As a preliminary result, for any function of <math>(q,\dot q,t)</math>,  

<math> \frac{\partial}{\partial \dot q^r}\frac{d}{dt}-\frac{d}{dt}\frac{\partial}{\partial \dot q^r}=\frac{\partial}{\partial q^r}+\frac{\partial \ddot q^s}{\partial \dot q^r}\frac{\partial}{\partial \dot q^s}</math>

which can be used to calculate the quantity:

<math> \frac{\partial}{\partial \dot q^r}\left(\frac {dF}{dt}\right)-p_s\left(\frac{\partial}{\partial \dot q^r}\left(\frac {d}{dt}\phi^s\right)\right)-\dot p_s\frac{\partial}{\partial \dot q^r}(\phi^s)=\frac{d}{dt}\cancel{\left(\frac{\partial}{\partial \dot q^r}F-p_s\frac{\partial}{\partial \dot q^r}\phi^s\right)}+\frac{\partial \ddot q^s}{\partial \dot q^r}\cancel{\left(\frac{\partial}{\partial \dot q^s}F-p_m\frac{\partial}{\partial \dot q^s}\phi^m\right)}-p_s\frac{\partial \phi^s}{\partial q^r}+\frac{\partial F}{\partial q^r}=\{p_r,(p\phi-F)\}</math>


This relation can be restated and combined with the formula for <math>\delta p_r</math>to give the required relation for momentum.

<math>\{ p_r,\epsilon(p_s\phi^s-F)\}=\frac{\partial}{\partial \dot q^r}(\delta L)-p_s\frac{\partial}{\partial \dot q^r}(\delta \dot q^s) - \dot p_s \frac{\partial}{\partial \dot q^r}(\delta q^s)=\delta p_r+\epsilon\left(\frac{\partial L}{\partial q^s}-\frac{d}{dt}\frac{\partial L}{\partial \dot q^s}\right)\frac{\partial \phi^s}{\partial \dot q^r} </math>

|}

==== Noether Invariant ====
Using Euler Lagrange relation for the provided Lagrangian, the invariants of motion can be derived as:<math display="block">\delta L-\epsilon\frac d {dt}F(q,\dot q,t)= \epsilon\phi\cancelto{=0}{\left(\frac{\partial}{\partial q}-\frac{d}{dt}\frac{\partial}{\partial \dot q}\right)L}+\epsilon\frac{d}{dt}\left(\phi\frac{\partial}{\partial \dot q}L- F\right)=\epsilon\frac{d}{dt}\left(\phi\frac{\partial}{\partial \dot q}L- F\right)=0</math>

Hence <math>\left(\phi\frac{\partial}{\partial \dot q}L-F\right)=p\phi-F</math> is a constant of motion. Hence, the derived Noether invariant also generates the same symmetry transformation as shown previously.

=== Examples of ICT ===

==== Time evolution ====
Taking <math>G(q,p,t)=H(q,p,t) </math> and <math>\alpha = dt </math>, then <math>\delta \eta = (J \nabla_\eta H) dt = \dot{\eta} dt = d\eta </math>. Thus the continuous application of such a transformation maps the coordinates <math>\eta(\tau) </math> to <math>\eta(\tau+t)   </math>. Hence if the Hamiltonian is time translation invariant, i.e. does not have explicit time dependence, its value is conserved for the motion.

==== Translation ====
Taking <math>G(q,p,t)=p_k  </math>, <math> \delta p_i = 0 </math> and <math> \delta q_i = \alpha \delta_{ik} </math>. Hence, the canonical momentum generates a shift in the corresponding generalized coordinate and if the Hamiltonian is invariant of translation, the momentum is a constant of motion.

==== Rotation ====
Consider an orthogonal system for an N-particle system:

<math display="block">\begin{array}{l}{{\mathbf q=\left(x_{1},y_{1},z_{1},\ldots,x_{n},y_{n},z_{n}\right),}}\\ {{\mathbf p=\left(p_{1x},p_{1y},p_{1z},\ldots,p_{n x},p_{n y},p_{n z}\right).}}\end{array}</math>

Choosing the generator to be: <math>G=L_{z}=\sum_{i=1}^{n}\left(x_{i}p_{i y}-y_{i}p_{i x}\right) </math> and the infinitesimal value of <math> \alpha = \delta \phi </math>, then the change in the coordinates is given for x by:

<math display="block">\begin{array}{c}
{\delta x_{i}=\{x_{i},G\}\delta\phi=\displaystyle\sum_{j} \{x_{i},x_{j}p_{j y}-y_{j}p_{j x}\}\delta\phi=\displaystyle\sum_{j}(\underbrace{\{x_{i},x_{j}p_{j y}\}}_{=0} -{ \{x_{i},y_{j}p_{j x}\}}})\delta\phi\\
{{=\displaystyle -\sum_{j} y_{j} \underbrace{\{x_i,p_{jx}\}}_{=\delta_{ij}}\delta\phi=- y_{i} \delta \phi}}  
\end{array}  </math>

and similarly for y:

<math display="block">\begin{array}{c}

\delta y_{i}=\{y_{i},G\}\delta\phi=\displaystyle\sum_{j}\{y_{i},x_{j}p_{j y}-y_{j}p_{j x}\}\delta\phi=\displaystyle\sum_{j}(\{y_{i},x_{j}p_{j y}\}-\underbrace{ \{y_{i},y_{j}p_{j x}\}}_{=0})\delta \phi\\ 
{=\displaystyle\sum_{j}x_{j}\underbrace{\{y_i,p_{jy}\}}_{=\delta_{ij}} \delta\phi=x_{i}\delta\phi\,,} 
\end{array} </math>

whereas the z component of all particles is unchanged: <math display="inline"> \delta z_{ i }=\left\{z_{i},G\right\}\delta\phi=\sum_{j}\left\{z_{i},x_{j}p_{j y}-y_{j}p_{j x}\right\}\delta \phi =0</math>.

These transformations correspond to rotation about the z axis by angle <math>\delta \phi </math> in its first order approximation. Hence, repeated application of the infinitesimal canonical transformation generates a rotation of system of particles about the z axis. If the Hamiltonian is invariant under rotation about the z axis, the generator, the component of angular momentum along the axis of rotation, is an invariant of motion.<ref name=":1" />