User:EditingPencil/sandbox/Canonical transformation

Allowing the values of $\backslash alpha$ to take continuous range of values in:$$\backslash begin\{align\}$$

& 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} which can be expressed as $\backslash epsilon^\backslash mu(\backslash eta,t;\backslash alpha\; )$ where $\backslash epsilon^\backslash mu(\backslash eta,t;0)=\backslash eta^\backslash mu$.

One parameter subgroup of canonical transformations concerns when the infinitesimal generator function is of the same functional form regardless of the coordinates, that is when

$\backslash delta\backslash epsilon^\backslash mu(\backslash eta,t;\backslash alpha)=\backslash delta\backslash alpha\backslash \{\backslash epsilon^\backslash nu,G\; \backslash \}=\backslash delta\backslash alpha\; J^\{\backslash mu\backslash nu\}\backslash frac\{\backslash partial\; G\}\{\backslash partial\; \backslash epsilon^\backslash nu\}(\backslash epsilon(\backslash eta,t;\backslash alpha\; ),t)$

for all $\backslash eta$ such that the generator has no explicit dependance on $\backslash eta$ or $\backslash alpha$. This forms a differential equation as follows

$\backslash frac\{d\backslash epsilon^\backslash mu(\backslash eta,t;\backslash alpha)\}\{d\; \backslash alpha\}=\; J^\{\backslash mu\backslash nu\}\backslash frac\{\backslash partial\; G\}\{\backslash partial\; \backslash epsilon^\backslash nu\}(\backslash epsilon(\backslash eta,t;\backslash alpha\; ),t)$

It follows that $\backslash epsilon^\backslash mu(\backslash epsilon(\backslash eta,t;\backslash alpha\_1);\backslash alpha\_2)=\backslash epsilon^\backslash mu(\backslash eta,t;\backslash alpha\_1+\backslash alpha\_2)$ since this equation is trivially satisfied when $\backslash alpha\_2=0$ which is considered initial values and the differential equations are of the same form implying the relation due to uniqueness of solutions with initial values given. Hence it also follows that such canonical transformations form a one parameter subgroup.

Another conclusion can follow from the nature of the generator

$\backslash frac\{d\; G(\backslash epsilon(\backslash eta;\backslash alpha),t)\}\{d\; \backslash alpha\}=\backslash \{G,G\backslash \}=0,\backslash ,\backslash forall\; \backslash alpha\; \backslash implies\; G(\backslash epsilon(\backslash eta;\backslash alpha),t)=G(\backslash eta,t)$

which implies that $\backslash frac\{d\backslash epsilon^\backslash mu(\backslash eta,t;\backslash alpha)\}\{d\; \backslash alpha\}=\; \backslash \{\backslash epsilon^\backslash mu(\backslash eta,t;\backslash alpha),G(\backslash eta,t)\backslash \}\_\backslash eta=:-\backslash tilde\; G\; \backslash epsilon^\backslash mu$ upon taking repeatedly in steps we get the solution as:

$\backslash epsilon(\backslash eta,t;\backslash alpha)=\backslash eta+\; \backslash alpha\backslash \{\backslash eta,G(\backslash eta,t)\backslash \}+\backslash frac\{1\}\{2!\}\backslash alpha^2\; \backslash \{\backslash \{\backslash eta,G(\backslash eta,t)\backslash \},G(\backslash eta,t)\backslash \}+\backslash cdots=e^\{-\; \backslash alpha\; \backslash tilde\; G\}\; \backslash eta$

where the differential operator is defined as $\backslash tilde\; G:=\; (\backslash nabla\_\backslash eta\; G)^T\; J\; \backslash nabla\_\backslash eta$.

Change in function values $\backslash frac\{df(\backslash epsilon(\backslash eta;\backslash alpha),t)\}\{d\; \backslash alpha\}=\; \backslash \{f(\backslash epsilon(\backslash eta;\backslash alpha),t),G(\backslash eta,t)\backslash \}\_\backslash eta=:-\backslash tilde\; G\; f(\backslash epsilon(\backslash eta;\backslash alpha),t)$ upon taking repeatedly in steps and using $\backslash epsilon(\backslash eta,t;0)=\backslash eta$ we get similarly

$f(e^\{-\backslash alpha\backslash tilde\; G\}\backslash eta,t)=f(\backslash epsilon(\backslash eta;\backslash alpha),t)=f(\backslash eta,t)+\; \backslash alpha\backslash \{f(\backslash eta,t),G(\backslash eta,t)\backslash \}+\backslash frac\{1\}\{2!\}\backslash alpha^2\; \backslash \{\backslash \{f(\backslash eta,t),G(\backslash eta,t)\backslash \},G(\backslash eta,t)\backslash \}+\backslash cdots=e^\{-\; \backslash alpha\; \backslash tilde\; G\}\; f(\backslash eta,t)$

Change in a function that preserves values on physical states in phase space as $f(\backslash epsilon,t)=f(\backslash epsilon(\backslash eta;\backslash alpha),t)=f\text{'}(\backslash epsilon(\backslash eta;\backslash alpha+\backslash delta\backslash alpha),t)=\; f\text{'}(\backslash epsilon\text{'},t)$ can be expressed as upto first order as:

$\backslash delta\text{'}\; f=f\text{'}(\backslash epsilon)-f(\backslash epsilon)=f\text{'}(\backslash epsilon)-f\text{'}(\backslash epsilon\text{'})\backslash approx\; f(\backslash epsilon(\backslash eta;\backslash alpha-\backslash delta\backslash alpha))-f(\backslash epsilon(\backslash eta;\backslash alpha))\; =-\backslash delta\; \backslash alpha\backslash \{f,G\backslash \}$

Including the change in the function, it can be expressed as $f(\backslash epsilon,t;\backslash alpha)$ where it is explicitly dependant on $\backslash alpha$ such that $\backslash frac\{\backslash partial\; f(\backslash epsilon,t;\backslash alpha)\}\{\backslash partial\; \backslash alpha\}\; =-\backslash \{f,G\backslash \}$ which indicates that the function transforms oppositely to that of the coordinates to preserve well defined mapping from a physical point in phase space to scalar values.

Let the set of generators of canonical transformations of the one parameter kind be closed under Poisson brackets i.e. $\backslash \{\backslash lambda\_i,\backslash lambda\_j\backslash \}=c\_\{ij\}^k\; \backslash lambda\_k+d\_\{ij\}$ then it also follows that$\backslash widetilde\{\backslash \{\backslash lambda\_i,\backslash lambda\_j\backslash \}\}=c\_\{ij\}^k\; \backslash tilde\; \backslash lambda\_k$. Using Jacobi identity of Poisson bracket, it can be shown that the exponential map of the group elements formed by $\backslash tilde\; \backslash lambda$'s forms a Lie group.

Using Baker-Campbell-Hausdroff formula on the exponential map, since $\backslash tilde\; \backslash lambda$ are linear operators, closure of the exponential set is given if the commutator, $\backslash tilde\; \backslash lambda\_i\backslash tilde\backslash lambda\_j-\backslash tilde\; \backslash lambda\_j\backslash tilde\backslash lambda\_i$ and hence all other terms arising in the BCH formula can be given as element of the generator set.

$(\backslash tilde\; \backslash lambda\_i\backslash tilde\backslash lambda\_j-\backslash tilde\; \backslash lambda\_j\backslash tilde\backslash lambda\_i)f=\backslash \{\backslash lambda\_i,\backslash \{\backslash lambda\_j,f\backslash \}\backslash \}-\backslash \{\backslash lambda\_j,\backslash \{\backslash lambda\_i,f\backslash \}\backslash \}=\backslash \{\backslash lambda\_i,\backslash \{\backslash lambda\_j,f\backslash \}\backslash \}+\backslash \{\backslash lambda\_j,\backslash \{f,\backslash lambda\_i\backslash \}\backslash \}=\backslash \{\backslash \{\backslash lambda\_i,\backslash lambda\_j\backslash \},f\backslash \}=\backslash widetilde\{\backslash \{\backslash lambda\_i,\backslash lambda\_j\backslash \}\}f=c\_\{ij\}^k\backslash tilde\backslash lambda\_k\; f$

Since the commutator is closed it follows that $e^\{a\_i\backslash tilde\; \backslash lambda\_i\}e^\{a\_j\backslash tilde\; \backslash lambda\_j\}=e^\{a\_\{ij\}^k\backslash tilde\; \backslash lambda\_k\}$.