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Infinitesimal transformations and the Hamiltonian as generator

  1. Apr 17, 2014 #1

    I don't understand how one arrives at the conclusion that the hamiltonian is a generating function.

    When you have an infinitesimal canonical transformation like:

    [itex]Q_{i}=q_{i}+ \delta q_{i}[/itex]
    [itex]P_{i}=p_{i}+\delta p_{i}[/itex]

    Then the generating function is:

    [itex]F_{2}=q_{i}P_{i}+ \epsilon G(q,P,t)[/itex]

    Then the transformation equations are:

    [itex]p_{j} = \frac{\partial F_{2} }{\partial q_{j} }= P_{j}+ \epsilon \frac{\partial G}{\partial q_{j}}[/itex]

    [itex]\delta p_{j} = P_{j}-p_{j} = -\epsilon \frac{\partial G}{\partial q_{j}}[/itex]


    [itex]Q_{j} = \frac{\partial F_{2} }{ \partial P_{j}}=q_{j}+ \epsilon \frac{\partial G}{\partial P_{j}}[/itex]
    [itex]\delta q_{j} = Q_{j}-q_{j} = \epsilon \frac{\partial G}{\partial P_{j}}[/itex]

    Leading to:

    [itex]\delta p_{j} = -\epsilon \frac{\partial G}{\partial q_{j}}[/itex]
    [itex]\delta q_{j} = \epsilon \frac{\partial G}{\partial p_{j}}[/itex]

    Which look super similar to Hamilton's equations of course.
    Then the poisson brackets, out of nowhere, refering only to the definition, tell you that:

    [itex]\dot{q_{i}} = [q_{i},H][/itex]
    [itex]\dot{p_{i}} = [p_{i},H][/itex]

    And then one says, OH GOD that looks like [itex]G[/itex] was secretly the Hamiltonian [itex]H[/itex], let's say [itex]H[/itex] is the generator of time evolution !

    And then I don't understand why the hamiltonian appears in something which needs the definition of the hamiltonian. Aren't canonical transformations definied in that they satisfy the principle of least action(among other things), which then includes the old hamiltonian and the new one. How can then the hamiltonian arise again out of that? How is this put together??

  2. jcsd
  3. Apr 17, 2014 #2


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    No, as they written, these equations say that [itex]G[/itex] is the (conserved) generator of infinitesimal transformation [itex]\epsilon[/itex] in phase space. However, one can easily prove that [itex]G = H[/itex] if and only if [itex]\epsilon[/itex] is an infinitesimal time translation [itex]\delta t[/itex]. One can also (even easier) prove that [itex]G = p[/itex] if and only if [itex]\epsilon = \delta q[/itex].

    No, the definition alone does not give you these equations. You get them when you insert Hamilton’s equations in the definition of the bracket.
  4. Apr 18, 2014 #3
    Hi, Thx for reply. It is really hard, I'm not sure how to formulate my question, the introduction of poisson brackets caused chaos in my head. Is the operation of taking the poisson brackets a canonical transformation? And if yes, how does this work? How can it be a transformation from (p,q) to (Q,P) ?

  5. Apr 18, 2014 #4


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    Yes, provided that the bracket is taken with the generator of the canonical transformation.
    You have already done that yourself when you wrote
    [tex]\delta p_{ j } = P_{ j } - p_{ j } = - \epsilon \frac{ \partial G }{ \partial q_{ j } } , \ \ (1)[/tex]
    [tex]\delta q_{ j } = Q_{ j } – q_{ j } = \epsilon \frac{ \partial G }{ \partial p_{ j } } . \ \ (2)[/tex]
    But, from the definition of Poisson bracket
    [tex]\{ F ( q , p ) , G ( q , p ) \} \equiv \frac{ \partial F }{ \partial q } \frac{ \partial G }{ \partial p } - \frac{ \partial F }{ \partial p } \frac{ \partial G }{ \partial q }[/tex]
    we find
    [tex]\{ p_{ j } , G \} = - \frac{ \partial G }{ \partial q_{ j } } ,[/tex]
    [tex]\{ q_{ j } , G \} = \frac{ \partial G }{ \partial p_{ j } } .[/tex]
    So, if you put these in eq(1) and (2), you find
    [tex]P_{ j } = p_{ j } + \epsilon \{ p_{ j } , G \}[/tex]
    [tex]Q_{ j } = q_{ j } + \epsilon \{ q_{ j } , G \}.[/tex]
    The importance of the Poisson bracket (between any two functions) relies on the fact that they are invariant under canonical transformations, i.e. you can test that a given transformations are canonical if they leave the bracket invariant. That is to say, you show that
    [tex]\{ F \ , \ G \}_{ (q , p) } = \{ F \ , \ G \}_{ (Q , P) } ,[/tex]
    if and only if the transformations are canonical.
  6. Apr 18, 2014 #5
    Hi, first of all, to clear things up, you should notice that what enters in the Poisson Brackets is functions (this is clear since the P.B involves differentiation). However, once also notices that the coordinates q and p enter the P.B. which "contradicts" the above statement.

    What I'm trying to stress out here is that the P.B. is an operation on the space of functions on the phase space, and not an operation on the phase space itself. You should think of q and p as coordinate functions which can be chosen as a basis of the space of functions, i.e. the dual space to the phase space - the dual phase space for short. When one evaluates q and p on a point living on the phase space then these will give you the position and momentum of your system. Similarly, all the functions, like the Hamiltonian H, written w.r.t the {q,p} basis, e.g. H=H(q,p), when evaluated at a point u on the phase space give you a real number, H(u)=H( q(u), p(u) )[itex]\in[/itex]R = ENERGY of your system. Thus the functions correspond to physical observables.

    Now the way that one may view a symmetry of a system is either through operations on phase space or through operations on the dual phase space. The two viewpoints are equivalent, with the former having the symmetry generators realized as vector fields on phases space, and the former having them realized as generating functions on the dual phase space.

    There is a nice mathematical story which reveals the interrelation between vector fields and functions on phase space, but for our purposes let me state one of the implications of this story. The components of a vector field on phase space may be written in terms of a function G as [itex]\vec{V}[/itex]=([itex]\frac{δG}{δp}[/itex],-[itex]\frac{δG}{δq}[/itex]). Moreover, an infinitesimal symmetry of a point [itex]\vec{x}[/itex]=(q,p) is then written as δ[itex]_{ε}[/itex](q,p)=([itex]\vec{V}[/itex][itex]\bullet[/itex][itex]\frac{δ}{δ\vec{y}}[/itex]) [itex]\vec{x}[/itex] ε = ([itex]\frac{δG}{δp}[/itex],-[itex]\frac{δG}{δq}[/itex]) ε. (When one seeks for the symmetries of an action is in fact looking at the infinitesimal transformation of the action w.r.t a vector field - a symmetry generator.)

    Example: The vector field associated to H generates time translations, in particular using the above equation with the infinitesimal parameter ε identified with the infinitesimal time parameter δt one ends up with Hanilton's e.o.m: ([itex]\dot{q}[/itex],[itex]\dot{p}[/itex])= ([itex]\frac{δH}{δp}[/itex],-[itex]\frac{δH}{δq}[/itex])

    The above calculation relied on the realization of symmetries on phase space. Now let us repeat the same calculation from the point of view of dual phase space. There one has generating functions and an infinitesimal symmetry is implemented with the the help of the P.B;
    δ[itex]_{ε}[/itex]q= ε{q,G}= ε[itex]\frac{δG}{δp}[/itex] and for time translations one recovers Hamilton's e.o.m.

    Equipped with these concepts, one can proceed with the understanding of canonical transformations.

    In simple words, we call a canonical transformation a transformation which respects this equivalence between vector fields and functions on phase space. To be a little bit more precise, this equivalence originates from the existence of a symplectic structure on phase whose existence, allows one to express vector fields in terms of generating functions and hence, a canonical transformation, which for practical purposes must leave the Hamiltonian invariant, must be such that does not destroy this symplectic structure.

    For example, consider the case where one would like to study a system in a different and more convenient coordinate system than the given one. Since the physics is captured in H, which is written w.r.t. the {q,p} basis, then the new Hamiltonian K which will be written in terms of some new basis {Q,P} must be such that when we evaluate H and K at the same phase space point u they must agree and give the same energy E=H(u)=K(u). How does H transforms under an infinitesimal symmetry δ[itex]^{(G)}_{ε}[/itex] on the coords? Clearly,

    δH(q,p)=[itex]\frac{δH}{δq}[/itex] δq + [itex]\frac{δH}{δp}[/itex]δp

    Plugging in δ[itex]^{(G)}_{ε}[/itex](q,p) we obtain:


    Now define K:=H(q,p)+δH(q,p) and check that it is a well defined Hamiltonian w.r.t. to the P.B., i.e. it time translates (Q,P)=(q,p)+δ(q,p) with the help of the P.B.. This implies that the value of the P.B. is the same for the two different sets of coordinate bases.

    At last, one can play the same game with other physical observables/generating functions and convince himself that a canonical transformation must be such that the P.B. is invariant, or if you prefer, the physical laws -e.g. energy and momentum conservation - do not change under canonical symmetry operations - associated to time and space translations.
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