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2

Hints to the question (v1): Let us parametrize the problem wrt. an arbitrary world-line parameter $\tau$ (which does not have to be the proper time). The Lagrange multiplier $\lambda=\lambda(\tau)$ depends on $\tau$, but it does not depend on the canonical variables $x^{\mu}$ and $p_{\mu}$. Similarly, $x^{\mu}$ and $p_{\mu}$ depend only on $\tau$. The ...

1

The Hamiltonian can be used to describe an evolution of the "density in phase" of a system of N bodies. The density in phase is a conserved quantity for a system in equilibrium by Liouville's Theorem. The position and momenta can describe any general intensive parameter. Gibbs used this approach to derive statistical mechanics. This approach of the concept ...

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Basically, the answer to both questions is yes. In the field theory case, however, this is true (optimistically speaking according to Witten-1) - in the case when the field equations are hyperbolic wave equations. In this case one needs the initial conditions to be defined on global Cauchy hypersurface. In this case there is a one to one correspondence ...

1

I can answer (1) and (2). The answer is: NO. Passing form classical mechanics to quantum one requires, in general, to add more information. There is no rigorous machinery allowing one to write the quantum corresponding of a classical object. Physically speaking, this is because quantum structures are more fundamental in Nature than classical ones. ...

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Yes. There is a standard way to generalize to field theory. Let a theory of $n\geq 1$ fields $\phi^i$ with a Lagrangian density $\mathcal L = \mathcal L(\phi^i, \partial_\mu\phi^i)$ be given. Here we use that standard abuse of notation in which $\phi^i$ denotes the vector whose components are the fields; $\phi^i = (\phi^1, \dots, \phi^n)$. To obtain the ...

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In my opinion, your professor is being liberal with terminology in a confusing way, and I think you've essentially already pointed out why in your comment on Kyle's answer. Let's examine a simple example. Consider the free particle moving in three spatial dimensions. The configuration space of the free particle is $\mathbb R^3$ and its momentum space is ...

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Recall that $$\dot{\mathbf{p}}=-\frac{\partial H}{\partial\mathbf{q}}$$ Since $H$ does not actually depend on $\phi$, then $$\dot{p}_\phi=0=-\frac{\partial H}{\partial\phi}$$ This will eliminate $p_\phi$ and $\phi$ from your coordinates: $H(p_r,p_\theta,p_\phi,r,\theta,\phi)\to H(p_r,p_\theta,r,\theta)$.

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This is a fact about the hamiltonian compared to the lagrangian which I find not trivial (and worth to keep in mind). Suppose that the lagrangian $L$ and hamiltonian $H$ are cyclic with respect to some coordinate $q_1$. Then we have a theorem (cfr. [1]): The evolution of the other coordinates $q_2,...,q_n$ is the one of a system with $n-2$ independent ...

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I) The restricted$^1$ transformation (RT) $$\tag{1} (q,p)~\longrightarrow~ (Q,P) ~:=~(q, \sqrt{p} - \sqrt{q})$$ of OP's professor with inverse RT $$\tag{2} (Q,P)~\longrightarrow~ (q,p) ~:=~(Q, (P+ \sqrt{Q})^2) ,$$ and with Hamiltonian $H=\frac{p^2}{2}$ and Kamiltonian $K=\frac{p^3}{3}$ is indeed interesting. Apparently we should assume that $p,q,Q\geq ... 2 One way to see the relationship of Hamilotian classical mechanics and Quantum mechanics is not to look for a direct translation of Hamiltionian -> Quantum Hamiltionian (which exists: Geometric Quantization), but consider the reverse relationship. Given a Hamiltion operator and evaluating it on wave functions of the form$e^{\frac{i}{\hbar} \phi}$(which can ... 2 The canonical (Hamiltonian) formalism offers one of the main paths for quantizing gravity. General Relativity can be expressed in terms of the ADM 3+1 decomposition of spacetime: http://en.wikipedia.org/wiki/ADM_formalism And Hamiltonian's underlie quantum mechanics: http://en.wikipedia.org/wiki/Hamiltonian_(quantum_mechanics) Not only does this provide ... 6 First of all, Lagrangian is a mathematical quantity which has no physical meaning but Hamiltonian is physical (for example, it is total energy of the system, in some case) and all quantities in Hamiltonian mechanics has physical meanings which makes easier to have physical intuition. In Hamiltonian mechanics you have canonical transformations which allows ... 14 Some more comments to add to user1504's response: For a system with configuration space of dimension$n$, Hamilton's equations are a set of$2n$, coupled, first-order ODEs while the Euler-Lagrange equations are a set of$n$uncoupled, second-order ODEs. In a given problem it might be easier to solve the first order Hamilton's equations (although sadly, I ... 4 Cool question! Thanks to user lionelbrits for his answer that prompted me to pull out my mechanics books and check the definitions of "canonical transformation" given by different authors. If you look in Goldstein's classical mechanics texts in the section on canonical transformations, then you'll find that canonical transformations are essentially defined ... 21 There are several reasons for using the Hamiltonian formalism: 1) Statistical physics. The standard thermal states weight pure states according to$Prob(state) \propto e^{-H(state)/k_BT}$. So you need to understand Hamiltonians to do stat mech in real generality. 2) Geometrical prettiness. Hamilton's equations say that flowing in time is equivalent to ... 1 The original coordinates satisfy the equations of motion when the integral of$p\, \dot{q} - H(p,q)$is minimized, and the new coordinates satisfy the equations of motion when the integral of$P\, \dot{Q} - K(P,Q)$is minimized. There is no requirement that$H$and$K$be numerically equal. The transformation is canonical if the Poisson bracket remains ... 1 The name seems appropriate if consider that it probably comes from the case when the manifold is the cotangent bundle of a manifold. Then a point on$T^*M$is a pair$(x,\alpha)$, where$x$is a point on$M$and$\alpha$a one form. The definition of the tautological one form is: the value of the form at a point$(x,\alpha)$on a tangent vector is obtained ... 1 I) On a general symplectic manifold$({\cal M},\omega)$(typically called phase space by physicists), one can locally choose a symplectic potential$\theta\in \Gamma(T^{*}{\cal M}|_{\cal U})$, which is a one-form such that $$\tag{1} \mathrm{d}\theta~=~\omega,$$ cf. Poincare Lemma. Here${\cal U}\subseteq {\cal M}$denotes a local neighborhood. Note ... 2 The problem with the phase space flow in Hamiltonian mechanics is that the flow itself is non-dynamical, that is, the flow is immediately defined for a given Hamiltonian, so there is no independent equation governing its evolution. Thus, Liouville equation is simply a transport of a scalar variable in a given flow. So, dimensional analysis of the flow ... 0 This approach to classical mechanics is named after Edward Routh (with ou rather than just u). In this approach the Legendre transform is only carried out for the cyclic coordinates. Goldstein covers it in section 8.3 (Routh's procedure). 4 Consider the time derivative of the Hamiltonian $$\frac{dH(q,p,t)}{dt}=\frac{\partial H}{\partial q}\dot{q}+\frac{\partial H}{\partial p}\dot{p}+\frac{\partial H}{\partial t}=-\dot{p}\dot{q}+\dot{q}\dot{p}+\frac{\partial H}{\partial t}$$ From this you see that the Hamiltonian is conserved if it does not depend on time,$t$, explicitly.$H$may or may not be ... 1 The Euler Lagrange equations just give the differential equations that determine the motion of the object. The end points are boundary conditions for the differential equation. The differential equation which the brachistochrone curve satisfies will have its constants fixed so that it reduces to a line when you give it the boundary conditions of the object ... 4 It's because they're based on the historical approach: Schroedinger's equation. Schroedinger's equation was discovered on its own before we knew about canonical quantization. Dirac came up with the canonical quantization rules which re-wrote (and generalized) Schroedinger's equation into the familiar one we have today,$\hat{H} \psi = i \dot{\psi}$. That ... 2 I don't have an answer for why there is no simple Lagrangian formulation, but I can explain some of why a Hamiltonian one is easy. Part of the way to go from Classical Mechanics to Quantum is by replacing Poisson brackets with commutators, and observables with operators on Hilbert space and their expectation values. So the equation$\frac{d}{dt} f(q, p, t) ...

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I can think of several reasons for why using Hamiltonians is preferred, but the most important, I'ld say, is that you need to use path integral formalism in order to formulate (non relativistic) QM in terms of the Lagrangian, which, for an undergrad course, is a bit of an overkill. Also, many of the most renowned equations in QM like, say, the Schrödinger ...

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