17

The reason the Hamiltonian is undefined is that converting a Lagrangian to a Hamiltonian requires: Finding $p$ Writing $H=p\dot q-L$ Expressing $H$ in terms of $p$ and $q$, eliminating all dependence on $\dot q$. For the third step to be possible, you need to be able to define a coordinate transformation between the $(q,\dot q)$ coordinate system and the $(...


13

OP's Lagrangian $$L~:=~\frac{\dot{q}}{2}\sin^2(q)~=~\frac{d}{dt}\left(\frac{q}{4}-\frac{1}{8}\sin(2q)\right)\tag{1}$$ is a total time derivative, so the Euler-Lagrange (EL) equation is a triviality $0=0$, cf. e.g. this Phys.SE post, and hence not suitable to determine physical trajectories. In other words, $q$ is a gauge degree of freedom. If $q$ is ...


11

In general, the mapping defined by $$ p_i(q,\dot{q}) = \frac{\partial L}{\partial \dot{q}^i}$$ is neither injective nor surjective. Theories in which it is not are constrained Hamiltonian theories and equivalently Lagrangian gauge theories in which the solutions to the equations of motion contain arbitrary functions of time. "Constrained" means that the $q$ ...


9

The full groupoid of all (possibly non-linear) canonical transformations (CTs) is infinite-dimensional. Infinitesimal CTs are Hamiltonian vector fields $$ \delta z^I ~=~{\{F(z,t),z^I\}}, \qquad I~\in~\{1,\ldots,2n\}.\tag{1}$$ The symplectic group $Sp(2n, \mathbb{R})$ of dimension $n(2n+1)$ is the group of all linear time-independent CTs. The corresponding ...


8

There are already some nice answers regarding intuitive interpretations of the Legendre transform. What I want to contribute here, is a more physical reason/motivation for why they appear. I.e., instead of focusing on their physical interpretation, I will focus on the physical requirements that uniquely define Legendre transforms. This explains why you would ...


6

I understand the "magic recipe" impression:) It is better to introduce quantization as an answer to a question. The latter is: given a classical Hamiltonian dynamical system, how to construct a family of quantum systems indexed by $\hbar$ such that, when taking the $\hbar\rightarrow 0$ limit, one recovers the given classical system. The magic recipe (...


6

Yes, absolutely. Copied and pasted from the end of this answer: Given any smooth function $F$, we can define a Hamiltonian vector field $X_F$, and a corresponding flow $\varphi_{X_F}$. Given some parameter value $t$, $\varphi_{X_F}(\Omega, t)$ constitutes a canonical transformation (i.e. a change of coordinates). We would then call $F$ the generator of ...


5

We need the complex conjugated variable $\bar{u}$ because the Hamiltonian $H(u,\bar{u})$ is typically not holomorphic in $u$. The normalization of the CCR$^1$ $$\{u,\bar{u}\} ~=~-i \tag{1}$$ explains the $\sqrt{2}$-normalization in $$ u=\frac{q+ip}{\sqrt{2}}.\tag{2}$$ Recall that Hamilton's equations can be written in terms of the Poisson bracket $$\dot{u}...


5

My favourite non-cotangent-bundle is the classical theory of spin where the Poisson brackets are $$ \{S_i,S_j\} =\epsilon_{ijk} S_k, $$ and the phase space is the two-sphere $S^2$. With spherical polar coordinates, the symplectic form is $$ \omega= J \sin\theta\, d\theta\wedge d\phi. $$


5

Note that in many applications, there is a simplification: the Hamiltonian is quadratic in the momenta, i.e. the path integral over the momenta is Gaussian, which can be performed exactly. On the other hand, if the Hamiltonian has higher-orders of momenta, then the path integral over the momenta can generically not be performed exactly, and we have to rely ...


5

The $i$th component of the angular momentum $L_i$ is the Noether charge for 3D rotation around the $i$th axis. It satisfies the $so(3)$ Lie algebra $$\{L_i,L_j\}~=~ \epsilon_{ijk}L_k.\tag{1}$$ The corresponding Hamiltonian vector field $X_{L_i}\equiv\{L_i,\cdot\}$ is the generator of 3D rotations around the $i$th axis, cf. this Phys.SE post. It satisfies ...


5

Considering a translated function: $$f(x)\rightarrow f(x+\Delta x)$$ We can expand in Taylor series around $x$ and write $$f(x+\Delta x)=\sum_{n=0}^\infty\frac1{n!}\left. \frac{\partial^n f}{\partial x^n}\right|_{x}\Delta x^n\equiv\sum_{n=0}^\infty\frac1{n!}\left(\Delta x\left. \frac{\partial}{\partial x}\right|_{x}\right)^nf(x)$$ Which formally could be ...


5

Consider a Lagrangian system with $n$ DOF. In the case where the Hessian matrix $\frac{\partial^2 L}{\partial v^i \partial v^j}$ has constant rank $r$, it is possible to replace $r$ velocities with $r$ momenta in the definition of the Hamiltonian. It is proven in theorem 2 of my Phys.SE answer here, that this Hamiltonian will not depend on the remaining $n-r$...


5

OP has already noted that the 2D harmonic oscillator is completely Liouville-integrable with 2 globally defined, Poisson-commuting, real integrals of motion $H_1$ and $H_2$. Since the phase space has 4 real dimensions, there can at most be 3 independent real integrals of motion, and 4 independent real constants of motion. By definition an integral of motion ...


5

In general, if you have a nonstandard kinetic term, it may be impossible to transform the equations of motion from Lagrangian to Hamilton form. Probably the simplest situation of this type (following an example from Nambu) is if the kinetic energy $K$ is a quartic function of the velocity. The fourth power makes $K$ bounded below (as it should be), and it ...


5

Lorentz force is an example of non-conservative conservative force that is discussed in pretty much any theoretical mechanics textbook. The Lagrangian is: $$L = \frac{m}{2}\dot{\mathbf{r}}\cdot\dot{\mathbf{r}} + q \mathbf{A}(\mathbf{r})\cdot\dot{\mathbf{r}} - q\phi(\mathbf{r}),$$ from which the momentum, the Hamiltonian, and the Hamilton equations follow as ...


5

Revised answer: The first one was wrong even if three people liked it! The "Lagrangian" derived in the cited paper is first order in the time derivative of $c_i$ and so the "Lagrangian" is really that of the Hamiltonian action principle. This starts from the action functional $$ S[p,q] = \int \{\sum_i p_i\dot q_i -H(p,q)\}dt $$ whose ...


4

I explain in general how the moment map leads to an actual map from $\mathfrak{g}$ to the algebra of classical observables $C^\infty(M)$ in this answer of mine, but I will exemplify the reverse direction for 1D translations in the following: The translation algebra $\mathfrak{g}\cong \mathbb{R}$ acts on $M\cong \mathbb{R}^2$ as $f_c : (q,p)\mapsto (q + c, p)...


4

$I_\zeta$ is called a homotopy operator because of eq. (1.60). The explicit form to all orders can e.g. be found in P.J. Olver's book Applications of Lie Groups to Differential Equations, 2nd edition, 1993, eq. (5.134).


4

This might not be the most general formulation, but maybe it's general enough to answer the question. I'll use some abbreviations that are hopefully clear, like omitting the time-argument and writing $\partial\phi$ instead of $\partial_\mu\phi$. I'll write $\dot\phi$ for the time-derivative of $\phi$. Start with any Lagrangian density $L(x)$ that can be ...


4

OP's question about existence of quantization is very broad and is literally the topic of whole books and current research. For a general classical system, there is no proof that a consistent quantization exists. Although many partial results in e.g. geometric quantization and deformation quantization have been obtained. The quantization procedure is not ...


4

There are two issues here. The first is that if $dx$ is expressed as a function of $x$, then that means that the coordinate change corresponds to adding a position dependent shift to the position coordinate. Adding a constant shift is what is prescribed here, so $dx$ has no $x$ dependence. If it would make you more comfortable, you could write it as $X = ...


3

Physicists conventionally normalize the square root Lagrangian a bit differently, namely as$^1$ $$L_0~:=~ -m\sqrt{-\dot{x}^2}, \qquad \dot{x}^2~:=~g_{\mu\nu}(x)~ \dot{x}^{\mu}\dot{x}^{\nu}~<~0,\tag{1}$$ where $m>0$ is the mass. (OP assumes that $m=1$.) It is important to note that the Legendre transformation of the square root Lagrangian (1) is ...


3

Obviously L&L's conclusion (45.9) is correct as always, but their 2nd proof in the paragraph between eqs. (45.9) & (45.10) seems not entirely convincing, although they do mention all the important ingredients along the way. In contrast, their 1st proof method by direct calculation is a straightforward exercise and quite convincing. In this answer, we ...


3

The space of all solutions of the Euler-Lagrange equations in the configuration space is equivalent to the phase space. Indeed, given an initial condition (consisting of a position and a velocity) there is a unique solution, and (assuming a fixed total mass) the velocity determines the momentum. Hence the points of phase space (i.e., pairs of position and ...


3

To derive the minimal coupling Hamiltonian, someone starts by calculating the Lagrangian $L$ of a particle in an electromagnetic field. The acting force is the Lorentz force $\boldsymbol{F} = q(\boldsymbol{E} + \boldsymbol{v}/c \times \boldsymbol{B})$. Since $\boldsymbol{F}$ depends on velocity, we have to find a generalized potential $U$ that satisfies the ...


3

All examples of phase spaces which are not cotangent bundles must come from systems with gauge symmetries (in the Lagrangian formalism) resp. systems with constraints (in the Hamiltonian formalism), since then the true phase space is not the cotangent bundle of the configuration space itself, but the surface inside it defined by the constraints, with the ...


3

Long story short: Delaunay variables are something which is called action-angle variables, which makes them extremely useful in perturbation theory. Let me first talk about what action-angle variables in general and then mention a few facts about Delaunay variables themselves. Let us consider a Hamiltonian system of $N$ degrees of freedom with generic ...


3

If $H$ is time independent and satisfies $\{\rho, H\}=0$, also $H(t)=f(t)H+ g(t)$ does for every pairs of smooth maps $f,g$. Hence the answer is negative. $H$ can be time dependent.


3

A more geometric approach is to consider the $(2n+1)$-dimensional contact manifold ${\cal M}$ with coordinates $(q^i,p_j,t)$. The Hamiltonian action functional is $$S_H[\gamma]~=~\int_I \gamma^{\ast} \Theta, \qquad \Theta~=~p_j \mathrm{d}q^j -H \mathrm{d}t, \tag{1}$$ where $\gamma:I\to {\cal M}$ is a curve. This action formulation (1) is world-line (WL) ...


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