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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 ...

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No, $SO(2n,\mathbb{F})\subseteq Sp(2n,\mathbb{F})$ [understood via their standard embedding into $GL(2n,\mathbb{F})$] is only true for $n=1$. This fact can be proved by considering the corresponding Lie algebras. Specifically, $Sp(4,\mathbb{R})$ is (the double cover of) the restricted anti de Sitter group $SO^+(3,2;\mathbb{R})$, cf. e.g. my Math.SE answer ...

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Yes, any smooth function on the phase space can be Hamiltonian. And to any Hamiltonian corresponds a Hamiltonian vector field $V_H$, such that $$i_{V_H} \omega = -dH$$ In the simple case of $\mathbb{R}^{2n}$ the symplectic form is $$\omega = \sum_i d q_i \wedge d p_i$$

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The hamiltonian of a system has to be an hermitian operator since it's associated to a measurable quantity, the energy. By symply taking the classical hamiltonian$$H=\omega xp$$ and converting it directly to an operatore $$\hat H = \omega\hat x\hat p$$ you can easily see that this operator is not hermitian since, given $\hat x$ and $\hat p$ hermitian, $$\hat ... 2 This is an interesting question because in general there is no unique answer, although there is a "natural" answer for the simplest observables, which include xp. The lack of uniqueness is linked to the ordering problem. The simplest quantization is due to Dirac; he tried to implement the idea that the commutator of two operators representing ... 2 Hint: Hamilton's equations of motion here are exactly the same as they are in classical mechanics, with ordinary derivatives replaced by functional derivatives. This is because in general, the Hamiltonian(not the hamiltonian density) H(t)=H[\psi(\cdot,t),\dot{\psi}(\cdot,t)] is a functional of the fields and conjugate momenta at a given time slice, and at ... 2 I am not entirely sure of what objections you have, but I will try to explain what Tong is getting at from a different angle, and perhaps then the route he takes through his notes will make more sense. The idea he's trying to get across is roughly that one can view time evolution as being a canonical transformation which is generated by the Hamiltonian ... 1 To familiarize your self with the subject you may consult the aforementioned Arnold's book, or Nakahara "Geometry, Topology and Physics" https://www.academia.edu/29696440/GEOMETRY_TOPOLOGY_AND_PHYSICS_SECOND_EDITION_Nakahara, which is a very pedagogical introduction for physicists with the mathematical machinery. In simple words, the symplectic ... 1 If OP is already familiar with Poisson brackets then it seems that the central piece of information relevant to OP's question is the following theorem. Theorem: Let there be given a 2n-dimensional manifold M. There is a canonical bijective correspondence between symplectic structures \omega\in\Omega^2(M) and non-degenerate Poisson structures \{\cdot,\... 1 No, when we consider the action principle S=\int\! dt(p \dot{q} -H(q,p)) for OP's Hamiltonian, then p acts as a Lagrange multiplier that imposes the EOM \dot{q} \approx \alpha q. This EOM does not have a regular Lagrangian formulation, i.e. a formulation without the use of other variables than q. 1 Your interpretation is correct. All you have to do is write the powers of S_z in matrix form, and add a unit matrix factor for the terms without S_z. 1 First of all, S is not a scalar, it is an operator! If \{\chi_{s,m_s}\} is the simultaneous eigenbasis of S_{z} and S(S+1), then$$\tag{2} S(S+1) \chi_{s,m_s}= s(s+1) \hbar^{2} \chi_{s,m_s} \tag{1} S_{z} \chi_{s,m_s} = m_{s}\hbar \chi_{s,m_s}However, as [S(S+1),S_{z}]=0 and the quantum number s is the same throughout this problem, the ... 1 What you want to do is use the \mathfrak{sp}(2,\mathbb{R})\sim \mathfrak{su}(1,1) as your Hamiltonian can be expressed as combinations of the generators \begin{align} \hat K_{+}=\frac{1}{4}\hat a^\dagger \hat a^\dagger\, ,\qquad \hat K_{-}=\frac{1}{4}\hat a \hat a\, ,\qquad \hat K_0=\frac{1}{4}(\hat a^\dagger\hat a +\hat a\hat a^\dagger)\, , \end{align} ... 1 In terms of standard differential geometry, the phase space of physics is a symplectic manifold with symplectic structure \omega = \mathrm{d}q^i \wedge \mathrm{d}p_i (with summation over repeated indices). For any observable f(q,p), the associated Hamiltonian vector field X_f is defined abstractly by \omega(X_f,V) = \mathrm{d}f(V) and in local ... 1 On a mathematical level, a Lagrange multiplier in the Lagrangian is no different from a "real" coordinate whose velocity does not appear in the Lagrangian, such as A_0 in the context of Maxwell field theory. One can therefore subject a Lagrangian containing a Lagrange multiplier to the standard Hamilton-Dirac procedure and obtain a corresponding ... 1 Ashok Das is not ignoring any factor of -i in formulas. The word "represent" in the sentence \overline{\psi} represents the conjugate to \psi is here used semantically in a weaker sense than "is equal", e.g. "is equal up to a multiplicative constant". 1 There is no need to eliminate the A_0 field^1. The long story short is that the Hamiltonian Lagrangian density^2\begin{array}{ccc} {\cal L}_H~=~\vec{\Pi}\cdot\dot{\vec{A}} - {\cal H}&\stackrel{\vec{\Pi}}{\longrightarrow} & {\cal L}~=~-\frac{1}{4}F_{\mu\nu}F^{\mu\nu}-\frac{1}{2}m^2 A_{\mu}A^{\mu} \cr\cr \downarrow A_0& &\downarrow A_0\...

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It seems to me that some of the notation in the mentioned textbook is implicit. Let's start with the phase space where coordinates are $(q,p)$. We have a canonical transformation to coordinates $(q',p')$. A point in the phase space will be denoted by $A$, and some other point by $B$. $\quad \bullet\quad$ In the passive viewpoint, all quantities at a point $A$...

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Consider canonical transformations $\Phi_\lambda:p_i,q^j \to p'_i,q'^j$ depending smoothly on a parameter $\lambda$ such that $\lambda=0$ reduces to an identity. Such a transformation is at least locally expressible (for $\lambda \approx 0$) as $$\Phi_\lambda(z) = \exp(\lambda \{z,G\})$$ where $G$ is some generating function and $z=(p_i,q^j)$. To linear ...

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These are very good questions. Refs. 1 & 2 are not entirely consistent on these issues. Let us analyze the situation. In general a Hamiltonian version of the stationary action principle is of form $$S_H[z]~=~\int_{t_i}^{t_f}\! dt~L_H(z,\dot{z},t),\tag{1}$$ where the $2n$-dimensional phase space has (not necessary canonical) coordinates $(z^1,\ldots,z^{... 1 There are three points I'd like to clarify. You can formulate the deterministic time-evolution laws of classical mechanics as first-order partial differential equations and thus, you wouldn't need to specify any time derivatives of quantities that specify the state of the system at a given instant of time. This is exactly what Hamiltonian formalism does. A ... 1 If we have the 4-vectors for position and momentum, the results are not frame-dependent in classical (relativistic) mechanics. In relativistic QM, if a spinor is the solution of an Hamiltonian, it is also not frame-dependent. 1 It might be clearer if we reduce to the one-dimensional case. Suppose we have just one coordinate$q$and one momentum$p\$. Hamilton's equations are: $$\dot{q}=\frac{\partial H}{\partial p}$$ $$\dot{p}=-\frac{\partial H}{\partial q}$$ Suppose we apply the following transformation: $$Q=q+\alpha F(q,p)$$ $$P=p+\alpha E(q,p)$$ Now let's suppose we want to write ...

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