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The fields satisfy the wave equation. We can therefore write \begin{split} \phi(x) = \int \frac{ d^3 p}{ (2\pi)^3} \frac{1}{2 \omega_{\bf p} } \left[ a({\bf p}) e^{i p \cdot x} + b^\dagger({\bf p} ) e^{- i p \cdot x} \right] \\ \phi^\dagger (x) = \int \frac{ d^3 p}{ (2\pi)^3} \frac{1}{2 \omega_{\bf p} } \left[ b({\bf p}) e^{i p \cdot x} + ...

1

I) First some terminology. Consider a symplectic manifold $(M;\omega)$. In a local chart $U\subseteq \mathbb{R}^{2n}$, the symplectic two-form reads $$\tag{1} \omega~=~\frac{1}{2} \omega_{ij}~\mathrm{d}x^i \wedge \mathrm{d}x^j ,$$ and the corresponding Poisson bi-vector $$\tag{2} \pi~=~\frac{1}{2} \pi^{ij}~\partial_i \wedge \partial_i,$$ where $$... 2 For a convex function you can do the following: For each point on the graph of the function, draw the line tangent to the function at that point. That point can now be identified by its original x and y=f(x) coordinates, or by specifying the slope of that tangent line and its corresponding y-intercept. Each point maps to one and only one line, and ... 1 Alternatively, there exists an extended approach to the Legendre transformation between the Lagrangian and Hamiltonian formalism, cf. e.g. Ref. 1. Let us suppress explicit time dependence t from the notation in the following. Consider the extended Lagrangian$$\tag{1} L_E(q,v,p)~:=~ p_i(\dot{q}^i-v^i)+L(q,v)~=~p_i\dot{q}^i-h(q,v,p), \tag{2} ...

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Ok, let us start from scratch. A function $g: \mathbb R^n \to \mathbb R$ with $f \in C^2(\mathbb R^n)$ is said to be convex if its Hessian matrix (i.e. the one with coefficients $\partial^2 f/\partial x_i \partial x_j$) is everywhere (strictly) positive defined. Let $\Omega \subset \mathbb R \times \mathbb R^n$ be an open set, and focus on a jointly ...

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Try using $U(x) = \Theta(x)$, $\Theta(x) \equiv \begin{cases} 1, & x>0\\ 0, & x<0\end{cases}$ . The force becomes $\dot{p} = - \delta(x)$, as anticipated. Because the force is infinite with direction opposite that of the momentum, the particle cannot cross the plane $x=0$. Another way to see that the particle won't cross $x=0$ is to integrate ...

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The gist of the response before the edit remains valid. The hamiltonian is defined as $$H(q,p,t) \equiv p \dot{q} - L(q,\dot{q},t),$$ the Legendre trasform of $L$. The Legendre transform takes $p$ to $\dot{q}$, because $L$ is convex, and this map is defined by $p = \partial L /\partial \dot{q}$. From the latter equation it is obvious that the map is ...

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I) Lagrangian formalism. Let us suppress position dependence $q^i$ and explicit time dependence $t$ in the following, and also assume that the Lagrangian $L=L(v)$ is a smooth function of the velocities $v^i$, where $i=1, \ldots, n$. Define functions $$\tag{1} g_i(v)~:=~\frac{\partial L(v)}{\partial v^i}, \qquad i=1, \ldots, n;$$ $$\tag{2} ... 0 Hints: Prove that a one-parameter group (\Phi_t)_{t\in I} of diffeomorphisms \Phi_t: M \to M is generated by a vector field X\in\Gamma(M). Prove that if the one-parameter group (\Phi_t)_{t\in I} preserves the a form \omega then {\cal L}_{X}\omega =0. Prove that {\cal L}_{X}\omega =0 together with the fact that \omega is a symplectic ... 0 See V.I.Arnold Mathematical Methods of Classical Mechanics chapter 44 E for definitions and proofs. Also check the footnotes up to p.241. In particular, Landau & Lifshitz conflates the two definitions of a canonical transformation and has several errors on the subject. If we define a canonical transformation as the diffeomorphism on a symplectic ... 4 Refs. 1 and 2 define a canonical transformation (CT)$$\tag{1} (q^i,p_i)~\longrightarrow~ (Q^i,P_i)$$[together with choices of Hamiltonian H(q,p,t) and Kamiltonian K(Q,P,t)] as satisfying$$ \tag{2} (p_i\mathrm{d}q^i-H\mathrm{d}t)-(P_i\mathrm{d}Q^i -K\mathrm{d}t) ~=~\mathrm{d}F$$for some generating function F. On the other hand, Wikipedia (March ... 7 Perhaps the simplest and most intuitive approach is to regularize the hard wall potential$$V_0(x)~=~\left\{ \begin{array}{rcl} 0 &\text{for}& x<0 \cr\cr \infty &\text{for}& x>0\end{array}\right. $$as$$ \lim_{\varepsilon \to 0^+} V_{\varepsilon}(x) ~=~V_0(x).$$For instance, one could choose the regularized potential as$$ ...

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Well, you better check your "pretty straightforward" calculation again, as $$\dot{x}\neq\frac{p_{x}+p_{y}}{4m}$$ and $$\dot{y}\neq\frac{p_{x}+p_{y}}{m}.$$

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The classical Lagrangian for the free electron field is, $$L=\int d^{3}x(i\psi^{\dagger}\frac{\partial \psi}{\partial t}+i\psi^{\dagger}\alpha_{r}\frac{\partial \psi}{\partial x^{r}}-m\psi^{\dagger}\beta \psi) \ .$$ The q's are $q^{i}(t)\rightarrow q^{(a,x)}\rightarrow \psi^{a}(t,x)$ and so the velocities are $\dot{q}^{i}(t)\rightarrow \frac{\partial ... 4 The Hamiltonian density for any classical field is given by: $$\mathcal{H} = \pi \dot{\phi} - \mathcal{L}$$ where$\pi$is the canonical momentum density: $$\pi(\mathbf{x},t) = \frac{\partial \mathcal{L}}{\partial \dot{\phi}(\mathbf{x},t)}$$ In classical point particle mechanics the Poisson brackets ... 4 I) At the classical level, there is no convexity condition. If an action functional$S$yields a stationary action principle, so will the negative action$-S$. (Under sign changes, a convex function turns in concave function and vice versa.) Or one could imagine a theory, which is convex in one section and concave in other sector. II) On the Lagrangian side ... 1 I) Well semiclassically, the Hamilton-Jacobi equation famously appears to lowest order in$\hbar\$ in a WKB expansion of the Schrödinger equation. II) The quantum concept of a canonical transformation (CT) $$\tag{1} z^I=(\hat{q}^i; \hat{p}_i) ~\longrightarrow~ Z^J=(\hat{Q}^j; \hat{P}_j)$$ (where the old and new canonical variables both satisfy the CCR) ...

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