# Tag Info

## Hot answers tagged hamiltonian-formalism

5

First-class constraints generate gauge transformations (assuming the Dirac conjecture), i.e. map physically equivalent states onto each other. Even if you do not assume the Dirac conjecture, then first-class constraints still generate transformations that map allowed states, i.e. those on the constraint surface, to other allowed states. Second-class ...

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Conservation laws I would start from the most basic question. Any operator evolves in time as: $$O(t) = e^{\frac{i}{\hbar} Ht} O(0) e^{-\frac{i}{\hbar} Ht}$$ (this is a consequence of the Hamiltonian being the generator of time translations) where $U(t) = e^{-\frac{i}{\hbar} Ht}$ is the time evolution operator. It follows straightforwardly that when we ...

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First allow me to type out some basic equations in a more pedestrian notation. The Gallilean symmetry acts as \begin{align} p &\mapsto p + m v \\ q &\mapsto q + vt. \end{align} This is generated by the time dependent boost charge $$K(t) = -mq + pt.$$ (In fact, if one realizes that $p_0 = \gamma m$ in relativistic mechanics, in the $v \ll c$ limit ...

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On one hand, $$0~=~[0,0]~=~[f(x),\vec{p}\cdot\vec{\nabla}f]~=~i\hbar (\vec{\nabla}f)^2.$$ On the other hand, a constraint function $f$ typically satisfies a regularity condition $$\left .\vec{\nabla}f \right|_{f=0}~\neq~\vec{0}.$$

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Would an example suffice? If so, consider the case $f(\vec x) = x_1$. Then (1) says $x_1=0$, which is already inconsistent with the commutation relation, and (3) says $p_1=0$, which is again inconsistent with the commutation relation. If $x_1$ or $p_1$ is zero, then we can't have $[x_1,p_1]\neq 0$.

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Sometimes it so happens that you have chosen your coordinate in such a way that Hamiltonian looks horrible (that is not separable) but you can choose good coordinates to make you're hamiltonian look good (that is separable). All you need to do is to look for such a coordinate. How are you gonna do it? You have to choose eigenvectors of the matrices you have ...

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Perhaps it is helpful to take a step back and review the definitions: In this answer, we will assume that the Lagrangian $L=T-U$ is the difference between kinetic and (possibly velocity-dependent) potential energy. Consider the (Lagrangian) energy function $$h(q,\dot{q},t)~=~\left(\sum_{j=1}^n\dot{q}^j\frac{\partial }{\partial \dot{q}^j}-1 \right)L(q,\dot{... 3 First, let us consider the situation described by the OP: an atom and a EM mode that are coupled and exchanging energy (as described, for example, by the Jaynes-Cummings model). As the OP says, once you take into account both the energy in the atom and in the field the total energy is conserved. However, this does not make it any closer to an eigenstate ... 3 The fundamental problem here is that many people, and also Pitts in his paper, are not careful about what theory they are currently talking about. "Quantization of Gauge Systems" by Henneaux and Teitelboim is actually careful about this, and their chapter 3 shows the correct resolution of this problem, even though Pitts cites it as an example for ... 2 The fact that,$$ \phi_{1}=p_{x}+\frac{qB}{2c}y\approx0 $$Does not imply that,$$ p_{x}=-\frac{qB}{2c}y $$But only,$$ p_{x}\approx-\frac{qB}{2c}y and conversely for y, x. That is, p_x is a function of y only on the constraint surface. Because the Dirac bracket is a derivative operator built by design to keep you on the constraint surface, while ... 2 For mechanical system you can use this: Euler Lagrange \begin{align*} &\mathcal{L} =T-U\\ &\frac{d}{dt}\left(\frac{\partial \mathcal{L}}{\partial \dot{\boldsymbol{q}}}\right)- \frac{\partial \mathcal{L}}{\partial \boldsymbol{q}} =\left[\frac{\partial \boldsymbol{R}}{\partial \boldsymbol{q}}\right]^T\boldsymbol{f}_a\qquad\qquad (1) \end{... 2 It seems like you're confused about the structure of quantum mechanics. When we say that X is conserved in quantum mechanics, we just mean that the operator X commutes with the hamiltonian. It doesn't mean that we can't have superpositions of different eigenstates of X. It is also in general not meaningful operationally to ask whether a system is in a ... 1 I have a distinct feeling you chose a huge number of interacting particles so you won't have to model them in your mind. Choose just two states, and consider a quantum flipflop, describable by a two-vector, |\psi(t)\rangle. In the most general unitary evolution, your energy is time-independent, \langle \psi(t)|H|\psi(t)\rangle = \langle \psi(0)| e^{...

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According to these lecture notes, the combination of the following Hamiltonian density and constraint gives rise to the Maxwell equations: $$\mathcal{H} = \frac{\varepsilon_0}{2}\mathbf{E}^2 + \frac{1}{2\mu_0}\mathbf{B}^2 - j_\mu A^\mu$$ and $$\nabla\cdot\mathbf{E}=\frac{\rho}{\varepsilon_0}$$

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OP's 2 different conditions arise from using 2 different definitions of the Legendre transformation. One definition uses supremum while another definition uses substitution, cf. e.g. my Phys.SE answer here.

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No, there is no "general conservation law" like what you imagine. The reason is simple: a variational principle is just that, a variational principle. It need not have anything to do with physics at all. It could be something as simple as finding the vertex of a parabola. What is considered a dynamic system is a matter of taste because time is just ...

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Concerning OP's first question: One can argue that a notion of volume/information is conserved in the Lagrangian formulation, cf. e.g. my Phys.SE answer here. Concerning OP's last question: The fact that the Hamiltonian formulation in phase space has a greater symmetry (symplectomorphism symmetry, Liouville's theorem) than the corresponding Lagrangian ...

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ACuriousMind has already given as good answer. Let me just add that it is typically the other way around, i.e. that first-class systems are simpler than second-class systems, at least at the formal level. This is e.g. because it is often difficult to invert the matrix of constraints to construct the Dirac bracket, especially in field theory, where it ...

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I think it is a matter of definitions: microcanonical, canonical, and grand canonical distributions are quite general and applicable to quantum and classical systems, interacting or not (the assumtpions involved in tehir derivation are of different nature). Boltzmann distribution is usually a synonym for the canonical distribution, although in some contexts ...

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The serial nature of the skewed statements and braided questions makes it very hard to come up with a decent answer without teaching the subject properly as mainstream texts do: Sakurai, Landau-Lifshits, Merzbacher, Messiah... We notice that, by construction, $[H_{rel},L_i]=0 \ \ ; \ \ i=1,2,3$ "Construction" is unfortunate. "Inspection"...

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