# Tag Info

27

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 ...

22

In order to use Lagrangians in QM, one has to use the path integral formalism. This is usually not covered in a undergrad QM course and therefore only Hamiltonians are used. In current research, Lagrangians are used a lot in non-relativistic QM. In relativistic QM, one uses both Hamiltonians and Lagrangians. The reason Lagrangians are more popular is that ...

21

Notice first that the phase space of any theory is nothing but the space of all its classical solutions. The traditional presentation of phase spaces by fields and their canonical momenta on a Cauchy surface is just a way of parameterizing all solutions by initial value data -- if possible. This is often possible, but comes with all the disadvantages that a ...

17

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 ...

13

See http://en.wikipedia.org/wiki/Legendre_transformation#Applications In theoretical physics, the basic or defining mathematical properties of the Legendre transformation are used to switch between one form of the energy - or "potential", as the generalized energies are called in thermodynamics - to another. This is important to switch between the ...

12

Vladimir's answer has the right essence but it is also misleading, so let me clarify. The formula $$H = \sum_i p_i\dot q_i - L$$ relating the Hamiltonian and the Lagrangian is completely general. It holds in all theories that admit both Lagrangians and Hamiltonians, whether they're relativistic or not, whether or not they have any other symmetry aside ...

11

In an ideal, holonomic and monogenic system (the usual one in classical mechanics), Hamiltonian equals total energy when and only when both the constraint and Lagrangian are time-independent and generalized potential is absent. So the condition for Hamiltonian equaling energy is quite stringent. Dan's example is one in which Lagrangian depends on time. A ...

11

A common mistake when students begin the study of the quantum harmonic oscillator is to try to convert everything to integrals. The thing is, in most curricula, the QHO is also used as a way to secretly acquaint you with bra-ket notation, and all the conveniences it offers. In reality, you shouldn't need any integrals at all here. $\lvert n \rangle$ is a ...

10

Legendre transformations are commonly used in thermodynamics (to switch between different independent variables) and classical mechanics (to switch between the Lagrange and Hamilton formalisms). But you rightly ask: what exactly is a Legendre transformation? Where does it come from? What makes it work? In (1D) classical mechanics, for example: if we have a ...

9

A fairly basic remark to make is that usually we can plainly identify $$L = T-U$$ where $T$ is the kinetic energy and $U$ is the potential energy, and $$H = T+U$$ Expressing these quantities for e.g. a Hooke-like spring (or any system where $U\neq 0$) would give you a problem with the sign of $U$ if you simply substitute the expression you find for ...

8

The Hamiltonian is in general not equal to the energy when the coordinates explicitly depend on time. For example, we can take the system of a bead of mass $m$ confined to a circular ring of radius $R$. If we define the $0$ for the angle $\theta$ to be the bottom of the ring, the Lagrangian $$L=\frac{mR^2\dot{\theta}^2}{2}-mgR(1-\cos{(\theta)}).$$ The ...

8

I) OP is given a problem of the form $$\tag{1} \dot{q}~=~f(q,p), \qquad \dot{p}~=~g(q,p),$$ where $f$ and $g$ are two given smooth functions. OP is asked to derive the integrability condition for the eqs. (1) to be Hamilton's eqs. $$\tag{2} \dot{q}~=~\frac{\partial H}{\partial p}, \qquad \dot{p}~=~-\frac{\partial H}{\partial q}.$$ OP correctly ...

8

The word quantization surface is not standard terminology. It apparently refer to a (generalized) Cauchy surface. A (generalized) Cauchy surface is a hypersurface on which the initial conditions are given for a well-posed initial value problem. Phrased differently, for given initial conditions on the Cauchy surface, there exists a unique solution for the ...

8

Comments to the question (v2): 1) The correspondence between Lagrangian (L) and Hamiltonian (H) theories is mired with subtleties. Some general tools for singular Legendre transformations are available, such as Dirac-Bergmann analysis, Faddeev-Jackiw method, etc. But rather than claiming complete understanding and existence of the L-H correspondence, it is ...

7

Well, perhaps one should consider reading The Hamiltonian formulation of General Relativity: myths and reality for further mathematical details. But I would like to remind to you with most constrained Hamiltonian systems, the Poisson bracket of the constraint generates gauge transformations. For General Relativity, foliating spacetime $\mathcal{M}$ as ...

7

The action functional and Hamilton's principal function are two different mathematical objects related to the same physical quantity. The action along a trajectory $\gamma:[t_1,t_2]\rightarrow Q$ is given by $$S[\gamma] = \int_{t_1}^{t_2}L(\gamma(t'),\dot\gamma(t'),t')dt'$$ whereas the pricipal function is the solution of the Hamilton-Jacobi equation $$... 7 Lubos Motl and Vladimir Kalitvianski have already provided correct conventional answers concerning the Legendre transformation from Lagrangian to Hamiltonian formalism. Nevertheless, it seems appropriate to mention that OP's second equation(v2)$$\mathcal{H} ~=~ \pi_{\mu}\partial^{\mu} \phi - \mathcal{L}$$is precisely the starting point for De ... 7 The ordering ambiguity is the statement – or the "problem" – that for a classical function f(x,p), or a function of analogous phase space variables, there may exist multiple operators \hat f(\hat x,\hat p) that represent it. In particular, the quantum Hamiltonian isn't uniquely determined by the classical limit. This ambiguity appears even if we require ... 7 Your interpretation is not quite right. One sharp interpretation one can give to this "cutting" of phase space into cubes of size h^{2N} (here N is the dimension of the system's configuration space), is that it allows one to use classical phase space to count the number of energy eigenstates of the corresponding quantum hamiltonian. Instead of trying ... 7 You're on the right track. Complete the square on x and you'll have some newly defined Harmonic oscillator whose position operator you have found. The additional constant that comes from completing the square will add to your ground state energy. (Once you have completed the square, you should have something of the form H = constant + \frac{P^2}{2M} + ... 7 First of all, Lagrangian is a mathematical quantity which has no physical meaning but Hamiltonian is physical (for example, it is total energy of the system, in some case) and all quantities in Hamiltonian mechanics has physical meanings which makes easier to have physical intuition. In Hamiltonian mechanics you have canonical transformations which allows ... 7 Perhaps the simplest and most intuitive approach is to regularize the hard wall potential$$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).$$For instance, one could choose the regularized potential as$$ ...

6

It seems there is a simply way to do this for polynomials with finite degree $d$. Since $a^\dagger a$ is the number operator, we can take $a^\dagger a = N$, where $N$ is the number of excitations corresponding to a particular level. Then if the Hamiltonian has the general form $H = \sum_{k=0}^d c_k (a^\dagger a)^k$, the energy corresponding to a particular ...

6

First, one must appreciate that the phase space is classically parameterized by $x,p$ and coordinates on an ordinary plane commute with each other, $xp=px$. However, in quantum mechanics, this ain't the case. Instead, we have the Heisenberg commutator $$xp - px = i\hbar.$$ This means that quantum mechanically, the phase space is not an ordinary plane (or ...

6

I) At least three different quantities in physics are customary called an action and denoted with the letter $S$. The (off-shell) action $$\tag{1}S[q]~:=~ \int_{t_i}^{t_f}\! dt \ L(q(t),\dot{q}(t),t)$$ is a functional of the full position curve/path $q^i:[t_i,t_f] \to \mathbb{R}$ for all times $t$ in the interval $[t_i,t_f]$. See also this question. ...

6

You have already mentioned the correct reason---the Lagrangian is manifestly Lorentz-invariant whereas the Hamiltonian is not. Since a relativistic field theory must be build of Lorentz-invariant quantities only the Lagrangian approach is good. Compare for example the expressions for a free real scalar field $\phi$  ...

6

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 ...

6

The overall idea is the following. As the symplectic manifold is affine (in the sense of affine spaces not in the sense of the existence of an affine connection), when you fix a point $O$, the manifold becomes a real vector space equipped with a non-degenerate symplectic form. A quantization procedure is nothing but the assignment of a (Hilbert-) Kahler ...

5

Coordinate invariance guarantees that the phase space $M$ can be endowed with a symplectic 2-form $\omega$ which locally is given by $\omega = dq^i \wedge dp_i$. This form is closed ($d\omega = 0$) and nondegenerate, i.e. $\omega^n$ is a volume form, where $2n$ is the dimension of $M$. The other conditions say that for any Hamiltonian function $H$, the ...

5

Nope, this operation is not a symmetry in the physics sense. A symmetry transformation is a transformation that changes or mixes the values of the basic degrees of freedom such as positions and momenta $x(t)$, $p(t)$ in mechanics or the values of fields such as $\vec E(\vec x,t)$ and $\vec B(\vec x,t)$ in electromagnetism. The Hamiltonian is not an ...

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