I have just started studying Classical Mechanics and I just can't wrap my head around why we legendre transform lagrangian to hamiltonian if it just reverses velocity and conjugate momentum.
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I've just finished a Classical Mechanics course, and looking back on it some things are not quite clear. In the first half we covered the Lagrangian formalism, which I thought was pretty cool. I specially appreciated the freedom you have when choosing coordinates, and the fact that you can basically ignore constraint forces. Of course, most simple situations you can solve using good old $F=ma$, but for more complicated stuff the whole formalism comes in pretty handy.  Then in the second half we switched to Hamiltonian mechanics, and that's where I began to lose sight of why we were doing things the way we were. I don't have any problem understanding the Hamiltonian, or Hamilton's equations, or the Hamilton-Jacobi equation, or what have you. My issue is that I don't understand why would someone bother developing all this to do the same things you did before but in a different way. In fact, in most cases you need to start with a Lagrangian and get the momenta from $p = \frac{\partial L}{\partial \dot{q}}$, and the Hamiltonian from $H = \sum \dot{q_i}p_i - L$. But if you already have the Lagrangian, why not just solve the Euler-Lagrange equations?  I guess maybe there are interesting uses of the Hamiltion formalism and we just didn't do a whole lot of examples (it was the harmonic oscillator the whole way, pretty much). I've also heard that it allows a somewhat smooth transition into quantum mechanics. We did work out a way to get Schrödinger's equation doing stuff with the action. But still something's not clicking.  My questions are the following: Why do people use the Hamiltonian formalism? Is it better for theoretical work? Are there problems that are more easily solved using Hamilton's mechanics instead of Lagrange's? What are some examples of that?
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I am trying to write the Lagrangian and Hamiltonian for the forced Harmonic oscillator before quantizing it to get to the quantum picture. For EOM $$m\ddot{q}+\beta\dot{q}+kq=f(t),$$ I write the Lagrangian  $$ L=\frac{1}{2}m\dot{q}^{2}-\frac{1}{2}kq^{2}+f(t)q$$ with Rayleigh dissipation function as   $$ D=\frac{1}{2}\beta\dot{q}^{2}$$ to put in Lagrangian EOM $$0 = \frac{\mathrm{d}}{\mathrm{d}t} \left ( \frac {\partial L}{\partial \dot{q}_j} \right ) - \frac {\partial L}{\partial q_j} + \frac {\partial D}{\partial \dot{q}_j}. $$  On Legendre transform of $L$, I get $$H=\frac{1}{2m}{p}^{2}+\frac{1}{2}kq^{2}-f(t)q.$$  How do I include the dissipative term to get the correct EOM from the Hamiltonian's EOM?
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I've seen the 'derivation' as to why momentum is an operator, but I still don't buy it. Momentum has always been just a product $m{\bf v}$. Why should it now be an operator. Why can't we just multiply the wave function by $\hbar{\bf k}$? Why should momentum be a derivative of a wave function?
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We know that angular momentum is defined as the cross product of position and linear momentum. By taking the time derivative, we can deduce that the time rate of change of the angular momentum equals the net torque.  Then why is the conservation of angular momentum considered a law, when we can easily show it is conserved when no net torque is applied, by using Newton's laws?  It truly confuses me.
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Here is a question that's been bothering me since I was a sophomore in university, and should have probably asked before graduating:  In analytic (Lagrangian) mechanics, the derivation of the Euler-Lagrange equations from the principle of least action assumes that the start and end coordinates at the initial and final times are known. As a consequence, any variation on the physical path must vanish at its boundaries. This conveniently cancels out the contributions of the boundary terms after integration by parts, and setting the requirement for minimal action, we obtain the E.L. equations.  This is all nice and dandy, but our intention is finding the location of a particle at a time in the future, which we do not know a priori; after we derive any equations of motion for a system, we solve them by applying initial values instead of boundary conditions.   How are these two approaches consistent?
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The least-action principle is a statement in classical physics saying that all bodies in a system follow a trajectory that minimize the following functional (ignoring explicit time dependence for now): $$ S[L] = \int dt L(x(t), \dot{x}(t)) \qquad\rightarrow\qquad  \frac{d}{dt}\Big(\frac{\partial L}{\partial \dot{x}}\Big) - \frac{\partial L}{\partial x} = 0. $$ The Hamiltonian of the same system can be constructed using the Legendre transform: $$ H(p, x) = \dot{x}\cdot p - L(x, \dot{x})\;;\quad p \equiv \frac{\partial L}{\partial\dot{x}}. $$ Pedagogically, one first studies Lagrangian Mechanics starting from the least-action principle, and eventually constructs an equivalent Hamiltonian Mechanics framework. But suppose I was a weird physicist who wanted to teach Hamiltonian mechanics first, and then later construct Lagrangian Mechanics. What would be the best way to do this?
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Lagrangian and Hamiltonian formulations are the bedrock of particle and field theories, produce the same equations of motion, and are related through a Legendre transform.  Are there more such mathematical objects that are equivalent, or are these two in some way unique?  If so, why are there two equivalent systems,  rather than a single (or more)?
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How is it possible to define the energy for a massless particle in a Freedman-Robertson-Walker metric if the Lagrangian is not invariant under time-translation?  I have postulated the simplest Lagrangian for a massless particle (a Lagrange multiplier times the constraint) and it is obviously not time-invariant. Not only: if I work since at the beginning using the proper time as a evolution parameter, the dynamics in the Lagrange parameter and the spatial components is physically inconsistent, giving meaningless results like a scale factor equal to the Lagrange multiplier. Is there something wrong in my assumption for the Lagrangian?
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I have a question concerning the time ordering operator. Let's suppose we have a time evolution generated by some Hamiltonian $H(t)$ given by $$ U(t)=T_\leftarrow\exp\left(-\mathrm{i}\int_0^t\mathrm{d}s\,H(s)\right)\tag{1}. $$ In Breuer and Petruccione, it's said that if the commutator of the Hamiltonian at some time $t$ with itself at some other time $t^\prime$ is a $c$-number function/ a complex function, i.e. $$\left[H(t),\,H(t^\prime)\right]=f(t,t'),$$ then the time evolution is given by $$ U(t)=\exp\left(-\frac{1}{2}\int_0^t\mathrm{d}s\int_0^t\mathrm{d}s^\prime\,\left[H(s),\,H(s^\prime)\right]\Theta(s-s^\prime)\right)$$ $$\times \exp\left(-\mathrm{i}\int_0^t\mathrm{ds}\,H(s)\right),\tag{2} $$ where $\Theta(s-s^\prime)$ is the Heaviside function. No proof or reference is given and I could not find any explanation anywhere, this is why I ask here. Any help would be very appreciated.
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I have heard that one can recover classical mechanics from quantum mechanics in the limit the $\hbar$ goes to zero.  How can this be done?  (Ideally, I would love to see something like: as $\hbar$ goes to zero, the position wavefunction reduces to a delta function and that the Schrodinger equation/Feynman path integral reduces to the Newtonian/Lagrangian/Hamiltonian equations of motion.)
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Is there a way of explaining why no differential equations in physics exceed order two without delving into Lagrangian and Hamiltonian mechanics - i.e. from Newtonian mechanics? Moreover, is there such an alternative proof, which focuses more on physical reasoning than abstract mathematics?
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If there is a system, described by an Lagrangian $\mathcal{L}$ of the form   $$\mathcal{L} = T-V = \frac{m}{2}\left(\dot{r}^2+r^2\dot{\phi}^2\right) + \frac{k}{r},\tag{1}$$   where $T$ is the kinetic energy and $V$ the potential energy, it is also possible to define the total energy $E$ of the system   $$ E = T + V =\frac{m}{2}\left(\dot{r}^2+r^2\dot{\phi}^2\right) - \frac{k}{r}.\tag{2}$$  If the angular momentum $M$ is defined by $$ M = m r^{2} \dot{\phi},\tag{3}$$ then $E$ can be written as  $$ E = \frac{m}{2}\dot{r}^2 + \underbrace{\frac{M^2}{2mr^2}-\frac{k}{r}}_{V_\textrm{eff}\left(r\right)},\tag{4}$$  where the last two terms are written as new, "effective" potential $V_\textrm{eff}\left(r\right)$.    In addition, using the definition of $M$, the Lagrangian $\mathcal{L}$ can be written as  $$ \mathcal{L} = \frac{m}{2}\dot{r}^2 + \underbrace{\frac{M^2}{2mr^2}+\frac{k}{r}}_{-V_\textrm{eff}\left(r\right)},\tag{5}$$  where the sign of $V_\textrm{eff}\left(r\right)$ has been changed, since $\mathcal{L} = T-V$. But from this argument, it appears that there are two different possible ways to construct the same effective potential. This seems to me contradictory. Where is my mistake?
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Is there a proof from the first principle that for the Lagrangian $L$,    $$L = T\text{(kinetic energy)} - V\text{(potential energy)}$$  in classical mechanics? Assume that Cartesian coordinates are used. Among the combinations, $L = T - nV$, only $n=1$ works. Is there a fundamental reason for it?  On the other hand, the variational principle used in deriving the equations of motion, Euler-Lagrange equation, is general enough (can be used to to find the optimum of any parametrized integral) and does not specify the form of Lagrangian. I appreciate for anyone who gives the  answer, and if possible, the primary source (who published the answer first in the literature).      Notes added on Sept 22: - Both answers are correct as far as I can find. Both answerers were not sure about what I meant by the term I used: 'first principle'. I like to elaborate what I was thinking, not meant to be condescending or anything near to that. Please have a little understanding if the words I use are not well-thought of. - We do science by collecting facts, forming empirical laws, building a theory which generalizes the laws, then we go back to the lab and find if the generalization part can stand up to the verification. Newton's laws are close to the end of empirical laws, meaning that they are easily verified in the lab. These laws are not limited to gravity, but are used mostly under the condition of gravity. When we generalize and express them in Lagrangian or Hamiltonian, they can be used where Newton's laws cannot, for example, on electromagnetism, or any other forces unknown to us. Lagrangian or Hamiltonian and the derived equations of motion are generalizations and more on the theory side, relatively speaking; at least those are a little more theoretical than Newton's laws. We still go to lab to verify these generalizations, but it's somewhat harder to do so, like we have to use Large Hadron Collider. - But here is a new problem, as @Jerry Schirmer pointed out in his comment and I agreed. Lagrangian is great tool if we know its expression. If we don't, then we are at lost. Lagrangian is almost as useless as Newton's laws for a new mysterious force. It's almost as useless but not quite, because we can try and error. We have much better luck to try and error on Lagrangian than on equations of motion. - Oh, variational principle is a 'first principle' in my mind and is used to derive Euler-Lagrange equation. But variational principle does not give a clue about the explicit expression of Lagrangian. This is the point I'm driving at. This is why I'm looking for help, say, in Physics SE. If someone knew the reason why n=1 in L=T-nV, then we could use this reasoning to find out about a mysterious force. It looks like that someone is in the future.
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In Hamiltonian Mechanics, a version of Hamilton's principle is shown to evolve a system according to the same equations of motion as the Lagrangian, and therefore Newtonian formalism. In particular, letting $\delta$ indicate a variation of the path through phase space, $$ \delta \int_{t_1}^{t_2} \big(p_i \dot{q}_i - H(p_i,q_i,t)\big)dt = 0 $$ is shown to generate the same motion equations as those found by the Legendre transform of the Lagrangian equations of motion. Essentially, when we calculate the Euler-Lagrange equations for the above integrand, we find $$ \dot{p}_i + \frac{\partial H}{\partial q_i} = 0\text{, } \dot{q}_i - \frac{\partial H}{\partial p_i} = 0. $$ Now, since these are both first-order equations of motion, we require $2n$ boundary conditions to obtain a solution, where $n$ is the number of particles. This is perfectly consistent with the Lagrangian formalism which had $2n$ boundary conditions due to its $n$ distinct second-order motion equations.   An important thing to note is that since $p_i\dot{q}_i - H(p_i,q_i,t)$ contains no $\dot{p}_i$ dependence, the variation in $p_i(t)$ need not be zero at the endpoints of the path. This is not the case with $q_i(t)$ because the $\dot{q}_i$ dependance results in the appearance of boundary terms that must be set to zero in order to obtain the motion equations.  Now, as far as the boundary conditions are concerned, this makes sense. For the path to be uniquely specified given Newtonian mechanics, we require $2n$ boundary conditions. These can either be the usual Newtonian choice of initial positions and velocities/momenta, or they can be the initial and final positions. Both are perfectly acceptable mathematically speaking. If we were required, in the modified Hamilton's principle variation, to enforce that the variations in the momenta also be zero at the initial and final times, this would mean $2n$ additional boundary conditions, which would generally over constrain a Newtonian system.   However, my confusion is this: when textbooks (Goldstein in particular) consider generating functions, they require that the integrand be invariant to the addition of a total time derivative of a function $F(q_i,p_i,t)$ of the phase space coordinates. However, adding such a function will in general add some $\dot{p}_i$ dependence to the integrand, which will then add boundary terms to the equations of motion unless we require that the $p_i(t)$ variations be zero at the boundaries. This is fine as far as it goes. We can always define the variation however we like. The point is just that it spits out the right motion equations at the end of the day. But my confusion lies in how this doesn't generally over constrain the system. If we specify all the positions and momenta at both the initial and the final times, might not the required path to connect those points in phase space be non-Newtonian? Does this only matter if you're actually trying to use the stationary action principle to find the paths, rather than just using it to find the motion equations?
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What exactly are  (and )?   I want to self-study QM, and I've heard from most people that Hamiltonian mechanics is a prereq. So I wikipedia'd it and the entry only confused me more. I don't know any differential equations yet, so maybe that's why.    But what's the difference between Hamiltonian (&amp; Lagrangian mechanics) and Newtonian mechanics?  And why is Hamiltonian mechanics used for QM instead of Newtonian?  Also, what would the prereqs for studying Hamiltonian mechanics be?
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In reading Weinstock's Calculus of Variations, on  he explains how Schrodinger apparently first derived the Schrodinger equation from variational principles.  Unfortunately I don't think page 262 is showing so I'll explain the gist of it:  "In his initial paper" he considers the reduced Hamilton-Jacobi equation   $$\frac{1}{2m}\left[\left(\frac{\partial S}{\partial x}\right)^2 \ + \ \left(\frac{\partial S}{\partial y}\right)^2 \ + \ \left(\frac{\partial S}{\partial z}\right)^2\right] \ + \ V(x,y,z) \ - \ E \ = \ 0$$   for a single particle of mass $m$ in an arbitrary force field described by a potential $V = V(x,y,z)$.  With a change of variables $S \ = \ K\log(\Psi)$, (where $K$ will turn out to be $\hbar=h/2\pi$) it reduces to  $$\frac{K^2}{2m}\left[\left(\frac{\partial \Psi}{\partial x}\right)^2 \ + \ \left(\frac{\partial \Psi}{\partial y}\right)^2 \ + \ \left(\frac{\partial \Psi}{\partial z}\right)^2\right] \ + \ (V \ - \ E)\Psi^2 \ = 0.$$  Now instead of solving this he, randomly from my point of view, chose to integrate over space  $$I  = \iiint_\mathcal{V}\left(\frac{K^2}{2m}\left[\left(\frac{\partial \Psi}{\partial x}\right)^2 \ + \ \left(\frac{\partial \Psi}{\partial y}\right)^2 \ + \ \left(\frac{\partial \Psi}{\partial z}\right)^2\right] + (V - E)\Psi^2\right)\,\mathrm{d}x\,\mathrm{d}y\,\mathrm{d}z$$   then Weinstock extremizes this integral which gives us the Schrodinger equation.  Apparently as the book then claims on page 264 it is only after this derivation that he sought to connect his idea's to deBroglie's wave-particle duality.  Thus I have three questions,  1) What is the justification for Feynman's famous quote:     Where did we get that [Schrödinger's equation] from? It's not possible to derive it from anything you know. It came out of the mind of Schrödinger.   The Feynman Lectures on Physics   in light of the above derivation. I note that all the derivations I've seen of the Schrodinger equation doing something like using operators such as $i\hbar\partial/\partial t = E$ to derive it always mention it's merely heuristic, yet what Schrodinger apparently originally did seems like a roundabout way of solving the Hamilton-Jacobi equation with no heuristic-ness in sight. What subtleties am I missing here? Why would I be a fool to arrogantly 'correct' someone who says Schrodinger is not derivable from anything you know?  Note I've read a ton of the threads on this topic on this forum &amp; none even go near the calculus of variations, in fact the above contradicts  that even refers to Schrodinger's Nobel prize lecture, so hopefully it's not a duplicate.  2) Is the mathematical trick Schrodinger has used something you can use to solve problems?   3) Why can't you use this exact derivation in the relativistic case?  Edit: On the topic of complex numbers:  On page 276, 14 pages after he explains what I posted, &amp; after he goes through how Schrodinger linked his work with DeBroglie's work, it is only then that Weinstock says:     In a more complete study of quantum mechanics than the present one the   admissibility of complex eigenfunctions $\Psi$ is generally shown to be   necessary. If $\Psi$ is complex, the quantity $|\Psi|^2$ is employed as the   position probability-density function inasmuch as $\Psi^2$ is not restricted   to real nonnegative values.   Apparently Schrodinger was able to do what I have posted using real-valued functions &amp; have K as I have defined it, without i. If you're following what Weinstock is saying he shows how the hydrogen atom energy levels are explainable without complex numbers, i.e. he is able to derive a physical interpretation of the eigenvalues (discrete energy levels) of the Schrodinger equation that were in accord with experiment (see Section 11.3 Page 279 on). As far as I understand it it is in trying to find a physical interpretation of the eigenfunctions that one is forced into complex numbers, though apparently, according to the book, it can be shown to be necessary. That is something to take into account!
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Basically, I just want to know the advantages of Lagrangian and Hamiltonian mechanics over Newtonian mechanics, that made it much more preferable and widely used!