Constructing a hamiltonian for a harmonic oscillator

In summary: Hamiltonian mechanics is a reformulation of classical mechanics that emphasizes the underlying symmetries and conservation laws of the system. It is particularly useful for systems with many degrees of freedom, such as in quantum mechanics.
  • #1
woodssnoop
10
0
Hello:

I am trying to understand how to build a hamiltonian for a general system and figure it is best to start with a simple system (e.g. a harmonic oscillator) first before moving on to a more abstract understanding. My end goal is to understand them enough so that I can move to symplectic transforms and then on to symplectic integration methods, but I plan on taking this one step at a time. From what I know and understand so far:

[itex]T = \frac{1}{2} m v^{2}[/itex]

[itex]V = \frac{1}{2} k q^{2}[/itex]

[itex]L(q,\dot{q},t) = T + V[/itex]

[itex]H(p,q,t) = p \dot{q} - L(q,\dot{q},t) [/itex]

[itex]\dot{q} =\frac{\partial H}{\partial p}[/itex]

[itex]\dot{p} = - \frac{\partial H}{\partial q}[/itex]


I have been replacing [itex] v [/itex] with [itex] \dot{q} [/itex], but I don't believe I am getting the right answer. So my first questions are:

1a. Are the terms [itex] p [/itex] and [itex] \dot{q} [/itex] the same thing, and if not why?
1b. Are the [itex] \dot{q}, \dot{p} [/itex] and other dotted terms I see in many texts referring to the time derivative of that term? If so, why is [itex] \dot{p} [/itex] not referred to as [itex] \ddot{q} [/itex]?​

While I was working though the problem I tried above, I noticed that given [itex] q = a \sin(2\pi f t) [/itex], [itex] H [/itex] could be expressed as just a function of just [itex] t [/itex].

2. So can the hamiltonian be function of only [itex] t [/itex]?​

Thank for your help in advance,
Dan
 
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  • #2
woodssnoop said:
1a. Are the terms [itex] p [/itex] and [itex] \dot{q} [/itex] the same thing, and if not why?
No, they are not. They happen to be easily related in this particular case since they are proportional to each other. But generalized velocity and generalized momentum are not generally proportional to each other (although it might be true for most if not all example you encounter in class). A simple case to the contrary is a relativistic particle.

When you really dwell into Hamiltonian mechanics, you realized that [itex]p[/itex] and [itex]q[/itex] (not [itex]\dot{q}[/itex]) are sort of interchangeable via what's called Canonical Transformation. In fact, you can already see that the two Hamiltonian equations are invariant under if you interchange [itex]p[/itex] and [itex]q[/itex] and make some clever sign change. Mathematically, [itex]\{p,q\}[/itex] are points of an geometric object called the Cotangent Bundle. But that's probably just formalism mumbo jumbo for you at this point.

1b. Are the [itex] \dot{q}, \dot{p} [/itex] and other dotted terms I see in many texts referring to the time derivative of that term? If so, why is [itex] \dot{p} [/itex] not referred to as [itex] \ddot{q} [/itex]?
Because they are different thing. Again, this only works when [itex]p \propto \dot{q}[/itex], which is not necessarily true.

While I was working though the problem I tried above, I noticed that given [itex] q = a \sin(2\pi f t) [/itex], [itex] H [/itex] could be expressed as just a function of just [itex] t [/itex].

2. So can the hamiltonian be function of only [itex] t [/itex]?
This is a specific trajectory of the motion, of a particular initial condition. Hamiltonian mechanics is more interested in the general structure of the system itself. So in fact, you are throwing away information by making such a substitution: instead of studying all trajectories govern by this Hamiltonian, you are studying a particular trajectory that solves the Hamiltonian.
 
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  • #3
mathfeel said:
No, they are not. They happen to be easily related in this particular case since they are proportional to each other. But generalized velocity and generalized momentum are not generally proportional to each other (although it might be true for most if not all example you encounter in class). A simple case to the contrary is a relativistic particle.

Thank you for replying. I made a mistake in my first question and I should probably rephrase as well. I meant to ask, can [itex] m \dot{q} [/itex] be replaced by [itex] p [/itex]? I understand that if the mass was not constant then this would not be possible, but in a non-relativistic calculation, such as this harmonic approximation, are these terms (i.e. [itex] q, \dot{q}, p [/itex], etc.) treated the same way as in Newtonian mechanics?

This is a specific trajectory of the motion, of a particular initial condition. Hamiltonian mechanics is more interested in the general structure of the system itself. So in fact, you are throwing away information by making such a substitution: instead of studying all trajectories govern by this Hamiltonian, you are studying a particular trajectory that solves the Hamiltonian.

So if I understand you right, Hamiltonian mechanics is not generally used for finding trajectories of particles, just the quantities of a general system such as energy, average velocity, average position, etc.?
 
  • #4
woodssnoop said:
[itex]L(q,\dot{q},t) = T + V[/itex]

Also, the quote above should be:

[itex]L(q,\dot{q},t) = T - V[/itex]
 
  • #5
woodssnoop said:
Thank you for replying. I made a mistake in my first question and I should probably rephrase as well. I meant to ask, can [itex] m \dot{q} [/itex] be replaced by [itex] p [/itex]? I understand that if the mass was not constant then this would not be possible, but in a non-relativistic calculation, such as this harmonic approximation, are these terms (i.e. [itex] q, \dot{q}, p [/itex], etc.) treated the same way as in Newtonian mechanics?
I suppose one can interpret generalized coordinate, its time derivative, and its conjugate momentum in Newtonian term if you want (and it is often good to do so). For you SHO problem, they are the same anyway. But in principle, there can be system with effective Hamiltonian in which this is not the case. For example, for a charged particle moving in a plane with a uniform magnetic field perpendicular to the plane. The conjugate momentum [itex]p[/itex] is not the same as the mechanical momentum [itex]m\dot{q}[/itex]

So if I understand you right, Hamiltonian mechanics is not generally used for finding trajectories of particles, just the quantities of a general system such as energy, average velocity, average position, etc.?
There is no less power to solve for trajectory in Hamiltonian form than in the Newtonian form, when it can be done in closed form. And no, it is not correct to say that Hamiltonian mechanics can only solve for "average" something. Hamiltonian mechanics visualizes motion as flow in the phase space. So you end up with a picture of motion of an incompressible fluid in [itex](q,p)[/itex] space and you can trace any trajectory corresponding to any initial condition from that.
 

1. What is a Hamiltonian in the context of a harmonic oscillator?

The Hamiltonian is a mathematical operator that describes the total energy of a system, including its potential and kinetic energy, in the context of a harmonic oscillator. It is often represented by the symbol H.

2. How is the Hamiltonian constructed for a harmonic oscillator?

The Hamiltonian for a harmonic oscillator is constructed by combining the potential energy function, which is proportional to the square of the displacement from equilibrium, with the kinetic energy function, which is proportional to the square of the velocity. This results in a quadratic equation that represents the total energy of the oscillator.

3. Why is the Hamiltonian important in the study of harmonic oscillators?

The Hamiltonian is important because it allows us to analyze the behavior of a harmonic oscillator and make predictions about its energy levels and dynamics. It also provides a mathematical framework for understanding the relationship between energy and motion in the oscillator.

4. Can the Hamiltonian be used to solve for the energy levels of a harmonic oscillator?

Yes, the Hamiltonian can be solved to determine the energy levels of a harmonic oscillator. By solving the Schrödinger equation, which uses the Hamiltonian operator, we can find the allowed energy levels and corresponding wavefunctions for the harmonic oscillator.

5. How does the Hamiltonian change if the harmonic oscillator is coupled to another system?

If the harmonic oscillator is coupled to another system, such as another oscillator or an external potential, the Hamiltonian will include terms that account for this coupling. These terms will affect the energy levels and dynamics of the oscillator and can be used to study the behavior of more complex systems.

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