# Can the hamiltonian be derived from phase space evolution?

Given the phase space evolution of a system, $$x(t)$$ and $$p(t)$$, is there any way of getting the hamiltonian to make a later study of the system under the hamiltonian formalism?

My first thought was to take the time derivatives of the phase space coordinates, $$\dot{x}(t)$$ and $$\dot{p}(t)$$, and then try to integrate the Hamilton's equations $$\dot{x} = \dfrac{\partial H}{\partial p} \quad\text{and}\quad \dot{p} =- \dfrac{\partial H}{\partial x}\ ,$$ which gives me a hamiltonian with the generic form $$H(x,p,t) = f(x,p,t) + g(t)\ ,$$ where $$f$$ is the result of direct integration of the phase space space coordinates and $$g$$ is an arbitrary function of time.

However I think that this is not a correct approach, since I think that I should get rid of the time dependence of $$x$$ and $$p$$ prior to integrating the Hamilton's equations.

How would you get a hamiltonian when the trajectories $$x(t)$$ and $$p(t)$$ are given without any extra information?

• Do you know the trajectories for every possible choice of initial condition, or do you only know a single trajectory? Oct 26, 2020 at 16:47
• @J.Murray I have single expressions for x(t) and p(t) only, so I suppose that they represent a single trajectory where the initial conditions were already applied. Oct 26, 2020 at 16:54

I want to answer the question "if you know the equations $$\dot{x}=\ldots~,\dot{p}=\ldots~$$ how you get the Hamiltonian?"

To obtain the Hamiltonian, you have to solve these partial differential equations :

$$-{\frac {\partial }{\partial x}}H \left( x,p \right) =f \left( x \right)$$ and

$${\frac {\partial }{\partial p}}H \left( x,p \right) =g \left( p \right)$$

which give you the solution:

$$H \left( x,p \right) ={ F_2} \left( x \right) +{ F_1} \left( p \right)$$

with :

$$f \left( x \right) =-{\frac {d}{dx}}{ F_2} \left( x \right)\tag 1$$ $$g \left( p \right) ={\frac {d}{dp}}{ F_1} \left( p \right)\tag 2$$

Example:

$$\dot{p}=-k\,x~,\dot{x}=\frac pm$$

$$\Rightarrow$$

Eq. (1)

$$-kx=-{\frac {d}{dx}}{F_2} \left( x \right) ~\Rightarrow~,F_2=\frac 12 k\,x^2+c_2$$

Eq. (2)

$$\frac pm=\frac{d}{dp}\,F_1~\Rightarrow~,F_1=\frac 12 \frac{p^2}{m}+c_1$$

and the Hamiltonian is:

$$H=\frac 12\left(k\,x^2+\frac {p^2}{m}\right)+c$$

Edit

in case that you know $$x=x(t)~,p=p(t)$$ you have to solve these equations

$$-{\frac {\partial }{\partial x}}H \left( x,p \right) =f \left( t \right)$$ and

$${\frac {\partial }{\partial p}}H \left( x,p \right) =g \left( t \right)$$ ($$f(t)=\dot{x}~,g(t)=\dot{p}$$)

which give you the solution:

$$H=F_3(t)\,x+F_4(t)\,p+F_5(t)$$

where :

$$f(t)=-F_3~,g(t)=F_4$$

• I know what you say but, in my case what I have is the time evolution of x and p. I mean, the functions x(t) and p(t) that one usually get after solving the differential equations of movement. I don't have the relations x = x(p) and p = p(x). Oct 26, 2020 at 20:07
• I will write you the solution soon
– Eli
Oct 26, 2020 at 20:13
• @Jaime_mc2 see new document
– Eli
Oct 26, 2020 at 20:22