Canonical Transformation (two degrees of freedom)

In summary: For the second equation, you can use the fact that $$Q_2 = P_2 = b \cdot t$$. So you can solve for $$q_1$$ and $$q_2$$ in terms of $$t$$ and $$P_1$$ and $$P_2$$. For the last two equations, $$P_1$$ and $$P_2$$ are constants, so $$p_1$$ and $$p_2$$ are also constants. Therefore, their equations in terms of $$t$$ and initial values are simply $$p_1 = p_{1,0}$$ and $$p_2 = p_{2,0}$$. In summary, the equations for $$q_1$$, $$
  • #1
sayebms
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Homework Statement


Point transformation in a system with 2 degrees of freedom is: $$Q_1=q_1^2\\Q_2=q_q+q_2$$
a) find the most general $P_1$ and $P_2$ such that overall transformation is canonical
b) Show that for some $P_1$ and $P_2$ the hamiltonain $$H=\frac{a}{2}(\frac{p_1-p_2}{wq_1})^2+bp_2+c(q_1+q_2)^2$$
(Note: a,b,c, constant) can be transformed in a way that $Q_1$ and $Q_2$ could be ignored.
c) Finally solve the problem and find equations for $q_1,q_2,p_1$ and $p_2$ in terms of t and their initial values.

Homework Equations


-Symplictic approach

The Attempt at a Solution


a) in order to find the transformation for $P$'s I use the symplectic approach:
in this approach we have the following condition for the canonical transformations $$\tilde{M}JM=J$$
where M is the matrix:
$$M=\begin{bmatrix} \frac{\partial Q_1}{\partial q_1}&\frac{\partial Q_1}{\partial q_2}&\frac{\partial Q_1}{\partial p_1}&\frac{\partial Q_1}{\partial p_2}\\\frac{\partial Q_2}{\partial q_1}&\frac{\partial Q_2}{\partial q_2}&\frac{\partial Q_2}{\partial p_1}&\frac{\partial Q_2}{\partial p_2}\\\frac{\partial P_1}{\partial q_1}&\frac{\partial P_1}{\partial q_2}&\frac{\partial P_1}{\partial p_1}&\frac{\partial P_1}{\partial p_2}\\\frac{\partial P_2}{\partial q_1}&\frac{\partial P_2}{\partial q_2}&\frac{\partial
P_2}{\partial p_1}&\frac{\partial P_2}{\partial p_2}\end{bmatrix}$$
and $$J=\begin{bmatrix} 0&0&1&0\\0&0&0&1\\-1&0&0&0\\0&-1&0&0\end{bmatrix}$$
so applying the symplectic methods conditions I get the following equations:$$\frac{\partial P_2}{\partial p_2}=1\\\frac{\partial P_2}{\partial p_1}=0\\\frac{\partial P_1}{\partial p_2}=\frac{-1}{2q_1}\\\frac{\partial P_1}{\partial p_1}=\frac{1}{2q_1}$$
which then gives me the following most general transformations for $P_1$ and $P_2$ :
$$P_1=\frac{1}{2q_1}(p_1-p_2)+g(q_1,q_2)\\P_2=p_2+f(q_1,q_2)\\2q_1 \frac{\partial g}{\partial q_2}+\frac{\partial f}{\partial q_2}-\frac{\partial f}{\partial q_1}$$
b) this part is very simple if we choose $$P_1=\frac{p_1-p_2}{2q_1}\\g(q_1,q_2)=0$$and $$P_2=p_2+\frac{c}{b}(q_1+q_2)^2\\f(q_1,q_2)=\frac{c}{b}(q_1+q_2)^2$$
we see that with these choices the three equations we found earlier are satisfied. and the Hamiltonian becomes:$$H=\frac{a}{2}P_1^2+bP_2$$
But I don't know how I can proceed for solving part c). Any help is appreciated. Thank you very much.[/B]
 
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  • #2
You have the Hamiltonian, so you can just write down the equations of motion and solve them.
 
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  • #3
so using the Hamiltonian equations of motion I have: $$\dot{Q_1}=\frac{\partial H}{\partial P_1}=aP_1 \Longrightarrow \frac{d(q_1^2)}{dt}=2q_1\dot{q_1}=aP_1\\\dot{Q_2}=\frac{\partial H}{\partial P_2}=b \Longrightarrow \dot{q_1}+\dot{q_2}=b\\\dot{P_1}=0\\\dot{P_2}=0$$ So I guess these are the equations that I am supposed to solve. But there is one more thing that I don't get; the problem hasn't given us anything on initial conditions but still wants the equations of q,p,t in terms of t and Initial values.Thank you very much again.
 
  • #4
$$P_1$$ is a constant of motion, so you can integrate the right side to get $$Q_1 = aP_1 \cdot t$$.
 

FAQ: Canonical Transformation (two degrees of freedom)

What is a canonical transformation?

A canonical transformation is a mathematical operation that transforms a set of canonical variables into a new set of canonical variables while preserving the Hamiltonian equations of motion. It is used to simplify the equations of motion in classical mechanics and to find new solutions to the equations of motion.

What are the two degrees of freedom in a canonical transformation?

The two degrees of freedom in a canonical transformation refer to the two types of canonical variables involved in the transformation. These are usually the generalized coordinates and their conjugate momenta.

What is the importance of a canonical transformation?

A canonical transformation is important because it allows us to simplify the equations of motion in classical mechanics and to find new solutions to these equations. It also helps us to uncover hidden symmetries in a physical system and to better understand its dynamics.

What is the difference between a canonical transformation and a change of coordinates?

A canonical transformation differs from a simple change of coordinates in that it preserves the Hamiltonian equations of motion, while a change of coordinates does not necessarily do so. A canonical transformation also allows for a more general transformation, including nonlinear transformations, while a change of coordinates is typically only a linear transformation.

Can a canonical transformation be used in quantum mechanics?

Yes, a canonical transformation can also be applied in quantum mechanics. In this case, it is used to transform the operators and wavefunctions that describe the system, while still preserving the commutation relations between them. This allows for a simpler description of the system and can reveal new insights into its dynamics.

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