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I have to find a constant of motion associated to this Hamiltonian but I don't know how to proceed.

$$H=\frac{\mathbf{p_0}^2}{2m}+\frac{\mathbf{p_1}^2}{2m}+\frac{\mathbf{p_2}^2}{2m}-2V(\mathbf{r_1}- \mathbf{r_0})+V(\mathbf{r_2}-\mathbf{r_1})$$

where $$V(\mathbf x)=\frac {e^2}{|\mathbf x|}.$$

I don't know what $\mathbf x$ is.

This Hamiltonian refers to a system of 3 particles $(0,1,2)$ with mass $m$ and charge $e$.

The coordinates are $r^\alpha_i$ and conjugate momenta $p^\beta_j$ with $\alpha, \beta=0,1,2$.

I have written all the information that I have.

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This may be of interest: – sujeet Jun 1 '13 at 18:12
@sujeet yes, sure, but I don't know what function considering in addition to the Hamiltonian.. have you got any idea? – sunrise Jun 1 '13 at 18:15
Just to be clear, since it looks like the potential is confusing you a little, the full Hamiltonian is like this: $$ H = \sum\limits_{i=0}^2 \frac{\mathbf{p_i}^2}{2m} - 2\frac{e^2}{|\mathbf{r_1} - \mathbf{r_0}|} + \frac{e^2}{|\mathbf{r_2} - \mathbf{r_1}|} ~. $$ – sujeet Jun 1 '13 at 18:32
@sujeet Oh, thank you! I'd have never understood if you didn't write it :) – sunrise Jun 1 '13 at 18:55

Use Poisson brackets. A constant of motion is some function $F(q_i,p_i)$ of phase space that is independent of time, i.e. such that $\frac{dF}{dt}=0$. Now, consider that $$\frac{dF}{dt}=\frac{\partial{F}}{\partial{q}_1}\frac{d{q_1}}{d{t}}+\ldots+\frac{\partial{F}}{\partial{q}_n}\frac{dq_n}{dt}+\frac{\partial{F}}{\partial{p}_1}\frac{d{p_1}}{d{t}}+\ldots+\frac{\partial{F}}{\partial{p}_n}\frac{dp_n}{dt}+\frac{\partial{F}}{\partial{t}}\\=\sum_{j=1}^n\left(\frac{\partial{F}}{\partial{q}_j}\dot{q}_j+\frac{\partial{F}}{\partial{p}_j}\dot{p}_j\right)+\frac{\partial{F}}{\partial{t}}$$ but since we defined $F=F(q_i,p_i)$ ($F$ doesn't depend explicitly on time), $\partial{F}/\partial{t}=0$, so that you may just calculate $$\sum_{j=1}^n\left(\frac{\partial{F}}{\partial{q}_j}\dot{q}_j+\frac{\partial{F}}{\partial{p}_j}\dot{p}_j\right)=\sum_{j=1}^n\left(\frac{\partial{F}}{\partial{q}_j}\frac{\partial{\mathcal{H}}}{\partial{p}_j}-\frac{\partial{F}}{\partial{p}_j}\frac{\partial{\mathcal{H}}}{\partial{q}_j}\right)\equiv\left\{F,\mathcal{H}\right\}$$ which is the way the Poisson bracket $\{F,\mathcal{H}\}$ is defined. Note that your hamiltonian does not depend on time, so that immediately $\{\mathcal{H},\mathcal{H}\}=0$ and energy is conserved. This way you may look for other conserved quantities like linear momentum $p_k$ or angular momentum $L_k$, for example.

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I'd like to not go at random... :) There is a way to analyze the simmetry of the system, of the Hamiltonian... and hypotize the constant of motion? – sunrise Jun 1 '13 at 18:36
A basic one is that if the Hamiltonian does not contain a particular coordinate $q_k$ explicitly, $p_k$ is a constant of motion ($q_k$ is called a cyclic coordinate), but again this may be justified by the Poisson bracket $\{p_k,\mathcal{H}\}$ (which you were supposed to see by yourself). There sure are other ways, but maybe even more contrived, as the Hamilton-Jacobi equations. Maybe you just have to get your hands dirty sometimes. – user24999 Jun 1 '13 at 19:12
Thanks a lot for your help. Now I try to explain better my problem: I'm not sure that the text of the exercise is complete or there's some missing information. (In this Hamiltonian there are all $q$s and so I can't find the constants..). This is a doubt that is driving me crazy! :) – sunrise Jun 1 '13 at 20:26

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