Difference between revisions of "User:Nik"
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I) \( \{Q_i(q,p), Q_j(q,p)\} = 0 \) | I) \( \{Q_i(q,p), Q_j(q,p)\} = 0 \) | ||
| − | II) \( \{P_i (q,p), P_j(q,p)\} = 0 \) | + | II) \( \{Q_i(q,p), P_j(q,p)\} = \delta_{ij} \) |
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| + | III) \( \{P_i (q,p), P_j(q,p)\} = 0 \) | ||
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==Solution== | ==Solution== | ||
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===Solution I)=== | ===Solution I)=== | ||
| − | + | This is the most complicated problem but we calculate it first because we use it in II) und III) | |
$$ \{Q_i, Q_j\} = \sum_{k=1}^{n}\left(\frac{\partial Q_i}{\partial q_k}\frac{\partial Q_j}{\partial p_k} - \frac{\partial Q_i}{\partial p_k}\frac{\partial Q_j}{\partial q_k}\right) $$ | $$ \{Q_i, Q_j\} = \sum_{k=1}^{n}\left(\frac{\partial Q_i}{\partial q_k}\frac{\partial Q_j}{\partial p_k} - \frac{\partial Q_i}{\partial p_k}\frac{\partial Q_j}{\partial q_k}\right) $$ | ||
| + | First we write equations (1) and (2) again an shift our eyeballs between them and look at it as matrix multiplication: | ||
| + | $$ \underbrace{-\frac{\partial^2 \Phi}{\partial q_i \partial q_k}(q, Q)}_{D_{ik}} = \sum_{s=1}^{n}\left(\underbrace{\frac{\partial^2 \Phi}{\partial q_i \partial Q_s}(q, Q)}_{A_{is}}\underbrace{\frac{\partial Q_s}{\partial q_k}}_{C_{sk}}\right) $$ | ||
| + | $$ \underbrace{\delta_{ik}}_{\mathbb{1}} = \sum_{s=1}^{n}\left(\underbrace{\frac{\partial^2 \Phi}{\partial q_i \partial Q_s}(q, Q)}_{A_{is}}\underbrace{\frac{\partial Q_s}{\partial p_k}}_{B_{sk}}\right) $$ | ||
| + | Now let's do some linear algebra: We know that \( \mathbb{1} = AB \Rightarrow A^{-1} = B \text{ and } D = AC \) | ||
| + | $$ \Rightarrow A^{-1}D = C $$ | ||
| + | $$\Rightarrow BD = C $$ | ||
===Solution II)=== | ===Solution II)=== | ||
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| + | $$ \{Q_i, P_j\} = \sum_{k=1}^{n}\left( \frac{\partial Q_i}{\partial q_k}\frac{\partial P_j}{\partial p_k} - \frac{\partial Q_i}{\partial p_k}\frac{\partial P_j}{\partial q_k} \right) $$ | ||
| + | Now we put in the results from equation (3) and (4) | ||
| + | $$ = \sum_{k=1}^{n}\left(\frac{\partial Q_i}{\partial q_k}\left(-\sum_{s=1}^{n}\left(\frac{\partial^2 \Phi}{\partial Q_j \partial Q_s}(q, Q)\frac{\partial Q_s}{\partial p_k}\right)\right) - \frac{\partial Q_i}{\partial p_k}\left(-\frac{\partial^2 \Phi}{\partial Q_j \partial q_k}(q, Q) - \sum_{s=1}^{n}\left(\frac{\partial^2 \Phi}{\partial Q_j \partial Q_s}(q, Q)\frac{\partial Q_s}{\partial q_k}\right)\right)\right) $$ | ||
| + | Now we multiply out and separate all the sums. We can rearrange it like this because all the parts we put inside are independent on the running variable. | ||
| + | $$ = \sum_{k=1}^{n}\left(\frac{\partial Q_i}{\partial p_k}\frac{\partial^2 \Phi}{\partial Q_j \partial q_k}(q, Q)\right) + \sum_{k=1}^{n}\sum_{s=1}^{n}\left(\frac{\partial Q_i}{\partial p_k}\frac{\partial^2 \Phi}{\partial Q_j \partial Q_s}(q, Q)\frac{\partial Q_s}{\partial q_k}\right) - \sum_{k=1}^{n}\sum_{s=1}^{n}\left(\frac{\partial Q_i}{\partial q_k}\frac{\partial^2 \Phi}{\partial Q_j \partial Q_s}(q, Q)\frac{\partial Q_s}{\partial p_k}\right) $$ | ||
| + | Now we rearrange again such that we can build a new Poisson bracket | ||
| + | $$ = \sum_{k=1}^{n}\left(\frac{\partial Q_i}{\partial p_k}\frac{\partial^2 \Phi}{\partial Q_j \partial q_k}(q, Q)\right) + \sum_{s=1}^{n}\left(\frac{\partial^2 \Phi}{\partial Q_j \partial Q_s}(q, Q) \underbrace{\sum_{k=1}^{n}\left(\frac{\partial Q_i}{\partial p_k}\frac{\partial Q_s}{\partial q_k} - \frac{\partial Q_i}{\partial q_k}\frac{\partial Q_s}{\partial p_k}\right)}_{=\{Q_i, Q_s\}=0}\right) $$ | ||
| + | Now we use the definition from the problem and then the chain rule | ||
| + | $$ = \sum_{k=1}^{n}\left(\frac{\partial Q_i}{\partial p_k}\frac{\partial p_k}{\partial Q_j} \right) = \frac{\partial Q_i}{\partial Q_j} = \delta_{ij} $$ | ||
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| + | ===Solution III)=== | ||
$$ \{P_i, P_j\} = \sum_{k=1}^{n}\left(\frac{\partial P_i}{\partial q_k}\frac{\partial P_j}{\partial p_k} - \frac{\partial P_i}{\partial p_k}\frac{\partial P_j}{\partial q_k}\right) $$ | $$ \{P_i, P_j\} = \sum_{k=1}^{n}\left(\frac{\partial P_i}{\partial q_k}\frac{\partial P_j}{\partial p_k} - \frac{\partial P_i}{\partial p_k}\frac{\partial P_j}{\partial q_k}\right) $$ | ||
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Now we use the chain rule and the definition again | Now we use the chain rule and the definition again | ||
$$ = \frac{\partial P_i}{\partial Q_j} + \frac{\partial P_i}{\partial Q_j} = 0 $$ | $$ = \frac{\partial P_i}{\partial Q_j} + \frac{\partial P_i}{\partial Q_j} = 0 $$ | ||
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Revision as of 20:28, 5 January 2015
Contents
Foreword
I use \(Q\:/\:P\) for the transformed system instead of \(\widetilde{q}\:/\:\widetilde{p}\) because it's easier to write in Latex.
Problem
Let \( \Phi \in C^\infty(\mathbb{R}^n) \) have the property that the system \( p_i = \frac{\partial}{\partial q_i} \Phi (q, Q) \) has a unique smooth solution \( Q = Q(q,p) \).
Define \( P_i(q,p) = - \frac{\partial}{\partial Q_i} \Phi (q, Q) | _{Q= Q(q,p)} \)
Let \( \{\cdot,\cdot\} \) be the Poisson bracket, such that \( \{f,g\} = \sum_{j=1}^n \frac{\partial f}{\partial q_j} \frac{\partial g}{\partial p_j} - \frac{\partial f}{\partial p_j} \frac{\partial g}{\partial q_j} \)
Show that:
I) \( \{Q_i(q,p), Q_j(q,p)\} = 0 \)
II) \( \{Q_i(q,p), P_j(q,p)\} = \delta_{ij} \)
III) \( \{P_i (q,p), P_j(q,p)\} = 0 \)
Solution
Important equations
$$ \underbrace{\frac{\partial p_i}{\partial q_k}}_{0} = \frac{\partial}{\partial q_k}\left(\frac{\partial \Phi}{\partial q_i}(q, Q)\right) $$ $$ = \sum_{s=1}^{n}\left(\frac{\partial^2 \Phi}{\partial q_i \partial q_s}(q, Q)\underbrace{\frac{\partial q_s}{\partial q_k}}_{\delta_{sk}} + \frac{\partial^2 \Phi}{\partial q_i \partial Q_s}(q, Q)\frac{\partial Q_s}{\partial q_k}\right) $$ $$ \Rightarrow 0 = \frac{\partial^2 \Phi}{\partial q_i \partial q_k}(q, Q) + \sum_{s=1}^{n}\left(\frac{\partial^2 \Phi}{\partial q_i \partial Q_s}(q, Q)\frac{\partial Q_s}{\partial q_k}\right) $$ $$ \tag{1} \Rightarrow \boxed{ -\frac{\partial^2 \Phi}{\partial q_i \partial q_k}(q, Q) = \sum_{s=1}^{n}\left(\frac{\partial^2 \Phi}{\partial q_i \partial Q_s}(q, Q)\frac{\partial Q_s}{\partial q_k}\right) } $$
$$ \underbrace{\frac{\partial p_i}{\partial p_k}}_{\delta_{ik}} = \frac{\partial}{\partial p_k}\left(\frac{\partial \Phi}{\partial q_i}(q, Q)\right) $$
$$ = \sum_{s=1}^{n}\left(\frac{\partial^2 \Phi}{\partial q_i \partial q_s}(q, Q)\underbrace{\frac{\partial q_s}{\partial p_k}}_{0} + \frac{\partial^2 \Phi}{\partial q_i \partial Q_s}(q, Q)\frac{\partial Q_s}{\partial p_k}\right) $$
$$ \tag{2} \boxed{ \Rightarrow \delta_{ik} = \sum_{s=1}^{n}\left(\frac{\partial^2 \Phi}{\partial q_i \partial Q_s}(q, Q)\frac{\partial Q_s}{\partial p_k}\right) } $$
$$ \frac{\partial P_i}{\partial q_k} = \frac{\partial}{\partial q_k}\left(-\frac{\partial \Phi}{\partial Q_i}(q, Q)\right) $$
$$ = \sum_{s=1}^{n}\left(-\frac{\partial^2 \Phi}{\partial Q_i \partial q_s}(q, Q)\underbrace{\frac{\partial q_s}{\partial q_k}}_{\delta_{sk}} - \frac{\partial^2 \Phi}{\partial Q_i \partial Q_s}(q, Q)\frac{\partial Q_s}{\partial q_k}\right) $$
$$ \tag{3} \boxed{ \Rightarrow \frac{\partial P_i}{\partial q_k} = -\frac{\partial^2 \Phi}{\partial Q_i \partial q_k}(q, Q) - \sum_{s=1}^{n}\left(\frac{\partial^2 \Phi}{\partial Q_i \partial Q_s}(q, Q)\frac{\partial Q_s}{\partial q_k}\right) } $$
$$ \frac{\partial P_i}{\partial p_k} = \frac{\partial}{\partial p_k}\left(-\frac{\partial \Phi}{\partial Q_i}(q, Q)\right) $$
$$ = \sum_{s=1}^{n}\left(-\frac{\partial^2 \Phi}{\partial Q_i \partial q_s}(q, Q)\underbrace{\frac{\partial q_s}{\partial p_k}}_{0} - \frac{\partial^2 \Phi}{\partial Q_i \partial Q_s}(q, Q)\frac{\partial Q_s}{\partial p_k}\right) $$
$$ \tag{4} \Rightarrow \boxed{ \frac{\partial P_i}{\partial p_k} = -\sum_{s=1}^{n}\left(\frac{\partial^2 \Phi}{\partial Q_i \partial Q_s}(q, Q)\frac{\partial Q_s}{\partial p_k}\right) } $$
Solution I)
This is the most complicated problem but we calculate it first because we use it in II) und III) $$ \{Q_i, Q_j\} = \sum_{k=1}^{n}\left(\frac{\partial Q_i}{\partial q_k}\frac{\partial Q_j}{\partial p_k} - \frac{\partial Q_i}{\partial p_k}\frac{\partial Q_j}{\partial q_k}\right) $$ First we write equations (1) and (2) again an shift our eyeballs between them and look at it as matrix multiplication: $$ \underbrace{-\frac{\partial^2 \Phi}{\partial q_i \partial q_k}(q, Q)}_{D_{ik}} = \sum_{s=1}^{n}\left(\underbrace{\frac{\partial^2 \Phi}{\partial q_i \partial Q_s}(q, Q)}_{A_{is}}\underbrace{\frac{\partial Q_s}{\partial q_k}}_{C_{sk}}\right) $$ $$ \underbrace{\delta_{ik}}_{\mathbb{1}} = \sum_{s=1}^{n}\left(\underbrace{\frac{\partial^2 \Phi}{\partial q_i \partial Q_s}(q, Q)}_{A_{is}}\underbrace{\frac{\partial Q_s}{\partial p_k}}_{B_{sk}}\right) $$ Now let's do some linear algebra: We know that \( \mathbb{1} = AB \Rightarrow A^{-1} = B \text{ and } D = AC \) $$ \Rightarrow A^{-1}D = C $$ $$\Rightarrow BD = C $$
Solution II)
$$ \{Q_i, P_j\} = \sum_{k=1}^{n}\left( \frac{\partial Q_i}{\partial q_k}\frac{\partial P_j}{\partial p_k} - \frac{\partial Q_i}{\partial p_k}\frac{\partial P_j}{\partial q_k} \right) $$ Now we put in the results from equation (3) and (4) $$ = \sum_{k=1}^{n}\left(\frac{\partial Q_i}{\partial q_k}\left(-\sum_{s=1}^{n}\left(\frac{\partial^2 \Phi}{\partial Q_j \partial Q_s}(q, Q)\frac{\partial Q_s}{\partial p_k}\right)\right) - \frac{\partial Q_i}{\partial p_k}\left(-\frac{\partial^2 \Phi}{\partial Q_j \partial q_k}(q, Q) - \sum_{s=1}^{n}\left(\frac{\partial^2 \Phi}{\partial Q_j \partial Q_s}(q, Q)\frac{\partial Q_s}{\partial q_k}\right)\right)\right) $$ Now we multiply out and separate all the sums. We can rearrange it like this because all the parts we put inside are independent on the running variable. $$ = \sum_{k=1}^{n}\left(\frac{\partial Q_i}{\partial p_k}\frac{\partial^2 \Phi}{\partial Q_j \partial q_k}(q, Q)\right) + \sum_{k=1}^{n}\sum_{s=1}^{n}\left(\frac{\partial Q_i}{\partial p_k}\frac{\partial^2 \Phi}{\partial Q_j \partial Q_s}(q, Q)\frac{\partial Q_s}{\partial q_k}\right) - \sum_{k=1}^{n}\sum_{s=1}^{n}\left(\frac{\partial Q_i}{\partial q_k}\frac{\partial^2 \Phi}{\partial Q_j \partial Q_s}(q, Q)\frac{\partial Q_s}{\partial p_k}\right) $$ Now we rearrange again such that we can build a new Poisson bracket $$ = \sum_{k=1}^{n}\left(\frac{\partial Q_i}{\partial p_k}\frac{\partial^2 \Phi}{\partial Q_j \partial q_k}(q, Q)\right) + \sum_{s=1}^{n}\left(\frac{\partial^2 \Phi}{\partial Q_j \partial Q_s}(q, Q) \underbrace{\sum_{k=1}^{n}\left(\frac{\partial Q_i}{\partial p_k}\frac{\partial Q_s}{\partial q_k} - \frac{\partial Q_i}{\partial q_k}\frac{\partial Q_s}{\partial p_k}\right)}_{=\{Q_i, Q_s\}=0}\right) $$ Now we use the definition from the problem and then the chain rule $$ = \sum_{k=1}^{n}\left(\frac{\partial Q_i}{\partial p_k}\frac{\partial p_k}{\partial Q_j} \right) = \frac{\partial Q_i}{\partial Q_j} = \delta_{ij} $$
Solution III)
$$ \{P_i, P_j\} = \sum_{k=1}^{n}\left(\frac{\partial P_i}{\partial q_k}\frac{\partial P_j}{\partial p_k} - \frac{\partial P_i}{\partial p_k}\frac{\partial P_j}{\partial q_k}\right) $$ Now we put in the results from equation (3) and (4) $$ = \sum_{k=1}^{n}\left(\frac{\partial P_i}{\partial q_k}\left(-\sum_{s=1}^{n}\left(\frac{\partial^2 \Phi}{\partial Q_j \partial Q_s}(q, Q)\frac{\partial Q_s}{\partial p_k}\right)\right) - \frac{\partial P_i}{\partial p_k}\left(-\frac{\partial^2 \Phi}{\partial Q_j \partial q_k}(q, Q) - \sum_{s=1}^{n}\left(\frac{\partial^2 \Phi}{\partial Q_j \partial Q_s}(q, Q)\frac{\partial Q_s}{\partial q_k}\right)\right)\right) $$ Now we multiply out and separate all the sums. We can rearrange it like this because all the parts we put inside are independent on the running variable. $$ = \sum_{k=1}^{n}\left(\frac{\partial P_i}{\partial p_k}\frac{\partial^2 \Phi}{\partial Q_j \partial q_k}(q, Q)\right) + \sum_{k=1}^{n}\sum_{s=1}^{n}\left(\frac{\partial P_i}{\partial p_k}\frac{\partial^2 \Phi}{\partial Q_j \partial Q_s}(q, Q)\frac{\partial Q_s}{\partial q_k}\right) - \sum_{k=1}^{n}\sum_{s=1}^{n}\left(\frac{\partial P_i}{\partial q_k}\frac{\partial^2 \Phi}{\partial Q_j \partial Q_s}(q, Q)\frac{\partial Q_s}{\partial p_k}\right) $$ Now we rearrange again such that we can build a new Poisson bracket $$ = \sum_{k=1}^{n}\left(\frac{\partial P_i}{\partial p_k}\frac{\partial^2 \Phi}{\partial Q_j \partial q_k}(q, Q)\right) + \sum_{s=1}^{n}\left(\frac{\partial^2 \Phi}{\partial Q_j \partial Q_s}(q, Q) \underbrace{\sum_{k=1}^{n}\left(\frac{\partial P_i}{\partial p_k}\frac{\partial Q_s}{\partial q_k} - \frac{\partial P_i}{\partial q_k}\frac{\partial Q_s}{\partial p_k}\right)}_{=\{P_i, Q_s\}=-\delta_{is}}\right) $$ Now we use the definition from the problem and take the \( \delta \) out. $$ = \sum_{k=1}^{n}\left(\frac{\partial P_i}{\partial p_k}\frac{\partial p_k}{\partial Q_j}\right) - \frac{\partial^2 \Phi}{\partial Q_j \partial Q_i}(q, Q) $$ Now we use the chain rule and the definition again $$ = \frac{\partial P_i}{\partial Q_j} + \frac{\partial P_i}{\partial Q_j} = 0 $$
