Conserved quantities via Poisson brackets

AI Thread Summary
The discussion revolves around the calculation of conserved quantities using Poisson brackets, specifically focusing on the Hamiltonian and angular momentum in scalar notation. The participant rewrites the Hamiltonian and angular momentum, then simplifies the Poisson bracket expressions, ultimately leading to a result that should equal zero for conservation. They explore the implications of index manipulation in the Levi-Civita symbol and confirm that the indices can be swapped due to the summation, reinforcing the need for careful notation. The conversation concludes with a reminder of the fundamental property that total angular momentum generates rotations for vector quantities, ensuring that certain scalar quantities commute with angular momentum components.
Lambda96
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Homework Statement
Show that ##L_j^{(1)}+L_j^{(2)}## is preserved
Relevant Equations
See screenshot
Hi,

Results from the previous task, which we may use
Bildschirmfoto 2023-01-15 um 19.12.52.png
I am unfortunately stuck with the following task
Bildschirmfoto 2023-01-12 um 10.48.04.png


Hi,

I have first started to rewrite the Hamiltonian and the angular momentum from vector notation to scalar notation:

$$H=\frac{1}{2m}\vec{p_1}^2+\frac{1}{2m}\vec{p_2}^2-\alpha|\vec{q_1}- \vec{q_2}|^2= \sum\limits_{i=1}^{3}\frac{1}{2m}p_{1i}^2+ \sum\limits_{i=1}^{3}\frac{1}{2m}p_{2i}^2-\alpha \sum\limits_{i=1}^{3}(q_{1i}-q_{2i})^2$$

$$L_j^{(1)}+L_j^{(2)}=\sum\limits_{k,l}\epsilon_{jkl}q_k^{(1)}p_l^{(1)}+\sum\limits_{k,l}\epsilon_{jkl}q_k^{(2)}p_l^{(2)}$$

Before inserting the above values into the Poisson bracket, I have rewritten the Poisson bracket as follows

$$\{ H,L_j^{(1)}+L_j^{(2)} \}=\{ H,L_j^{(1)}\}+\{H,L_j^{(2)} \}$$

To save paperwork, I will now only calculate ##\{ H,L_j^{(1)}\}##. the calculation for ##L_j^{(2)## would run analogously

$$\{ \sum\limits_{i=1}^{3}\frac{1}{2m}p_{1i}^2+ \sum\limits_{i=1}^{3}\frac{1}{2m}p_{2i}^2-\alpha \sum\limits_{i=1}^{3}(q_{1i}-q_{2i})^2, \sum\limits_{k,l}\epsilon_{jkl}q_k^{(1)}p_l^{(1)}\}$$
$$\{ \sum\limits_{i=1}^{3}\frac{1}{2m}p_{1i}^2,\sum\limits_{k,l}\epsilon_{jkl}q_k^{(1)}p_l^{(1)}\}+ \{\sum\limits_{i=1}^{3}\frac{1}{2m}p_{2i}^2,\sum\limits_{k,l}\epsilon_{jkl}q_k^{(1)}p_l^{(1)}\}+ \{-\alpha \sum\limits_{i=1}^{3}(q_{1i}-q_{2i})^2, \sum\limits_{k,l}\epsilon_{jkl}q_k^{(1)}p_l^{(1)}\}$$
 

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Because of the previous problem, the terms with ##q^2## become zero, so ##\{ \vec{q}^2,L_j \}=0## also the terms with the different variables become zero and only the following term remains

$$\{-\alpha \sum\limits_{i=1}^{3}(q_{1i}-q_{2i})^2, \sum\limits_{k,l}\epsilon_{jkl}q_k^{(1)}p_l^{(1)}\}$$

Now you can square the bracket off and get ##q^2## terms again, which again add up to zero, leaving only the following term.

$$\{\alpha \sum\limits_{i=1}^{3}q_{1i}q_{2i}, \sum\limits_{k,l}\epsilon_{jkl}q_k^{(1)}p_l^{(1)}\}$$

For the left-hand side I have now used Leibniz's rule and the following applies to the term then ##\{q_{2i},q_k^{(1)}p_l^{(1)} \}=0## and the following term remains

$$\{\alpha \sum\limits_{i=1}^{3}q_{1i}, \sum\limits_{k,l}\epsilon_{jkl}q_k^{(1)}p_l^{(1)}\}q_{2i}$$

Now you can factor out some terms and get the following representation

$$\alpha \sum\limits_{i=1}^{3}\sum\limits_{k,l}\epsilon_{jkl}\{q_{1i}, q_k^{(1)}p_l^{(1)}\}q_{2i}$$

Now you can use the result from the previous task and get

$$\epsilon_{jki}q_k^{(1)}q_i^{(2)}$$

for the calculation ##\{H,L_j^{(2)} \}## I got the following ##\epsilon_{jki}q_k^{(2)}q_i^{(1)}##, so I get as final result:

$$\epsilon_{jki}q_k^{(1)}q_i^{(2)}+\epsilon_{jki}q_k^{(2)}q_i^{(1)}$$

Actually, the result should be 0 in order to conserve the observable, have I miscalculated somewhere or can I still zero the above term with an index manipulation?

Sorry to split my calculation into two posts, but when I do the whole calculation in one post, it is not displayed correctly.
 
Lambda96 said:
so I get as final result:
$$\epsilon_{jki}q_k^{(1)}q_i^{(2)}+\epsilon_{jki}q_k^{(2)}q_i^{(1)}$$
Actually, the result should be 0
I think you are essentially there. Play with the expression above and see if you can see why it equals 0. Note that ##k## and ##i## are dummy summation indices.
 
Thank you for your help TSny 👍👍

I have now proceeded as follows

$$\epsilon_{jki}q_k^{(1)}q_i^{(2)}+\epsilon_{jki}q_k^{(2)}q_i^{(1)}$$

Now in the second term I have swapped the indices of ##q_k^{(2)}q_i^{(1)}## to ##q_i^{(2)}q_k^{(1)}##.

$$\epsilon_{jki}q_k^{(1)}q_i^{(2)}+\epsilon_{jik}q_i^{(2)}q_k^{(1)}$$

By doing this, I have made an anticyclic permutation to the Levi-Civita symbol, which makes it negative and I can write the above equation as follows.

$$\epsilon_{jki}q_k^{(1)}q_i^{(2)}-\epsilon_{jik}q_i^{(2)}q_j^{(1)}=0$$
 
Lambda96 said:
I have now proceeded as follows

$$\epsilon_{jki}q_k^{(1)}q_i^{(2)}+\epsilon_{jki}q_k^{(2)}q_i^{(1)}$$

Now in the second term I have swapped the indices of ##q_k^{(2)}q_i^{(1)}## to ##q_i^{(2)}q_k^{(1)}##.

$$\epsilon_{jki}q_k^{(1)}q_i^{(2)}+\epsilon_{jik}q_i^{(2)}q_k^{(1)}$$
This looks good. Note that the index ##j## is a fixed index that is not summed. It corresponds to the fixed index ##j## that you started with in the expression ##\{H, L_j^{(1)}\}##. The reason that you were allowed to swap the indices ##i## and ##k## in the second term is that you are summing over ##i## and ##k##. That is, the equality $$\epsilon_{jki}q_k^{(1)}q_i^{(2)} + \epsilon_{jki}q_k^{(2)}q_i^{(1)} =\epsilon_{jki}q_k^{(1)}q_i^{(2)} + \epsilon_{jik}q_i^{(2)}q_k^{(1)}$$ should actually be written as $$\sum_{k, i}\left(\epsilon_{jki}q_k^{(1)}q_i^{(2)} + \epsilon_{jki}q_k^{(2)}q_i^{(1)} \right) =\sum_{k, i}\left(\epsilon_{jki}q_k^{(1)}q_i^{(2)} + \epsilon_{jik}q_i^{(2)}q_k^{(1)}\right)$$ It's only because of the summation over ##i## and ##k## that we have equality of these two expressions.

Let me know if this is not clear.

Lambda96 said:
By doing this, I have made an anticyclic permutation to the Levi-Civita symbol, which makes it negative and I can write the above equation as follows.

$$\epsilon_{jki}q_k^{(1)}q_i^{(2)}-\epsilon_{jik}q_i^{(2)}q_j^{(1)}=0$$
The second term is not correct. Double-check this term to make sure this is what you intended to type. The index ##j## should still be a fixed index that occurs only in the levi-civita symbol and a summation over ##k## and ##i## should be included.
 
Thank you for your help and explanation TSny 👍👍👍

In my notes I had included the sum over k and had unfortunately forgotten to include this in my post 4, but in my notes I had forgotten the sum over i in the course of the calculation, thanks to your explanation I have now noticed this, thank you very much.

Unfortunately I can't quite understand why I wrote j instead of k in my last equation, probably I just did too much index notation the last couple of days :smile:

Thanks again for your help 👍
 
One can also argue without any calculation ;-)). The total angular momentum ##\vec{J}=\vec{L}^{(1)}+\vec{L}^{(2)}## generate rotations for all vector quantities of both systems, from this you get that ##\vec{V}^{(j)} \cdot \vec{W}^{(k)}## as scalars commute with all components of ##\vec{J}##.
 
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