Semi-classical quantum/Poisson Brackets

  • Thread starter barnflakes
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  • #1
barnflakes
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4
I'm trying to workout the poisson bracket [itex]\{L_n, L_m\}[/itex] n,m = 1,2,3.

In the answers, my lecturer has written [itex]\{L_1, L_2\} = \{q_2 p_3 - q_3 p_2 , q_3 p_1 - q_1 p_3\} [/itex] which of course I understand. Then the very next line he writes [itex] = q_2 \{p_3 , q_3\} p_1 + p_2 \{q_3, p_3\} q_1[/itex]

I don't quite see how he has jumped so quickly from one to the other. Has he missed several lines of calculations here, or is he using some property of the Poisson Bracket that I can't see?

Thank you
 
Last edited:

Answers and Replies

  • #2
meopemuk
1,761
53
You only need the property that there are just three non-zero Poisson brackets involving q and p

[tex] \{q_1,p_1\} = \{q_2,p_2\} = \{q_3,p_3\} = 1 [/tex]

All other brackets vanish. Then

[itex]\{L_1, L_2\} = \{q_2 p_3 - q_3 p_2 , q_3 p_1 - q_1 p_3\} =
\{q_2 p_3, q_3 p_1\} - \{q_3 p_2 , q_3 p_1\} - \{q_2 p_3, q_1 p_3\} + \{q_3 p_2 , q_1 p_3\} = \{q_2 p_3, q_3 p_1\} + \{q_3 p_2 , q_1 p_3\}[/itex]

[itex] = q_2 \{p_3 , q_3\} p_1 + p_2 \{q_3, p_3\} q_1 = q_1 p_2 - q_2 p_1 = L_3[/itex]

Eugene.
 
  • #3
barnflakes
156
4
You only need the property that there are just three non-zero Poisson brackets involving q and p

[tex] \{q_1,p_1\} = \{q_2,p_2\} = \{q_3,p_3\} = 1 [/tex]

All other brackets vanish. Then

[itex]\{L_1, L_2\} = \{q_2 p_3 - q_3 p_2 , q_3 p_1 - q_1 p_3\} =
\{q_2 p_3, q_3 p_1\} - \{q_3 p_2 , q_3 p_1\} - \{q_2 p_3, q_1 p_3\} + \{q_3 p_2 , q_1 p_3\} = \{q_2 p_3, q_3 p_1\} + \{q_3 p_2 , q_1 p_3\}[/itex]

[itex] = q_2 \{p_3 , q_3\} p_1 + p_2 \{q_3, p_3\} q_1 = q_1 p_2 - q_2 p_1 = L_3[/itex]

Eugene.

Thanks Eugene, I'm a still confused exactly how you've used the [tex] \{q_1,p_1\} = \{q_2,p_2\} = \{q_3,p_3\} = 1 [/tex] condition in this [tex]\{q_2 p_3, q_3 p_1\} - \{q_3 p_2 , q_3 p_1\} - \{q_2 p_3, q_1 p_3\} + \{q_3 p_2 , q_1 p_3\} [/tex] expression though? Could you be a bit more explicit? Sorry for the stupid questions, I have a feeling I really don't understand poisson brackets very well at all yet.
 
  • #4
meopemuk
1,761
53
Take, for example, the second term

[tex] \{q_3 p_2 , q_3 p_1\} [/tex]

and consider the left part of the bracket [tex] q_3 p_2 [/tex]. The first factor q3 has zero brackets with both factors on the right side (q3 and p1). The second factor p2 also has zero brackets with q3 and p1. Therefore, the product [tex] q_3 p_2 [/tex] has zero bracket with the product [tex] q_3 p_1 [/tex].

You can get this result also by the direct use of the Poisson bracket definition

[tex] \{f, g\} \equiv \frac{\partial f}{\partial \mathbf{q}} \cdot \frac{\partial g}{\partial \mathbf{p}} - \frac{\partial f}{\partial \mathbf{p}} \cdot \frac{\partial g}{\partial \mathbf{q}} [/tex]

Eugene.
 
  • #5
meopemuk
1,761
53
In my preceding post I assumed the following identity

[tex] \{ab , c\} = a \{b , c\} + \{a , c\} b[/tex]

which can be proved for all a,b,c from the Poisson bracket definition.

Eugene.
 
  • #6
barnflakes
156
4
Thanks again Eugene. It is clear you have a much more intuitive understand of poisson brackets than I do, so instead I have tried using the direction definition of the Poisson bracket.

If we take the first term:

[tex]
\{q_2 p_3, q_3 p_1\}
[/tex]

Then applying the above definition we have:

[tex](q_2 p_3)_p (q_3 p_1)_q - (q_2 p_3)_q (q_3 p_1)_p[/tex]

Where subscript p & q denote differentiation wrt that momentum & coordinate.

So considering just the very first term [tex](q_2 p_3)_p (q_3 p_1)_q [/tex] and using that definition I obtain:

[tex](\frac{\partial q_2}{\partial p_3} \cdot p_3 + \frac{\partial p_3}{\partial p_3} \cdot q_2)(\frac{\partial q_3}{\partial q_3} \cdot p_1 + \frac{\partial p_1}{\partial q_3}) \cdot q_3 = p_1 q_2[/tex]

Is this correct? If so, when I work out the entire expression, I obtain:

[tex]q_2 p_1 - p_3 q_3 - q_3 p_1 + p_2 q_3 -q_2 p_3 + p_3 q_1 + q_3 p_3 - p_2 q_1 = q_2 p_1 - q_3 p_1 + p_2 q_3 -q_2 p_3 + p_3 q_1 - p_2 q_1[/tex]

and I cannot see how that relates to the final answer?
 
  • #7
meopemuk
1,761
53
This is not correct. In the definition of the Poisson bracket the derivative [tex] \partial f/\partial \mathbf{q} [/tex] means a 3-vector [tex] (\partial f/\partial q_1, \partial f/\partial q_2, \partial f/\partial q_3) [/tex] and [tex] \partial f/\partial \mathbf{p} \equiv (\partial f/\partial p_1, \partial f/\partial p_2, \partial f/\partial p_3) [/tex]. Using this notation we obtain

[tex] \{q_2 p_3, q_3 p_1 \} = \left( \frac{\partial (q_2 p_3)}{\partial q_1 }, \frac{\partial (q_2 p_3)}{\partial q_2 }, \frac{\partial (q_2 p_3)}{\partial q_3} \right) \cdot \left( \frac{\partial (q_3 p_1)}{\partial p_1 }, \frac{\partial (q_3 p_1)}{\partial p_2 }, \frac{\partial( q_3 p_1)}{\partial p_3}\right) - \left( \frac{\partial (q_2 p_3)}{\partial p_1 }, \frac{\partial (q_2 p_3)}{\partial p_2 }, \frac{\partial (q_2 p_3)}{\partial p_3} \right) \cdot \left( \frac{\partial (q_3 p_1)}{\partial q_1 }, \frac{\partial (q_3 p_1)}{\partial q_2 }, \frac{\partial (q_3 p_1)}{\partial q_3} \right) [/tex]

[tex]= (0, p_3, 0) \cdot (q_3, 0, 0) - (0, 0, q_2) \cdot (0, 0, p_1) = -q_2p_1[/tex]

Eugene.
 
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