Solution for assignment 14 set 4 Poisson Brackets

Let us consider functions g = g(q_{k}, p_{k}, t) and h = h(q_{k}, p_{k}, t). The Poisson bracket is defined by

[g, h] \begin{matrix}def \\ = \end{matrix} ∑_{k} (\frac{∂g}{∂q_{k}} \frac{∂h}{∂p_{k}} − \frac{∂h}{∂q_{k}} \frac{∂g}{∂p_{k}} ) .

The properties of the assignment are shown in the following:

1. Resulting in the definition of the total time derivative:

\frac{dg}{dt} = [g, H] + \frac{∂g}{∂t} = ∑_{k} (\frac{∂g}{∂q_{k}} \frac{∂H}{∂p_{k}} − \frac{∂H}{∂q_{k}} \frac{∂g}{∂p_{k}} ) + \frac{∂g}{∂t}
= ∑_{k} (\frac{∂g}{∂q_{k}} q̇_{k} + \frac{∂g}{∂p_{k}} ṗ_{k}) + \frac{∂g}{∂t} .

2. Using ∂q_{j}/∂p_{k} = 0:

q̇_{j} = [q_{j}, H] = ∑_{k} (\frac{∂q_{j}}{∂q_{k}} \frac{∂H}{∂p_{k}} − \frac{∂H}{∂q_{k}} \frac{∂q_{j}}{∂p_{k}} ) = ∑_{k} δ_{jk} q̇_{k} = q̇_{j} ,

3. Using ∂p_{j}/∂q_{k} = 0:

ṗ_{j} = [p_{j}, H] = ∑_{k} (\frac{∂p_{j}}{∂q_{k}} \frac{∂H}{∂p_{k}} − \frac{∂H}{∂q_{k}} \frac{∂p_{j}}{∂p_{k}} ) = ∑_{k} δ_{jk} ṗ_{k} = ṗ_{j} ,

4. Commutators of generalized coordinates and momenta:

[q_{i}, q_{j}] = ∑_{k} (\frac{∂q_{i}}{∂q_{k}} \frac{∂q_{j}}{∂p_{k}} − \frac{∂q_{j}}{∂q_{k}} \frac{∂q_{i}}{∂p_{k}} ) = 0

as ∂q_{j}/∂p_{k} = 0 and ∂q_{i}/∂p_{k} = 0. Similarly

[p_{i}, p_{j}] = ∑_{k} (\frac{∂p_{i}}{∂q_{k}} \frac{∂p_{j}}{∂p_{k}} − \frac{∂p_{j}}{∂q_{k}} \frac{∂p_{i}}{∂p_{k}} ) = 0 ,

while

[q_{i}, p_{j}] = ∑_{k} (\frac{∂q_{i}}{∂q_{k}} \frac{∂p_{j}}{∂p_{k}} − \frac{∂p_{j}}{∂q_{k}} \frac{∂q_{i}}{∂p_{k}} ) = ∑_{k} (δ_{ik} δ_{j_{k}} − 0) = δ_{ij} .

5. Angular momentum commutator:

[x_{i}, L_{j}] = ∑_{k} ∑_{l} ∑_{m} ϵ_{jlm} (\frac{∂x_{i}}{∂x_{k}} \frac{∂(x_{l}p_{m})}{∂p_{k}} − \frac{∂(x_{l}p_{m})}{∂x_{k}} \frac{∂x_{i}}{∂p_{k}} )
= ∑_{k} ∑_{l} ∑_{m} ϵ_{jlm} δ_{ik} x_{l} δ_{mk} = ∑_{l} ∑_{m} ϵ_{jlm} δ_{im} x_{l} = ∑_{l} ϵ_{jli} x_{l} = ∑_{l} ϵ_{ijl} x_{l} .

[p_{i}, L_{j}] = ∑_{k} ∑_{l} ∑_{m} ϵ_{jlm} (\frac{∂p_{i}}{∂x_{k}} \frac{∂(x_{l}p_{m})}{∂p_{k}} − \frac{∂(x_{l}p_{m})}{∂x_{k}} \frac{∂p_{i}}{∂p_{k}} )
= − ∑_{k} ∑_{l} ∑_{m} ϵ_{jlm} δ_{lk} p_{m} δ_{ik} = − ∑_{l} ∑_{m} ϵ_{jlm} δ_{li} p_{m} = − ∑_{m} ϵ_{jim} p_{m} = ∑_{m} ϵ_{ijm} x_{m} .

[L_{i}, L_{j}] = ∑_{k} ∑_{l} ∑_{m} ∑_{n} ∑_{r} ϵ_{ikl}ϵ_{jmn} (\frac{∂(x_{k}p_{l})}{∂x_{r}} \frac{∂(x_{m}p_{n})}{∂p_{r}} − \frac{∂(x_{m}p_{n})}{∂x_{r}} \frac{∂(x_{k}p_{l})}{∂p_{r}})
= ∑_{k} ∑_{l} ∑_{m} ∑_{n} ∑_{r} ϵ_{ikl}ϵ_{jmn} (δ_{kr}δ_{nr}x_{m}p_{l} − δ_{mr}δ_{lr}x_{k}p_{n})
= ∑_{k} ∑_{l} ∑_{m} ∑_{n} ϵ_{ikl}ϵ_{jmn} (δ_{kn}x_{m}p_{l} − δ_{ml}x_{k}p_{n})
= ∑_{k} ∑_{l} (∑_{m} ϵ_{ikl}ϵ_{jmk}x_{m}p_{l} − ∑_{n} ϵ_{ikl}ϵ_{jln}x_{k}p_{n})
= ∑_{k} ∑_{l} (∑_{m} ϵ_{kli}ϵ_{kjm}x_{m}p_{l} − ∑_{n} ϵ_{lik}ϵ_{lnj}x_{k}p_{n})
= ∑_{l} ∑_{m}(δ_{lj}δ_{im} − δ_{lm}δ_{ij})x_{m}p_{l} − ∑_{k} ∑_{n}(δ_{in}δ_{kj} − δ_{ij}δ_{kn})x_{k}p_{n}
= x_{i}p_{j} − δ_{ij} x⃗ · p⃗ − x_{j}p_{i} + δ_{ij} x⃗ · p⃗ = x_{i}p_{j} − x_{j}p_{i} .

This agrees with

∑_{k} ϵ_{ijk}L_{k} = ∑_{k} ∑_{l} ∑_{m} ϵ_{ijk}ϵ_{klm}x_{l}p_{m} = ∑_{k} ∑_{l} ∑_{m} ϵ_{kij}ϵ_{klm}x_{l}p_{m}
= ∑_{l} ∑_{m}(δ_{il}δ_{jm} − δ_{im}δ_{jl}) x_{l}p_{m} = x_{i}p_{j} − x_{j}p_{i} .