Poisson brackets - Aetherwisp

The Poisson brackets map a pair of functions in phase space to another function in phase space:

f(\mathbf{q},\mathbf{p},t),\ g(\mathbf{q},\mathbf{p},t)\quad\mapsto \quad \{ f,g \}(\mathbf{q},\mathbf{p},t)=\sum_{i=1}^{n} \left( \frac{ \partial f }{ \partial q_{i} } \frac{ \partial g }{ \partial p_{i} } - \frac{ \partial f }{ \partial p_{i} } \frac{ \partial g }{ \partial q_{i} } \right)

They can be written in a more compact form as follows:

\{ f,g \}(\mathbf{x},t)=\sum_{i,j=1}^{2n} \frac{ \partial f }{ \partial x_{i} } E_{ij}\frac{ \partial g }{ \partial x_{j} } =\nabla_{\mathbf{x}}f\cdot \mathrm{E}\nabla_{\mathbf{x}}g

where $\mathbf{x}=(\mathbf{q},\mathbf{p})=(q_{1},\ldots,q_{n},p_{1},\ldots,p_{n})$ are the canonical coordinates and $\mathrm{E}$ is the standard symplectic matrix.

Properties#

The brackets are bilinear: $\{ f,\alpha g+\beta h \}=\alpha \{ f,g \}+\beta \{ f,h \}$ .
They are antisymmetric: $\{ f,g \}=-\{ g,f \}$ .
They obey the Jacobi identity: $\{ f,\{ g,h \} \}+\{ g, \{ h,f \} \}+\{ h, \{ f,g \} \}=0$
$\{ f,g\cdot h \}=g\{ f,h \}+\{ f,g \}h$ .
$\{f,f\}=0$ (due to antisymmetry).
$\{f(x),x\}=0$ .

The first three are the most important, as a Vector space equipped with an operation that satisfies them is a Lie algebra.

Fundamental Poisson brackets#

Some instances of Poisson brackets are particularly important, specifically the brackets on the generalized coordinates and conjugate momenta of the phase space:

\{ q_{i},q_{j} \}=\sum_{k=1}^{n} \frac{ \partial q_{i} }{ \partial q_{k} } \underbrace{ \frac{ \partial q_{j} }{ \partial p_{k} } }_{ 0 } - \underbrace{ \frac{ \partial q_{i} }{ \partial p_{k} } }_{ 0 } \frac{ \partial q_{j} }{ \partial q_{k} }=0

\{ p_{i},p_{j} \}=\sum_{k=1}^{n} \underbrace{ \frac{ \partial p_{i} }{ \partial q_{k} } }_{ 0 } \frac{ \partial p_{j} }{ \partial p_{k} } - \frac{ \partial p_{i} }{ \partial p_{k} } \underbrace{ \frac{ \partial p_{j} }{ \partial q_{k} } }_{ 0 }=0

\{ q_{i},p_{j} \}=\sum_{k=1}^{n} \left( \underbrace{ \frac{ \partial q_{i} }{ \partial q_{k} } }_{ \delta_{ik} } \underbrace{ \frac{ \partial p_{j} }{ \partial p_{k} } }_{ \delta_{jk} } - \cancel{ \frac{ \partial q_{i} }{ \partial p_{k} } \frac{ \partial p_{j} }{ \partial q_{k} } } \right)=\sum_{k=1}^{n} \delta_{ik}\delta_{jk}=\delta_{ij}

In short:

\boxed{\{ q_{i},q_{j} \}=0,\quad \{ p_{i},p_{j} \}=0,\quad \{ q_{i},p_{j} \}=\delta_{ij}}

Also using compact formalism:

\{ x_{i},x_{j} \}=\sum_{k,l=1}^{2n} \underbrace{ \frac{ \partial x_{i} }{ \partial x_{k} } }_{ \delta_{ik} } \mathrm{E}_{kl}\underbrace{ \frac{ \partial x_{j} }{ \partial x_{l} } }_{ \delta_{jl} } =\sum_{k,l=1}^{2n} \delta_{ik}\delta_{jl}\mathrm{E}_{kl}=\mathrm{E}_{ij}

And so

\boxed{\{ x_{i},x_{j} \}=E_{ij}}

We can also check the Poisson brackets of $x_{i}$ with a function $G$ defined in phase space:

\{ x_{i},G \}(\mathbf{x},t)=\sum_{j,k=1}^{2n} \underbrace{ \frac{ \partial x_{i} }{ \partial x_{j} } }_{ \delta_{ij} } \mathrm{E}_{jk}\frac{ \partial G }{ \partial x_{k} } =\sum_{k=1}^{2n} \mathrm{E}_{ik}\frac{ \partial G }{ \partial x_{k} }

which is equivalent to

\boxed{\{ \mathbf{x},G \}=\mathrm{E}\nabla_{x}G}

This is notable, as we know that the Hamilton equations can be written as $\dot{\mathbf{x}}=\mathrm{E}\nabla_{x}H$ , for some Hamiltonian $H$ . Because of this, we can write the Hamilton equations as Poisson brackets:

\boxed{\dot{\mathbf{x}}=\{ \mathbf{x},H \}}

Regarding constants of motion#

Poisson brackets have some interesting behavior with respect to constants of motion. Say now $f(q,p,t)$ is a constant of motion over the motion $q(t)$ and $p(t)$ , so

\frac{d}{dt} f(q(t),p(t),t)=0

for $q(t)$ and $p(t)$ that solve the Hamilton equations. If we calculate the total derivative explicitly and apply the Hamilton equations, we get

\begin{align} \frac{d}{dt} f(q(t),p(t),t)&=\sum_{i=1}^{n} \left( \frac{ \partial f }{ \partial p_{i} } \dot{p}_{i}+ \frac{ \partial f }{ \partial q_{i} } \dot{q}_{i} \right)+\frac{ \partial f }{ \partial t } \\ &=\sum_{i=1}^{n} \left( -\frac{ \partial f }{ \partial p_{i} } \frac{ \partial H }{ \partial q_{i} } +\frac{ \partial f }{ \partial q_{i} } \frac{ \partial H }{ \partial p_{i} } \right)+\frac{ \partial f }{ \partial t } \\ &=\{ f,H \}(q(t),p(t),t)+\frac{ \partial f }{ \partial t } \end{align}

This lets us say that $f$ is a constant of motion if and only if

\{ f,H \}+\frac{ \partial f }{ \partial t } =0

In the special (but common) case where $f$ does not explicitly depend on $t$ , we get

\boxed{f\text{ is a constant of motion}\quad\Leftrightarrow\quad \{ f,H \}=0}

As a side note, the Hamiltonian itself is a constant if

\{ H,H \}+\frac{ \partial H }{ \partial t } =\frac{ \partial H }{ \partial t } =0

In other words, it's always constant unless it explicitly changes over time. This is what we want though, as it is just another formulation of conservation of energy: if $H$ , which is total mechanical energy, changed on its own without being explicitly time-dependent, it would mean that energy is being created or destroyed, which cannot be possible.

Examples#

> Between position and angular momentum they are essentially the same: > $$\{ q_{l},L_{i} \}=\sum_{k=1}^{3} \epsilon_{lik}q_{k}

The Poisson brackets between components of the angular momentum are among the most relevant in physics (and later on, quantum mechanics):
$\{ L_{i},L_{j} \}&=\left\{ \sum_{m,s=1}^{3} \epsilon_{ims}q_{m}p_{s}, L_{j} \right\} \\ \text{(bilinear)}&=\sum_{m,s=1}^{3} \epsilon_{ims}\{ q_{m}p_{s},L_{j} \} \\ \text{(commutative)}&=\sum_{m,s=1}^{3} \epsilon_{ims}(q_{m}\{ p_{s},L_{j} \}+\{ q_{m},L_{j} \}p_{j}) \\ &=\sum_{m,s=1}^{3} \epsilon_{ims}\left( \sum_{r=1}^{3} q_{m}\epsilon_{sjr}p_{r}+\sum_{r=1}^{3} \epsilon_{mjr}q_{r}p_{s} \right) \\ &=\sum_{m,r=1}^{3} q_{m}p_{r}\underbrace{ \sum_{s=1}^{3} \epsilon_{ims}\epsilon_{jrs} }_{ \delta_{ij}\delta_{mr}-\delta_{ir}\delta_{mj} }-\sum_{r,s=1}^{3} q_{r}p_{s}\underbrace{ \sum_{m=1}^{3} \epsilon_{mis}\epsilon_{mjr} }_{ \delta_{ij}\delta_{sr}-\delta_{ir}\delta_{sj} } \\ &=-q_{j}p_{i}+q_{i}p_{j} \\ &=\sum_{k=1}^{3} \epsilon_{ijk}(\mathbf{q}\times \mathbf{p})_{k} \\ &=\sum_{k=1}^{3} \epsilon_{ijk}L_{k} \end{align}$

> where at the end we used the Levi-Civita tensor expression for the [[Vector product]] of $\mathbf{q}$ and $\mathbf{p}$ (which is angular momentum). So, the Poisson brackets on angular momentum component is > $$\boxed{\{ L_{i},L_{j} \}=\sum_{k=1}^{3} \epsilon_{ijk}L_{k}}

Specializing to specific indexes, we see the following fundamental relations:

> Another important set of brackets is between the square [[norm]] of the angular momentum and any of its components (for brevity, $\lvert \mathbf{L} \rvert^{2}\equiv L^{2}$): > $$\begin{align} > \{ L^{2},L_{i} \}&=\sum_{r=1}^{3} \{ L_{r}^{2},L_{i} \} \\ > \text{(commutative)}&=\sum_{r=1}^{3} (L_{r}\{ L_{r},L_{i} \}+\{ L_{r},L_{i} \}L_{r}) \\ > &=2\sum_{r=1}^{3} L_{r}\{ L_{r},L_{i} \} \\ > &=2\sum_{r=1}^{3} L_{r}\sum_{s=1}^{3} \epsilon_{ris}L_{s} \\ > &=-2\sum_{r,s=1}^{3} \epsilon_{irs}L_{r}L_{s} \\ > &=\ldots > \end{align}

We'll use a useful property of tensors. Consider two tensors, an antisymmetric tensor $a_{lk}$ and a symmetric one $s_{lk}$ . The contraction of their sum is zero: $\sum_{n,m}a_{nm}s_{nm}=0$ . This is because due to antisymmetry, as $\sum_{n,m}a_{nm}b_{nm}=-\sum_{nm}a_{mn}b_{mn}$ and the only number that is equal to its negative is zero. In our case, $\epsilon_{irs}$ is antisymmetric, and the product $L_{r}L_{s}$ is a symmetric tensor. So, we can use this little trick to conclude: