Invariance of the Poisson Bracket

[Luke Tan]

TL;DR

When it is mentioned that the poisson bracket is invariant under a canonical transformation, does this mean the functional form of the poisson bracket, or the numerical value?

I've recently been starting to get really confused with the meaning of equality in multivariable calculus in general.

When we say that the poisson bracket is invariant under a canonical transformation $q, p \rightarrow Q,P$, what does it actually mean?

If the poisson bracket $[u,v]_{q,p}$ were, say, $[u,v]_{q,p}=q-p$, does the invariance mean that the poisson bracket is $[u,v]_{Q,P}=Q-P$?

This would seem to make the least sense to me.

However, the only other definition I can think of would be that the numerical value is conserved, say if we had $P=-q$, $Q=p$, the poisson bracket $[u,v]_{Q,P}=-P-Q$, and this would make sense for the most part to me.

However, this would then raise the confusing question as to what the significance of this invariance is. From what I can see, any transformation equations $Q=Q(q,p)$ and $P=P(q,p)$ can easily be inverted to get $q=q(Q,P)$ and $p=p(Q,P)$, which we then substitute into the poisson bracket, or any other function as a matter of fact, and this will naturally satisfy the condition that the numerical value is invariant.

Which is the correct definition of invariance, and if it's that the numerical value doesn't change, why then is this invariance so significant?

 

[anuttarasammyak]

What are your u and v with relation to p and q ? Poisson bracket you mean is classical one or quantum ?

 

[Luke Tan]

What are your u and v with relation to p and q ? Poisson bracket you mean is classical one or quantum ?

u and v arent really any definite functions, I just want to get an idea of how the poisson bracket transforms under a canonical transformation and what exactly is invariant.

Classical poisson bracket

 

[wrobel]

Let $M$ be a symplectic manifold that is on $M$ a 2-form $\omega=\sum_{i<j}\omega_{ij}(x)dx^i\wedge dx^j$ is defined. Here $x=(x^1,\ldots, x^r)$ are arbitrary local coordinates in $M$.
The form $\omega$ must obey two conditions
1) it is non degenerate: $\det(\omega_{ij}(x))\ne 0,\quad \forall x\in M$
2) it is closed: $d\omega=0$.
The first condition implies that $\dim M$ is an even number.
Let $f,g:M\to\mathbb{R}$ be two smooth functions. The Poisson bracket by definition is a function
$$\{f,g\}:=\omega^{ij}\frac{\partial f}{\partial x^i}\frac{\partial g}{\partial x^j}.$$
$\omega^{ij}$ is the matrix inverse to the matrix $\omega_{ij}$.

Coordinates $(x^1,\ldots,x^{2m})=(q^1,\ldots,q^m,p_1,\ldots,p_m)$ are said to be canonical coordinates (or symplectic coodinates) if $\omega=dq^i\wedge dp_i.$

In canonical coordinates the Poisson bracket has especially simple form
$$\{f,g\}=\frac{\partial f}{\partial p_i}\frac{\partial g}{\partial q^i}-\frac{\partial g}{\partial p_i}\frac{\partial f}{\partial q^i}.$$

A change of variables $(q,p)\mapsto (Q,P)$ is said to be canonical if it preserves canonical shape of the 2-form:
$$\omega=dq^i\wedge d p_i=dQ^i\wedge dP_i.$$

 

[wrobel]

Summary:: When it is mentioned that the poisson bracket is invariant under a canonical transformation, does this mean the functional form of the poisson bracket, or the numerical value?

If the poisson bracket [u,v]q,p[u,v]_{q,p} were, say, [u,v]q,p=q−p[u,v]_{q,p}=q-p, does the invariance mean that the poisson bracket is [u,v]Q,P=Q−P[u,v]_{Q,P}=Q-P?

consider a canonical change $p=-Q,\quad q=P$

 

[Luke Tan]

Let $M$ be a symplectic manifold that is on $M$ a 2-form $\omega=\sum_{i<j}\omega_{ij}(x)dx^i\wedge dx^j$ is defined. Here $x=(x^1,\ldots, x^r)$ are arbitrary local coordinates in $M$.
The form $\omega$ must obey two conditions
1) it is non degenerate: $\det(\omega_{ij}(x))\ne 0,\quad \forall x\in M$
2) it is closed: $d\omega=0$.
The first condition implies that $\dim M$ is an even number.
Let $f,g:M\to\mathbb{R}$ be two smooth functions. The Poisson bracket by definition is a function
$$\{f,g\}:=\omega^{ij}\frac{\partial f}{\partial x^i}\frac{\partial g}{\partial x^j}.$$
$\omega^{ij}$ is the matrix inverse to the matrix $\omega_{ij}$.

Coordinates $(x^1,\ldots,x^{2m})=(q^1,\ldots,q^m,p_1,\ldots,p_m)$ are said to be canonical coordinates (or symplectic coodinates) if $\omega=dq^i\wedge dp_i.$

In canonical coordinates the Poisson bracket has especially simple form
$$\{f,g\}=\frac{\partial f}{\partial p_i}\frac{\partial g}{\partial q^i}-\frac{\partial g}{\partial p_i}\frac{\partial f}{\partial q^i}.$$

A change of variables $(q,p)\mapsto (Q,P)$ is said to be canonical if it preserves canonical shape of the 2-form:
$$\omega=dq^i\wedge d p_i=dQ^i\wedge dP_i.$$

Um sorry I'm not familiar with differential geometry, is there any other way I can understand this?

 

[anuttarasammyak]

If the poisson bracket [u,v]q,p[u,v]q,p[u,v]_{q,p} were, say, [u,v]q,p=q−p[u,v]q,p=q−p[u,v]_{q,p}=q-p, does the invariance mean that the poisson bracket is [u,v]Q,P=Q−P[u,v]Q,P=Q−P[u,v]_{Q,P}=Q-P?

Say $u=p^2,v=q$ and $Q=p,P=-q$,

\{p^2,q\}_{p.q}=2p
\{p^2,q\}_{P,Q}=2p =2Q \neq 2P

 

[Luke Tan]

Say $u=p^2,v=q$ and $Q=p,P=-q$,

\{p^2,q\}_{p.q}=2p
\{p^2,q\}_{P,Q}=2p =2Q \neq 2P

So would it be correct to say that it is the numerical value that is invariant?

 

[anuttarasammyak]

In the case of post #7 the invariant result 2p is a function of p. Also the function would be expressed as 2p(P,Q)=2Q which is the function of canonical transformation variables.

Of course when you input value p=p', function 2p becomes the value 2p'.