Summary of the question:
-
How can I prove the equal-time Poisson bracket relations for the classical Hamiltonian field theory? I.e
$$[q(x,t),H(t)]_\mathrm{PB}=\dot{q}(x,t)\tag{1}$$ for a field $q$ and $$[\Pi(x,t),H(t)]_\mathrm{PB}=\dot{\Pi}(x,t)\tag{2}$$ for the conjugate momentum field $\Pi$. -
Since we should have the right physics, I think these 'abstract' relations themselves are not enough, but their specific compatibility with known physical laws (e.g. Euler-Lagrange equation for fields) should also be addressed. Do they appear while addressing the previous question? Or are those 'abstract' relations just enough to ensure that we are doing the right physics?
Hello,
As a preliminary step of studying quantum field theory, I am looking at some classical field theories, especially focusing on their formulations.
During that, I am suffering a problem regarding the Poisson bracket relations in Hamiltonian field theory.
First, let's start from the Euler-Lagrange equation
$$\frac{\partial}{\partial t}\left(\frac{\partial \mathcal{L}}{\partial(\partial_t q)}\right)+\frac{\partial}{\partial x}\left(\frac{\partial \mathcal{L}}{\partial(\partial_x q)}\right)-\frac{\partial \mathcal{L}}{\partial q}=0\tag{3}$$
for a Lagrangian density $\mathcal{L}$ involving a field variable $q(x,t)$.
For a canonical momentum density
$$\Pi(x,t):=\partial\mathcal{L}/\partial(\partial_t q)\tag{4}$$
conjugate to $q$, we define the Hamiltonian density
$$\mathcal{H}(x,t):=\Pi(x,t)\frac{\partial q}{\partial t} – \mathcal{L}.\tag{5}$$
Since we are formulating a Hamiltonian mechanics, we desire the following equal-time Poisson bracket relations hold:
(A) $\quad[q(x,t),\,\Pi(x',t)]_\mathrm{PB}=\delta(x-x')$
(B) $\quad[q(x,t),\,q(x',t)]_\mathrm{PB}=[\Pi(x,t),\,\Pi(x',t)]_\mathrm{PB}=0.$
According to some lecture notes and web pages, I found a definition of Poisson bracket
$$[A(q,\Pi),\,B(q,\Pi)]_{\mathrm{PB}}:=\int dx \left(
\frac{\delta A}{\delta q(x)}\frac{\delta B}{\delta \Pi(x)}-\frac{\delta A}{\delta \Pi(x)}\frac{\delta B}{\delta q(x)}
\right)\tag{6}$$
where $\delta$ is the functional derivative.
Meanwhile, based on the definition of functional derivative, I derived some formulae like
$$\frac{\delta q (x)}{\delta q(y)} = \frac{\delta \Pi (x)}{\delta \Pi(y)}=\delta(x-y),\tag{7}$$
$$\frac{\delta A(x)}{\delta q(y)}=\frac{\partial A(x)}{\partial q(x)} \delta(x-y),\quad \frac{\delta A(x)}{\delta \Pi(y)}=\frac{\partial A(x)}{\partial \Pi(x)} \delta(x-y)\tag{8}$$
(I'm not sure the accuracy of those formulae, and I don't have a clear intuitive picture about those formulae in my mind; to me, they are just bunch of symbols providing some rules).
Starting form the suggested definition of the Poisson bracket and using the formulae above, I think I somehow successfully checked that the basic Poisson bracket relations (A), (B) hold, given the definition of the PB.
But here's my first problem: the definition of the Poisson bracket involves only one argument $x$, but we should deal with two arguments $x$ and $x'$ while calculating the PB. This is very confusing to me, and I cannot make sure that my derivation below is rigorous enough.
$$
[q(x,t),\Pi(x',t)]_\mathrm{PB}=\left[\int dx' q(x',t)\delta(x-x'),\,\,\Pi(x',t)\right]_\mathrm{PB}\\
=\int dy \left(\int dx' \frac{\delta q(x')}{\delta q(y)}\delta(x-x')\frac{\delta \Pi(x')}{\delta \Pi(y)} – \int dx' {\frac{\delta q(x')}{\delta \Pi(y)}}\delta(x-x'){\frac{\delta \Pi(x')}{\delta q(y)}}\right) \\
= \int dy \int dx' \delta(x'-y) \delta(x-x') \delta(x'-y) \\
=\delta(x-x').\tag{9}
$$
To avoid the confusion above, in the second line, I introduced a new variable $y$ for the dummy index defining the Poisson bracket. Also, in the final step, dealing with the delta functions seems unsatisfactory. Is every step okay here?
Next, I want to look at the PB relations involving the Hamiltonian $H=\int dx \mathcal{H}$ as an argument. I started from here:
$$
[q(x,t),H(t)]_\mathrm{PB}=\int dx'\,\left[
q(x,t),\,\Pi(x',t)\frac{\partial q}{\partial t}(x',t)-\mathcal{L}\left(
\partial_t q,\,\partial_x q,\,q
\right)
\right]_\mathrm{PB},\tag{10}
$$
written from the definition of the Hamiltonian density. Using the distributive law of Poisson bracket and the formula $$[A,\,BC] = [A,B]C + B[A,C],\tag{11}$$ I could reach
$$
=\int dx' [q(x,t),\Pi(x',t)]_\mathrm{PB} \frac{\partial q}{\partial t}(x',t)
+ \int dx' \Pi(x',t) \left[q(x,t),\frac{\partial q}{\partial t}(x',t) \right]_\mathrm{PB} \\
\quad\quad-\int dx'\left[
q(x,t),\mathcal{L}\left(\partial_t q(x',t),\partial_x q(x',t),q(x',t)\right)
\right]_\mathrm{PB}.\tag{12}
$$
In the first term, using the relation $[q(x,t),\Pi(x',t)]=\delta(x-x'),$ I think I can obtain $\partial q(x,t) / \partial t$. However, I cannot go further with the other terms. So here comes my second question: how can I derive the final PB relation for $q$, expectedly something like $[q,H]=\dot{q}$? In addition, how can I derive the similar PB relation for $\Pi$?
Finally, in the whole step above, we only dealt with the 'abstract' algebraic properties of Poisson bracket relations, not specifically checking the compatibility of those with already-known physical laws (such as Euler-Lagrange equations). Do they appear while addressing the previous questions? Or are those 'abstract' relations just enough to ensure that we are doing the right physics?
Best Answer
I think that your issues are not specifically due to classical field theory, but rather Hamiltonian mechanics in general. Formally, there is no conceptual difference between the two, you just replace the Kronecker deltas of the finite dimensional phase space with Dirac deltas in field theory.
The Hamilton's equations of motion are typically obtained by a variational principle. Your real trajectories are stationary points of the action: $$ S = \int pdq-Hdt $$ Using calculus of variation, you can deduce Hamilton's equations of motion the Euler-Lagrange equation gives: $$ \dot q = \frac{\partial H}{\partial p} \\ \dot p = -\frac{\partial H}{\partial q} $$ which you can then rewrite more conveniently using Poisson brackets. Note that this is equivalent to the Euler-Lagrange equation of the corresponding Lagrangian thanks to the property of the Legendre transform, even though the candidate trajectories in the Hamiltonian formalism are more diverse than in the Lagrangian setting. Therefore, if you capture the "right" physics in your Lagrangian (typically by symmetry considerations), then the Hamilton equations of motion will also capture the "right" physics due to this equivalence.
Back to your field theory setting, the same method applies. From the variational principle using the action $S$: $$ S = \int \Pi\cdot dq -H dt \\ \Pi \cdot \dot q = \int \Pi(x) \dot q(x) d^nx $$ you obtain the Hamilton's equations of motion: $$ \begin{align} \dot q(x) &= \frac{\delta H}{\delta \Pi(x)} \\ &= \frac{\partial \mathcal H}{\partial \Pi}(x) \\ \dot\Pi(x,) &= -\frac{\delta H}{\delta q(x)} \\ &= -\frac{\partial \mathcal H}{\partial q}(x) \end{align} $$ where the second equations assume that (as it is often the case) the Hamiltonian is of the form: $$ H = \int \mathcal H(q,\Pi)(x) d^nx $$ The correct definition of the Poisson bracket two functionals $A,B$ of $q,\Pi$ is: $$ [A,B] = \frac{\delta A}{\delta q}\frac{\delta B}{\delta \Pi}-\frac{\delta A}{\delta \Pi}\frac{\delta B}{\delta q} $$ The Poisson bracket is still a functional of $q,\Pi$, but is implicitly a function of $x$. Things simplify if you assume that they are of the form: $$ A = \int [\mathcal A(\Pi,q)](x) d^nx\\ B = \int [\mathcal B(\Pi,q)](x) d^nx $$ In which case: $$ \begin{align} \frac{\delta A}{\delta q(x)} &= \frac{\partial \mathcal A}{\partial q}(x) & \frac{\delta B}{\delta q(x)} &= \frac{\partial\mathcal B}{\partial q}(x) \\ \frac{\delta A}{\delta \Pi(x)} &= \frac{\partial\mathcal A}{\partial \Pi}(x) & \frac{\delta\mathcal B}{\delta \Pi(x)} &= \frac{\partial\mathcal B}{\partial \Pi}(x) \\ [A,B] &= \frac{\partial \mathcal A}{\partial q}\frac{\partial \mathcal B}{\partial \Pi}-\frac{\partial \mathcal B}{\partial q}\frac{\partial \mathcal A}{\partial \Pi} \end{align} $$ Even the elementary fields can be written this way: $$ q(x) = \int q(y)\delta(y-x)d^ny \\ \Pi(x) = \int \Pi(y)\delta(y-x)d^ny $$ Or more generally, this applies to local observables as well, which justify the equations whose accuracy you were not sure: $$ \mathcal A(x) = \int [\mathcal A(q,\Pi)] (y)\delta(y-x)d^ny \\ \mathcal B(x) = \int [\mathcal B(q,\Pi)] (y)\delta(y-x)d^ny $$ This is why you have: $$ \begin{align} [q(x),\Pi(y)] &= \delta(x-y) \\ [q(x),A] &= \frac{\partial \mathcal A}{\partial \Pi}(x) \\ [\Pi(x),A] &= -\frac{\partial \mathcal A}{\partial q}(x) \\ [\mathcal A(x),\mathcal B(y)] &= \frac{\partial \mathcal A}{\partial q}(x)\frac{\partial \mathcal B}{\partial \Pi}(x)\delta(x-y)-\frac{\partial \mathcal B}{\partial q}(x)\frac{\partial \mathcal A}{\partial \Pi}(x)\delta(x-y) \end{align} $$
Hope this helps.