Let $\mathscr{B}$ be a space of physics we have and $\mathscr{T}$ be the duration. Let $\mathscr{L}$ be a lagrangian density of the field such that the action is a functional of $\phi:\mathbb{R}^4\rightarrow\mathbb{R}$: $$S=\int_{\mathscr{B} \times\mathscr{T}}d^4x\mathscr{L}(\phi(t,\vec{x}),\partial_\mu\phi(t,\vec{x})).\tag{1}$$

We can then derive the equations of motion: $$\frac{\delta S}{\delta \phi}=0.\tag{2}$$

Otherwise we can define the hamiltonian density $$\mathscr{H}=\pi\dot\phi-\mathscr{L}=\mathscr{H}(\pi,\phi,\partial_i\phi)\tag{3}$$ whereas $$\pi=\frac{\partial\mathscr{L}}{\partial\dot\phi}\tag{4}$$ and $i=1.2.3$. Then the hamiltonian is a functional of $(\mathbb{R}^3\rightarrow\mathbb{R})$:

$$H(t)=\int_\mathscr{B} d^3x\mathscr{H}.\tag{5}$$ Let $\phi(t)$ and $\pi(t)$ be 2 functions $\mathbb{R}^3\rightarrow \mathbb{R}$. We define the Poisson bracket for 2 functionals $A[\phi,\pi]$ and $B[\phi,\pi]$: $$\{A,B\}=\frac{\delta A}{\delta\pi}\frac{\delta B}{\delta\phi}-\frac{\delta A}{\delta\phi}\frac{\delta B}{\delta\pi}\tag{6}$$ and we have the canonical relation (for $t,\vec{x},\vec{y}$ fixed): $$\{\pi(t,\vec{x}),\phi(t,\vec{y})\}=\delta^{(3)}(\vec{x}-\vec{y}).\tag{7}$$

How can we show that the equation of motions is now


  • $\begingroup$ Comments to the post (v4): 1. Note that the two formulas (6) & (7) for the Poisson bracket are incompatible, cf. e.g. this related Phys.SE post. 2. For the Legendre transformation in point mechanics, see this Phys.SE question and links therein. $\endgroup$
    – Qmechanic
    Dec 7, 2015 at 19:06

1 Answer 1


First, note that the EOM are the Euler-Lagrange equations:

$$ \frac{\delta S}{\delta \phi}=\partial_\mu\left(\frac{\partial \mathscr L}{\partial\phi_{,\mu}}\right)-\frac{\partial \mathscr L}{\partial \phi}=\dot\pi+\partial_i\left(\frac{\partial \mathscr L}{\partial\phi_{,i}}\right)-\frac{\partial \mathscr L}{\partial \phi} $$ where I isolated the $\mu=0$ term, and used the definition of $\pi$.

Next, use $\mathscr L=\pi\dot\phi-\mathscr H$: $$ \frac{\delta S}{\delta \phi}=\dot\pi+\partial_i\left(\frac{\partial \mathscr L}{\partial\phi_{,i}}\right)-\frac{\partial \mathscr L}{\partial \phi}=\dot\pi-\partial_i\left(\frac{\partial \mathscr H}{\partial\phi_{,i}}\right)+\frac{\partial \mathscr H}{\partial \phi}=\dot\pi+\frac{\delta H}{\delta \phi} $$ because $\pi$ and $\dot\phi$ are not functions of $\phi$.

Thus, we get $\dot\pi=-\frac{\delta H}{\delta\phi}=\{H,\pi\}$.

The other relation is easier: $$ \{H,\phi\}=\frac{\delta \mathscr H}{\delta \pi} $$ Next, use $\mathscr H=\pi\dot\phi-\mathscr L$: $$ \{H,\phi\}=\frac{\delta \mathscr H}{\delta \pi}=\dot\phi $$ because $\mathscr L$ is not a function of $\pi$.


Let $f(\phi,\pi)$ and $g(\phi,\pi)$ be two function on "Phase space". Then, by definition, $$ \{f(\boldsymbol x),g(\boldsymbol y)\}\equiv\int\mathrm d\boldsymbol z\ \frac{\delta f(\boldsymbol x)}{\delta \phi(\boldsymbol z)}\frac{\delta g(\boldsymbol y)}{\delta \pi(\boldsymbol z)}-\frac{\delta g(\boldsymbol y)}{\delta \phi(\boldsymbol z)}\frac{\delta f(\boldsymbol x)}{\delta \pi(\boldsymbol z)} $$ where everything is evaluated at the same time $t$ (not writen for clarity). This means that $$ \{H(t),\pi(t,\boldsymbol x)\}=\int\mathrm d\boldsymbol z\ \frac{\delta H(t)}{\delta \phi(\boldsymbol z)}\frac{\delta \pi(\boldsymbol x)}{\delta \pi(\boldsymbol z)}-\frac{\delta \pi(\boldsymbol x)}{\delta \phi(\boldsymbol z)}\frac{\delta H(t)}{\delta \pi(\boldsymbol z)}=\int\mathrm d\boldsymbol z\ \frac{\delta H(t)}{\delta \phi(\boldsymbol z)}\delta(\boldsymbol x-\boldsymbol z) $$ which equals $\frac{\delta H}{\delta \phi}$, as expected.


Proof of $\frac{\delta H}{\delta \phi}=\frac{\partial \mathscr H}{\partial \phi}-\partial_i\left(\frac{\partial\mathscr H}{\partial\phi_{,i}}\right)$:

The dynamical variables of the Lagrangian are $\phi$ and $\partial_\mu\phi$. In the Hamiltonian formulation, we change $\partial_0\phi\leftrightarrow\pi$, so that the dynamical variables of the Hamiltonian are $\phi,\,\partial_i\phi$ and $\pi$.

With this in mind, the Hamiltonian is, by definition, $$ H=\int\mathrm d\boldsymbol x\ \mathscr H(\phi(\boldsymbol x),\phi_{,i}(\boldsymbol x),\pi(\boldsymbol x)) $$

Therefore, $$ \delta H=\int\mathrm d\boldsymbol x\ \delta\mathscr H(\phi(\boldsymbol x),\phi_{,i}(\boldsymbol x),\pi(\boldsymbol x)) $$ where $$ \delta\mathscr H(\phi,\phi_{,i},\pi)=\frac{\partial \mathscr H}{\partial \phi}\delta\phi+\frac{\partial \mathscr H}{\partial \phi_{,i}}\delta\phi_{,i}+\frac{\partial \mathscr H}{\partial\pi}\delta\pi $$

Next, as we want $\frac{\delta H}{\delta \phi}$, we want to leave $\pi$ unchanged, so $\delta\pi=0$ (this is analogous to ordinary partial derivatives: when you calculate $\frac{\partial f(x,y)}{\partial x}$ you want to make a small displacement of $x$, while leaving $y$ unchanged)$\phantom{}^1$.

Anyway, in the integral over $\mathrm d\boldsymbol x$, we can integrate by parts the $\delta \phi_{,i}=\partial_i\delta \phi$ to make the derivative act on $\frac{\partial\mathscr H}{\partial\phi_{,i}}$:

$$ \delta\mathscr H(\phi,\phi_{,i},\pi)=\frac{\partial \mathscr H}{\partial \phi}\delta\phi-\partial_i\left(\frac{\partial \mathscr H}{\partial \phi_{,i}}\right)\delta\phi+\text{surface terms} $$

Finally, back to $H$: $$ \delta H=\int\mathrm d\boldsymbol x\ \left(\frac{\partial \mathscr H}{\partial \phi}-\partial_i\left(\frac{\partial \mathscr H}{\partial \phi_{,i}}\right)\right)\delta\phi $$ where I assumed that surface terms dont contribute. The expression in the parentheses is, by definition, the functional derivative of $H$ w.r.t. $\phi$.

$\phantom{}^1$: If we took $\delta\pi\neq 0$ and $\delta\phi=0$, we would get $\frac{\delta H}{\delta \pi}$ instead. All this is possible because the system is unconstrained, which is not true in some theories (such as the Dirac Lagrangian); in these cases, you can't use Poisson brackets, but Dirac brackets instead.

  • $\begingroup$ $$\{H,\pi\}=\frac{\delta H}{\delta\pi}\frac{\delta \pi}{\delta\phi}-\frac{\delta H}{\delta\phi}\frac{\delta \pi}{\delta\pi}$$. What does it mean exactly $\frac{\delta \pi}{\delta\pi}$? Normally I know that $\frac{\delta\pi(a)}{\delta\pi(x)}=\delta(x-a)$ but it seems it is not the same here? $\endgroup$ Dec 6, 2015 at 20:11
  • $\begingroup$ @Mr.T I believe it should be more clear now. If you have any other doubt, feel free to ask. $\endgroup$ Dec 6, 2015 at 20:27
  • $\begingroup$ So $$\{A,B\}=\frac{\delta A}{\delta\pi}\frac{\delta B}{\delta\phi}-\frac{\delta A}{\delta\phi}\frac{\delta B}{\delta\pi}$$ is not a good definition? We have to add the integral $\int$? $\endgroup$ Dec 6, 2015 at 20:30
  • $\begingroup$ Can you show me clearly what are the functionals $\pi$ and $\phi$? (as opposite to the functions $\pi$ and $\phi$) $\endgroup$ Dec 6, 2015 at 20:33
  • 1
    $\begingroup$ @Mr.T I edited my answer to include a proof of that statement. PS: IMHO you should change the title of your post into something like "Field theory: equivalence between Hamiltonian and Lagrangian formulation" as this is really the topic of your question. This way, it will be useful for other people if they run into the same doubts as you. $\endgroup$ Dec 6, 2015 at 22:33

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