# Finite transformations: correspondence between classical and quantum

Shankar, R. (1994), Principles of Quantum Mechanics, Plenum Press. Chapter 11, Exercise 11.2.3.

The translation operator for an infinitesimal translation ${\varepsilon}$ is, to first order in ${\varepsilon}$:

$\displaystyle T\left(\varepsilon\right)=I-\frac{i\varepsilon}{\hbar}P \ \ \ \ \ (1)$

where ${P}$, the momentum operator, serves as the generator of translations. To derive a formula for a finite (non-infinitesimal) translation over a distance ${a}$, we divide the interval ${a}$ into ${N}$ segments, each of width ${a/N}$, so that for very large ${N}$, the width becomes infinitesimal. Then we have

$\displaystyle T\left(a\right)=\left(I-\frac{ia}{\hbar N}P\right)^{N} \ \ \ \ \ (2)$

This formula is reminiscent of one definition of the exponential function (which can be found in most introductory calculus texts):

$\displaystyle e^{-ax}=\lim_{N\rightarrow\infty}\left(1-\frac{ax}{N}\right)^{N} \ \ \ \ \ (3)$

When we try to apply a formula that is valid for ordinary numbers to a case containing operators, we need to take care that any commutation relations involving the operators are taken into account. In this case, 2 contains only the momentum operator and the identity operator, which commute with each other, so we can in fact apply the limit formula directly to the operator case. We therefore have

$\displaystyle T\left(a\right)=\lim_{N\rightarrow\infty}\left(I-\frac{ia}{\hbar N}P\right)^{N}=e^{-iaP/\hbar} \ \ \ \ \ (4)$

In the position basis, ${P=-i\hbar\frac{d}{dx}}$, so if we apply ${T\left(a\right)}$ to a state vector ${\psi\left(x\right)=\left\langle x\left|\psi\right.\right\rangle }$ we can expand the exponential in a Taylor series to get

$\displaystyle \left\langle x\left|T\left(a\right)\right|\psi\right\rangle =\psi\left(x\right)-a\frac{d\psi}{dx}+\frac{a}{2!}\frac{d^{2}\psi}{dx^{2}}+\ldots \ \ \ \ \ (5)$

We can extend our analysis of the correspondence between classical and quantum versions of translations. In the passive transformation model, the transformation is applied to operators rather than state vectors, so for a finite translation of an operator ${\Omega}$ we have

$\displaystyle \Omega\rightarrow T^{\dagger}\left(a\right)\Omega T\left(a\right)=e^{iaP/\hbar}\Omega e^{-iaP/\hbar} \ \ \ \ \ (6)$

The operator expression on the RHS can be expanded using Hadamard’s lemma, which for two operators ${A}$ and ${B}$ is

$\displaystyle e^{-A}Be^{A}=B+\left[B,A\right]+\frac{1}{2!}\left[\left[B,A\right],A\right]+\ldots \ \ \ \ \ (7)$

where each term contains the commutator of the previous term’s commutator with ${A}$.

In this case gives us

$\displaystyle e^{iaP/\hbar}\Omega e^{-iaP/\hbar}=\Omega+a\left(-\frac{i}{\hbar}\right)\left[\Omega,P\right]+\frac{a^{2}}{2!}\left(-\frac{i}{\hbar}\right)^{2}\left[\left[\Omega,P\right],P\right]+\ldots \ \ \ \ \ (8)$

For example, in the case ${\Omega=X}$, ${\left[X,P\right]=i\hbar I}$ and all higher commutators are zero (since they involve the commutator of a constant with ${P}$), so we get

$\displaystyle e^{iaP/\hbar}Xe^{-iaP/\hbar}=X+aI \ \ \ \ \ (9)$

so the system is translated by a distance ${a}$, as we’d expect.

For higher powers of ${X}$, we can use the result

$\displaystyle \left[X^{n},P\right]=i\hbar nX^{n-1} \ \ \ \ \ (10)$

We therefore get

 $\displaystyle e^{iaP/\hbar}X^{n}e^{-iaP/\hbar}$ $\displaystyle =$ $\displaystyle X^{n}+anX^{n-1}+\frac{a^{2}}{2!}n\left(n-1\right)X^{n-2}+\ldots+\frac{a^{n}}{n!}n!I\ \ \ \ \ (11)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \sum_{m=0}^{n}\binom{n}{m}X^{n-m}\left(aI\right)^{m}\ \ \ \ \ (12)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left(X+aI\right)^{n} \ \ \ \ \ (13)$

We’re allowed to treat ${X}$ as an ordinary number in these equations since it is (apart from ${I}$), the only operator present so all terms commute.

In the classical case, the infinitesimal change ${\delta\omega}$ of a variable ${\omega}$ under an infinitesimal displacement ${\delta a}$ generated by the momentum ${p}$ is given by the Poisson bracket

$\displaystyle \delta\omega=\delta a\left\{ \omega,p\right\} \ \ \ \ \ (14)$

We can write this as a derivative:

$\displaystyle \frac{d\omega}{da}=\left\{ \omega,p\right\} \ \ \ \ \ (15)$

For a finite translation by an amount ${a}$, we can write the value of ${\omega}$ as a Taylor series relative to some starting point ${a_{0}}$ as

$\displaystyle \omega\left(a_{0}+a\right)=\omega\left(a_{0}\right)+a\frac{d\omega}{da}+\frac{a^{2}}{2!}\frac{d^{2}\omega}{da^{2}}+\ldots \ \ \ \ \ (16)$

where all derivatives are evaluated at ${a=a_{0}}$.

We can write all the derivatives in terms of Poisson brackets by using 15. For example

$\displaystyle \frac{d^{2}\omega}{da^{2}}=\frac{d}{da}\left(\frac{d\omega}{da}\right)=\left\{ \frac{d\omega}{da},p\right\} =\left\{ \left\{ \omega,p\right\} ,p\right\} \ \ \ \ \ (17)$

Thus the variable ${\omega}$ transforms according to

$\displaystyle \omega\left(a_{0}+a\right)=\omega+a\left\{ \omega,p\right\} +\frac{a^{2}}{2!}\left\{ \left\{ \omega,p\right\} ,p\right\} +\ldots \ \ \ \ \ (18)$

Comparing this with 8, we see that the two expressions match if we use the usual recipe for converting classical Poisson brackets to quantum commutators, namely ${\left\{ a,b\right\} =-\frac{i}{\hbar}\left[A,B\right]}$.

Although we’ve worked this out for the special case of translations, the same principle can be used for other transformations. For example, the angular momentum about the ${z}$ axis is

$\displaystyle \ell_{z}=xp_{y}-yp_{x} \ \ \ \ \ (19)$

and serves as the generator of rotations about the ${z}$ axis. Suppose we have a rotation through an angle ${\theta}$ and we want to see how the two coordinates ${x}$ and ${y}$ transform. The expansion 18 becomes

$\displaystyle \bar{x}=x+\theta\left\{ x,\ell_{z}\right\} +\frac{\theta^{2}}{2!}\left\{ \left\{ x,\ell_{z}\right\} ,\ell_{z}\right\} +\ldots \ \ \ \ \ (20)$

The relevant Poisson brackets are (using the generic term ${q_{i}}$ to represent the two coordinates ${x}$ and ${y}$):

 $\displaystyle \left\{ x,\ell_{z}\right\}$ $\displaystyle =$ $\displaystyle \sum_{i}\left[\frac{\partial x}{\partial q_{i}}\frac{\partial\ell_{z}}{\partial p_{i}}-\frac{\partial x}{\partial p_{i}}\frac{\partial\ell_{z}}{\partial q_{i}}\right]\ \ \ \ \ (21)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -y\ \ \ \ \ (22)$ $\displaystyle \left\{ y,\ell_{z}\right\}$ $\displaystyle =$ $\displaystyle \sum_{i}\left[\frac{\partial y}{\partial q_{i}}\frac{\partial\ell_{z}}{\partial p_{i}}-\frac{\partial y}{\partial p_{i}}\frac{\partial\ell_{z}}{\partial q_{i}}\right]\ \ \ \ \ (23)$ $\displaystyle$ $\displaystyle =$ $\displaystyle x \ \ \ \ \ (24)$

Looking at how ${x}$ transforms, we see that the Poisson brackets in 20 will cycle through the four values

 $\displaystyle \left\{ x,\ell_{z}\right\}$ $\displaystyle =$ $\displaystyle -y\ \ \ \ \ (25)$ $\displaystyle \left\{ \left\{ x,\ell_{z}\right\} ,\ell_{z}\right\}$ $\displaystyle =$ $\displaystyle -\left\{ y,\ell_{z}\right\} =-x\ \ \ \ \ (26)$ $\displaystyle \left\{ \left\{ \left\{ x,\ell_{z}\right\} ,\ell_{z}\right\} ,\ell_{z}\right\}$ $\displaystyle =$ $\displaystyle -\left\{ x,\ell_{z}\right\} =y\ \ \ \ \ (27)$ $\displaystyle \left\{ \left\{ \left\{ \left\{ x,\ell_{z}\right\} ,\ell_{z}\right\} ,\ell_{z}\right\} ,\ell_{z}\right\}$ $\displaystyle =$ $\displaystyle \left\{ y,\ell_{z}\right\} =x \ \ \ \ \ (28)$

The series 20 thus expands to

 $\displaystyle \bar{x}$ $\displaystyle =$ $\displaystyle x\left[1-\frac{\theta^{2}}{2!}+\frac{\theta^{4}}{4!}-\ldots\right]-y\left[\theta-\frac{\theta^{3}}{3!}+\frac{\theta^{5}}{5!}-\ldots\right]\ \ \ \ \ (29)$ $\displaystyle$ $\displaystyle =$ $\displaystyle x\cos\theta-y\sin\theta \ \ \ \ \ (30)$

We can do the same calculation for ${\bar{y}}$ to get

$\displaystyle \bar{y}=x\sin\theta+y\cos\theta \ \ \ \ \ (31)$

# Angular momentum – Poisson bracket to commutator

Shankar, R. (1994), Principles of Quantum Mechanics, Plenum Press. Section 7.4, Exercise 7.4.8.

The classical angular momentum components are

 $\displaystyle \ell_{x}$ $\displaystyle =$ $\displaystyle yp_{z}-zp_{y}\ \ \ \ \ (1)$ $\displaystyle \ell_{y}$ $\displaystyle =$ $\displaystyle zp_{x}-xp_{z}\ \ \ \ \ (2)$ $\displaystyle \ell_{z}$ $\displaystyle =$ $\displaystyle xp_{y}-yp_{x} \ \ \ \ \ (3)$

In the position basis, we can replace each coordinate by its quantum operator ${x\rightarrow X}$, ${y\rightarrow Y}$ and ${z\rightarrow Z}$, and each momentum component by the derivative ${p_{i}\rightarrow-i\hbar\partial/\partial q_{i}}$, where ${q_{i}}$ is the ${i}$th coordinate. This gives

 $\displaystyle L_{x}$ $\displaystyle =$ $\displaystyle -i\hbar\left(y\frac{\partial}{\partial z}-z\frac{\partial}{\partial y}\right)\ \ \ \ \ (4)$ $\displaystyle L_{y}$ $\displaystyle =$ $\displaystyle -i\hbar\left(z\frac{\partial}{\partial x}-x\frac{\partial}{\partial z}\right)\ \ \ \ \ (5)$ $\displaystyle L_{z}$ $\displaystyle =$ $\displaystyle -i\hbar\left(x\frac{\partial}{\partial y}-y\frac{\partial}{\partial x}\right) \ \ \ \ \ (6)$

Because coordinates always commute with momentum components of other coordinates (${x}$ commutes with ${p_{y}}$ and ${p_{z}}$, etc), there is no ordering ambiguity in making the transition from classical to quantum mechanics. That is, we could place the coordinate on either side of the momentum in each term for all components ${L_{i}}$.

Classically, we can calculate the Poisson brackets for the angular momentum components. For example

 $\displaystyle \left\{ \ell_{x},\ell_{y}\right\}$ $\displaystyle =$ $\displaystyle \sum_{i}\left(\frac{\partial\ell_{x}}{\partial q_{i}}\frac{\partial\ell_{y}}{\partial p_{i}}-\frac{\partial\ell_{x}}{\partial p_{i}}\frac{\partial\ell_{y}}{\partial q_{i}}\right)\ \ \ \ \ (7)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -p_{y}\left(-x\right)-yp_{z}\ \ \ \ \ (8)$ $\displaystyle$ $\displaystyle =$ $\displaystyle xp_{y}-yp_{z}\ \ \ \ \ (9)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \ell_{z} \ \ \ \ \ (10)$

According to the rule for converting classical Poisson brackets to quantum commutators, we should get (since there is no ordering ambiguity)

$\displaystyle \left[L_{x},L_{y}\right]=i\hbar L_{z} \ \ \ \ \ (11)$

As we’ve seen earlier, this is verified by direct calculation using the position-momentum commutator

$\displaystyle \left[q_{i},p_{j}\right]=i\hbar\delta_{ij} \ \ \ \ \ (12)$

# Poisson brackets to commutators: classical to quantum

Shankar, R. (1994), Principles of Quantum Mechanics, Plenum Press. Section 7.4, Exercise 7.4.7.

The postulates of quantum mechanics that we described earlier included specifications for the matrix elements of position ${X}$ and momentum ${P}$ in position space:

 $\displaystyle \left\langle x\left|X\right|x^{\prime}\right\rangle$ $\displaystyle =$ $\displaystyle x\delta\left(x-x^{\prime}\right)\ \ \ \ \ (1)$ $\displaystyle \left\langle x\left|P\right|x^{\prime}\right\rangle$ $\displaystyle =$ $\displaystyle -i\hbar\delta^{\prime}\left(x-x^{\prime}\right) \ \ \ \ \ (2)$

A more fundamental form of this postulate is to specify the commutation relation between ${X}$ and ${P}$, which is independent of the basis and is

$\displaystyle \left[X,P\right]=i\hbar \ \ \ \ \ (3)$

This allows the construction of explicit forms of the operators in other bases, such as the momentum basis, where

 $\displaystyle X$ $\displaystyle =$ $\displaystyle i\hbar\frac{d}{dp}\ \ \ \ \ (4)$ $\displaystyle P$ $\displaystyle =$ $\displaystyle p \ \ \ \ \ (5)$

We can verify this by calculating the commutator by applying it to a function ${f\left(p\right)}$:

 $\displaystyle \left[X,P\right]f$ $\displaystyle =$ $\displaystyle i\hbar\frac{d}{dp}\left(pf\left(p\right)\right)-i\hbar p\frac{d}{dp}f\left(p\right)\ \ \ \ \ (6)$ $\displaystyle$ $\displaystyle =$ $\displaystyle i\hbar f\left(p\right)+i\hbar p\frac{d}{dp}f\left(p\right)-i\hbar p\frac{d}{dp}f\left(p\right)\ \ \ \ \ (7)$ $\displaystyle$ $\displaystyle =$ $\displaystyle i\hbar f\left(p\right) \ \ \ \ \ (8)$

Thus 3 is satisfied in the momentum basis as well.

The standard recipe for converting a classical system to a quantum one is to first calculate the Poisson bracket for two physical quantities in the classical system, which gives

$\displaystyle \left\{ \omega,\lambda\right\} =\sum_{i}\left(\frac{\partial\omega}{\partial q_{i}}\frac{\partial\lambda}{\partial p_{i}}-\frac{\partial\omega}{\partial p_{i}}\frac{\partial\lambda}{\partial q_{i}}\right) \ \ \ \ \ (9)$

where ${q_{i}}$ and ${p_{i}}$ are the canonical coordinates and momenta. To convert to a quantum commutator, we replace the classical quantities by their quantum operator equivalents and the Poisson bracket by ${i\hbar}$ times the corresponding commutator. That is

$\displaystyle \left[\Omega,\Lambda\right]=i\hbar\left\{ \omega,\lambda\right\} \ \ \ \ \ (10)$

For the case of ${X}$ and ${P}$, we have, in classical mechanics in one dimension

$\displaystyle \left\{ x,p\right\} =\frac{\partial x}{\partial x}\frac{\partial p}{\partial p}-\frac{\partial x}{\partial p}\frac{\partial p}{\partial x}=1 \ \ \ \ \ (11)$

so the quantum commutator is given by 3.

For other quantities, we can use the theorems on the Poisson brackets to reduce them:

 $\displaystyle \left\{ \omega,\lambda\right\}$ $\displaystyle =$ $\displaystyle -\left\{ \lambda,\omega\right\} \ \ \ \ \ (12)$ $\displaystyle \left\{ \omega,\lambda+\sigma\right\}$ $\displaystyle =$ $\displaystyle \left\{ \omega,\lambda\right\} +\left\{ \omega,\sigma\right\} \ \ \ \ \ (13)$ $\displaystyle \left\{ \omega,\lambda\sigma\right\}$ $\displaystyle =$ $\displaystyle \left\{ \omega,\lambda\right\} \sigma+\left\{ \omega,\sigma\right\} \lambda \ \ \ \ \ (14)$

Quantum commutators obey similar rules

 $\displaystyle \left[\Omega,\Lambda\right]$ $\displaystyle =$ $\displaystyle -\left[\Lambda,\Omega\right]\ \ \ \ \ (15)$ $\displaystyle \left[\Omega,\Lambda+\Gamma\right]$ $\displaystyle =$ $\displaystyle \left[\Omega,\Lambda\right]+\left[\Omega,\Gamma\right]\ \ \ \ \ (16)$ $\displaystyle \left[\Omega\Lambda,\Gamma\right]$ $\displaystyle =$ $\displaystyle \Omega\left[\Lambda,\Gamma\right]+\left[\Omega,\Gamma\right]\Lambda \ \ \ \ \ (17)$

The main difference between Poisson brackets and commutators is that, for the latter, the order of the operators in the last equation can make a difference. That is, in 14 we could also have written

$\displaystyle \left\{ \omega,\lambda\sigma\right\} =\sigma\left\{ \omega,\lambda\right\} +\lambda\left\{ \omega,\sigma\right\} \ \ \ \ \ (18)$

since all three quantities are numerical (not operators), so multiplication commutes. In 17 it is not true in general that, for example

$\displaystyle \Omega\left[\Lambda,\Gamma\right]+\left[\Omega,\Gamma\right]\Lambda=\left[\Lambda,\Gamma\right]\Omega+\left[\Omega,\Gamma\right]\Lambda \ \ \ \ \ (19)$

The conversion from classical to quantum mechanics can then be achieved in general by replacing

$\displaystyle \left\{ \omega\left(x,p\right),\lambda\left(x,p\right)\right\} =\gamma\left(x,p\right) \ \ \ \ \ (20)$

by

$\displaystyle \left[\Omega\left(X,P\right),\Lambda\left(X,P\right)\right]=i\hbar\Gamma\left(X,P\right) \ \ \ \ \ (21)$

where each of the operators in the last equation is obtained by replacing ${x}$ in the first equation by ${X}$ and ${p}$ by ${P}$. We do need to be careful with the ordering of the operators in the quantum version, however.

As an example, suppose we have

 $\displaystyle \Omega$ $\displaystyle =$ $\displaystyle X\ \ \ \ \ (22)$ $\displaystyle \Lambda$ $\displaystyle =$ $\displaystyle X^{2}+P^{2} \ \ \ \ \ (23)$

In the classical version, we calculate the Poisson bracket

 $\displaystyle \left\{ \omega,\lambda\right\}$ $\displaystyle =$ $\displaystyle \left\{ x,x^{2}+p^{2}\right\} \ \ \ \ \ (24)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left\{ x,x^{2}\right\} +\left\{ x,p^{2}\right\} \ \ \ \ \ (25)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 0+2\left\{ x,p\right\} p\ \ \ \ \ (26)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 2p \ \ \ \ \ (27)$

Thus, by our rule above, the quantum version should be

$\displaystyle \left[\Omega,\Lambda\right]=2i\hbar P \ \ \ \ \ (28)$

We can verify this using 17

 $\displaystyle \left[X,X^{2}+P^{2}\right]$ $\displaystyle =$ $\displaystyle \left[X,X^{2}\right]+\left[X,P^{2}\right]\ \ \ \ \ (29)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 0-\left[P^{2},X\right]\ \ \ \ \ (30)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -P\left[P,X\right]-\left[P,X\right]P\ \ \ \ \ (31)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -P\left(-i\hbar\right)-\left(-i\hbar\right)P\ \ \ \ \ (32)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 2i\hbar P \ \ \ \ \ (33)$

In this case, there is no ordering ambiguity in the quantum version, since ${\left[X,P\right]=i\hbar}$ is just a number.

For a second example, suppose we have

 $\displaystyle \Omega$ $\displaystyle =$ $\displaystyle X^{2}\ \ \ \ \ (34)$ $\displaystyle \Lambda$ $\displaystyle =$ $\displaystyle P^{2} \ \ \ \ \ (35)$

The classical version gives us, using the relations 14, 11 and 27

 $\displaystyle \left\{ x^{2},p^{2}\right\}$ $\displaystyle =$ $\displaystyle -\left\{ p^{2},x^{2}\right\} \ \ \ \ \ (36)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -2\left\{ p^{2},x\right\} x\ \ \ \ \ (37)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 2\left\{ x,p^{2}\right\} x\ \ \ \ \ (38)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 4px \ \ \ \ \ (39)$

In the classical case, this result is the same as ${4xp}$, but because ${X}$ and ${P}$ don’t commute in the quantum form, we need to be careful about the ordering.

We can do the calculation:

$\displaystyle \left[X^{2},P^{2}\right]=X\left[X,P^{2}\right]+\left[X,P^{2}\right]X \ \ \ \ \ (40)$

From 33 we have

$\displaystyle \left[X,P^{2}\right]=2i\hbar P \ \ \ \ \ (41)$

so we get

$\displaystyle \left[X^{2},P^{2}\right]=2i\hbar\left(XP+PX\right) \ \ \ \ \ (42)$

Thus if the Poisson bracket involves a product of ${p}$ and ${x}$, this should be replaced by

$\displaystyle xp\mbox{ or }px\rightarrow\frac{1}{2}\left(XP+PX\right) \ \ \ \ \ (43)$

in the quantum version.

# Passive, regular and active transformations. Invariance of the Hamiltonian and generators of transformations

References: Shankar, R. (1994), Principles of Quantum Mechanics, Plenum Press. Sections 2.7 & 2.8; Exercises 2.8.1 – 2.8.2.

The canonical transformations we’ve considered so far are of the form

 $\displaystyle \overline{q}_{i}$ $\displaystyle =$ $\displaystyle \overline{q}_{i}\left(q,p\right)\ \ \ \ \ (1)$ $\displaystyle \overline{p}_{i}$ $\displaystyle =$ $\displaystyle \overline{p}_{i}\left(q,p\right) \ \ \ \ \ (2)$

The interpretation of these transformations is that we are using a new set of coordinates and momenta to describe the same point in phase space. For example, in 2-d we can describe the point one unit along the ${y}$ axis by the coordinates ${x=0,y=1}$ if we use rectangular coordinates, or by ${r=1,\theta=\frac{\pi}{2}}$ if we use polar coordinates. The numerical values of the coordinates are different in the two systems, but the geometric point being described is the same. Such a transformation is called a passive transformation. In a passive transformation, any function ${\omega}$ always has the same value at a given point in phase space no matter which coordinate system we’re using, so we can say that

$\displaystyle \omega\left(q,p\right)=\omega\left(\bar{q},\bar{p}\right) \ \ \ \ \ (3)$

where it is understood that ${\left(q,p\right)}$ and ${\left(\bar{q},\bar{p}\right)}$ both refer to the same point, but in different representations.

One characteristic of a passive transformation is that the ranges of the variables used to represent a point in phase space need not be the same in the two systems. For example, in 2-d rectangular coordinates, both ${x}$ and ${y}$ can range from ${-\infty}$ to ${+\infty}$, while in polar coordinates ${r}$ ranges between 0 and ${+\infty}$ while the angle ${\theta}$ runs between 0 and ${2\pi}$.

A special type of transformation is a regular transformation, in which the variables in the two systems have the same ranges. For example, if we translate a 2-d system by 1 unit along the ${x}$ axis, the new coordinates are related to the old ones by

 $\displaystyle \bar{x}$ $\displaystyle =$ $\displaystyle x-1\ \ \ \ \ (4)$ $\displaystyle \bar{y}$ $\displaystyle =$ $\displaystyle y \ \ \ \ \ (5)$

Both the original and barred systems have the same range (${-\infty}$ to ${+\infty}$).

Although we can interpret a regular transformation as a passive transformation, we can also think of it in a different way. We can image that instead of just providing a different label for the same point that the transformed coordinate has actually shifted the system to a new location in phase space. In the above example, this would mean that we have physically moved the system by 1 unit along the ${x}$ axis. This interpretation is known as an active transformation.

If a function ${\omega}$ is invariant under an active transformation, then it satisfies the condition

$\displaystyle \omega\left(q,p\right)=\omega\left(\bar{q},\bar{p}\right) \ \ \ \ \ (6)$

Although mathematically this is the same as 3, physically it means something quite different, since now the points ${\left(q,p\right)}$ and ${\left(\bar{q},\bar{p}\right)}$ refer to different points in phase space, so we’re saying that the function ${\omega}$ does not change when we move the physical system in the way specified by the active transformation.

We now restrict ourselves to talking about regular canonical transformations. Consider some dynamical variable (it could be momentum or angular momentum, for example) ${g\left(q,p\right)}$ and suppose we define the transformations

 $\displaystyle \bar{q}_{i}$ $\displaystyle =$ $\displaystyle q_{i}+\epsilon\frac{\partial g}{\partial p_{i}}\equiv q_{i}+\delta q_{i}\ \ \ \ \ (7)$ $\displaystyle \bar{p}_{i}$ $\displaystyle =$ $\displaystyle p_{i}-\epsilon\frac{\partial g}{\partial q_{i}}\equiv p_{i}+\delta p_{i} \ \ \ \ \ (8)$

where ${\epsilon}$ is some infinitesimal quantity.

First, we need to show that, to first order in ${\epsilon}$, this is a canonical transformation. The required conditions for this are

 $\displaystyle \left\{ \overline{q}_{i},\overline{q}_{j}\right\}$ $\displaystyle =$ $\displaystyle \left\{ \overline{p}_{i},\overline{p}_{j}\right\} =0\ \ \ \ \ (9)$ $\displaystyle \left\{ \overline{q}_{i},\overline{p}_{j}\right\}$ $\displaystyle =$ $\displaystyle \delta_{ij} \ \ \ \ \ (10)$

Consider first (we’ll use the summation convention, so the index ${k}$ is summed in what follows):

 $\displaystyle \left\{ \overline{q}_{i},\overline{p}_{j}\right\}$ $\displaystyle =$ $\displaystyle \frac{\partial}{\partial q_{k}}\left(q_{i}+\epsilon\frac{\partial g}{\partial p_{i}}\right)\frac{\partial}{\partial p_{k}}\left(p_{j}-\epsilon\frac{\partial g}{\partial q_{j}}\right)-\nonumber$ $\displaystyle$ $\displaystyle$ $\displaystyle \frac{\partial}{\partial p_{k}}\left(q_{i}+\epsilon\frac{\partial g}{\partial p_{i}}\right)\frac{\partial}{\partial q_{k}}\left(p_{j}-\epsilon\frac{\partial g}{\partial q_{j}}\right)\ \ \ \ \ (11)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left(\delta_{ik}+\epsilon\frac{\partial^{2}g}{\partial p_{i}q_{k}}\right)\left(\delta_{jk}-\epsilon\frac{\partial^{2}g}{\partial p_{k}q_{j}}\right)-\nonumber$ $\displaystyle$ $\displaystyle$ $\displaystyle \left(0+\epsilon\frac{\partial^{2}g}{\partial p_{i}p_{k}}\right)\left(0-\epsilon\frac{\partial^{2}g}{\partial q_{j}q_{k}}\right) \ \ \ \ \ (12)$

The zeroes in the last line follow from the fact that ${q_{k}}$ and ${p_{k}}$ are independent variables. We can now keep terms only up to first order in ${\epsilon}$ to get

 $\displaystyle \left\{ \overline{q}_{i},\overline{p}_{j}\right\}$ $\displaystyle =$ $\displaystyle \delta_{ik}\delta_{jk}+\epsilon\left(\frac{\partial^{2}g}{\partial p_{i}q_{k}}\delta_{jk}-\frac{\partial^{2}g}{\partial p_{k}q_{j}}\delta_{ik}\right)\ \ \ \ \ (13)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \delta_{ij}+\epsilon\left(\frac{\partial^{2}g}{\partial p_{i}q_{j}}-\frac{\partial^{2}g}{\partial p_{i}q_{j}}\right)\ \ \ \ \ (14)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \delta_{ij} \ \ \ \ \ (15)$

The other two brackets work out similarly:

 $\displaystyle \left\{ \overline{q}_{i},\overline{q}_{j}\right\}$ $\displaystyle =$ $\displaystyle \frac{\partial}{\partial q_{k}}\left(q_{i}+\epsilon\frac{\partial g}{\partial p_{i}}\right)\frac{\partial}{\partial p_{k}}\left(q_{j}+\epsilon\frac{\partial g}{\partial p_{j}}\right)-\nonumber$ $\displaystyle$ $\displaystyle$ $\displaystyle \frac{\partial}{\partial p_{k}}\left(q_{i}+\epsilon\frac{\partial g}{\partial p_{i}}\right)\frac{\partial}{\partial q_{k}}\left(q_{j}+\epsilon\frac{\partial g}{\partial p_{j}}\right)\ \ \ \ \ (16)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left(\delta_{ik}+\epsilon\frac{\partial^{2}g}{\partial p_{i}q_{k}}\right)\left(0+\epsilon\frac{\partial^{2}g}{\partial p_{k}p_{j}}\right)-\nonumber$ $\displaystyle$ $\displaystyle$ $\displaystyle \left(0+\epsilon\frac{\partial^{2}g}{\partial p_{i}p_{k}}\right)\left(\delta_{jk}+\epsilon\frac{\partial^{2}g}{\partial p_{j}q_{k}}\right)\ \ \ \ \ (17)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \delta_{ik}\epsilon\frac{\partial^{2}g}{\partial p_{k}p_{j}}-\delta_{jk}\epsilon\frac{\partial^{2}g}{\partial p_{i}p_{k}}\ \ \ \ \ (18)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \epsilon\left(\frac{\partial^{2}g}{\partial p_{i}p_{j}}-\frac{\partial^{2}g}{\partial p_{i}p_{j}}\right)\ \ \ \ \ (19)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 0 \ \ \ \ \ (20)$
 $\displaystyle \left\{ \overline{p}_{i},\overline{p}_{j}\right\}$ $\displaystyle =$ $\displaystyle \frac{\partial}{\partial q_{k}}\left(p_{i}-\epsilon\frac{\partial g}{\partial q_{i}}\right)\frac{\partial}{\partial p_{k}}\left(p_{j}-\epsilon\frac{\partial g}{\partial q_{j}}\right)-\nonumber$ $\displaystyle$ $\displaystyle$ $\displaystyle \frac{\partial}{\partial p_{k}}\left(p_{i}-\epsilon\frac{\partial g}{\partial q_{i}}\right)\frac{\partial}{\partial q_{k}}\left(p_{j}-\epsilon\frac{\partial g}{\partial q_{j}}\right)\ \ \ \ \ (21)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left(0-\epsilon\frac{\partial^{2}g}{\partial q_{i}q_{k}}\right)\left(\delta_{jk}-\epsilon\frac{\partial^{2}g}{\partial p_{k}q_{j}}\right)-\nonumber$ $\displaystyle$ $\displaystyle$ $\displaystyle \left(\delta_{ik}-\epsilon\frac{\partial^{2}g}{\partial q_{i}p_{k}}\right)\left(0-\epsilon\frac{\partial^{2}g}{\partial q_{j}q_{k}}\right)\ \ \ \ \ (22)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\delta_{jk}\epsilon\frac{\partial^{2}g}{\partial q_{i}q_{k}}+\delta_{ik}\epsilon\frac{\partial^{2}g}{\partial q_{k}q_{j}}\ \ \ \ \ (23)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\epsilon\left(\frac{\partial^{2}g}{\partial q_{i}q_{j}}-\frac{\partial^{2}g}{\partial q_{i}q_{j}}\right)\ \ \ \ \ (24)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 0 \ \ \ \ \ (25)$

Thus all the brackets check out, so the transformation is canonical.

The point of all this is that, if the Hamiltonian is invariant under the transformations 7 and 8 then the variable ${g}$ is conserved (that is, doesn’t change with time). ${g}$ is called the generator of the transformation. We can verify this by using the chain rule to calculate the variation in ${H}$:

 $\displaystyle \delta H$ $\displaystyle =$ $\displaystyle \frac{\partial H}{\partial q_{i}}\delta q_{i}+\frac{\partial H}{\partial p_{i}}\delta p_{i}\ \ \ \ \ (26)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \epsilon\left[\frac{\partial H}{\partial q_{i}}\frac{\partial g}{\partial p_{i}}-\frac{\partial H}{\partial p_{i}}\frac{\partial g}{\partial q_{i}}\right]\ \ \ \ \ (27)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \epsilon\left\{ H,g\right\} \ \ \ \ \ (28)$

Since ${H}$ is invariant, we must have ${\delta H=0}$, so

$\displaystyle \left\{ H,g\right\} =0 \ \ \ \ \ (29)$

However, this is the condition for ${g}$ to be conserved. QED.

Example Suppose we have a two particle system moving in one dimension, with positions ${q_{1},q_{2}}$ and momenta ${p_{1},p_{2}}$. If we take

$\displaystyle g=p_{1}+p_{2} \ \ \ \ \ (30)$

we get

 $\displaystyle \delta q_{i}$ $\displaystyle =$ $\displaystyle \epsilon\frac{\partial g}{\partial p_{i}}=\epsilon\ \ \ \ \ (31)$ $\displaystyle \delta p_{i}$ $\displaystyle =$ $\displaystyle -\epsilon\frac{\partial g}{\partial q_{i}}=0 \ \ \ \ \ (32)$

That is, each particle gets shifted by the same amount ${\epsilon}$ but the momentum of each particle remains unchanged. Thus the total momentum is the generator of infinitesimal translations. The physical interpretation of this is that, since the momentum of each particle is conserved, the total kinetic energy

$\displaystyle T=\frac{p_{1}^{2}}{2m_{1}}+\frac{p_{2}^{2}}{2m_{2}} \ \ \ \ \ (33)$

remains unchanged. Since the total energy is invariant, the total potential energy of the system is unaffected by a translation, which means that there is no external force on the system.

# Poisson brackets are invariant under a canonical transformation

References: Shankar, R. (1994), Principles of Quantum Mechanics, Plenum Press. Section 2.7; Exercise 2.7.9.

The Poisson bracket of two functions is defined as

$\displaystyle \left\{ \omega,\sigma\right\} =\sum_{i}\left(\frac{\partial\omega}{\partial q_{i}}\frac{\partial\sigma}{\partial p_{i}}-\frac{\partial\omega}{\partial p_{i}}\frac{\partial\sigma}{\partial q_{i}}\right) \ \ \ \ \ (1)$

Calculating the Poisson bracket requires knowing ${\omega}$ and ${\sigma}$ as functions of the coordinates ${q_{i}}$ and momenta ${p_{i}}$ in the particular coordinate system we’re using. However, we’ve seen that the Euler-Lagrange and Hamilton’s equations are invariant under a canonical transformation and since the Poisson bracket is a fundamental quantity in classical mechanics, in particular because the time derivative of a function ${\omega}$ is the Poisson bracket ${\left\{ \omega,H\right\} }$ with the Hamiltonian, it’s natural to ask how the Poisson bracket of two functions transforms under a canonical transformation.

The simplest way of finding out (although not the most elegant) is to write the canonical transformation as

 $\displaystyle \bar{q}_{i}$ $\displaystyle =$ $\displaystyle \bar{q}_{i}\left(q,p\right)\ \ \ \ \ (2)$ $\displaystyle \bar{p}_{i}$ $\displaystyle =$ $\displaystyle \bar{p}\left(q,p\right) \ \ \ \ \ (3)$

We can then write the Poisson bracket in the new coordinates as

$\displaystyle \left\{ \omega,\sigma\right\} _{\bar{q},\bar{p}}=\sum_{j}\left(\frac{\partial\omega}{\partial\bar{q}_{j}}\frac{\partial\sigma}{\partial\bar{p}_{j}}-\frac{\partial\omega}{\partial\bar{p}_{j}}\frac{\partial\sigma}{\partial\bar{q}_{j}}\right) \ \ \ \ \ (4)$

Assuming the transformation is invertible, we can use the chain rule to calculate the derivatives with respect to the barred coordinates. This gives the following (we’ve used the summation convention in which any index repeated twice in a product is summed; thus in the following, there are implied sums over ${i,j}$ and ${k}$):

 $\displaystyle \left\{ \omega,\sigma\right\} _{\bar{q},\bar{p}}$ $\displaystyle =$ $\displaystyle \left(\frac{\partial\omega}{\partial q_{i}}\frac{\partial q_{i}}{\partial\bar{q}_{j}}+\frac{\partial\omega}{\partial p_{i}}\frac{\partial p_{i}}{\partial\bar{q}_{j}}\right)\left(\frac{\partial\sigma}{\partial q_{k}}\frac{\partial q_{k}}{\partial\bar{p}_{j}}+\frac{\partial\sigma}{\partial p_{k}}\frac{\partial p_{k}}{\partial\bar{p}_{j}}\right)-\nonumber$ $\displaystyle$ $\displaystyle$ $\displaystyle \left(\frac{\partial\omega}{\partial q_{i}}\frac{\partial q_{i}}{\partial\bar{p}_{j}}+\frac{\partial\omega}{\partial p_{i}}\frac{\partial p_{i}}{\partial\bar{p}_{j}}\right)\left(\frac{\partial\sigma}{\partial q_{k}}\frac{\partial q_{k}}{\partial\bar{q}_{j}}+\frac{\partial\sigma}{\partial p_{k}}\frac{\partial p_{k}}{\partial\bar{q}_{j}}\right)\ \ \ \ \ (5)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\partial\omega}{\partial q_{i}}\frac{\partial\sigma}{\partial p_{k}}\left(\frac{\partial q_{i}}{\partial\bar{q}_{j}}\frac{\partial p_{k}}{\partial\bar{p}_{j}}-\frac{\partial q_{i}}{\partial\bar{p}_{j}}\frac{\partial p_{k}}{\partial\bar{q}_{j}}\right)+\nonumber$ $\displaystyle$ $\displaystyle$ $\displaystyle \frac{\partial\omega}{\partial p_{i}}\frac{\partial\sigma}{\partial q_{k}}\left(\frac{\partial p_{i}}{\partial\bar{q}_{j}}\frac{\partial q_{k}}{\partial\bar{p}_{j}}-\frac{\partial p_{i}}{\partial\bar{p}_{j}}\frac{\partial q_{k}}{\partial\bar{q}_{j}}\right)+\nonumber$ $\displaystyle$ $\displaystyle$ $\displaystyle \frac{\partial\omega}{\partial q_{i}}\frac{\partial\sigma}{\partial q_{k}}\left(\frac{\partial q_{i}}{\partial\bar{q}_{j}}\frac{\partial q_{k}}{\partial\bar{p}_{j}}-\frac{\partial q_{i}}{\partial\bar{p}_{j}}\frac{\partial q_{k}}{\partial\bar{q}_{j}}\right)+\nonumber$ $\displaystyle$ $\displaystyle$ $\displaystyle \frac{\partial\omega}{\partial p_{i}}\frac{\partial\sigma}{\partial p_{k}}\left(\frac{\partial p_{i}}{\partial\bar{q}_{j}}\frac{\partial p_{k}}{\partial\bar{p}_{j}}-\frac{\partial p_{i}}{\partial\bar{p}_{j}}\frac{\partial p_{k}}{\partial\bar{q}_{j}}\right)\ \ \ \ \ (6)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\partial\omega}{\partial q_{i}}\frac{\partial\sigma}{\partial p_{k}}\left\{ q_{i},p_{k}\right\} +\frac{\partial\omega}{\partial p_{i}}\frac{\partial\sigma}{\partial q_{k}}\left\{ p_{i},q_{k}\right\} +\nonumber$ $\displaystyle$ $\displaystyle$ $\displaystyle \frac{\partial\omega}{\partial q_{i}}\frac{\partial\sigma}{\partial q_{k}}\left\{ q_{i},q_{k}\right\} +\frac{\partial\omega}{\partial p_{i}}\frac{\partial\sigma}{\partial p_{k}}\left\{ p_{i},p_{k}\right\} \ \ \ \ \ (7)$

For a canonical transformation, the Poisson brackets in the last equation satisfy

 $\displaystyle \left\{ q_{i},p_{k}\right\}$ $\displaystyle =$ $\displaystyle -\left\{ p_{i},q_{k}\right\} =\delta_{ik}\ \ \ \ \ (8)$ $\displaystyle \left\{ q_{i},q_{k}\right\}$ $\displaystyle =$ $\displaystyle \left\{ p_{i},p_{k}\right\} =0 \ \ \ \ \ (9)$

[Actually, we had worked out these conditions for the barred coordinates in terms of the original coordinates, but since the transformation is invertible and both sets of coordinates are canonical, the Poisson brackets work either way.] Applying these conditions to the above, we find

 $\displaystyle \left\{ \omega,\sigma\right\} _{\bar{q},\bar{p}}$ $\displaystyle =$ $\displaystyle \left(\frac{\partial\omega}{\partial q_{i}}\frac{\partial\sigma}{\partial p_{k}}-\frac{\partial\omega}{\partial p_{i}}\frac{\partial\sigma}{\partial p_{k}}\right)\delta_{ik}\ \ \ \ \ (10)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\partial\omega}{\partial q_{i}}\frac{\partial\sigma}{\partial p_{i}}-\frac{\partial\omega}{\partial p_{i}}\frac{\partial\sigma}{\partial p_{i}}\ \ \ \ \ (11)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left\{ \omega,\sigma\right\} _{q,p} \ \ \ \ \ (12)$

Thus the Poisson bracket is invariant under a canonical transformation.

# Canonical transformations: a few more examples

References: Shankar, R. (1994), Principles of Quantum Mechanics, Plenum Press. Section 2.7; Exercises 02.07.06 – 02.07.07, 02.07.08(4).

Here are a few more examples of canonical variable transformations.

Example 1 First, we revisit the two-body problem, in which we simplified the problem by transforming from the coordinates ${\mathbf{r}_{1}}$ and ${\mathbf{r}_{2}}$ of the masses ${m_{1}}$ and ${m_{2}}$ to two new position vectors:

 $\displaystyle \mathbf{r}$ $\displaystyle \equiv$ $\displaystyle \mathbf{r}_{1}-\mathbf{r}_{2}\ \ \ \ \ (1)$ $\displaystyle \mathbf{r}_{CM}$ $\displaystyle \equiv$ $\displaystyle \frac{m_{1}\mathbf{r}_{1}+m_{2}\mathbf{r}_{2}}{M} \ \ \ \ \ (2)$

Here ${M\equiv m_{1}+m_{2}}$ is the total mass, ${\mathbf{r}}$ is the relative position, and ${\mathbf{r}_{CM}}$ is the position of the centre of mass. The conjugate momenta in the original system are

$\displaystyle \mathbf{p}_{i}=m\dot{\mathbf{r}}_{i} \ \ \ \ \ (3)$

The conjugate momenta transform according to

 $\displaystyle \mathbf{p}_{CM}$ $\displaystyle =$ $\displaystyle M\mathbf{r}_{CM}=\mathbf{p}_{1}+\mathbf{p}_{2}\ \ \ \ \ (4)$ $\displaystyle \mathbf{p}$ $\displaystyle =$ $\displaystyle \mu\dot{\mathbf{r}}\ \ \ \ \ (5)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{m_{2}\mathbf{p}_{1}-m_{1}\mathbf{p}_{2}}{M} \ \ \ \ \ (6)$

where ${\mu=m_{1}m_{2}/M}$ is the reduced mass.

To check that this is a canonical transformation, we need to calculate the Poisson brackets. To make things easier, note that the new coordinates depend only on the old coordinates (and not on the momenta), and conversely, the new momenta depend only on the old momenta (and not on the coordinates). Since the Poisson brackets ${\left\{ \overline{q}_{i},\overline{q}_{j}\right\} }$and ${\left\{ \overline{p}_{i},\overline{p}_{j}\right\} }$all involve taking derivatives of coordinates with respect to momenta (in the first case) or momenta with respect to coordinates (in the second case), all these brackets are zero. We need, therefore, to check only the mixed brackets between coordinates and momenta.

Because we’re dealing with 3-d vector equations, there are 3 components to each vector and to be thorough, we need to calculate all possible brackets between all pairs of components. However, if we do the ${x}$ component of each, it should be obvious that the ${y}$ and ${z}$ components behave in the same way.

First, consider

$\displaystyle \left\{ r_{x},p_{x}\right\} =\sum_{i}\left(\frac{\partial r_{x}}{\partial q_{i}}\frac{\partial p_{x}}{\partial p_{i}}-\frac{\partial r_{x}}{\partial p_{i}}\frac{\partial p_{x}}{\partial q_{i}}\right) \ \ \ \ \ (7)$

In the RHS, the term ${q_{i}}$ stands for all 6 components of the original position vectors, that is ${q_{i}=\left\{ r_{1x},r_{1y},\ldots,r_{2z}\right\} }$ and the term ${p_{i}}$ in the denominators refers to all 6 components of the original momentum vectors. The ${p_{x}}$ in the numerators refers to the ${x}$ component of ${\mathbf{p}}$ in 6. Hopefully this won’t cause too much confusion.

The second term on the RHS is zero because it involves derivatives of coordinates with respect to momenta (and vice versa). In the first term, ${r_{x}}$ depends only the ${x}$ components of ${\mathbf{r}_{1}}$ and ${\mathbf{r}_{2}}$, and ${p_{x}}$ depends only on the ${x}$ components of ${\mathbf{p}_{1}}$and ${\mathbf{p}_{2}}$, so we have

 $\displaystyle \left\{ r_{x},p_{x}\right\}$ $\displaystyle =$ $\displaystyle \frac{\partial r_{x}}{\partial r_{1x}}\frac{\partial p_{x}}{\partial p_{1x}}+\frac{\partial r_{x}}{\partial r_{2x}}\frac{\partial p_{x}}{\partial p_{2x}}\ \ \ \ \ (8)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left(1\right)\frac{m_{2}}{M}+\left(-1\right)\left(-\frac{m_{1}}{M}\right)\ \ \ \ \ (9)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{m_{1}+m_{2}}{M}\ \ \ \ \ (10)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 1 \ \ \ \ \ (11)$

The same result is obtained for the ${y}$ and ${z}$ components. If we look at mixing two different components, we have, for example

$\displaystyle \left\{ r_{x},p_{y}\right\} =\frac{\partial r_{x}}{\partial r_{1x}}\frac{\partial p_{y}}{\partial p_{1x}}+\frac{\partial r_{x}}{\partial r_{2x}}\frac{\partial p_{y}}{\partial p_{2x}}+\frac{\partial r_{x}}{\partial r_{1y}}\frac{\partial p_{y}}{\partial p_{1y}}+\frac{\partial r_{x}}{\partial r_{2y}}\frac{\partial p_{y}}{\partial p_{2y}}=0 \ \ \ \ \ (12)$

This is zero because each term in the sum contains a derivative of an ${x}$ component with respect to a ${y}$ component (or vice versa), all of which are zero.

For the centre of mass components, we have

 $\displaystyle \left\{ r_{CMx},p_{CMx}\right\}$ $\displaystyle =$ $\displaystyle \frac{\partial r_{CMx}}{\partial r_{1x}}\frac{\partial p_{CMx}}{\partial p_{1x}}+\frac{\partial r_{CMx}}{\partial r_{2x}}\frac{\partial p_{CMx}}{\partial p_{2x}}\ \ \ \ \ (13)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{m_{1}}{M}\left(1\right)+\frac{m_{2}}{M}\left(1\right)\ \ \ \ \ (14)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 1\ \ \ \ \ (15)$ $\displaystyle \left\{ r_{CMx},p_{CMy}\right\}$ $\displaystyle =$ $\displaystyle \frac{\partial r_{CMx}}{\partial r_{1x}}\frac{\partial p_{CMy}}{\partial p_{1x}}+\frac{\partial r_{CMx}}{\partial r_{2x}}\frac{\partial p_{CMy}}{\partial p_{2x}}+\frac{\partial r_{CMx}}{\partial r_{1y}}\frac{\partial p_{CMy}}{\partial p_{1y}}+\frac{\partial r_{CMx}}{\partial r_{2y}}\frac{\partial p_{CMy}}{\partial p_{2y}}\ \ \ \ \ (16)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 0 \ \ \ \ \ (17)$

where the last bracket is zero for the same reason as ${\left\{ r_{x},p_{y}\right\} }$: we’re mixing ${x}$ and ${y}$ in the derivatives. Again, it should be obvious that the brackets for the other combinations of ${x}$, ${y}$ and ${z}$ components work out the same way.

Example 2 A bizarre transformation of variables in one dimension is given by

 $\displaystyle \overline{q}$ $\displaystyle =$ $\displaystyle \ln\frac{\sin p}{q}=\ln\sin p-\ln q\ \ \ \ \ (18)$ $\displaystyle \overline{p}$ $\displaystyle =$ $\displaystyle q\cot p \ \ \ \ \ (19)$

To show this is canonical, we need calculate only ${\left\{ \overline{q},\overline{p}\right\} }$ (since the Poisson bracket of a function with itself is always zero, we have ${\left\{ \overline{q},\overline{q}\right\} =\left\{ \overline{p},\overline{p}\right\} =0}$). We need one rather obscure derivative of a trig function.

 $\displaystyle \frac{d}{dp}\cot p$ $\displaystyle =$ $\displaystyle \frac{d}{dp}\left(\frac{\cos p}{\sin p}\right)\ \ \ \ \ (20)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{-\sin^{2}p-\cos^{2}p}{\sin^{2}p}\ \ \ \ \ (21)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -1-\cot^{2}p \ \ \ \ \ (22)$

We get

 $\displaystyle \left\{ \overline{q},\overline{p}\right\}$ $\displaystyle =$ $\displaystyle \frac{\partial\overline{q}}{\partial q}\frac{\partial\overline{p}}{\partial p}-\frac{\partial\overline{q}}{\partial p}\frac{\partial\overline{p}}{\partial q}\ \ \ \ \ (23)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left(-\frac{1}{q}\right)\left(q\left(-1-\cot^{2}p\right)\right)-\frac{\cos p}{\sin p}\cot p\ \ \ \ \ (24)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 1+\cot^{2}p-\cot^{2}p\ \ \ \ \ (25)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 1 \ \ \ \ \ (26)$

Thus the transformation is canonical.

Example 3 Finally, we return to the point transformation, which is given in general by

 $\displaystyle \overline{q}_{i}$ $\displaystyle =$ $\displaystyle \overline{q}_{i}\left(q_{1},\ldots,q_{n}\right)\ \ \ \ \ (27)$ $\displaystyle \overline{p}_{i}$ $\displaystyle =$ $\displaystyle \sum_{j}\frac{\partial q_{j}}{\partial\overline{q}_{i}}p_{j} \ \ \ \ \ (28)$

In this case, the coordinate transformation to ${\overline{q}}$ is completely arbitrary, but the momentum transformation must follow the formula given. The derivatives ${\frac{\partial q_{i}}{\partial\overline{q}_{j}}}$ in the formula for ${\overline{p}_{i}}$ are taken at constant ${\overline{q}}$. As in the earlier examples, since the coordinate formulas depend only on the old coordinates, and the momentum formulas depend only on the old momenta, the Poisson brackets satisfy

$\displaystyle \left\{ \overline{q}_{i},\overline{q}_{j}\right\} =\left\{ \overline{p}_{i},\overline{p}_{j}\right\} =0 \ \ \ \ \ (29)$

For the mixed brackets, we have

 $\displaystyle \left\{ \overline{q}_{i},\overline{p}_{j}\right\}$ $\displaystyle =$ $\displaystyle \sum_{k}\left(\frac{\partial\overline{q}_{i}}{\partial q_{k}}\frac{\partial\overline{p}_{j}}{\partial p_{k}}-\frac{\partial\overline{q}_{i}}{\partial p_{k}}\frac{\partial\overline{p}_{j}}{\partial q_{k}}\right)\ \ \ \ \ (30)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \sum_{k}\frac{\partial\overline{q}_{i}}{\partial q_{k}}\frac{\partial q_{k}}{\partial\overline{q}_{j}}\ \ \ \ \ (31)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\partial\overline{q}_{i}}{\partial\overline{q}_{j}}\ \ \ \ \ (32)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \delta_{ij} \ \ \ \ \ (33)$

The second term in the first line is zero (mixed derivatives again). We used 28 to calculate the derivative ${\frac{\partial\overline{p}_{j}}{\partial p_{k}}}$ and get the second line and then notice that the sum is an expansion of the chain rule for the derivative in line 3. Since ${\overline{q}_{i}}$ and ${\overline{q}_{j}}$ are independent variables, the result is that given in the last line. Thus a point transformation is a canonical transformation.

# Conditions for a transformation to be canonical

References: Shankar, R. (1994), Principles of Quantum Mechanics, Plenum Press. Section 2.7; Exercise 2.7.3.

We’ve seen that the Euler-Lagrange equations are invariant under canonical transformations, but in the Hamiltonian formalism where the system moves in a ${2n}$-dimensional phase space with ${n}$ coordinates ${q}$ and ${n}$ momenta ${p}$, more general transformations are possible:

 $\displaystyle \overline{q}_{i}$ $\displaystyle =$ $\displaystyle \overline{q}_{i}\left(q,p\right)\ \ \ \ \ (1)$ $\displaystyle \overline{p}_{i}$ $\displaystyle =$ $\displaystyle \overline{p}_{i}\left(q,p\right) \ \ \ \ \ (2)$

In order for such a transformation to be canonical, we require that the new variables ${\overline{q}}$ and ${\overline{p}}$ satisfy Hamilton’s equations, that is

 $\displaystyle \frac{\partial H}{\partial\overline{p}_{i}}$ $\displaystyle =$ $\displaystyle \dot{\overline{q}}_{i}\ \ \ \ \ (3)$ $\displaystyle -\frac{\partial H}{\partial\overline{q}_{i}}$ $\displaystyle =$ $\displaystyle \dot{\overline{p}}_{i} \ \ \ \ \ (4)$

In principle, then, we could check the Hamiltonian in the new coordinates to see if these equations are valid, but it would seem that whether or not a set of coordinates and momenta is canonical should be determinable from the variables themselves, and not depend on the specific Hamiltonian. Here we derive a set of conditions on the ${\overline{q}}$ and ${\overline{p}}$ that determine whether or not the transformation is canonical.

The time derivative of any function ${\omega}$ can be written as a Poisson bracket:

$\displaystyle \dot{\omega}=\left\{ \omega,H\right\} \ \ \ \ \ (5)$

For the transformed velocities, we have

 $\displaystyle \dot{\overline{q}}_{j}$ $\displaystyle =$ $\displaystyle \left\{ \overline{q}_{j},H\right\} \ \ \ \ \ (6)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \sum_{i}\left(\frac{\partial\overline{q}_{j}}{\partial q_{i}}\frac{\partial H}{\partial p_{i}}-\frac{\partial\overline{q}_{j}}{\partial p_{i}}\frac{\partial H}{\partial q_{i}}\right) \ \ \ \ \ (7)$

Here, ${H}$ is written as a function ${H\left(q,p\right)}$ of the original variables. If we write it as a function of the transformed variables, we can find the two derivatives of ${H}$ in 7 by using the chain rule:

 $\displaystyle \frac{\partial H\left(\overline{q},\overline{p}\right)}{\partial p_{i}}$ $\displaystyle =$ $\displaystyle \sum_{k}\left(\frac{\partial H}{\partial\overline{q}_{k}}\frac{\partial\overline{q}_{k}}{\partial p_{i}}+\frac{\partial H}{\partial\overline{p}_{k}}\frac{\partial\overline{p}_{k}}{\partial p_{i}}\right)\ \ \ \ \ (8)$ $\displaystyle \frac{\partial H\left(\overline{q},\overline{p}\right)}{\partial q_{i}}$ $\displaystyle =$ $\displaystyle \sum_{k}\left(\frac{\partial H}{\partial\overline{q}_{k}}\frac{\partial\overline{q}_{k}}{\partial q_{i}}+\frac{\partial H}{\partial\overline{p}_{k}}\frac{\partial\overline{p}_{k}}{\partial q_{i}}\right) \ \ \ \ \ (9)$

Inserting these into 7 we get

 $\displaystyle \dot{\overline{q}}_{j}$ $\displaystyle =$ $\displaystyle \sum_{i}\sum_{k}\left[\frac{\partial\overline{q}_{j}}{\partial q_{i}}\left(\frac{\partial H}{\partial\overline{q}_{k}}\frac{\partial\overline{q}_{k}}{\partial p_{i}}+\frac{\partial H}{\partial\overline{p}_{k}}\frac{\partial\overline{p}_{k}}{\partial p_{i}}\right)-\frac{\partial\overline{q}_{j}}{\partial p_{i}}\left(\frac{\partial H}{\partial\overline{q}_{k}}\frac{\partial\overline{q}_{k}}{\partial q_{i}}+\frac{\partial H}{\partial\overline{p}_{k}}\frac{\partial\overline{p}_{k}}{\partial q_{i}}\right)\right]\ \ \ \ \ (10)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \sum_{k}\frac{\partial H}{\partial\overline{q}_{k}}\sum_{i}\left(\frac{\partial\overline{q}_{j}}{\partial q_{i}}\frac{\partial\overline{q}_{k}}{\partial p_{i}}-\frac{\partial\overline{q}_{j}}{\partial p_{i}}\frac{\partial\overline{q}_{k}}{\partial q_{i}}\right)+\sum_{k}\frac{\partial H}{\partial\overline{p}_{k}}\sum_{i}\left(\frac{\partial\overline{q}_{j}}{\partial q_{i}}\frac{\partial\overline{p}_{k}}{\partial p_{i}}-\frac{\partial\overline{q}_{j}}{\partial p_{i}}\frac{\partial\overline{p}_{k}}{\partial q_{i}}\right)\ \ \ \ \ (11)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \sum_{k}\frac{\partial H}{\partial\overline{q}_{k}}\left\{ \overline{q}_{j},\overline{q}_{k}\right\} +\sum_{k}\frac{\partial H}{\partial\overline{p}_{k}}\left\{ \overline{q}_{j},\overline{p}_{k}\right\} \ \ \ \ \ (12)$

In order for this result to satisfy 3, we must have

 $\displaystyle \left\{ \overline{q}_{j},\overline{q}_{k}\right\}$ $\displaystyle =$ $\displaystyle 0\ \ \ \ \ (13)$ $\displaystyle \left\{ \overline{q}_{j},\overline{p}_{k}\right\}$ $\displaystyle =$ $\displaystyle \delta_{jk} \ \ \ \ \ (14)$

We can repeat the calculation for ${\dot{\overline{p}}_{i}}$:

 $\displaystyle \dot{\overline{p}}_{j}$ $\displaystyle =$ $\displaystyle \left\{ \overline{p}_{j},H\right\} \ \ \ \ \ (15)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \sum_{i}\left(\frac{\partial\overline{p}_{j}}{\partial q_{i}}\frac{\partial H}{\partial p_{i}}-\frac{\partial\overline{p}_{j}}{\partial p_{i}}\frac{\partial H}{\partial q_{i}}\right)\ \ \ \ \ (16)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \sum_{i}\sum_{k}\left[\frac{\partial\overline{p}_{j}}{\partial q_{i}}\left(\frac{\partial H}{\partial\overline{q}_{k}}\frac{\partial\overline{q}_{k}}{\partial p_{i}}+\frac{\partial H}{\partial\overline{p}_{k}}\frac{\partial\overline{p}_{k}}{\partial p_{i}}\right)-\frac{\partial\overline{p}_{j}}{\partial p_{i}}\left(\frac{\partial H}{\partial\overline{q}_{k}}\frac{\partial\overline{q}_{k}}{\partial q_{i}}+\frac{\partial H}{\partial\overline{p}_{k}}\frac{\partial\overline{p}_{k}}{\partial q_{i}}\right)\right]\ \ \ \ \ (17)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \sum_{k}\frac{\partial H}{\partial\overline{q}_{k}}\sum_{i}\left(\frac{\partial\overline{p}_{j}}{\partial q_{i}}\frac{\partial\overline{q}_{k}}{\partial p_{i}}-\frac{\partial\overline{p}_{j}}{\partial p_{i}}\frac{\partial\overline{q}_{k}}{\partial q_{i}}\right)+\sum_{k}\frac{\partial H}{\partial\overline{p}_{k}}\sum_{i}\left(\frac{\partial\overline{p}_{j}}{\partial q_{i}}\frac{\partial\overline{p}_{k}}{\partial p_{i}}-\frac{\partial\overline{p}_{j}}{\partial p_{i}}\frac{\partial\overline{p}_{k}}{\partial q_{i}}\right)\ \ \ \ \ (18)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \sum_{k}\frac{\partial H}{\partial\overline{q}_{k}}\left\{ \overline{p}_{j},\overline{q}_{k}\right\} +\sum_{k}\frac{\partial H}{\partial\overline{p}_{k}}\left\{ \overline{p}_{j},\overline{p}_{k}\right\} \ \ \ \ \ (19)$

Requiring this to satsify 4, we have

 $\displaystyle \left\{ \overline{p}_{j},\overline{p}_{k}\right\}$ $\displaystyle =$ $\displaystyle 0\ \ \ \ \ (20)$ $\displaystyle \left\{ \overline{p}_{j},\overline{q}_{k}\right\}$ $\displaystyle =$ $\displaystyle -\delta_{jk} \ \ \ \ \ (21)$

The last equation is equivalent to

$\displaystyle \left\{ \overline{q}_{j},\overline{p}_{k}\right\} =\delta_{jk} \ \ \ \ \ (22)$

which agrees with 14. Thus in order for the transformation to be canonical, the conditions are

 $\displaystyle \left\{ \overline{q}_{j},\overline{q}_{k}\right\}$ $\displaystyle =$ $\displaystyle \left\{ \overline{p}_{j},\overline{p}_{k}\right\} =0\ \ \ \ \ (23)$ $\displaystyle \left\{ \overline{q}_{j},\overline{p}_{k}\right\}$ $\displaystyle =$ $\displaystyle \delta_{jk} \ \ \ \ \ (24)$

Note that these Poisson brackets require calculating the derivatives of the new variables ${\overline{q}}$ and ${\overline{p}}$ with respect to the original ones ${q}$ and ${p}$, but they don’t involve any particular Hamiltonian. Thus it’s possible to determine whether or not a transformation is canonical entirely from the transformation equations 1 and 2.

# Cyclic coordinates and Poisson brackets

References: Shankar, R. (1994), Principles of Quantum Mechanics, Plenum Press. Section 2.7; Exercises 2.7.1 – 2.7.2.

Hamilton’s canonical equations are:

 $\displaystyle \frac{\partial H}{\partial p_{i}}$ $\displaystyle =$ $\displaystyle \dot{q}_{i}\ \ \ \ \ (1)$ $\displaystyle -\frac{\partial H}{\partial q_{i}}$ $\displaystyle =$ $\displaystyle \dot{p}_{i} \ \ \ \ \ (2)$

If a coordinate ${q_{i}}$ is missing in the Hamiltonian (that is, ${H}$ is indepedent of ${q_{i}}$), then

$\displaystyle \dot{p}_{i}=-\frac{\partial H}{\partial q_{i}}=0 \ \ \ \ \ (3)$

Thus the conjugate momentum ${p_{i}}$ is conserved. Such a missing coordinate ${q_{i}}$ is known as a cyclic coordinate. [I’m not sure of the origin of this term. Again Google doesn’t provide a definitive answer.]

There is a general method for calculating the rate of change of some function ${\omega\left(p,q\right)}$ that depends on the momenta and coordinates, but not explicitly on the time (${\omega}$ is allowed to depend implicitly on time since ${p}$ and/or ${q}$ can depend on time). The time derivative can then be written using the chain rule:

 $\displaystyle \frac{d\omega}{dt}$ $\displaystyle =$ $\displaystyle \sum_{i}\left(\frac{\partial\omega}{\partial q_{i}}\dot{q}_{i}+\frac{\partial\omega}{\partial p_{i}}\dot{p}_{i}\right)\ \ \ \ \ (4)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \sum_{i}\left(\frac{\partial\omega}{\partial q_{i}}\frac{\partial H}{\partial p_{i}}-\frac{\partial\omega}{\partial p_{i}}\frac{\partial H}{\partial q_{i}}\right)\ \ \ \ \ (5)$ $\displaystyle$ $\displaystyle \equiv$ $\displaystyle \left\{ \omega,H\right\} \ \ \ \ \ (6)$

where in the second line we used Hamilton’s equations 1 and 2. The last line defines the Poisson bracket of the function ${\omega}$ with the Hamiltonian ${H}$. We can see that if ${\left\{ \omega,H\right\} =0}$, the function ${\omega}$ is conserved.

Since ${\left\{ H,H\right\} =0}$ automatically, the total energy (represented by the Hamiltonian) is conserved, provided there is no explicit time dependence. Such a time dependence can arise if the system is subject to some external force, for example.

From the definition 5 we can derive a few fundamental properties of Poisson brackets. We’ll consider a general Poisson bracket between two arbitrary functions ${\omega\left(p,q\right)}$ and ${\lambda\left(p,q\right)}$. Then

 $\displaystyle \left\{ \omega,\lambda\right\}$ $\displaystyle =$ $\displaystyle \sum_{i}\left(\frac{\partial\omega}{\partial q_{i}}\frac{\partial\lambda}{\partial p_{i}}-\frac{\partial\omega}{\partial p_{i}}\frac{\partial\lambda}{\partial q_{i}}\right)\ \ \ \ \ (7)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\sum_{i}\left(\frac{\partial\omega}{\partial p_{i}}\frac{\partial\lambda}{\partial q_{i}}-\frac{\partial\omega}{\partial q_{i}}\frac{\partial\lambda}{\partial p_{i}}\right)\ \ \ \ \ (8)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\sum_{i}\left(\frac{\partial\lambda}{\partial q_{i}}\frac{\partial\omega}{\partial p_{i}}-\frac{\partial\lambda}{\partial p_{i}}\frac{\partial\omega}{\partial q_{i}}\right)\ \ \ \ \ (9)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\left\{ \lambda,\omega\right\} \ \ \ \ \ (10)$

A Poisson bracket is distributive, in the sense that

 $\displaystyle \left\{ \omega,\lambda+\sigma\right\}$ $\displaystyle =$ $\displaystyle \sum_{i}\left(\frac{\partial\omega}{\partial q_{i}}\frac{\partial\left(\lambda+\sigma\right)}{\partial p_{i}}-\frac{\partial\omega}{\partial p_{i}}\frac{\partial\left(\lambda+\sigma\right)}{\partial q_{i}}\right)\ \ \ \ \ (11)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \sum_{i}\left(\frac{\partial\omega}{\partial q_{i}}\left[\frac{\partial\lambda}{\partial p_{i}}+\frac{\partial\sigma}{\partial p_{i}}\right]-\frac{\partial\omega}{\partial p_{i}}\left[\frac{\partial\lambda}{\partial q_{i}}+\frac{\partial\sigma}{\partial q_{i}}\right]\right)\ \ \ \ \ (12)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \sum_{i}\left(\frac{\partial\omega}{\partial q_{i}}\frac{\partial\lambda}{\partial p_{i}}-\frac{\partial\omega}{\partial p_{i}}\frac{\partial\lambda}{\partial q_{i}}\right)+\sum_{i}\left(\frac{\partial\omega}{\partial q_{i}}\frac{\partial\sigma}{\partial p_{i}}-\frac{\partial\omega}{\partial p_{i}}\frac{\partial\sigma}{\partial q_{i}}\right)\ \ \ \ \ (13)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left\{ \omega,\lambda\right\} +\left\{ \omega,\sigma\right\} \ \ \ \ \ (14)$

One more identity is useful, which we can derive using the product rule:

 $\displaystyle \left\{ \omega,\lambda\sigma\right\}$ $\displaystyle =$ $\displaystyle \sum_{i}\left(\frac{\partial\omega}{\partial q_{i}}\frac{\partial\left(\lambda\sigma\right)}{\partial p_{i}}-\frac{\partial\omega}{\partial p_{i}}\frac{\partial\left(\lambda\sigma\right)}{\partial q_{i}}\right)\ \ \ \ \ (15)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \sum_{i}\sigma\left(\frac{\partial\omega}{\partial q_{i}}\frac{\partial\lambda}{\partial p_{i}}-\frac{\partial\omega}{\partial p_{i}}\frac{\partial\lambda}{\partial q_{i}}\right)+\sum_{i}\lambda\left(\frac{\partial\omega}{\partial q_{i}}\frac{\partial\sigma}{\partial p_{i}}-\frac{\partial\omega}{\partial p_{i}}\frac{\partial\sigma}{\partial q_{i}}\right)\ \ \ \ \ (16)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left\{ \omega,\lambda\right\} \sigma+\left\{ \omega,\sigma\right\} \lambda \ \ \ \ \ (17)$

The Poisson brackets involving the coordinates ${q_{i}}$ and momenta ${p_{i}}$ turn up frequently, so it’s worth deriving them in detail. We have

$\displaystyle \left\{ q_{i},q_{j}\right\} =\sum_{k}\left(\frac{\partial q_{i}}{\partial q_{k}}\frac{\partial q_{j}}{\partial p_{k}}-\frac{\partial q_{i}}{\partial p_{k}}\frac{\partial q_{j}}{\partial q_{k}}\right)=0 \ \ \ \ \ (18)$

This follows because, in the Hamiltonian formalism, the ${q_{i}}$s and ${p_{i}}$s are independent variables, so ${\frac{\partial q_{j}}{\partial p_{k}}=\frac{\partial p_{j}}{\partial q_{k}}=0}$ for all ${j}$ and ${k}$. For the same reason, we have

$\displaystyle \left\{ p_{i},p_{j}\right\} =\sum_{k}\left(\frac{\partial p_{i}}{\partial q_{k}}\frac{\partial p_{j}}{\partial p_{k}}-\frac{\partial p_{i}}{\partial p_{k}}\frac{\partial p_{j}}{\partial q_{k}}\right)=0 \ \ \ \ \ (19)$

The mixed Poisson bracket is a different story, however:

 $\displaystyle \left\{ q_{i},p_{j}\right\}$ $\displaystyle =$ $\displaystyle \sum_{k}\left(\frac{\partial q_{i}}{\partial q_{k}}\frac{\partial p_{j}}{\partial p_{k}}-\frac{\partial q_{i}}{\partial p_{k}}\frac{\partial p_{j}}{\partial q_{k}}\right)\ \ \ \ \ (20)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \sum_{k}\delta_{ik}\delta_{jk}-0\ \ \ \ \ (21)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \delta_{ij} \ \ \ \ \ (22)$

Hamilton’s equations 1 and 2 can be written using Poisson brackets by setting ${\omega}$ equal to ${q_{i}}$ and ${p_{i}}$ respectively in 6:

 $\displaystyle \dot{q}_{i}$ $\displaystyle =$ $\displaystyle \left\{ q_{i},H\right\} \ \ \ \ \ (23)$ $\displaystyle \dot{p}_{i}$ $\displaystyle =$ $\displaystyle \left\{ p_{i},H\right\} \ \ \ \ \ (24)$

Example In two dimensions, we have a Hamiltonian:

$\displaystyle H=p_{x}^{2}+p_{y}^{2}+ax^{2}+by^{2} \ \ \ \ \ (25)$

If ${a=b}$, then in polar coordiantes, the only coordinate appearing in ${H}$ is the radial distance from the origin ${r=\sqrt{x^{2}+y^{2}}}$, which means that the polar angle ${\theta}$ is a cyclic coordinate. This means that the conjugate momentum ${p_{\theta}}$ must be conserved. That is,

$\displaystyle \dot{p}_{\theta}=\left\{ p_{\theta},H\right\} =0 \ \ \ \ \ (26)$

However, ${p_{\theta}}$ is the angular momentum ${\ell_{z}}$, so this just says that angular momentum is conserved.

To see this explicitly, it’s easier to convert to polar coordinates. From Hamilton’s equations

 $\displaystyle \dot{x}$ $\displaystyle =$ $\displaystyle \frac{\partial H}{\partial p_{x}}=2p_{x}\ \ \ \ \ (27)$ $\displaystyle \dot{y}$ $\displaystyle =$ $\displaystyle 2p_{y}\ \ \ \ \ (28)$ $\displaystyle p_{x}^{2}+p_{y}^{2}$ $\displaystyle =$ $\displaystyle \frac{1}{4}\left(\dot{x}^{2}+\dot{y}^{2}\right)\ \ \ \ \ (29)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{v^{2}}{4}\ \ \ \ \ (30)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{4}\left(\dot{r}^{2}+r^{2}\dot{\theta}^{2}\right) \ \ \ \ \ (31)$

where in the fourth line, ${v}$ is the linear velocity and in the fifth line we converted this to polar coordinates. Thus the Hamiltonian becomes, in the case where ${a=b}$:

$\displaystyle H=\frac{1}{4}\left(\dot{r}^{2}+r^{2}\dot{\theta}^{2}\right)+ar^{2} \ \ \ \ \ (32)$

To find the conjugate momenta in polar coordinates, we can write out the Lagrangian. We use ${p_{x}\dot{x}=\frac{\dot{x}^{2}}{2}}$ and ${p_{y}\dot{y}=\frac{\dot{y}^{2}}{2}}$ and get

 $\displaystyle L$ $\displaystyle =$ $\displaystyle \sum_{i}p_{i}\dot{q}_{i}-H\ \ \ \ \ (33)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{2}\left(\dot{x}^{2}+\dot{y}^{2}\right)-\frac{1}{4}\left(\dot{r}^{2}+r^{2}\dot{\theta}^{2}\right)-ar^{2}\ \ \ \ \ (34)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{4}\left(\dot{r}^{2}+r^{2}\dot{\theta}^{2}\right)-ar^{2} \ \ \ \ \ (35)$

The conjugate momenta are thus

 $\displaystyle p_{\theta}$ $\displaystyle =$ $\displaystyle \frac{\partial L}{\partial\dot{\theta}}=\frac{1}{2}r^{2}\dot{\theta}\ \ \ \ \ (36)$ $\displaystyle p_{r}$ $\displaystyle =$ $\displaystyle \frac{\partial L}{\partial\dot{r}}=\frac{\dot{r}}{2} \ \ \ \ \ (37)$

From this we can see that ${p_{\theta}}$ is indeed angular momentum as it’s proportional to the product of ${r}$ and the tangential velocity ${v_{\theta}=r\dot{\theta}}$. (‘Real’ momentum and angular momentum must, of course, also contain a factor of a mass, but from the definition of the Hamiltonian above, we see that the mass has been incorporated into the momentum parameters.)

Plugging these back into 32 we get

$\displaystyle H=p_{r}^{2}+p_{\theta}^{2}+ar^{2} \ \ \ \ \ (38)$

We can now calculate the Poisson brackets easily:

 $\displaystyle \left\{ p_{\theta},H\right\}$ $\displaystyle =$ $\displaystyle \sum_{i}\left(\frac{\partial p_{\theta}}{\partial q_{i}}\frac{\partial H}{\partial p_{i}}-\frac{\partial p_{\theta}}{\partial p_{i}}\frac{\partial H}{\partial q_{i}}\right)\ \ \ \ \ (39)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 0-\frac{\partial p_{\theta}}{\partial p_{\theta}}\frac{\partial H}{\partial\theta}=0\ \ \ \ \ (40)$ $\displaystyle \left\{ p_{r},H\right\}$ $\displaystyle =$ $\displaystyle \sum_{i}\left(\frac{\partial p_{r}}{\partial q_{i}}\frac{\partial H}{\partial p_{i}}-\frac{\partial p_{r}}{\partial p_{i}}\frac{\partial H}{\partial q_{i}}\right)\ \ \ \ \ (41)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 0-\frac{\partial p_{r}}{\partial p_{r}}\frac{\partial H}{\partial r}\ \ \ \ \ (42)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -2ar \ \ \ \ \ (43)$

Thus ${p_{\theta}}$ (the angular momentum) is conserved, while ${p_{r}<0}$, so that the object is always being pulled in towards the origin.

# Poisson brackets and Hamilton’s equations of motion

Reference: W. Greiner & J. Reinhardt, Field Quantization, Springer-Verlag (1996), Chapter 2, Section 2.2.

Although I’ve looked at Poisson brackets before, it’s worth going through G & R’s treatment as it is a fair bit simpler and gives clearer results.

First, we need the time derivative of a functional. In the simplest case, a functional ${F\left[\phi\right]}$ depends on a function ${\phi}$which in turn depends on an independent variable ${x}$. The functional itself does not depend on ${x}$, however, usually because ${F}$ is defined as the integral of ${\phi\left(x\right)}$ over some range of ${x}$ values, so the dependence on ${x}$ disappears in the integration.

We can generalize things a bit by taking ${\phi}$ as a function of two variables, say ${x}$ and ${t}$. If ${F}$ is defined in the same way (say, as an integral of ${\phi}$ over ${x}$), then the variable ${t}$ also appears in the functional, so we can write this as ${F\left(t\right)}$, which is

$\displaystyle F\left(t\right)=\int dx\;g\left(\phi\left(x,t\right)\right) \ \ \ \ \ (1)$

where ${g\left(\phi\right)}$ is some function of ${\phi}$. Since ${F}$ now depends on ${t}$, we can take the derivative ${dF/dt}$ which comes out to

$\displaystyle \dot{F}\equiv\frac{dF}{dt}=\int dx\frac{dg}{d\phi}\frac{\partial\phi}{\partial t}=\int dx\frac{dg}{d\phi}\dot{\phi}\left(x,t\right) \ \ \ \ \ (2)$

As we’ve seen before, the functional derivative of ${F}$ in this case is

$\displaystyle \frac{\delta F\left(t\right)}{\delta\phi\left(y,t\right)}=\frac{dg\left(\phi\left(y,t\right)\right)}{d\phi} \ \ \ \ \ (3)$

where the notation means that we evaluate the derivative on the RHS at the point ${\left(y,t\right)}$. Using this result, we can therefore write ${\dot{F}}$ as

$\displaystyle \dot{F}\left(t\right)=\int dx\frac{\delta F\left(t\right)}{\delta\phi\left(x,t\right)}\dot{\phi}\left(x,t\right) \ \ \ \ \ (4)$

We can generalize this to 4-d space time, so that ${x}$ now indicates the four-vector ${x=\left(\mathbf{x},t\right)}$, and the integral is over 3-d space:

$\displaystyle \dot{F}\left(t\right)=\int d^{3}\mathbf{x}\frac{\delta F\left(t\right)}{\delta\phi\left(x\right)}\dot{\phi}\left(x\right) \ \ \ \ \ (5)$

Generalizing even further, we can make ${F}$ a functional of two fields, ${\phi}$ and ${\pi}$, so we get

$\displaystyle \dot{F}\left(t\right)=\int d^{3}\mathbf{x}\left[\frac{\delta F\left(t\right)}{\delta\phi\left(x\right)}\dot{\phi}\left(x\right)+\frac{\delta F\left(t\right)}{\delta\pi\left(x\right)}\dot{\pi}\left(x\right)\right] \ \ \ \ \ (6)$

Interpreting ${\phi}$ as the field and ${\pi}$ as its conjugate momentum, we can now use Hamilton’s equations of motion

 $\displaystyle \dot{\phi}$ $\displaystyle =$ $\displaystyle \frac{\delta H}{\delta\pi}\ \ \ \ \ (7)$ $\displaystyle \dot{\pi}$ $\displaystyle =$ $\displaystyle -\frac{\delta H}{\delta\phi} \ \ \ \ \ (8)$

We get

$\displaystyle \dot{F}\left(t\right)=\int d^{3}\mathbf{x}\left[\frac{\delta F\left(t\right)}{\delta\phi\left(x\right)}\frac{\delta H}{\delta\pi}-\frac{\delta F\left(t\right)}{\delta\pi\left(x\right)}\frac{\delta H}{\delta\phi}\right] \ \ \ \ \ (9)$

The quantity on the RHS is defined to be the Poisson bracket:

$\displaystyle \left\{ F,H\right\} _{PB}\equiv\int d^{3}\mathbf{x}\left[\frac{\delta F\left(t\right)}{\delta\phi\left(x\right)}\frac{\delta H}{\delta\pi}-\frac{\delta F\left(t\right)}{\delta\pi\left(x\right)}\frac{\delta H}{\delta\phi}\right] \ \ \ \ \ (10)$

We thus have the general result that the time derivative of a functional is equal to its Poisson bracket with the Hamiltonian:

$\displaystyle \boxed{\dot{F}\left(t\right)=\left\{ F,H\right\} _{PB}} \ \ \ \ \ (11)$

We can use this result in a rather curious way to re-derive Hamilton’s equations of motion. We first observe that we can write the field ${\phi}$ as an integral:

$\displaystyle \phi\left(\mathbf{x},t\right)=\int d^{3}\mathbf{x}^{\prime}\;\phi\left(\mathbf{x}^{\prime},t\right)\delta^{3}\left(\mathbf{x}-\mathbf{x}^{\prime}\right) \ \ \ \ \ (12)$

This effectively defines an ordinary function ${\phi}$ as a functional depending on itself. In this case, both ${\mathbf{x}}$ and ${t}$ are parameters that are present on both sides of the equation; it is the dummy variable ${\mathbf{x}^{\prime}}$ that is the variable of integration.

Taking the variation on both sides, we get

$\displaystyle \delta\phi\left(\mathbf{x},t\right)=\int d^{3}\mathbf{x}^{\prime}\;\delta\phi\left(\mathbf{x}^{\prime},t\right)\delta^{3}\left(\mathbf{x}-\mathbf{x}^{\prime}\right) \ \ \ \ \ (13)$

[Be careful not to get the ${\delta}$s confused here: ${\delta\phi}$ is a variation of the function ${\phi}$ while ${\delta^{3}}$ is the 3-d delta function.] Comparing this to the definition of the functional derivative

$\displaystyle \delta F\left[\phi\right]\equiv\int d^{3}\mathbf{x}\frac{\delta F\left[\phi\right]}{\delta\phi\left(x\right)}\delta\phi\left(x\right) \ \ \ \ \ (14)$

we see that we have the functional derivative of ${\phi}$ with respect to itself:

$\displaystyle \frac{\delta\phi\left(\mathbf{x},t\right)}{\delta\phi\left(\mathbf{x}^{\prime},t\right)}=\delta^{3}\left(\mathbf{x}-\mathbf{x}^{\prime}\right) \ \ \ \ \ (15)$

We could use the same argument on the conjugate momentum, so we also have

$\displaystyle \frac{\delta\pi\left(\mathbf{x},t\right)}{\delta\pi\left(\mathbf{x}^{\prime},t\right)}=\delta^{3}\left(\mathbf{x}-\mathbf{x}^{\prime}\right) \ \ \ \ \ (16)$

Since ${\phi}$ and ${\pi}$ are independent fields

$\displaystyle \frac{\delta\pi\left(\mathbf{x},t\right)}{\delta\phi\left(\mathbf{x}^{\prime},t\right)}=\frac{\delta\phi\left(\mathbf{x},t\right)}{\delta\pi\left(\mathbf{x}^{\prime},t\right)}=0 \ \ \ \ \ (17)$

We can now use 11 to find the time derivatives of ${\phi}$ and ${\pi}$ by treating them as functionals:

 $\displaystyle \dot{\phi}\left(\mathbf{x},t\right)$ $\displaystyle =$ $\displaystyle \left\{ \phi\left(\mathbf{x},t\right),H\right\} \ \ \ \ \ (18)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \int d^{3}\mathbf{x}^{\prime}\left[\frac{\delta\phi\left(\mathbf{x},t\right)}{\delta\phi\left(\mathbf{x}^{\prime},t\right)}\frac{\delta H\left(t\right)}{\delta\pi\left(\mathbf{x}^{\prime},t\right)}-\frac{\delta\phi\left(\mathbf{x},t\right)}{\delta\pi\left(\mathbf{x}^{\prime},t\right)}\frac{\delta H\left(t\right)}{\delta\phi\left(\mathbf{x}^{\prime},t\right)}\right]\ \ \ \ \ (19)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \int d^{3}\mathbf{x}^{\prime}\left[\delta^{3}\left(\mathbf{x}-\mathbf{x}^{\prime}\right)\frac{\delta H\left(t\right)}{\delta\pi\left(\mathbf{x}^{\prime},t\right)}-0\right]\ \ \ \ \ (20)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\delta H\left(t\right)}{\delta\pi\left(\mathbf{x},t\right)} \ \ \ \ \ (21)$

This gives the first Hamilton equation of motion 7. We can work out the second equation similarly:

 $\displaystyle \dot{\pi}\left(\mathbf{x},t\right)$ $\displaystyle =$ $\displaystyle \left\{ \pi\left(\mathbf{x},t\right),H\right\} \ \ \ \ \ (22)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \int d^{3}\mathbf{x}^{\prime}\left[\frac{\delta\pi\left(\mathbf{x},t\right)}{\delta\phi\left(\mathbf{x}^{\prime},t\right)}\frac{\delta H\left(t\right)}{\delta\pi\left(\mathbf{x}^{\prime},t\right)}-\frac{\delta\pi\left(\mathbf{x},t\right)}{\delta\pi\left(\mathbf{x}^{\prime},t\right)}\frac{\delta H\left(t\right)}{\delta\phi\left(\mathbf{x}^{\prime},t\right)}\right]\ \ \ \ \ (23)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \int d^{3}\mathbf{x}^{\prime}\left[0-\delta^{3}\left(\mathbf{x}-\mathbf{x}^{\prime}\right)\frac{\delta H\left(t\right)}{\delta\phi\left(\mathbf{x}^{\prime},t\right)}\right]\ \ \ \ \ (24)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\frac{\delta H\left(t\right)}{\delta\phi\left(\mathbf{x},t\right)} \ \ \ \ \ (25)$

Finally, we can work out the Poisson brackets of the fields with each other, using the definition 10 and the results above.

 $\displaystyle \left\{ \phi\left(\mathbf{x},t\right),\pi\left(\mathbf{x}^{\prime},t\right)\right\} _{PB}$ $\displaystyle \equiv$ $\displaystyle \int d^{3}\mathbf{x}^{\prime\prime}\left[\frac{\delta\phi\left(\mathbf{x},t\right)}{\delta\phi\left(\mathbf{x}^{\prime\prime},t\right)}\frac{\delta\pi\left(\mathbf{x}^{\prime},t\right)}{\delta\pi\left(\mathbf{x}^{\prime\prime},t\right)}-\frac{\delta\phi\left(\mathbf{x},t\right)}{\delta\pi\left(\mathbf{x}^{\prime\prime},t\right)}\frac{\delta\pi\left(\mathbf{x}^{\prime},t\right)}{\delta\phi\left(\mathbf{x}^{\prime\prime},t\right)}\right]\ \ \ \ \ (26)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \int d^{3}\mathbf{x}^{\prime\prime}\left[\delta^{3}\left(\mathbf{x}-\mathbf{x}^{\prime\prime}\right)\delta^{3}\left(\mathbf{x}^{\prime}-\mathbf{x}^{\prime\prime}\right)-0\right]\ \ \ \ \ (27)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \delta^{3}\left(\mathbf{x}-\mathbf{x}^{\prime}\right) \ \ \ \ \ (28)$

The other two Poisson brackets are zero because of 17:

 $\displaystyle \left\{ \phi\left(\mathbf{x},t\right),\phi\left(\mathbf{x}^{\prime},t\right)\right\} _{PB}$ $\displaystyle \equiv$ $\displaystyle \int d^{3}\mathbf{x}^{\prime\prime}\left[\frac{\delta\phi\left(\mathbf{x},t\right)}{\delta\phi\left(\mathbf{x}^{\prime\prime},t\right)}\frac{\delta\phi\left(\mathbf{x}^{\prime},t\right)}{\delta\pi\left(\mathbf{x}^{\prime\prime},t\right)}-\frac{\delta\phi\left(\mathbf{x},t\right)}{\delta\pi\left(\mathbf{x}^{\prime\prime},t\right)}\frac{\delta\phi\left(\mathbf{x}^{\prime},t\right)}{\delta\phi\left(\mathbf{x}^{\prime\prime},t\right)}\right]\ \ \ \ \ (29)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \int d^{3}\mathbf{x}^{\prime\prime}\left[0-0\right]\ \ \ \ \ (30)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 0\ \ \ \ \ (31)$ $\displaystyle \left\{ \pi\left(\mathbf{x},t\right),\pi\left(\mathbf{x}^{\prime},t\right)\right\} _{PB}$ $\displaystyle \equiv$ $\displaystyle \int d^{3}\mathbf{x}^{\prime\prime}\left[\frac{\delta\pi\left(\mathbf{x},t\right)}{\delta\phi\left(\mathbf{x}^{\prime\prime},t\right)}\frac{\delta\pi\left(\mathbf{x}^{\prime},t\right)}{\delta\pi\left(\mathbf{x}^{\prime\prime},t\right)}-\frac{\delta\pi\left(\mathbf{x},t\right)}{\delta\pi\left(\mathbf{x}^{\prime\prime},t\right)}\frac{\delta\pi\left(\mathbf{x}^{\prime},t\right)}{\delta\phi\left(\mathbf{x}^{\prime\prime},t\right)}\right]\ \ \ \ \ (32)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \int d^{3}\mathbf{x}^{\prime\prime}\left[0-0\right]\ \ \ \ \ (33)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 0 \ \ \ \ \ (34)$

# Poisson brackets in classical field theory

References: Robert D. Klauber, Student Friendly Quantum Field Theory, (Sandtrove Press, 2013) – Chapter 2.

We’ve seen that the equations of motion of a particle in classical particle theory can be written in terms of Poisson brackets. For a general function ${u\left(q_{i},p_{i},t\right)}$ of the generalized coordinates ${q_{i}}$, conjugate momenta ${p_{i}}$ and time ${t}$, its time derivative is

 $\displaystyle \frac{du}{dt}$ $\displaystyle =$ $\displaystyle \sum_{k}\left(\frac{\partial u}{\partial q_{k}}\frac{\partial H}{\partial p_{k}}-\frac{\partial u}{\partial p_{k}}\frac{\partial H}{\partial q_{k}}\right)+\frac{\partial u}{\partial t}\ \ \ \ \ (1)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left\{ u,H\right\} +\frac{\partial u}{\partial t} \ \ \ \ \ (2)$

where the Poisson bracket is defined as

$\displaystyle \left\{ u,H\right\} \equiv\sum_{k}\left(\frac{\partial u}{\partial q_{k}}\frac{\partial H}{\partial p_{k}}-\frac{\partial u}{\partial p_{k}}\frac{\partial H}{\partial q_{k}}\right) \ \ \ \ \ (3)$

To extend this sort of analysis to classical field theory, we replace the generalized coordinates and momenta by the fields ${\phi^{r}}$ and their conjugate momenta ${\pi_{r}}$. Then, a general function ${u}$ becomes a function of these new variables ${u\left(\phi^{r},\pi_{r}t\right)}$ so its time derivative is now (implied sum over ${r}$):

$\displaystyle \frac{du}{dt}=\frac{\partial u}{\partial\phi^{r}}\dot{\phi}^{r}+\frac{\partial u}{\partial\pi_{r}}\dot{\pi}_{r}+\frac{\partial u}{\partial t} \ \ \ \ \ (4)$

Actually, since we’re taking the derivative of ${u}$ with respect to functions ${\phi^{r}}$ and ${\pi_{r}}$, these are really functional derivatives, so we should write this as

$\displaystyle \frac{du}{dt}=\frac{\delta u}{\delta\phi^{r}}\dot{\phi}^{r}+\frac{\delta u}{\delta\pi_{r}}\dot{\pi}_{r}+\frac{\partial u}{\partial t} \ \ \ \ \ (5)$

 $\displaystyle \dot{\phi}^{r}$ $\displaystyle =$ $\displaystyle \frac{\partial\mathcal{H}}{\partial\pi_{r}}\equiv\frac{\delta\mathcal{H}}{\delta\pi_{r}}\ \ \ \ \ (6)$ $\displaystyle \dot{\pi}_{r}$ $\displaystyle =$ $\displaystyle -\frac{\partial\mathcal{H}}{\partial\phi^{r}}+\partial_{i}\left(\frac{\partial\mathcal{H}}{\partial\phi_{,i}^{r}}\right)\equiv-\frac{\delta\mathcal{H}}{\delta\phi^{r}} \ \ \ \ \ (7)$

we have the equation of motion for ${u}$:

 $\displaystyle \frac{du}{dt}$ $\displaystyle =$ $\displaystyle \frac{\delta u}{\delta\phi^{r}}\frac{\delta\mathcal{H}}{\delta\pi_{r}}-\frac{\delta u}{\delta\pi_{r}}\frac{\delta\mathcal{H}}{\delta\phi^{r}}+\frac{\partial u}{\partial t} \ \ \ \ \ (8)$

The derivatives ${\frac{\partial u}{\partial\phi^{r}}}$ and ${\frac{\partial u}{\partial\pi_{r}}}$ require a bit of care. The field ${\phi^{r}}$ and its conjugate momentum ${\pi_{r}}$ are each defined at every point in space, and when we take the derivative of a function ${u}$ with respect to a field ${\phi^{r}}$, we get a (possibly) non-zero result only if we take the position ${\mathbf{x}}$ of the field in ${u}$ to be the same as the position of the field with respect to which we’re taking the derivative. This is easier to see if we consider 3-space to be divided up into a bunch of infinitesimal volume elements ${\Delta V_{j}}$ where ${j}$ is an index specifying which volume element we’re in. The field ${\phi^{r}}$ has a separate value in each volume element (and if we consider the elements to be small enough, the field value can be considered constant within each element). We name the field ${\phi_{j}^{r}}$ to be the field in element ${\Delta V_{j}}$. Then we can write ${u_{i}}$ as the function ${u}$ within volume element ${\Delta V_{i}}$. Then

$\displaystyle \frac{\delta u_{i}}{\delta\phi_{j}^{r}}=\delta_{ij}\frac{\delta u_{i}}{\delta\phi_{i}^{r}} \ \ \ \ \ (9)$

Given this, we can write the derivative as a sum over volume elements:

 $\displaystyle \frac{\delta u_{i}}{\delta\phi_{i}^{r}}$ $\displaystyle =$ $\displaystyle \sum_{j}\frac{\delta u_{i}}{\delta\phi_{j}^{r}}\ \ \ \ \ (10)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \sum_{j}\Delta V_{j}\left[\frac{1}{\Delta V_{j}}\frac{\delta u_{i}}{\delta\phi_{j}^{r}}\right] \ \ \ \ \ (11)$

In the limit of infinitesimal volume elements, the LHS becomes the derivative evaluated at a particular point ${\mathbf{x}}$, and the volume element in the sum on the RHS becomes the integration element ${d^{3}\mathbf{x}^{\prime}}$. Therefore, the quantity in square brackets must tend to

$\displaystyle \frac{1}{\Delta V_{j}}\frac{\delta u_{i}}{\delta\phi_{j}^{r}}\rightarrow\delta^{3}\left(\mathbf{x}-\mathbf{x}^{\prime}\right)\left.\frac{\delta u\left(\phi^{r},\pi_{r},t\right)}{\delta\phi^{r}}\right|_{\phi^{r}=\phi\left(\mathbf{x},t\right)}\equiv\frac{\delta u\left(\phi^{r}\left(\mathbf{x},t\right),\pi_{r}\left(\mathbf{x}^{\prime},t\right)\right)}{\delta\phi^{r}\left(\mathbf{x}^{\prime},t\right)} \ \ \ \ \ (12)$

We can do a similar calculation for ${\frac{\delta u}{\delta\pi_{r}}}$ with the result

$\displaystyle \frac{\delta u}{\delta\pi_{r}}=\delta^{3}\left(\mathbf{x}-\mathbf{x}^{\prime}\right)\left.\frac{\delta u\left(\phi^{r},\pi_{r},t\right)}{\delta\pi^{r}}\right|_{\pi^{r}=\pi\left(\mathbf{x},t\right)}\equiv\frac{\delta u\left(\phi^{r}\left(\mathbf{x}^{\prime},t\right),\pi_{r}\left(\mathbf{x},t\right)\right)}{\delta\pi^{r}\left(\mathbf{x}^{\prime},t\right)} \ \ \ \ \ (13)$

So we get

 $\displaystyle \frac{du\left(\mathbf{x}\right)}{dt}$ $\displaystyle =$ $\displaystyle \int d^{3}\mathbf{x}^{\prime\prime}\left(\frac{\delta u}{\delta\phi^{r}}\frac{\delta\mathcal{H}}{\delta\pi_{r}}-\frac{\delta u}{\delta\pi_{r}}\frac{\delta\mathcal{H}}{\delta\phi^{r}}\right)+\frac{\partial u}{\partial t}\ \ \ \ \ (14)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \int d^{3}\mathbf{x}^{\prime\prime}\delta^{3}\left(\mathbf{x}-\mathbf{x}^{\prime\prime}\right)\delta^{3}\left(\mathbf{x}^{\prime}-\mathbf{x}^{\prime\prime}\right)\left(\frac{\delta u}{\delta\phi^{r}}\frac{\delta\mathcal{H}}{\delta\pi_{r}}-\frac{\delta u}{\delta\pi_{r}}\frac{\delta\mathcal{H}}{\delta\phi^{r}}\right)+\frac{\partial u}{\partial t}\ \ \ \ \ (15)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left(\frac{\delta u}{\delta\phi^{r}}\frac{\delta\mathcal{H}}{\delta\pi_{r}}-\frac{\delta u}{\delta\pi_{r}}\frac{\delta\mathcal{H}}{\delta\phi^{r}}\right)\delta^{3}\left(\mathbf{x}-\mathbf{x}^{\prime}\right)+\frac{\partial u}{\partial t} \ \ \ \ \ (16)$

The Poisson bracket for fields is the first term:

 $\displaystyle \left\{ u,\mathcal{H}\right\}$ $\displaystyle =$ $\displaystyle \left(\frac{\delta u}{\delta\phi^{r}}\frac{\delta\mathcal{H}}{\delta\pi_{r}}-\frac{\delta u}{\delta\pi_{r}}\frac{\delta\mathcal{H}}{\delta\phi^{r}}\right)\delta^{3}\left(\mathbf{x}-\mathbf{x}^{\prime}\right) \ \ \ \ \ (17)$

For a field and its conjugate momentum, we have

 $\displaystyle \left\{ \phi^{r}\left(\mathbf{x}^{\prime},t\right),\pi_{r}\left(\mathbf{x},t\right)\right\}$ $\displaystyle =$ $\displaystyle \int d^{3}\mathbf{x}^{\prime\prime}\delta^{3}\left(\mathbf{x}-\mathbf{x}^{\prime\prime}\right)\delta^{3}\left(\mathbf{x}^{\prime}-\mathbf{x}^{\prime\prime}\right)\left(\frac{\delta\phi^{r}}{\delta\phi^{r}}\frac{\delta\pi_{r}}{\delta\pi_{r}}-\frac{\delta\phi^{r}}{\delta\pi_{r}}\frac{\delta\pi_{r}}{\delta\phi^{r}}\right)\ \ \ \ \ (18)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \int d^{3}\mathbf{x}^{\prime\prime}\delta^{3}\left(\mathbf{x}-\mathbf{x}^{\prime\prime}\right)\delta^{3}\left(\mathbf{x}^{\prime}-\mathbf{x}^{\prime\prime}\right)\left(1-0\right)\ \ \ \ \ (19)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \delta^{3}\left(\mathbf{x}-\mathbf{x}^{\prime}\right) \ \ \ \ \ (20)$

If we consider two different fields ${\phi^{r}}$ and ${\phi^{s}}$, these are independent variables so ${\frac{\delta\phi^{r}}{\delta\phi^{s}}=\frac{\delta\pi_{s}}{\delta\pi_{r}}=\delta_{s}^{r}}$ and we can generalize the bracket to

$\displaystyle \left\{ \phi^{r}\left(\mathbf{x}^{\prime},t\right),\pi_{r}\left(\mathbf{x},t\right)\right\} =\delta_{s}^{r}\delta^{3}\left(\mathbf{x}-\mathbf{x}^{\prime}\right) \ \ \ \ \ (21)$

Other brackets are zero, for example

 $\displaystyle \left\{ \phi^{r}\left(\mathbf{x}^{\prime},t\right),\phi^{r}\left(\mathbf{x},t\right)\right\}$ $\displaystyle =$ $\displaystyle \int d^{3}\mathbf{x}^{\prime\prime}\delta^{3}\left(\mathbf{x}-\mathbf{x}^{\prime\prime}\right)\delta^{3}\left(\mathbf{x}^{\prime}-\mathbf{x}^{\prime\prime}\right)\left(\frac{\delta\phi^{r}}{\delta\phi^{r}}\frac{\delta\phi^{r}}{\delta\pi_{r}}-\frac{\delta\phi^{r}}{\delta\pi_{r}}\frac{\delta\phi^{r}}{\delta\phi^{r}}\right)\ \ \ \ \ (22)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 0 \ \ \ \ \ (23)$

because ${\frac{\partial\phi^{r}}{\partial\pi_{r}}=0}$.

Hamilton’s equations of motion for fields can be written using Poisson brackets. We get

 $\displaystyle \frac{d\phi^{r}}{dt}$ $\displaystyle =$ $\displaystyle \dot{\phi}^{r}=\left(\frac{\delta\phi^{r}}{\delta\phi^{r}}\frac{\delta\mathcal{H}}{\delta\pi_{r}}-\frac{\delta\phi^{r}}{\delta\pi_{r}}\frac{\delta\mathcal{H}}{\delta\phi^{r}}\right)\delta^{3}\left(\mathbf{x}-\mathbf{x}^{\prime}\right)\ \ \ \ \ (24)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\delta\mathcal{H}}{\delta\pi_{r}}\delta^{3}\left(\mathbf{x}-\mathbf{x}^{\prime}\right)\ \ \ \ \ (25)$ $\displaystyle \frac{d\pi^{r}}{dt}$ $\displaystyle =$ $\displaystyle \dot{\pi}^{r}=\left(\frac{\delta\pi^{r}}{\delta\phi^{r}}\frac{\delta\mathcal{H}}{\delta\pi_{r}}-\frac{\delta\pi^{r}}{\delta\pi_{r}}\frac{\delta\mathcal{H}}{\delta\phi^{r}}\right)\delta^{3}\left(\mathbf{x}-\mathbf{x}^{\prime}\right)\ \ \ \ \ (26)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\frac{\delta\mathcal{H}}{\delta\phi_{r}}\delta^{3}\left(\mathbf{x}-\mathbf{x}^{\prime}\right) \ \ \ \ \ (27)$

I’m not too sure about the extra delta function that shows up here. Presumably it just serves to localize the result to one particular place, so that ${\phi^{r}}$ on the LHS and ${\frac{\delta\mathcal{H}}{\delta\pi_{r}}}$ must both be evaluated at the same location.

The commutators for quantum field theory are obtained by postulating the same relation between Poisson brackets and quantum commutators that we had in particle theory. Namely, we propose a commutator equivalent to each Poisson bracket (multiplied by ${i\hbar}$), so we have

 $\displaystyle \left[\phi^{r},\pi_{s}\right]$ $\displaystyle =$ $\displaystyle i\hbar\delta_{s}^{r}\delta^{3}\left(\mathbf{x}-\mathbf{x}^{\prime}\right)\ \ \ \ \ (28)$ $\displaystyle \left[\phi^{r},\phi^{s}\right]$ $\displaystyle =$ $\displaystyle 0\ \ \ \ \ (29)$ $\displaystyle \left[\pi_{r},\pi_{s}\right]$ $\displaystyle =$ $\displaystyle 0 \ \ \ \ \ (30)$