# Correspondence between classical and quantum transformations

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

When we consider infinitesimal transformations of some dynamical variable, there is a correspondence between classical and quantum mechanics which we can see as follows. First, we’ll summarize the results from classical mechanics. We can define a canonical transformation generated by a variable ${g}$ as

 $\displaystyle \bar{q}_{i}$ $\displaystyle =$ $\displaystyle q_{i}+\varepsilon\frac{\partial g}{\partial p_{i}}\equiv q_{i}+\delta q_{i}\ \ \ \ \ (1)$ $\displaystyle \bar{p}_{i}$ $\displaystyle =$ $\displaystyle p_{i}-\varepsilon\frac{\partial g}{\partial q_{i}}\equiv p_{i}+\delta p_{i} \ \ \ \ \ (2)$

Here, ${\varepsilon}$ is an infintesimal amount and ${\delta q_{i}}$ and ${\delta p_{i}}$ are the infinitesimal amounts by which the coordinates and momenta vary. It follows from these definitions that, for any dynamical variable ${\omega}$, its variation ${\delta\omega}$ is given by a Poisson bracket

$\displaystyle \delta\omega=\omega\left(\bar{q}_{i},\bar{p}_{i}\right)-\omega\left(q_{i},p_{i}\right)=\varepsilon\left\{ \omega,g\right\} \ \ \ \ \ (3)$

For the special cases of coordinates and momenta, this is

 $\displaystyle \delta q_{i}$ $\displaystyle =$ $\displaystyle \varepsilon\left\{ q_{i},g\right\} \ \ \ \ \ (4)$ $\displaystyle \delta p_{i}$ $\displaystyle =$ $\displaystyle \varepsilon\left\{ p_{i},g\right\} \ \ \ \ \ (5)$

If the generator is the momentum ${p_{j}}$, then

 $\displaystyle \delta q_{i}$ $\displaystyle =$ $\displaystyle \varepsilon\left\{ q_{i},p_{j}\right\} =\varepsilon\delta_{ij}\ \ \ \ \ (6)$ $\displaystyle \delta p_{i}$ $\displaystyle =$ $\displaystyle \varepsilon\left\{ p_{i},p_{j}\right\} =0 \ \ \ \ \ (7)$

Thus, in classical mechanics, ${p_{j}}$ is the generator of translations in direction ${j}$.

If ${\omega=H}$ (the Hamiltonian) and if ${\left\{ H,g\right\} =0}$, then ${g}$ is conserved (it doesn’t vary with time). Because the transformation 1 and 2 is canonical, it preserves the Poisson brackets so that

 $\displaystyle \left\{ \overline{q}_{i},\overline{q}_{j}\right\}$ $\displaystyle =$ $\displaystyle \left\{ \overline{p}_{i},\overline{p}_{j}\right\} =0\ \ \ \ \ (8)$ $\displaystyle \left\{ \overline{q}_{i},\overline{p}_{j}\right\}$ $\displaystyle =$ $\displaystyle \delta_{ij} \ \ \ \ \ (9)$

What do these things correspond to in quantum mechanics? [I find Shankar’s treatment in section 11.2 to be almost tautological, since it merely repeats the derivation given earlier. I’ll try to be a bit more general.]

Suppose we have some infinitesimal transformation given by a unitary operator ${U\left(\varepsilon\right)}$. We can then define the changes in ${X}$ and ${P}$ by

 $\displaystyle \delta X$ $\displaystyle =$ $\displaystyle U^{\dagger}\left(\varepsilon\right)XU\left(\varepsilon\right)-X\ \ \ \ \ (10)$ $\displaystyle \delta P$ $\displaystyle =$ $\displaystyle U^{\dagger}\left(\varepsilon\right)PU\left(\varepsilon\right)-P \ \ \ \ \ (11)$

Since ${U\left(\varepsilon\right)}$ describes an infinitesimal transformation, we can expand it to first order in ${\varepsilon}$:

$\displaystyle U\left(\varepsilon\right)=I-\frac{i\varepsilon}{\hbar}G \ \ \ \ \ (12)$

where ${G=G^{\dagger}}$ is some Hermitian operator known as the generator of the transformation. (We’ve seen a proof that the translation operator ${T\left(\varepsilon\right)}$ (a special case of ${U\left(\varepsilon\right)}$) is unitary and that its generator is Hermitian earlier, and the current case follows the same reasoning.) Using this form we have from 10 and 11, to order ${\varepsilon}$:

 $\displaystyle \delta X$ $\displaystyle =$ $\displaystyle \left(I+\frac{i\varepsilon}{\hbar}G\right)X\left(I-\frac{i\varepsilon}{\hbar}G\right)-X\ \ \ \ \ (13)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\frac{i\varepsilon}{\hbar}\left[X,G\right]\ \ \ \ \ (14)$ $\displaystyle \delta P$ $\displaystyle =$ $\displaystyle \left(I+\frac{i\varepsilon}{\hbar}G\right)P\left(I-\frac{i\varepsilon}{\hbar}G\right)-P\ \ \ \ \ (15)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\frac{i\varepsilon}{\hbar}\left[P,G\right] \ \ \ \ \ (16)$

If ${G=P}$, then

 $\displaystyle \delta X$ $\displaystyle =$ $\displaystyle -\frac{i\varepsilon}{\hbar}\left[X,P\right]=\varepsilon I\ \ \ \ \ (17)$ $\displaystyle \delta P$ $\displaystyle =$ $\displaystyle -\frac{i\varepsilon}{\hbar}\left[P,P\right]=0 \ \ \ \ \ (18)$

Comparing this with 6 and 7 we see that (in one dimension, where the classical coordinate is given by ${x}$ and momentum by ${p}$) there is a correspondence between the classical Poisson bracket and quantum commutator:

$\displaystyle \left\{ x,p\right\} \leftrightarrow-\frac{i}{\hbar}\left[X,P\right] \ \ \ \ \ (19)$

The momentum operator ${P}$ in quantum mechanics is thus the generator of translations, just as ${p}$ generates translations in classical mechanics.

More generally, we can define the variation in some arbitrary dynamical operator ${\Omega}$ in a similar way, using 12 to expand the RHS:

 $\displaystyle \delta\Omega$ $\displaystyle =$ $\displaystyle U^{\dagger}\left(\varepsilon\right)\Omega U\left(\varepsilon\right)-\Omega\ \ \ \ \ (20)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\frac{i\varepsilon}{\hbar}\left[\Omega,G\right] \ \ \ \ \ (21)$

The correspondence with classical mechanics is then

$\displaystyle \left\{ \omega,g\right\} \leftrightarrow-\frac{i}{\hbar}\left[\Omega,G\right] \ \ \ \ \ (22)$

The general rule is that a quantum commutator is ${i\hbar}$ times the corresponding classical Poisson bracket.

If ${\Omega=H}$ and ${\left[H,G\right]=0}$, then by Ehrenfest’s theorem, ${\left\langle \dot{G}\right\rangle =0}$ and the average value of ${G}$ is conserved.

The correspondence is a bit odd in that the generator ${g}$ in classical mechanics enters as a derivative in 1 and 2 while the generator ${G}$ in quantum mechanics enters as an operator (no derivatives) in 12.

One other feature is worth noting. A canonical transformation preserves the Poisson brackets 8 in the new coordinate system. In quantum mechanics, it is the commutators that get preserved. For example, using the fact that ${U}$ is unitary so that ${UU^{\dagger}=I}$:

 $\displaystyle U^{\dagger}\left[X,P\right]U$ $\displaystyle =$ $\displaystyle U^{\dagger}XPU-U^{\dagger}PXU\ \ \ \ \ (23)$ $\displaystyle$ $\displaystyle =$ $\displaystyle U^{\dagger}XUU^{\dagger}PU-U^{\dagger}PUU^{\dagger}XU\ \ \ \ \ (24)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left[U^{\dagger}XU,U^{\dagger}PU\right] \ \ \ \ \ (25)$