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)

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