Shankar, R. (1994), Principles of Quantum Mechanics, Plenum Press. Chapter 11, Exercise 11.2.3.
The translation operator for an infinitesimal translation is, to first order in :
where , the momentum operator, serves as the generator of translations. To derive a formula for a finite (non-infinitesimal) translation over a distance , we divide the interval into segments, each of width , so that for very large , the width becomes infinitesimal. Then we have
This formula is reminiscent of one definition of the exponential function (which can be found in most introductory calculus texts):
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
In the position basis, , so if we apply to a state vector we can expand the exponential in a Taylor series to get
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 we have
The operator expression on the RHS can be expanded using Hadamard’s lemma, which for two operators and is
where each term contains the commutator of the previous term’s commutator with .
In this case gives us
For example, in the case , and all higher commutators are zero (since they involve the commutator of a constant with ), so we get
so the system is translated by a distance , as we’d expect.
For higher powers of , we can use the result
We therefore get
We’re allowed to treat as an ordinary number in these equations since it is (apart from ), the only operator present so all terms commute.
In the classical case, the infinitesimal change of a variable under an infinitesimal displacement generated by the momentum is given by the Poisson bracket
We can write this as a derivative:
For a finite translation by an amount , we can write the value of as a Taylor series relative to some starting point as
where all derivatives are evaluated at .
We can write all the derivatives in terms of Poisson brackets by using 15. For example
Thus the variable transforms according to
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 .
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 axis is
and serves as the generator of rotations about the axis. Suppose we have a rotation through an angle and we want to see how the two coordinates and transform. The expansion 18 becomes
The relevant Poisson brackets are (using the generic term to represent the two coordinates and ):
Looking at how transforms, we see that the Poisson brackets in 20 will cycle through the four values
The series 20 thus expands to
We can do the same calculation for to get