# Does it make sense to define operators like r,theta, phi

I'm pondering, since we've introduced formalism, all operators are either scalars or vector components, does it make sense to define operators like r, theta, phi (as in spherical coordinates) which are neither?
In classical mechanics we can easily transform observables fro cartesian to spherical polar coordinate, but in QM it seems difficult(at least for me),for example for r=sqrt(x^2+y^2+z^2), it involves square root of operators;also,I have trouble evaluating relevant commutation relations.
Well I asked this question because I remember calculating <r> in wave mechanics, but we never introduce this kind of operator in formalism.

Related Quantum Physics News on Phys.org
tom.stoer
You can introduce these operators but you have to be careful regarding domain, commutation relation, etc.

While there are no problems with taking square roots of nonnegative operators like r=sqrt(x^2+y^2+z^2) - that is standard in functional analysis, operators like $$i\partial /\partial \phi$$ and their commutation relations really require a good and deep understanding of what you are doing.

Hi.
I'm pondering, since we've introduced formalism, all operators are either scalars or vector components, does it make sense to define operators like r, theta, phi (as in spherical coordinates) which are neither?
In classical mechanics we can easily transform observables fro cartesian to spherical polar coordinate, but in QM it seems difficult(at least for me),for example for r=sqrt(x^2+y^2+z^2), it involves square root of operators;also,I have trouble evaluating relevant commutation relations.
As for spherical coordinates, operators r and Pr = -ih' (d/dr + 1/r) satisfy commutation relation.
Due to boundedness at origin, both r and Pr are not self-adjoint operators, I hear in mathematical discussion. I should appreciate if someone explain it plainly and what's wrong with it in physics.

Regards.

tom.stoer
Nothing is wrong with them because they are still symmetric, which means that on the level of matrix elements (and this is what counts in physics) everything is fine.

The subtle problems with the radial momentum operator are discussed, for instance, in http://cdsweb.cern.ch/record/644435/files/0309170.ps.gz" [Broken] (after Eq. (8)).

Last edited by a moderator:
The answer to this question can be found by a deep study of the Lie Algebra of the group SO(3).

SO(3) is the group of rotations in three space. According to the principle axis theorem, there will be three principle axis for these rotations, corresponding to the three orthogonal axis of rotations. It must be noted that rotations in three space do not commute.

The algebra of infinitesimal rotations (lie algebra) along the principle axis is defined, as are all algebra's, by the commutation relations of the various elements.

The lie algebra elements of SO(3) are (linear) operator's which rotate a point in three space infinitesimally along one of the three principle axis. These infinitesimal rotations have the form of derivatives, which you should know are the most important class of linear operators.

It turns out that the commutation relations of the various derivative functions used in the theory of angular momentum [d/dr, d/d(theta), d/d(phi)], are the same commutation relations possessed by the lie algebra of SO(3). There is no mystery to this:

The lie algebra of SO(3) has the characteristic property that if A, B, C are the lie algebra for rotations about the principle axis, then any rotation in three space R (i.e. any element of SO(3) ), can be written as:

R = exp(xA)exp(yB)exp(zC), for some real numbers x,y,z.

Now it's very important to remark that the exponential function of a lie algebra does not posses the same properties as the exponential of say a complex number. This is because these exponential functions do not commute. By the Baker-Hausdorff formula we learn their commutation relation.

It should be noted that the equation which defines lie algebra's, the Jacobi identity, is also satisfied by the curvature tensor of the gravitational field. This helps to explain the fundamental importance of lie algebra's in theoretical physics. A general lie group is defined as a manifold with a group structure, and most lie groups do posses what could be called a "curvature". The lie algebra elements correspond to vectors which define the tangent plane to the manifold at a given point.

The nice thing about our lie algebra exponential is that it satisfies the same derivative rules, i.e. we can find the lie algebra by taking the derivative of the exponential function as illustrated below:

A = d/dx[exp(xA)] evaluated at x = 0.

In other words, for an arbitrary finite rotation along a principle axis, R(w), we can find the lie algebra for that axis by the formula.

A = d/dw[R(w)] evaluated at w = 0.

The foundation of quantum theory is that measurable physical quantities are represented by operators which do not commute. Therefore, the basic problem of quantum mechanics is to find the commutation relations for operators that correspond to physical quantities.

It turns out that the groups U(1) (EM), SU(2) (weak), and SU(3) (strong) are fundamental for high energy physics because their commutation relations are the same as for the most important fundamental physical quantities; these are all lie groups. We say, with some hand waving, that the fundamental particles ARE the lie algebra of these groups (i.e. for each lie algebra element of these groups, there corresponds a certain fundamental particle)

The significance of Lie Groups/Algebra's can be appreciated by considering the geometry of U(1), whose lie algebra corresponds to the photon. As is well known, the photon is it's own anti-particle; but it does have two spin states. These spin states correspond to the fact that the group U(1) is not simply connected, but contains two disconnected subgroups which are characterized by orientation, or in other words, by the sign of the determinant of the matrix which represents a given element of U(1).

It is now clear why the commutation relations for the derivative's (angular momentum operators) have the same form as for the lie algebra of SO(3); as well as why it "makes sense" to introduce derivatives as operators.

Last edited: