Why Does Commuting Operators Require Test Functions in Quantum Mechanics?

[Shackleford]

Why is it that I have to use a test function for commuting a position operator with either momentum or angular momentum, but I don't necessarily have to use a test function for commuting the momentum operator with angular momentum?

In the first case, it doesn't make sense if you don't have a test function. You need a test function to prove that [x_i, p_j] = i*h-bar*kronecker delta_ij.

Why is the test function f = f(x_i, x_j)? Where i is not equal to j.

I hope that makes sense.

 

[arkajad]

You need a test function when you want to calculate the commutation relations "from scratch". If you calculate commutation relations of some functions (polynomials) of operators of which the commutation relations you already know, then algebra is enough.

As for you last question - each angular momentum component involves two different coordinates.

 

[Shackleford]

You need a test function when you want to calculate the commutation relations "from scratch". If you calculate commutation relations of some functions (polynomials) of operators of which the commutation relations you already know, then algebra is enough.

As for you last question - each angular momentum component involves two different coordinates.

Well, for the momentum, I worked it out with a test function and without a test function. They both worked out correctly, so I assume you can correctly work out the momentum/angular momentum commutation relations without using a test function.

 

[arkajad]

Once you got momentum-positions commutation relations the rest can be reduced to the pure algebra, possibly with series expansions if you want to calculate commutators like [p,f(x)]

But if you want to calculate a general formula for,say, [p,f(x)], then using a test function is handy.

 

[caloygwafo]

well..n physics, angular momentum, moment of momentum, or rotational momentum[1][2] is a conserved vector quantity that can be used to describe the overall state of a physical system. The angular momentum L of a particle with respect to some point of origin is

\mathbf{L} = \mathbf{r} \times \mathbf{p} = \mathbf{r} \times m\mathbf{v}\, ,

where r is the particle's position from the origin, p = mv is its linear momentum, and × denotes the cross product.

The angular momentum of a system of particles (e.g. a rigid body) is the sum of angular momenta of the individual particles. For a rigid body rotating around an axis of symmetry (e.g. the fins of a ceiling fan), the angular momentum can be expressed as the product of the body's moment of inertia I (a measure of an object's resistance to changes in its rotation rate) and its angular velocity ω: