Angular Momentum eigenstates, and tensor products

In summary, the angular momentum of a particle can be found by summing up the product of a function of r and a function of theta. However, this sum is not always a product of 3 functions that depend separately on the 3 variables.
  • #1
darkSun
53
0
This is taken from a text problem, but I am putting it in this section because I think my question goes beyond the problem itself:

If a particle has a wave function psi = A*R(r)*cos2 (theta), for example, then if I want to find the probability that its angular momentum is l I would find the absolute value squared of

P(l) = <L=l|psi> .

But the eigenstates of the angular momentum operator are products of exp(i*l*theta/h) and an unspecified function of r. If I do the above integral, don't I have to know what the function of r in the eigenfunction for L is? (I mean Lz, to be precise).

This got me thinking--in this case, |psi> is a tensor product of kets from two Hilbert Spaces, a space with elements depending on r only and a space with elements depending on theta only, correct?

When such a product is put into the r, theta basis, the resulting wave function is a product of a function of r and a function of theta. So does this mean that all wave functions are like this, i.e. products of functions of one variable? In other words, it would be impossible to have a wave function like exp(i*r*theta), for instance? (Aside from the normalization problem)

Thanks for your time.
 
Physics news on Phys.org
  • #2
Yes, you can regard the (spinless, for simplicity) particle's state space as a tensor product of functions of r, theta and phi. This doesn't mean that every state function is a product of 3 functions that depend separately on the 3 variables. Rather, you can expand each state function in something like

[tex]\psi(r,\theta,\phi)=\sum_{abc}c^{abc}f_a(r)g_b(\theta)h_c(\phi)[/tex]

this is precisely the definition of tensor product.
 
  • #3
Ah, and in the case of a function like psi = cos(r*theta*phi) or something, the expansion would be the Taylor Series. thanks dude
 
  • #4
Or would it? I'm not sure I get it, actually.
 
  • #5
It's not quite a Taylor series, its a different type of expansion. For the theta and phi you take the spherical harmonics, which are a base of the functions defined on the unit spherical surface. For the r part you can take a whole set of functions that are a base of the (0, infinity) interval (Legendere functions, Bessel functions, etc), and you get something like

[tex]\psi(r,\theta,\phi)=\sum_{nlm}c^{nlm}J_n(r)Y_{lm}(\theta,\phi)[/tex]

The way to calculate the coefficients is to performs certain integrals. The meaning is essentially the same of Taylor expansion, only the base is different. In the case of taylor expansion the base are polinomials

[tex]1[/tex]
[tex]r,\theta,\phi[/tex]
[tex]r^2,\theta^2,\phi^2,r\theta,r\phi,\theta\phi[/tex]

[tex]\dots[/tex]

in other cases you have more complicated yet more useful functions.
 

FAQ: Angular Momentum eigenstates, and tensor products

What is an Angular Momentum eigenstate?

An Angular Momentum eigenstate is a state in which the angular momentum of a quantum particle is well-defined and can take on only one specific value. This value is determined by the quantum numbers associated with the particle's motion, such as its orbital angular momentum and spin.

How are Angular Momentum eigenstates related to tensor products?

Angular Momentum eigenstates are related to tensor products through the mathematical representation of the angular momentum operator. The angular momentum operator is a tensor product of the position and momentum operators, and its eigenstates are also tensor products of the position and momentum eigenstates. This relationship allows for the calculation of the angular momentum eigenstates in terms of the position and momentum eigenstates.

Can Angular Momentum eigenstates have multiple quantum numbers?

Yes, Angular Momentum eigenstates can have multiple quantum numbers. The total angular momentum of a quantum particle is the sum of its orbital angular momentum and spin, and each of these components can have its own set of quantum numbers. This results in a set of eigenstates for the total angular momentum, each with a unique combination of quantum numbers.

How do Angular Momentum eigenstates behave under rotations?

Angular Momentum eigenstates behave in a specific way under rotations, known as the rotation transformation rule. This rule states that under a rotation, the eigenstate will transform into a linear combination of itself and other eigenstates, with the coefficients of the linear combination determined by the rotation matrix. This allows for the prediction of the behavior of a quantum particle in an external magnetic field, for example.

Are Angular Momentum eigenstates always degenerate?

No, Angular Momentum eigenstates are not always degenerate. Degeneracy occurs when multiple eigenstates have the same eigenvalue, but this is not always the case for angular momentum. For example, in a free particle system, the eigenstates of the total angular momentum can have different eigenvalues and are therefore non-degenerate. However, in a system with additional symmetries, such as a spherically symmetric potential, the eigenstates can become degenerate due to the constraints on the possible values of the quantum numbers.

Similar threads

Back
Top