Past exam of the mathematics course of the University of Cambridge / 2008 / ib / Paper 3 / 7A (Quantum Mechanics)

Table of contents
- Solution 7A

Solution

In the position representation, the orbital angular momentum operator is

L = r \times p = - i ℏ r \times \nabla, L_{i} = ϵ_{iab} x_{a} p_{b} .

On a common smooth domain use

[x_{a}, p_{b}] = i ℏ δ_{ab}

,

[x_{a}, x_{b}] = [p_{a}, p_{b}] = 0

. Expanding the product commutator gives

[x_{a} p_{b}, x_{c} p_{d}] = - i ℏ δ_{b c} x_{a} p_{d} + i ℏ δ_{a d} x_{c} p_{b} .

Contracting with the two alternating tensors gives the orbital angular momentum commutation relations:

[L_{i}, L_{j}] = i ℏ (x_{i} p_{j} - x_{j} p_{i}) = i ℏ ϵ_{ijk} L_{k} .

For

L_{\pm} = L_{1} \pm i L_{2}

, these imply

[L_{3}, L_{\pm}] = \pm ℏ L_{\pm}

. Therefore

L_{3} ψ = 0

yields

L_{3} (L_{\pm} ψ) = L_{\pm} L_{3} ψ + [L_{3}, L_{\pm}] ψ = \pm ℏ L_{\pm} ψ .

Thus

L_{\pm} ψ has eigenvalue \pm ℏ whenever L_{\pm} ψ \neq = 0.

The nonzero qualification is necessary: a rotationally invariant wavefunction is annihilated by all three components, and the zero vector is not an eigenvector.

 Ancestors (9)

 View article source

 Discussion (0) New discussion

There are no discussions about this article yet.

 Articles by others on the same topic (0)

There are currently no matching articles.

  See all articles in the same topic Create my own version