Angular momentum and expectation value

[jaejoon89]

My teacher said that angular momentum doesn't have orientation in space - but how can that be? Isn't cos(theta) = L_z / |L vector| ?

Also (an unrelated question) could somebody give an example of how the integration process goes when you are trying to get an expectation value for something which isn't an eigenfunction? i.e. I know if it's an eigenfunction, everything normalizes so you just get the observable as the expectation value. But what if it isn't?

 

[kuruman]

My teacher said that angular momentum doesn't have orientation in space - but how can that be? Isn't cos(theta) = L_z / |L vector| ?

Although you can define L_z as the projection of the angular momentum along an external magnetic field, that's just the z-component. The angular momentum vector precesses about the magnetic field, so there is no fixed direction. Maybe that's what your teacher meant.

Also (an unrelated question) could somebody give an example of how the integration process goes when you are trying to get an expectation value for something which isn't an eigenfunction? i.e. I know if it's an eigenfunction, everything normalizes so you just get the observable as the expectation value. But what if it isn't?

Consider two energy eigenstates, |ψ₁> and |ψ₂>. Then

H|ψ₁>=E₁|ψ₁> and H|ψ₂>=E₂|ψ₂>. Now make a new wavefunction that is a normalized linear combination of these, to be specific

$$|\varphi>=\sqrt{\frac{1}{3}}|\psi_{1}>+\sqrt{\frac{2}{3}}|\psi_{2}>$$

Then

$$H|\varphi>=\sqrt{\frac{1}{3}}E_{1}|\psi_{1}>+\sqrt{\frac{2}{3}}E_{2}|\psi_{2}>$$

Clearly |φ> is not an eigenstate of the Hamiltonian H. The expectation value of the energy however, is

$$<E>= <\varphi|H|\varphi>=\frac{1}{3}E_{1}+\frac{2}{3}E_{2}$$

In other words the expectation value is neither of the eigenvalues, but a weighted average where the weights are the probability coefficients of the eigenstates.