0
$\begingroup$

I do not understand why it's supposed to be vanishing. Rather than discussing the question in its full generality I prefer to consider the following scenario, which I think sums up anything that's actually important for my level and my purpose, but keep in mind the real point of the question.

Consider a system of six identical particles with spin one-half, each one of them in a eigenstate with azimuthal quantum number $ l = 1 $ and identical quantum number $ n $ given by the following single-particle hamiltonian (no interaction then):

$H_a := \frac{\vec{p_a}^2}{2m} + V(r_a) $

Due to the indistinguishability of the particles, the total wavefunction, orbital part plus spin part, must be antisymmetric with respect to the exchange between any two particles. Given six states, one for each particle, the Slater determinant identifies such a function. The six states must be different from one another, otherwise the Slater determinant vanishes and there's no wavefunction at all. More correctly, not only do they have to be different, they have to be linearly independent as well.

If all particles are in an eigenstate of a projection of $ \vec{L_a} $ and $ \vec{S_a} $ then I can see clearly that the sum of these values add up to zero. Since $ l = 1 $, then $ m = -1, 0, 1. $ Each state with a definite $ m $ splits in two independent states, with spin up and down. I finally have six different state and the value of the projections clearly add up to zero.

Yet by hypothesis I have only defined the azimuthal quantum number, meaning the above is only a particular case. The particles can actually be in any state, as long as they are independent. For example I could have some particles in an eigenstate of $ L_x $ and some particles in an eigenstate of $ L_z $ and in this situation I really can't see how the projections would add up to zero.

Even by considering only the special case above I don't see how this implies that the total angular momentum, not only the the projection of the total angular momentum, is vanishing. A typical argument I have encountered says there's nothing special about a particular component, but I don't seem to understand it: if I say each particle is in an eigenstate of $ L_z $, the $ L_{ztot} = 0 $ and clearly this applies to $ L_x $ and $ L_y $ for instance but nothing says $ L_{ztot} $ , $ L_{xtot} $ and $ L_{ytot} $ are simultaneously zero. In fact it seems to me that I prepare the system in two different ways from the beginning because I would prepare eigenstates of $ L_z $ in the first case and eigenstates of $ L_x $ later. These are again particular cases and potentially different, I can't help but thinking a proper argument shouldn't rely on statements like 'Each particle is in an eigenstate of a projection' because all I know is that particles are in the eigenspace where $ l = 1 $, nothing else.

I apologize if the post sounds too confusing.

$\endgroup$
1
$\begingroup$

The answer is that the antisymmetry of the state implies that the total angular momentum vanishes. Let's start with the simplest example, two spin-1/2 states. We can form a two-particle state with $S_z=0$ as $|1/2\rangle|-1/2\rangle$, or as $|-1/2\rangle|1/2\rangle$, but neither of these has vanishing total angular momentum. Only the antisymmetric linear combination does, $(|1/2\rangle|-1/2\rangle-|-1/2\rangle|1/2\rangle)/\sqrt{2}$. To see why this works in generally, consider the totally antisymmetric combination of products of the $2j+1$ states of a spin-$j$ representation (i.e. the Slater determinant of these states). Now act with any rotation on this state. Each of the states in the product will be acted on by the same rotation operator, and the resulting state will still be totally antisymmetric. But there is only one totally antisymmetric tensor product of $n$ vectors in an $n$-dimensional vector space, so after the rotation the state is the same, up to a possible scalar multiple. In fact the scalar multiple is nothing but the determinant of the 1-particle rotation matrix, which is 1. This means the antisymmetric state is invariant under all rotations. The angular momentum components are the infinitesimal generators of rotations, so the fact that the state is rotationally invariant implies that $\vec J$ annihilates it.

$\endgroup$
  • $\begingroup$ Another way to see that, in addition to $J_z$, both $J_x$ and $J_y$ annihilate the totally antisymmetric state, is to act with $J_\pm=J_x\pm i J_y$ on the state. Ever term either vanishes because $m=j$ can't be raised or because $m=-j$ can't be lowered, or because the result of the action is to make two $m$ values the same, which yields zero because of the antisymmetry of the state. $\endgroup$ – Ted Jacobson Mar 5 '16 at 20:24
1
$\begingroup$

I'm not totally understanding the question, but I think the critical aspect here is that in atoms, the energy gap between successive levels is much greater than the "thermal" energy of the electrons. So the energy levels really do have to fill up pairwise, with each pair of electrons forming a singlet state with spin zero. I think it might be otherwise in a situation where you had energy levels very close together, like a small chunk of metal with several hundred electrons in the conduction band. In this case, the thermal energy of the electrons is much greater than the separation between energy levels, so there is no need for the electrons to fill up each mode pairwise. You can then have electrons with random spins this way and that way, and you can end up with a total spin of 11 or 12 with a z-projection of whatever.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.