1
$\begingroup$

I'm trying to understand what Landau and Lifshitz mean in their $\S31$ of "Quantum mechanics. Non-relativistic theory" about vector model of addition of angular momenta:

... This result can be illustrated by means of what is called the vector model. If we take two vectors $\mathbf L_1,\mathbf L_2$ of lengths $L_1$ and $L_2$, then the values of $L$ are represented by the integral lengths of the vectors $\mathbf L$ which are obtained by vector addition of $\mathbf L_1$ and $\mathbf L_2$; the greatest value of $L$ is $L_1+L_2$, which is obtained when $\mathbf L_1$ and $\mathbf L_2$ are parallel, and the least value is $|L_1-L_2|$, when $\mathbf L_1$ and $\mathbf L_2$ are antiparallel.

As I understand, I should take vectors $\mathbf L_1=(\sqrt{L_1^2-M_1^2}\;\;\;M_1)^T$ and $\mathbf L_2=(\sqrt{L_2^2-M_2^2}\;\;\;M_2)^T$, and get an integral length $L=|\mathbf L_1+\mathbf L_2|$. But when I take e.g. $L_1=3$, $M_1=1$, $L_2=5$, $M_2=-3$, I get

$$L=2\sqrt{7+4\sqrt2}\approx7.12,$$

which is by no means integral. What am I missing? Should I round the result to integer? Or does the vector model actually work only qualitatively?

$\endgroup$
1
$\begingroup$

which is by no means integral. What am I missing? Should I round the result to integer? Or does the vector model actually work only qualitatively?

They are saying that you should take $L_1$ and $L_2$ to be non-negative integer values. And then $L$ ranges over the non-negative integer values from $L_1+L_2$ to $|L_1-L_2|$.

For example, if $L_1=3$ and $L_2=5$ then $L$ can range from 8 to 2.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Hmm, so this model only gives us the range, like in triangle inequality, and is otherwise not useful, right? $\endgroup$ – Ruslan Mar 18 '15 at 18:51
  • $\begingroup$ Yes, this simple picture is really just used to help remember the correct range for $L$ given $L_1$ and $L_2$. As you have shown, you can't interpret the resultant quantum angular momentum $\vec L$ correctly just in terms of this simple picture. $\endgroup$ – hft Mar 18 '15 at 20:24
1
$\begingroup$

I don't quite understand the construction you are using for $\vec{L_1}$ and $\vec{L_2}$. Normally, $M$ is used to designated the projection of the angular momentum vector along one axis.

The idea of his model, consisting here of integer angular momentum, is to consider only the integer values obtained in all the possible vector addition of $\vec{L_1}$ and $\vec{L_2}$. Starting with both vectors aligned, you obtain the biggest possible value, being $L=L_1+L_2$, down to the smallest possible value when the vectors are contra-aligned, $L=|L_1-L_2|$, along with all the intermediate values, keeping only integers. These are the possible values for $L$ in addition of the angular momentum.

I myself think that approaching these topics algebraically is more complete, without refering to geometric analogies. In any case, you shouldn't bother too much with this part if the preceding demonstration is clear to you.

| cite | improve this answer | |
$\endgroup$
0
$\begingroup$

Please keep in mind that "quantum mechanics" (QM) begins with the word "quantum". Angular momentum in QM is quantized, i.e., values like 7.12 are never allowed for L. The rules for angular momenta arise from corresponding operators and their group theory.

I try to keep in mind when combining two angular momenta that much of the stuff in textbooks is a bit misleading. You will find statements such as "in the ... model $\vec L_1$ and $\vec L_2$ couple" or "in the ... representation $\vec L$ and $\vec S$ couple" [Here the ... descriptors could be "coupled", "uncoupled", "L-S", "jj", "irreducible", "direct product", etc.]

While they tell you which representation to use, one gets the impression that there is a "choice". Yes, there is a choice, but the "good" quantum numbers, i.e., eigenvalues of operators that commute with the Hamiltonian ($H$) and are thus useful labels for states because they do not change with time, are really determined by the physics. This means that knowledge of perturbative terms in the Hamiltonian for the system is important - it tells us which choice to make. For instance, since $\vec L\cdot\vec S$ can be expressed in terms of $J^2$, $L^2$, and $S^2$, if $H$ contains a term proportional to $\vec L\cdot\vec S$ we should use the $|j,m\rangle$ representation, and not the $|m_l, m_s\rangle$ representation.

The "vector model" or "semi-classical description" is just that: a semi-classical description. This is a useful description when you know which model you must use based on the Hamiltonian, but then you have to refer to a text for the resulting rules for the angular momenta and states. If you want to derive such things though, it's better to stick with quantum mechanics and group theory and directly obtain the result.

| cite | improve this answer | |
$\endgroup$

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.