In the book by Arthur Beiser, Concepts of modern physics, in the chapter LS coupling there is this image:

enter image description here


How do we get total orbital angular momentum $L=3$ (image (a)) out of quantum numbers $\ell_1=1 \Rightarrow m_{\ell 1}=-1,0,1$ and $\ell_2=2 \Longrightarrow m_{\ell 2} = -2,-1,0,1,2$? What about $L=2$ and $L=1$?

My attempt:

I tried to combine the $z$ components using equations $L_z=m_\ell \hbar$ and got these results for $\boxed{L\equiv L_z}$:

\begin{align} L_{1z}=& \left\{ \begin{aligned} -1&\hbar\\ 0&\\ 1&\hbar \end{aligned} \right.\\ L_{2z}=& \left\{ \begin{aligned} -2&\hbar\\ -1&\hbar\\ 0&\\ 1&\hbar\\ 2&\hbar \end{aligned} \right. \end{align}

If I sum these up one of the results is $-3\hbar$, which is weird and isn't listed as one of the possible values in the picture. Also the picture states that one of the possible total orbital angular momenta is $L=3$ and not $L=3\hbar$. What is this a mistake or what?


Firstly, let's drop the factors of $\hbar$ for the sake of simplicity, as is often done (see also Emilio Pisanty's answer). By the vector coupling model for the combination of angular momenta, we know that the combination of two angular momentum quantum numbers $j_1$ and $j_2$ will yield allowed total angular momentum quantum numbers $j$ determined by $|j_1-j_2| \leq j \leq j_1+j_2$.$^1$

In this case, we have

$$|l_1-l_2| \leq l \leq l_1+l_2$$ $$|2-1| \leq l \leq 2+1$$ $$1 \leq l \leq 3$$

and similarly

$$|s_1-s_2| \leq s \leq s_1+s_2$$ $$|1/2-1/2| \leq s \leq 1/2+1/2$$ $$0 \leq s \leq 1$$

$^1$ The way to derive this triangle inequality is by a process of elimination, starting from the condition that $m = m_1+m_2$ where the $m$'s are the eigenvalues (quantum numbers) of the $z$-components of the respective angular momenta. This condition can, in turn, be derived by expressing the coupled basis as linear combinations of the product vector basis and letting $j_z = j_{z1}+j_{z2}$ act on both sides of the equality (for more detail, see e.g. the wikipedia page on Clebsch-Gordan coefficients).

Now, we know that the largest possible value of $m$ is $j$. We also know that the largest possible $m$ is obtained if (and only if) $m_1$ and $m_2$ are also as large as possible, i.e. $m_i = j_i, i=1,2$. So we know that $m = j_1+j_2$ is the largest possible value for $m$ and therefore $j = j_1+j_2$ must exist. There are $2j+1$ values for $m$ associated with this value of $j$, given by $m = -j, \cdots, j$.

The second largest value of $m$ is $j_1+j_2-1$, which can be obtained in two ways. These are $(m_1 = j_1, m_2 = j_2-1)$ and $(m_1 = j_1-1, m_2 = j_2)$. However, one of them corresponds to the previous value of $j$ ($j_1+j_2$). This still means we have one possibility left to get $m = j_1+j_2-1$ which doesn't correspond to an earlier value of $j$. Therefore, $j = j_1+j_2-1$ must also exist.

We continue this reasoning until we get to the smallest allowed value of $j$. We can say that this value is reached after $n$ steps if $j_k-n = -j_k$, where $j_k = \min{\{j_1,j_2\}}$. (think about why this is true) Say $j_k = j_2$, then $n = 2j_2$ and the smallest allowed value of $j$, $j_{min}$ is found to be

$$j_{min} = j_1+j_2-n = j_1-j_2,$$

or in the general case (taking into account the possibility of $j_k=j_1$):

$$j_{min} = |j_1-j_2|.$$

Summarizing, we have found that all the allowed values for $j$ are:

$$j_1+j_2, j_1+j_2-1, j_1+j_2-2, \cdots, |j_1-j_2|$$

which is often written as

$$|j_1-j_2| \leq j \leq j_1+j_2.$$

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Factors of $\hbar$ are routinely dropped from angular momenta in quantum mechanics as there is very seldom a chance of confusion. Angular momentum is intrinsically restricted to values of the form $m\hbar$ (for specific components) or $j(j+1)\hbar^2$ (for squared magnitudes), where $j$ and $m$ must be either integers or half-integers. Knowing $j$ gives you all the informaiton you need about the system - so you might as well drop the $\hbar$.

The minus sign you are getting is a magnetic quantum number, $L_z=m\hbar$ with $m=-3$. This is perfectly allowed and it corresponds to a total angular momentum of $l=3$.

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  • $\begingroup$ Why do you all use $l$ (orbital quantum number) in place of $L$ (orbital angular momentum)? In the book it only says that $L$ can be 3 2 or 1 and not -3, -2, -1... $\endgroup$ – 71GA Sep 8 '13 at 18:07
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    $\begingroup$ Because it's more convenient. It is hard to know what $L$ can be if you do not define what you mean by it. But if you're wondering why the angular momentum magnitude can't be negative, you need to sit back and think. $\endgroup$ – Emilio Pisanty Sep 8 '13 at 18:27
  • $\begingroup$ So i did this like i should but forgot that there is no such thing as negative magnitude... so from all the solutions for $L=3\hbar, 2\hbar, 1\hbar, 0, -1\hbar, -2\hbar, -3\hbar$ i have to simply remove negative ones and 0. $\endgroup$ – 71GA Sep 8 '13 at 18:51
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    $\begingroup$ No, the procedure is not quite so simple. $L=0$ may or may not be present, and some of the smaller positive integers may be missing as well. The final quantum number $l$ must obey $|l_1-l_2|\leq l\leq l_2+l_2$. $\endgroup$ – Emilio Pisanty Sep 8 '13 at 19:03

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