Suppose I have two quantum systems $(1)$ and $(2)$, for each of them the angular momenta $J^{(1)}$ and $J^{(2)}$ are defined. Our purpose is to create a total angular momentum $ J$ which can describe simultaneously $(1)$ and $(2)$.

Since $J^{(1)}$ and $J^{(2)}$ can be acting on completely diffent Hilbert spaces, the only way to do this is to "promote" both of them in the tensor product: $$ { J^{(1)} \longrightarrow J^{(1)} \otimes {\mathbb I}^{(2)} \\ J^{(2)} \longrightarrow {\mathbb I}^{(1)} \otimes J^{(2)} } $$

and then define a total $J$ on ${\cal H}^{(1)} \otimes {\cal H}^{(2)} $ as follows: $$ J := J^{(1)} \otimes {\mathbb I}^{(2)} + {\mathbb I}^{(1)} \otimes J^{(2)} $$

If $(1)$ and $(2)$ are independent each other, a complete set of commuting observables CSCO is given by $\{J^{2(1)}, J^{2(2)}, J_z^{(1)}, J_z^{(2)} \}$. Another choice, particularly useful when there is interaction between the two systems is this alternative CSCO : $\{ J^{2(1)}, J^{2(2)}, J^{2}, J_z \}$. Their respective basis vectors are sometimes written as $| j_1,j_2, m_1, m_2 \rangle $ and $ | j_1, j_2, j, m \rangle$

Now, since they're both valid bases for the same Hilbert space ${\cal H}^{(1)} \otimes {\cal H}^{(2)} $ it should be possible to change representation from one base to another.

Here is my first question: why I want that? Mathematically it makes sense, but physically speaking... why should I care? Two non-interacting systems behave very differently from two interacting ones, and that's exactly why I chose two different sets of commuting observables. So where is the interest in describing one system with the "wrong" base?

But anyway, let's do this. The final result gives the so-called Clebsch-Gordan coefficients. It all starts with some preliminary observations:

Remark. Since we're dealing with bases, i.e. complete orthonormal sets, we have the following identites:

$$\sum_{j_1, j_2, m_1, m_2} |j_1, j_2, m_1, m_2 \rangle \langle j_1, j_2, m_1, m_2 | = \mathbb I \\ \sum_{j_1, j_2, j , m } |j_1, j_2,j,m \rangle \langle j_1, j_2, j, m | = \mathbb I $$

The same holds even if we keep $j_1$ and $j_2$ fixed. Why? A base is by definition the minimum amount of linear independent elements that generate the entire space. If I restrict it (by keeping $j_1$ and $j_2$ fixed) I shouldn't obtain a base...

  • $\begingroup$ What do you know about group theory? You know the difference between reducible and irriducible representations? Do you know what a Casimir operator is? $\endgroup$ Commented Jul 15, 2021 at 19:19
  • $\begingroup$ Your last equation is incorrect: there is no sum on $j_1,j_2$ since $\mathbb{I}=\sum_{jm}\vert jm\rangle\langle jm\vert$. In the context of this equation $j_1,j_2$ are fixed. $\endgroup$ Commented Jul 18, 2021 at 0:00
  • $\begingroup$ @ZeroTheHero does that mean I cannot say if the set $\displaystyle \sum_{j1_,j_2,m_1,m_2} |j_1,j_2,m_1,m_2⟩⟨j_1,j_2,m_1,m_2|$ is complete or not, even if it is for each fixed value of $j_1$ and $j_2$? $\endgroup$
    – ric.san
    Commented Jul 18, 2021 at 18:16
  • $\begingroup$ No. For fixed $j_1,j_2$ you have $\sum_{m_1m_2} \vert j_1m_1;j_2m_2\rangle \langle j_1m_1;j_2m_2\vert =\mathbb{I}$ and, for fixed $j_1m_1,j_2m_2$, you have $\sum_{j} \vert j m\rangle\langle j m \vert =\mathbb{I}$ where $m=m_1+m_2$. $\endgroup$ Commented Jul 18, 2021 at 19:49
  • $\begingroup$ Ok. That's really strange though. The atomic states are given by three quantum numbers $n , l$ and $m$, and $l$ goes from $0$ to $n-1$... $\endgroup$
    – ric.san
    Commented Jul 18, 2021 at 20:11

1 Answer 1

  • Obviously, if the two systems are not interacting with each other ("they are not coupled") there is no point in the coupled basis: it's not wrong, it's pointless. However, when you couple them by a rotationally invariant coupling, your uncoupled basis is useless, since neither $J_1$ nor $J_2$ are good constants of the motion (conserved). However, $J$ would be conserved, so a good quantum number. Quoting WP,

For instance, the orbit and spin of a single particle can interact through spin–orbit interaction, in which case the complete physical picture must include spin–orbit coupling. Or two charged particles, each with a well-defined angular momentum, may interact by Coulomb forces, in which case coupling of the two one-particle angular momenta to a total angular momentum is a useful step in the solution of the two-particle Schrödinger equation. In both cases the separate angular momenta are no longer constants of motion, but the sum of the two angular momenta usually still is.

  • The two freaky completeness relations you wrote are terribly misguided; where on earth did they come from? Normally you work within the Hilbert space of dimension $(2j_1+1)(2j_2+1) $ for a fixed $j_1$ and $j_2$, mere labels, and then you only sum $$ \sum_{m_1 = -j_1}^{j_1} \sum_{m_2 = -j_2}^{j_2} |j_1 \, m_1 \, j_2 \, m_2\rangle \langle j_1 \, m_1 \, j_2 \, m_2 |= {\mathbb I}. $$ As a consequence, $$ |j \, m\rangle = \sum_{m_1 = -j_1}^{j_1} \sum_{m_2 = -j_2}^{j_2} |j_1 \, m_1 \, j_2 \, m_2\rangle \langle j_1 \, m_1 \, j_2 \, m_2 | j \, m\rangle . $$

A similar relation holds for the coupled basis states for fixed $j_1$ and $j_2$, summing over $j,m$, where $j_1+j_2\geq j\geq |j_1-j_2|$.

I have no idea why you believe you don't get a basis for fixed $j_1$ and $j_2$. Have you tried a simple example? Adding spin 1/2 to spin 1 is illustrated on this site. Have you checked out the PDG or any basic QM text?

Add-on on OP's comment

Your lecturer flakes out and slips in 58:03, but quickly recovers in 58:19 ...59:30 and writes the identity I write for fixed $𝑗_1$ and $𝑗_2$, for the "smaller Hilbert space", utilized throughout the C-G discussion, of dimension $(2𝑗_1+1)(2𝑗_2+1)$ and never looks back. Unfortunate pedagogy.

I have personally never seen the broader construction summing over each orthogonal finite-dimensional Hilbert space, since these states are orthogonal, and do not share different Hilbert spaces: they are disjoint superselection sectors, and a classic case of notational overkill.

Both coupled and uncoupled bases connect to each other in each finite-dimensional "small Hilbert space" separately, since they share $𝑗_1$ and $𝑗_2$, and the different Hilbert spaces never get to "talk to each other". Hence, this broader superfluous and pointless reduced array can hardly yield anything useful (I'll stick my neck out here). I fear your lecturer was negligently hidebound.

  • $\begingroup$ This isn't quite right. The angular momentum operators $\vec{J}_{1}$ and $\vec{J}_{2}$ are not constants of the motion when you have two coupled angular momenta. This means that there are not well-defined eigenvalues $m_{1}\hbar$ and $m_{2}\hbar$ of $J_{1z}$ and $J_{2z}$ in an energy eigenstate. However, the total angular momentum squared is still conserved, as mentioned in the question; $\vec{J}^{2}_{i}$ still has an eigenvalue $j_{i}(j_{i}+1)\hbar^{2}$. $\endgroup$
    – Buzz
    Commented Jul 18, 2021 at 1:37
  • $\begingroup$ @Buzz ? Isn’t this precisely what I’m saying? $\endgroup$ Commented Jul 18, 2021 at 2:33
  • $\begingroup$ With the edits, yes. $\endgroup$
    – Buzz
    Commented Jul 18, 2021 at 4:35
  • $\begingroup$ @CosmasZachos I found these two identities on this MIT lecture: ocw.mit.edu/courses/physics/8-05-quantum-physics-ii-fall-2013/… min.58 $\endgroup$
    – ric.san
    Commented Jul 18, 2021 at 18:25

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