I'm following R. Shankar's Quantum mechanics and a certain statement, I just don't understand.

Suppose the effect of acting on a state $|\alpha lm\rangle $ with $T^q_k$. Let us rotate the resulting state and see what happens: $$U[R]T^q_k|jm\rangle=U[R]T^q_k U^\dagger[R]U[R]|jm\rangle $$ $$=\sum_{q'}D^{(k)}_{q'q}T^{q'}_k\sum_{m'}D^{(j)}_{m'm}|jm'\rangle=\sum_{q',m'}D^{(k)}_{q'q}D^{(j)}_{m'm}T^{q'}_k|jm'\rangle$$

We find that $T^q_k|jm\rangle$ responds to rotations like the product ket $|kq\rangle\otimes|jm\rangle$.

I understand this as $$U[R](|kq\rangle \otimes|jm\rangle)=\sum_{q'}D^{(k)}_{q'q}|kq'\rangle\otimes\sum_{m'}D^{(j)}_{m'm}|jm'\rangle$$

But then He said,

Thus, when we act on a state with $T^q_k$, we add angular momentum $(k,q)$ to the state.

In other words, an irreducible tensor operator $T^q_k$ imparts a definite amount of angular momentum $(k,q)$ to the state it acts on.

Can you please explain this paragraph? It will be great if you can give an example sort of thing for lower dimensions.

I attempt to test this statement backward As

$$T^q_k|jm\rangle =|j+k,q+m\rangle$$ As claimed by the author (apart from constant). Now If we act the rotation on this state then $$U[R]T^q_k|jm\rangle =U[R]|j+k,q+m\rangle=\sum_{m'}\mathcal{D}^{(k+j)}_{m',m+q}|k+j,m'\rangle$$

Now, this must be equal to the right-hand side of the author's calculation. But I don't understand How this supposed to be.

  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$
    – rob
    Commented Dec 13, 2020 at 21:55

2 Answers 2


The states $T^q_k\vert \alpha \ell m\rangle$ can have angular momentum in the range $$ k\otimes \ell=(\ell+k)\oplus (\ell+k-1)\ldots \oplus \vert \ell -k\vert\, , \tag{1} $$ thus changing the angular momentum of the initial state.

In other words, $$ T^q_k\vert \alpha \ell m\rangle =\sum_{\beta L} \vert \beta L, m+q\rangle \langle \beta L,m+q\vert T^q_k\vert \alpha \ell m\rangle\, , $$ where $L$ is in the range given in (1). The operator $U[R]$ is just a rotation so it will not change $L$: \begin{align} U[R]T^q_k\vert \alpha \ell m\rangle &=\sum_{\beta L} U[R]\vert \beta L, m+q\rangle \langle \beta L,m+q\vert T^q_k\vert \alpha \ell m\rangle\, , \\ &=\sum_{\beta L M'}\vert \beta L, M'\rangle D^L_{M',m+q}[R]\langle \beta L,m+k\vert T^q_k\vert \alpha \ell m\rangle\, . \tag{2} \end{align} Alternatively, since by definition \begin{align} U[R]T^q_k U^{-1}[R]&=\sum_{k'} T^{q'}_{k} D^k_{q'q}[R] \, ,\\ U[R]\vert jm\rangle &= \sum_{m'} \vert jm'\rangle D^j_{m'm}[R] \end{align} your first two equations follow immediately. The combination $$ D^k_{q'q}[R] D^j_{m'm}[R]=\sum_{L} c^{k L}_{q'q;m'm} D^{L}_{M'M}[R] \tag{3} $$ with $c^{k L}_{q'q;m'm}$ a known coefficient is of the same form as (2), with the range of $L$ the same as (1). Pushing through the calculation you'd get of course that (3) would eventually agree with (2).

Note that the resulting angular momentum can be increased or decreased as per (1).

  • $\begingroup$ Thank you for your response but I don't understand the first line. In calculation, you have $U$ on the left hand How you didn't consider that? $\endgroup$ Commented Dec 11, 2020 at 3:46
  • $\begingroup$ @CosmasZachos good catch... I will fix it. $\endgroup$ Commented Dec 11, 2020 at 23:46

Just going to illustrate usage, since asked, without covering the dispositive rotational features (tossing the baby with the bathwater!) well-covered in the other answer.

So, take $k=j=1/2$ , combining to $L=1$ or else 0, and $M=m+q$. The four vectors in the tensor product, then, break up into a 3d and a 1d subspace, $$ T^q_{(1/2)}|1/2 ~m\rangle =|1~m+q\rangle \langle 1~m+q| T^q_{(1/2)}|1/2 ~m\rangle + |0~0\rangle \langle 0~0| T^{q}_{(1/2)}|1/2 ~m\rangle \delta_{m,-q} . $$

Consider the matrix elements $$ \langle 1~m+q| T_{(1/2)}^q|1/2~m\rangle \\ = \langle 1/2 ~m; ~1/2 ~q|1~m+q\rangle \langle 1||T_{(1/2)}|| 1/2\rangle . $$ The second factor on the r.h.s is just a number, the reduced matrix elt, common to all m and q. The first factor is a Clebsch that you may use to correlate all four matrix elements, here somewhat trivial. But imagine a larger example...

As for intuition, I fear its is mostly math... Messiah v2 ch XIII Β§32 proves it wonderfully. My "mnemonic" is the way you already understand it as a sort of direct product of two states.

Note after the edit of the question It becomes clear you misread what the author is saying by "adding angular momentum to the state". He means exactly what @ZerotheHero points our in his answer, between his (1) and (2), which is what I am illustrating for the simplest case above, to wit, in your comment's language, $$ 𝑇^{π‘š_1}_{(𝑗_1)}|𝑗_2 ~ π‘š_2\rangle = \sum_j |j~~~m_1+m_2\rangle \langle j~~~m_1+m_2 | 𝑇^{π‘š_1}_{(𝑗_1)}|𝑗_2 ~ π‘š_2\rangle, $$ where j ranges over the allowable values in the Kronecker product (~ vector sum) of $j_1$ and $j_2$. It's understood that entries with $|m_1+m_2| > j$ are zero. By the Wigner-Eckart theorem, the coefficients in the sum on the right are proportional to Clebsches.

Conserning your Clebsch question, illustrated, starting from your conventional $$ |00\rangle= |1/2~1/2,1/2~-1/2\rangle \frac{1}{\sqrt 2 } +|1/2~-1/2,1/2~1/2\rangle \frac{-1}{\sqrt 2 }\\ |11\rangle= |1/2~1/2,1/2~1/2\rangle \\ |10\rangle= |1/2~1/2,1/2~-1/2\rangle \frac{1}{\sqrt 2 } +|1/2~-1/2,1/2~1/2\rangle \frac{1}{\sqrt 2 }\\ |1~-1\rangle= |1/2~-1/2,1/2~-1/2\rangle , $$ utilize the orthogonality of the Clebsch matrix to invert to the form we use, $$ |1/2~1/2,1/2~-1/2\rangle =|10\rangle \frac{1}{\sqrt 2 }+|00\rangle \frac{1}{\sqrt 2 } \\ |1/2~1/2,1/2~1/2\rangle =|11\rangle \\ |1/2~-1/2,1/2~1/2\rangle = |10\rangle \frac{1}{\sqrt 2 } +|00\rangle \frac{-1}{\sqrt 2 }\\ |1/2~-1/2,1/2~-1/2\rangle =|1~-1\rangle . $$

I'm not sure whether writing down the explicit tensor products in $\otimes$ notation might help you, but it can be easily done. Proble 4 in these notes might be helpful.

  • $\begingroup$ I tried to prove backward but still, I'm in mess. Can you help me through this? I edited the question. Please see my attempt. Thank you for your answer. But still, I'm having trouble. $\endgroup$ Commented Dec 12, 2020 at 16:18
  • $\begingroup$ Thank you for writing the Kronecker product, I don't about that in matrix terms so It's hard for me to think about the product state and sum state. So $\oplus $ of state refer to Kronecker sum? $$A\oplus B=A\otimes I_b+I_a\otimes B$$ $\endgroup$ Commented Dec 12, 2020 at 18:17
  • $\begingroup$ $\oplus$ means direct sum, so the separate blocks the direct product reduces to after a similarity transformation (Clebsching). What you write there is is not a direct sum, it is the coproduct, i.e. the composition of two reps , A and B, of the same Lie algebra. When you exponentiate it, it yields a group element $e^{i\theta A}\otimes e^{i\theta B}$, like the rotation matrix acting on the respective components here. $\endgroup$ Commented Dec 12, 2020 at 18:48
  • $\begingroup$ @CosmasZacos Thanks You for the attached notes, It makes my concept much more clearer. As for me, matrix makes much more sense than abstract notation. I want to learn more. Any reference book, Where I can find this sort of details. $\endgroup$ Commented Dec 13, 2020 at 19:36
  • $\begingroup$ These questions and also this one use the matrix notation. I have posted a few myself, but can't find them. Don't know about a book. This is what I have been teaching and illustrated for different examples on this site, but I don't know of a book... it is sort of self explanatory... $\endgroup$ Commented Dec 13, 2020 at 21:26

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