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It is known that Clebsch-Gordan coefficients are those of a linear transformation from the product basis $\{|j_1,j_2;m_1,m_2\rangle\}_{m_1\in \{-j_1,...,j_1\},m_2\in \{-j_2,...,j_2\}}$ to the coupled basis $\{|j_1,j_2;J,M\rangle\rangle\}_{J\in \{|j_1-j_2|,...,j_1+j_2\},M\in \{-J,...,J\}}$ (the double ket to distinguish between the two basis). These two basis are orthonormal basis in the space $\mathscr{E}(j_1)\otimes\mathscr{E}(j_2)$ and that means that there is a unitatry operator $\hat U$ that transforms one basis to the other. The matrix elements of this operator are by construction the CG-Coefficients.

On the other hand a physical rotation $R$ is implemented on a ket in $\mathscr{E}(j_1)\otimes\mathscr{E}(j_2)$ by a unitary operator $\mathscr D (R)$.

Question:

Is there a physical rotation $R$ such that $\mathscr D (R) = \hat U$, where $\hat U$ is the change-of-basis operator $\hat U: \{|j_1,j_2;m_1,m_2\rangle\} \mapsto \{|j_1,j_2;J,M\rangle\rangle\}$ ?

My answer is no and here are my two arguments

  1. Diagonalisation of observables (namely $\mathbf J^2$ & $J_z$) is not possible by physical rotations

  2. The map $\mathscr D :SO(3) \to SU((2j_1+1)(2j_2+1))$ may not be surjective, i.e. there are unitary operators $\hat{ \mathcal O} \in SU((2j_1+1)(2j_2+1))$ which do not correspond to physical rotations $R \in SO(3)$

These two arguments do not exclude each other but I hope that someone can make this mathematically more precise, I hope this question makes sense.

Added:

in other words are there e.g. Euler angles $\alpha, \beta,\gamma$ such that matrix elements are something like this $$ \mathscr D^{(J)}_{M',M}(\alpha, \beta,\gamma) = \langle j_1,j_2;m_1,m_2|j_1,j_2;J,M\rangle $$ I know the indices are not consistent, they do not make sense! but this may not be a reason to beleive that there is no $R(\alpha, \beta,\gamma)$ whose representation $\mathscr D(R)$ does as

$$\hat U: \{|j_1,j_2;m_1,m_2\rangle\} \mapsto \{|j_1,j_2;J,M\rangle\rangle\}$$

Physor
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  • Please confirm I have edited your question properly, you mean for any $\hat U$ right? – ohneVal Apr 08 '21 at 14:57
  • Possibly helpful. The Clebsch matrix changes basis on states and matrices, hence coproduces. – Cosmas Zachos Apr 08 '21 at 15:00
  • The Clebsch matrix can be made real and orthogonal, so it is a rotation. Did you try/look at the spin 1/2 ⊗ 1/2 paradigm? – Cosmas Zachos Apr 08 '21 at 15:14
  • @ohneVal I meant for the $\hat U$ that performs the change-of-basis from the $${|j_1,j_2;m_1,m_2\rangle} \mapsto {|j_1,j_2;J,M\rangle\rangle}$$ – Physor Apr 08 '21 at 15:15
  • @CosmasZachos your comments are not quite clear to me, but let me explain in the post above, I'll add something – Physor Apr 08 '21 at 15:26
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    The change-of-basis matrix $\hat U$ is the Clebsch matrix, which can be made orthogonal, so it is a rotation matrix in the full space, and transforms the reducible rotation matrices on your reduced Hilbert space, by acting on both sides. I am equally baffled by your "arguments". Just write down the 1/2 ⊗ 1/2 case and behold. I think you are entrapping yourself by excessively general notation. – Cosmas Zachos Apr 08 '21 at 15:35
  • Post "added" note: your anticipation is malformed. The Clebsch matrix, e.g. in the above 1/2 ⊗ 1/2 case, is a 4x4 orthogonal matrix, and not a rotation matrix on a reduced subspace! – Cosmas Zachos Apr 08 '21 at 15:41
  • The Clebsch Matrix is a rotation..Ok in some sense in Hilbert space, but the question is: does it correspond to some physical rotation of the system ? I thought about acting some arbitrary unitary $(2l+1)\times(2l+1)$-matrix on spherical harmonics to see if the result is a rotated spherical harmonics in the physical 3D-space, that works fine if I use D-Matrices but I don't know if it works for any (haven'T tried that yet)...but using ClebschGordan matrix would be special. and Please think about something more than that 1/2-spin, – Physor Apr 08 '21 at 15:42
  • Oh... "physical" denotes angles of the rotations on your space, identical for each Kronecker factor. No, the formal "angles" of the Clebsch matrix are fixed, and deal with the representation series, not the arbitrary rotation angles which they cover in their entirety: one Clebsch matrix covers all physical rotations. – Cosmas Zachos Apr 08 '21 at 15:46
  • I'm sorry that my question is very misunderstood by you, but in any case the CG-Matrix is only one rotation one operator it doesn't cover more than one rotation, namely from the old product basis to the coupled. I'm sorry again that my question is misunderstood like that – Physor Apr 08 '21 at 15:49
  • Here it is, if you wished to use it. It is but $I_2 \oplus \exp (-i\frac{\pi}{4}\sigma_2 )$, but in a space where the 4th components were moved over next to the 1st, distinctly not the reduction of the added spins! – Cosmas Zachos Apr 08 '21 at 16:03
  • ...and by the way, my statements on spin 1/2 work on all spins; I invited you to check the trivial case to appreciate the systematics of the spaces you have misconfigured.... – Cosmas Zachos Apr 09 '21 at 00:45

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