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Consider a system $A$ whose basis states are $|\phi_1\rangle_{(i)}\in H^{(1)}$ and a system $B$ whose basis states are $|\phi_2\rangle_{(j)}\in H^{(2)}$.
Then the basis states of the combined system is the tensor product of basis states of the individual system, i.e., $|\phi_1\rangle_{(i)}\otimes |\phi_2\rangle_{(j)}\in H^{(1)}\otimes H^{(2)}$.

We know that rotation of wavefunction about any axis in space is generated by angular momentum in that direction.
Suppose $L_z^{(1)\otimes (2)}$ is the angular momentum bector in z direction acting in $H^{(1)}\otimes H^{(2)}$, $L_z^{(1)}$ in $H^{(1)}$ and $L_z^{(2)}$ in $H^{(2)}$

We know that $L_z^{(1)\otimes (2)}=(L_z^{(1)}\otimes I^{(2)})+(I^{(1)}\otimes L_z^{(2)})$

Now consider $|\phi_1\rangle_{(i)}=|j_1,m_1\rangle$ and $|\phi_2\rangle_{(j)}=|j_2,m_2\rangle$

$L_z^{(1)\otimes (2)}(|j_1,m_1\rangle\otimes |j_2,m_2\rangle)=(L_z^{(1)}\otimes I^{(2)})(|j_1,m_1\rangle\otimes |j_2,m_2\rangle)+(I^{(1)}\otimes L_z^{(2)})(|j_1,m_1\rangle\otimes |j_2,m_2\rangle)$
$\implies L_z^{(1)\otimes (2)}(|j_1,m_1\rangle\otimes |j_2,m_2\rangle)=(L_z^{(1)}|j_1,m_1\rangle)\otimes (I^{(2)}|j_2,m_2\rangle)+(I^{(1)}|j_1,m_1\rangle)\otimes (L_z^{(2)}|j_2,m_2\rangle)\tag{1}$
$\implies L_z^{(1)\otimes (2)}(|j_1,m_1\rangle\otimes |j_2,m_2\rangle)=(m_1\hbar|j_1,m_1\rangle)\otimes |j_2,m_2\rangle+|j_1,m_1\rangle\otimes (m_2\hbar|j_2,m_2\rangle)\tag{2}$
$\implies L_z^{(1)\otimes (2)}(|j_1,m_1\rangle\otimes |j_2,m_2\rangle)=m_1\hbar(|j_1,m_1\rangle\otimes |j_2,m_2\rangle)+m_2\hbar(|j_1,m_1\rangle\otimes |j_2,m_2\rangle)\tag{3}$
$\implies L_z^{(1)\otimes (2)}(|j_1,m_1\rangle\otimes |j_2,m_2\rangle)=(m_1+m_2)\hbar(|j_1,m_1\rangle\otimes |j_2,m_2\rangle)\tag{4}$

In going from $(2)$ to $(3)$, I am not able to understand how.
$(m_1\hbar|j_1,m_1\rangle)\otimes |j_2,m_2\rangle=m_1\hbar(|j_1,m_1\rangle\otimes |j_2,m_2\rangle)$?

Because if the above identity is true then
$m_1\hbar(|j_1,m_1\rangle\otimes |j_2,m_2\rangle)= (m_1\hbar|j_1,m_1\rangle)\otimes |j_2,m_2\rangle=|j_1,m_1\rangle\otimes (m_1\hbar|j_2,m_2\rangle)$

Isn't this incorrect?
I am not able to justify to take the scalar out from the first state and multiply it to the whole composite state.
After all the tensor product of stayes is introduced to describe the states of multipartite system. Isn't that identity leads to ambiguity because.
$\lambda(|j_1,m_1\rangle\otimes |j_2,m_2\rangle)=(\lambda|j_1,m_1\rangle)\otimes |j_2,m_2\rangle=|j_1,m_1\rangle\otimes \lambda(|j_2,m_2\rangle)$?

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  • $\begingroup$ Related : Total spin of two spin-1/2 particles. $\endgroup$
    – Frobenius
    Commented Sep 8, 2022 at 15:21
  • $\begingroup$ Correct your garbled formula above your "Isn't this incorrect?" ... $\endgroup$ Commented Sep 12, 2022 at 16:51
  • $\begingroup$ Sorry for the inconvenience. I have updated it. $\endgroup$
    – Manu
    Commented Sep 12, 2022 at 17:18

1 Answer 1

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The ability to relocate numbers $(\lambda {\bf a})\otimes {\bf b}={\bf a}\otimes (\lambda {\bf b})= \lambda({\bf a}\otimes {\bf b})$ is part of the definition of the tensor product.

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  • $\begingroup$ Thanks for the answer, I have read about it. Bit what is the physical intuition behind this identity. After all the tensor product of stayes is introduced to describe the states of multipartite system. Isn't that identity leads to ambiguity because $\lambda(|a\rangle\otimes |b\rangle)=(\lambda |a\rangle)\otimes|b\rangle=|a\rangle\otimes(\lambda|b\rangle)$? $\endgroup$
    – Manu
    Commented Sep 9, 2022 at 2:06
  • $\begingroup$ @Manu Illustrate by tensor multiplying two two-vectors explicitly. $\endgroup$ Commented Sep 11, 2022 at 12:33

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