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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.
$\lambda(|j_1,m_1\rangle\otimes |j_2,m_2\rangle)=(\lambda|j_1,m_1\rangle)\otimes |j_2,m_2\rangle=|j_1,m_1\rangle\otimes \lambda(|j_2,m_2\rangle)$?

Manu
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1 Answers1

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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.

mike stone
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  • 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 $\lambda(|a\rangle\otimes |b\rangle)=(\lambda |a\rangle)\otimes|b\rangle=|a\rangle\otimes(\lambda|b\rangle)$? – Manu Sep 09 '22 at 02:06
  • @Manu Illustrate by tensor multiplying two two-vectors explicitly. – Cosmas Zachos Sep 11 '22 at 12:33