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Suppose I have two quantum systems associated with angular momenta $J_1$ and $J_2$ respectively.

  • I can define the angular momentum of the whole system with the operator $J$ acting on $\scr H:={\scr H_1 \otimes H_2}$, defined as $J := J_1 \otimes \mathbb I_2 + \mathbb I_1 \otimes J_2$, where $\mathbb I$ is the identity operator.

  • The inner product $\langle .,. \!\rangle : {\scr H \otimes H} \to \mathbb C $ is defined as $\langle\psi, \phi\rangle = \langle \psi_1 \otimes \psi_2, \phi_1 \otimes \phi_2\rangle = \langle \psi_1, \phi_1 \!\rangle _1 \langle \psi_2, \phi_2 \!\rangle _2$, where $\langle .,. \!\rangle_1 $ and $ \langle ., .\!\rangle_2$ are the inner products defined on $\scr H_1$ and $\scr H_2$ respectively.

Question. How is $J^2$ defined?

According to the previous definition $J^2 = J \cdot J = \left ( J_1 \otimes \mathbb I_2 + \mathbb I_1 \otimes J_2 \right ) \cdot \left ( J_1 \otimes \mathbb I_2 + \mathbb I_1 \otimes J_2 \right ) = \\ \underbrace{(J_1 \otimes \mathbb I_2) \cdot (J_1 \otimes \mathbb I_2)}_{J_1^2} + \underbrace{(J_1 \otimes \mathbb I_2) \cdot (\mathbb I_1 \otimes J_2)}_{J_1 \cdot J_2} + \underbrace{(\mathbb I_1 \otimes J_2) \cdot (J_1 \otimes \mathbb I_2)}_{J_2 \cdot J_1} + \underbrace{(\mathbb I_1 \otimes J_2) \cdot (\mathbb I_1 \otimes J_2)}_{J_2^2} $

What does it all mean? I know how the inner product behaves with vectors, but not with linear operators...

ric.san
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1 Answers1

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Perhaps it helps to write down all these things more explicitly:

To start, let us define $$J\equiv J_1 \otimes \mathbb{I}_2 + \mathbb{I}_1 \otimes J_2 $$ and $$J^2 \equiv J_x^2 + J_y^2 + J_z^2 \quad ,$$ where the analogous definition should hold for $J_1$ and $J_2$. Further, define for $k=x,y,z$: $$ J_k \equiv (J_1)_k \otimes \mathbb I_2 + \mathbb I_1 \otimes (J_2)_k$$

and let us compute \begin{align} J_k^2 \equiv J_k \, J_k &= \left((J_1)_k \otimes \mathbb{I}_2 + \mathbb{I}_1 \otimes (J_2)_k\right) \, \left((J_1)_k \otimes \mathbb{I}_2 + \mathbb{I}_1 \otimes (J_2)_k\right) \\ &= (J_1)_k \, (J_1)_k \otimes \mathbb{I}_2 + \mathbb{I}_1 \otimes (J_2)_k\, (J_2)_k + 2\, (J_1)_k \otimes (J_2)_k \quad . \end{align}

Adding the three contributions from $k=x,y,z$ yields

$$J^2 = J_1^2 \otimes \mathbb{I}_2 + \mathbb{I}_1 \otimes J_2^2 + 2\, J_1 \cdot J_2 \quad , $$

where we have defined $$J_1 \cdot J_2 \equiv (J_1)_x \otimes (J_2)_x + (J_1)_y \otimes (J_2)_y+ (J_1)_z \otimes (J_2)_z \quad .$$

Tobias Fünke
  • 8,777