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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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    Note that $(A \otimes B) (C \otimes D) = (AC) \otimes (BD)$. Could you specify your question? 'What does it all mean?' is a very broad question. – Tobias Fünke Jul 15 '21 at 11:49
  • Could you specify what is your doubt? – aneet kumar Jul 15 '21 at 11:50
  • $J^2=J_x^2+J_y^2+J_z^2$ where $J_x=J_x\otimes \mathbb{I} + \mathbb{I}\otimes J_x$ etc in your notation. – ZeroTheHero Jul 15 '21 at 11:51
  • @Jacob you've written the composition of two operators, not their inner product, I think. I'm wondering how to compute $(A \otimes B) \cdot (C \otimes D)$ – ric.san Jul 15 '21 at 11:52
  • @ZeroTheHero yes, I'd just see how to derive it from the definition of $J$. – ric.san Jul 15 '21 at 11:55
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    See for example this or this related post and the definition that @ZeroTheHero gave. I wanted to point out that you have made a mistake in the calculation, e.g. the first term of your last line is wrong (it must be an operator on the tensor Hilbert space, so $J_1^2 \otimes \mathbb{I}_2$). – Tobias Fünke Jul 15 '21 at 11:56
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    Note that it does boil down to $J_1^2+J_2^2+2 J_1\cdot J_2$ where $J_1\cdot J_2=J_{1x}J_{2x}+J_{1y}J_{2y}+J_{1z}J_{2z}$ with $J_{1x}=J_x\otimes \mathbb{I}$ etc. – ZeroTheHero Jul 15 '21 at 12:00
  • Thanks, that was actually my second doubt. You've written $2 J_1 \cdot J_2$, so the commutator $[J_1, J_2]$ should be zero. But is it true? Even for two interacting systems? – ric.san Jul 15 '21 at 12:02
  • I've added a comment in the related post @Jakob suggested: https://physics.stackexchange.com/questions/422369/how-is-the-product-l-cdot-s-between-orbital-and-spin-angular-momentum-operator – ric.san Jul 15 '21 at 12:21
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    See my F O U R T H___ A N S W E R here : Total spin of two spin- 1/2 particles. Especially all after equation (68). – Frobenius Jul 15 '21 at 12:23
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    yes the commutator is $0$ since $J_{1x}$ and $J_{2x}$ etc act in different spaces. – ZeroTheHero Jul 15 '21 at 15:37

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
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  • isn't the definition of $J^2$ colliding with that of $J_1 \cdot J_2$? Specifically, setting $1=2$, we obtain $J \cdot J = J_x \otimes J_x + J_y \otimes J_y + J_z \otimes J_z$, but each of them is not equal to $J_x^2 + J_y^2 + J_z^2 = J^2$. – ric.san Aug 13 '21 at 13:59
  • I don't understand. Could you please elaborate? The equation is also valid for the case of $J_1=J_2\equiv L$. Then $J^2 = L^2 \otimes \mathbb I + \mathbb I \otimes L^2 + 2, L \cdot L$. – Tobias Fünke Aug 13 '21 at 14:06
  • $J_x \otimes J_x \neq J_x J_x = J_x^2$ and similarly for the other components. It follows $J \cdot J \neq J^2$. – ric.san Aug 13 '21 at 14:12
  • I think your problem is one due to notation. For $L\equiv J_1 =J_2$, i.e. the case that both angular momenta are the same, we have that $(J_x)^2 = (L_x)^2 \otimes \mathbb I + \mathbb I \otimes (L_x)^2 + 2 L_x \otimes L_x$, as the above calculations show. – Tobias Fünke Aug 13 '21 at 14:16