Let's say we have two Hilbertspaces $H_1$ and $H_2$, with angular momentum operators $j_1$ and $j_2$. On their product space $H_1 \otimes H_2$ we define the total angular momentum operator as:

$J = j_1 \otimes id^{(2)} + id^{(1)} \otimes {j_2}$

Now i tried to calculate the commutator in the product space $[J_x,J_y]$ (it should equal to $i \hslash J_z$):

$$[J_x,J_y] = [j_{1x}\otimes id^{(2)} + id^{(1)} \otimes j_{2x},j_{1y}\otimes id^{(2)} + id^{(1)} \otimes j_{2y}]\tag{1}$$

This splits up into a sum of 4 commutators, the first being:

$$[j_{1x}\otimes id^{(2)},j_{1y}\otimes id^{(2)}] = (j_{1x}\otimes id^{(2)})(j_{1y}\otimes id^{(2)}) - (j_{1y}\otimes id^{(2)})(j_{1x}\otimes id^{(2)})\tag{2}$$.

Now lets multiply a vector $|\psi \rangle |\phi \rangle \in H_1 \otimes H_2$ onto this commutator:

$$((j_{1x}\otimes id^{(2)})(j_{1y}\otimes id^{(2)}) - (j_{1y}\otimes id^{(2)})(j_{1x}\otimes id^{(2)})) |\psi \rangle |\phi \rangle = (j_{1x}j_{1y}-j_{1y}j_{1x})|\psi \rangle (id^{(2)}id^{(2)}-id^{(2)}id^{(2)})|\phi \rangle = i\hslash j_{1z}|\psi \rangle \ 0|\phi \rangle = 0\tag{3}$$

And i think by the same logic the complete commutator equals to zero, which clearly can't be true.

Can somebody identify the mistake or explain how the commutator on the product space is defined? Thank you.

  • $\begingroup$ Related : Total spin of two spin- 1/2 particles. F O U R T H___ A N S W E R, SECTION C : Angular Momentum Coupling and Product Transformations, equations from (68) to (77). $\endgroup$
    – Frobenius
    Jun 14 at 9:35

1 Answer 1


Your equation $3$ is not right. You seem to generate 4 terms from 2 terms. To see how to fix it, let's continue from equation $2$ as this is correct. For general tensor product operators the multiplication is defined in the following way.

$$ (A \otimes B)(C \otimes D) = (AB) \otimes(CD) \tag{1}$$

We will now use this to solve your problem. Consider your equation $2$. Using the formula above, we get

$$(j_{1x}\otimes id^{(2)})(j_{1y}\otimes id^{(2)}) - (j_{1y}\otimes id^{(2)})(j_{1x}\otimes id^{(2)}) = (j_{1x}j_{1y}\otimes id^{(2)}id^{(2)}) - (j_{1y}j_{1x}\otimes id^{(2)}id^{(2)}) = (j_{1x}j_{1y}\otimes id^{(2)}) - (j_{1y}j_{1x}\otimes id^{(2)}) = (j_{1x}j_{1y} - j_{1y}j_{1x})\otimes id^{(2)} = [j_{1x},j_{1y}] \otimes id^{(2)}$$

as required.

  • $\begingroup$ Thank you, but why does $(j_{1x}j_{1y} \otimes id^{(2)}) - (j_{1y}j_{1x} \otimes id^{(2)}) \neq (j_{1x}j_{1y} - j_{1y}j_{1x}) \otimes (id^{(2)}- id^{(2)})$? $\endgroup$
    – Jahi02
    Jun 14 at 9:32
  • $\begingroup$ I think you are confounding tensor product and direct sum. The equation you wrote holds for direct sums rather than tensor product. You can see this if you "expand" the tensor product in my equation and you get the right answer. $\endgroup$
    – emir sezik
    Jun 14 at 9:36
  • $\begingroup$ I got it now: My problem was that i thougt:$|\alpha \rangle |\psi\rangle - |\beta \rangle |\psi \rangle = |\alpha - \beta \rangle |\psi - \psi \rangle$ but correct is $|\alpha\rangle |\psi\rangle - |\beta \rangle |\psi \rangle = |\alpha - \beta \rangle |\psi \rangle$, thank you so much $\endgroup$
    – Jahi02
    Jun 14 at 10:54

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