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I recently learned that position vectors and spin vectors lie in different spaces, and the complete wave is the tensor product of both. I wanted to know that whether we can talk about commutation of spin and position operators.

Can we talk about commutation of operators that act on different spaces? If so, how?

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3 Answers 3

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When two operators act on different spaces, they will always commute.

For instance: if $\hat A$ acts only on $\vert\psi\rangle_a$ so that and $\hat B$ only on $\vert \phi\rangle_b$ so that $$ \hat A\vert\psi\rangle_a \vert \phi\rangle_b=\vert\psi^\prime\rangle_a\vert \phi\rangle_b\, ,\quad \hat B\vert\psi\rangle_a\vert \phi\rangle_b = \vert\psi\rangle_a\vert\phi^\prime\rangle_b $$ then \begin{align} \hat B\left[\hat A\vert\psi\rangle_a\vert\phi\rangle_b\right]&= \hat B \vert\psi^\prime\rangle_a\vert\phi_b\rangle= \vert\psi^\prime\rangle_a\vert\phi^\prime\rangle_b\, ,\tag{1}\\ \hat A\left[\hat B\vert\psi\rangle_a\vert\phi\rangle_b\right]&= \hat A \vert\psi\rangle_a\vert\phi^\prime_b\rangle= \vert\psi^\prime\rangle_a\vert\phi^\prime\rangle_b\tag{2} \end{align} and clearly Eqs.(1) and (2) are equal.

In the specific case of spin, it turns out that spin operators cannot be expressed in terms of $x,y,z$ and $p_x,p_y,p_z$. As a result, an operator like $\hat p_x$ which contains derivatives w/r to $x$ cannot see any factors of $x$ in a spin operator, so that for instance, $$ \hat p_x \hat S_z\psi(x)= \left[\hat p_x\hat S_x\right]\psi(x)+\hat S_z\left[\hat p_x\psi(x)\right]=S_z\hat p_x\psi(x) $$ since $\partial \hat S_z/\partial x$ is necessarily $0$ as $\hat S_z$ cannot depend on $x$.

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$$(A\otimes I)(I\otimes B)=(A\otimes B) = (I\otimes B)(A\otimes I)$$

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You have already got correct answers proving it with abstract mathematics. But I would like to complement these with a concrete example, to show that the reasoning is actually very simple.

Let us take for example a single particle with spin $\frac 12$. This can be described by the tensor product of a simple wavefunction $\psi(x,y,z)$ with a $2$-component spinor $\begin{pmatrix}\psi_+\\\psi_-\end{pmatrix}$. This means you have a $2$-component spinor wavefunction. $$\psi = \begin{pmatrix} \psi_+(x,y,z)\\\psi_-(x,y,z)\end{pmatrix}$$

Then the $x$-operator is the multiplication by $x$ which obviously acts different for different positions $(x,y,z)$, but in the same way on $\psi_+$ and $\psi_+$.

And the $S_y$-operator is a matrix multiplication operator $$S_y=\frac 12 \hbar \begin{pmatrix}0&-i\\i&0\end{pmatrix}$$ which scrambles $\psi_+$ and $\psi_-$, but does so independent of $(x,y,z)$.

Now you can calculate $xS_y\psi$ and $S_yx\psi$ and see you get the same result.

$$\begin{align} xS_y\psi &= x \frac 12 \hbar \begin{pmatrix}0&-i\\i&0\end{pmatrix} \begin{pmatrix}\psi_+(x,y,z)\\\psi_-(x,y,z)\end{pmatrix} \\ &= x \frac 12 \hbar \begin{pmatrix}-i\psi_-(x,y,z)\\i\psi_+(x,y,z)\end{pmatrix} \\ &= \frac 12 \hbar \begin{pmatrix}-ix\psi_-(x,y,z)\\ix\psi_+(x,y,z)\end{pmatrix} \end{align}$$ and $$\begin{align} S_yx\psi &= \frac 12 \hbar \begin{pmatrix}0&-i\\i&0\end{pmatrix} x\begin{pmatrix}\psi_+(x,y,z)\\\psi_-(x,y,z)\end{pmatrix} \\ &= \frac 12 \hbar \begin{pmatrix}0&-i\\i&0\end{pmatrix} \begin{pmatrix}x\psi_+(x,y,z)\\x\psi_-(x,y,z)\end{pmatrix} \\ &= \frac 12 \hbar \begin{pmatrix}-ix\psi_-(x,y,z)\\ix\psi_+(x,y,z)\end{pmatrix} \end{align}$$

So it is $xS_y\psi=S_yx\psi$ for every $\psi$, and hence $[x,S_y]=0$.

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