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?


3 Answers 3


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 $x$.


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


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 $[x,S_y]=0$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.