# Spin-Angular Momentum Coupling: What component do $\hat{H}, \hat{x}$ and $\hat{p}$ act on?

Consider a QM system with Hamiltonian $$\hat{H}$$, position operator $$\hat{x}$$, momentum operator $$\hat{p}$$ and angular momentum operator $$\hat{L}$$, all acting on a Hilbert space $$\mathcal{H}$$. Assume now that the system has spin, for example if we consider a hydrogen atom, and introduce the spin operator $$\hat{S}$$.

If we want to take into account the interaction of the two, we study the Hilbert space $$\mathcal{H}\otimes\mathcal{H}$$, as far as I understand. Here angular momentum acts on the first component and spin acts on the second.

Question: What component do $$\hat{H}, \hat{x}$$ and $$\hat{p}$$ act on?

• The Hilbert space for a particle with spin $s=1/2$ is $H=L^2(\mathbb{R}) \otimes \mathbb{C}^2$. The position, momentum and angular momentum operator act on vectors of $L^2$, while the spin operators act on $\mathbb{C}^2$. The Hamiltonian is an operator on $H$. Is that your question? Jul 22, 2021 at 14:49
• I meant $L^2(\mathbb{R}^3)$ in the first line. The crucial point however is that the Hilbert space is not $\mathcal{H} \otimes \mathcal{H}$, whatever $\mathcal{H}$ is (you did not define it). For instance, $L^2$ is infinite-dimensional, whereas $\mathbb{C}^2$ is finite-dimensional, so both Hilbert spaces cannot be the same. Jul 22, 2021 at 15:11
• @Jakob that looks like an answer Jul 22, 2021 at 15:53

Consider first a spinless particle and denote, as you did, by $$\cal H$$ its Hilbert space. The operators $$\hat x$$ (position), $$\hat p$$ (momentum), $$\hat H$$ (hamiltonian) act on states of $$\cal H$$. The wavefunction of the particle in the state $$|\psi\rangle$$ is $$\psi(\vec r)=\langle \vec r|\psi\rangle$$
Consider now a spin $$1/2$$ without any other degree of freedom. Its Hilbert space, say $${\cal H}_{1/2}$$ is spanned by the two states $$\{|\uparrow\rangle,|\downarrow\rangle\}$$. The operators $$\hat S_i$$ ($$i=x,y,z$$) act on $${\cal H}_{1/2}$$. In the basis $$\{|\uparrow\rangle,|\downarrow\rangle\}$$, it can be written as a $$2\times 2$$ matrix. If the spin is in the state $$|\phi\rangle=a|\uparrow\rangle+b|\downarrow\rangle$$, the probability amplitude of finding it in the $$\uparrow$$ state is $$\langle\uparrow|\phi\rangle=a$$
Now, for a spin $$1/2$$-particle, i.e. having both translation and spin degrees of freedom, the quantum state belongs to the full Hilbert state $${\cal H}\otimes{\cal H}_{1/2}$$. The probability amplitude of finding the particle at point $$\vec r$$ with a spin $$\uparrow$$ is $$(\langle\vec r|\otimes\langle\uparrow|)(|\psi\rangle\otimes|\phi\rangle) =\langle\vec r|\psi\rangle\ \!\langle\uparrow|\phi\rangle$$ The position operator $$\hat x$$ acts only on $${\cal H}$$ (and should now be written $$\hat x\otimes\mathbb I$$): \eqalign{ \hat x(|\psi\rangle\otimes|\phi\rangle) &=(\hat x\otimes\mathbb I)|\psi\rangle\otimes|\phi\rangle\cr &=(\hat x|\psi\rangle)\otimes(\mathbb I|\phi\rangle)\cr &=(\hat x|\psi\rangle)\otimes|\phi\rangle\cr } The spin operators $$\hat S_i$$ act only on $${\cal H}_{1/2}$$: \eqalign{ \hat S_i(|\psi\rangle\otimes|\phi\rangle) &=(\mathbb I\otimes\hat S_i)|\psi\rangle\otimes|\phi\rangle\cr &=|\psi\rangle\otimes\hat S_i|\phi\rangle\cr } Some operators may act on both Hilbert state simultanously. It is the case of the spin-orbit interaction $$\hat W\sim \vec L.\vec S$$ that couples the spin and the angular momentum of the electron in an atom for example: \eqalign{ \vec L.\vec S|\psi\rangle\otimes|\phi\rangle &=(\vec L\otimes\mathbb I).(\mathbb I\otimes\vec S)|\psi\rangle\otimes|\phi\rangle\cr &=(\vec L|\psi\rangle)\otimes(\vec S|\phi\rangle)\cr } Note that such interaction causes the entanglement of translation and spin degrees of freedom. The eigenstates of $$H$$ are no longer a tensor product $$|\psi\rangle\otimes|\phi\rangle$$ but a linear superposition of such products.