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?

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    $\begingroup$ 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? $\endgroup$ Jul 22, 2021 at 14:49
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    $\begingroup$ 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. $\endgroup$ Jul 22, 2021 at 15:11
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    $\begingroup$ @Jakob that looks like an answer $\endgroup$
    – jacob1729
    Jul 22, 2021 at 15:53

1 Answer 1


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.

PS: see also the answer How is the product $L\cdot S$ between orbital and spin angular momentum operators defined? Do they act on the same or different Hilbert spaces?


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