I'm currently taking a course on quantum computing and we've just introduced the concept of spin in a not so very formal way and I'm no physicist, so please be gentle. Apparently, the solution to the time-independent Schrödinger's equation for, I guess, an electron-like particle has the following form:

$$\psi(\vec{x},s,t) = \psi(\vec{x})\otimes|s\rangle e^{-iEt/\hbar}$$

Where $s$ denotes the spin. My question is: on the right-hand side, why can we factorize the initial condition $\psi(\vec{x},s)$ as a tensor product of a spatial part and a spin part? I think the only assumption was that the spin contribution in the Hamiltonian is an additive term which depends only on $s$, and not on $\vec{x}$, e.g. in the case the particle is immersed in a (uniform?) magnetic field.

Possibly a duplicate of this, although I can't get much out of it.

  • $\begingroup$ If you care about QComp. and don't have a physics background, I'd recommend just ignoring where this comes from, rather than first learning "normal" quantum theory and then ignoring it again! $\endgroup$ Commented Apr 14, 2021 at 15:39

1 Answer 1


Let's say we consider a Hilbert space $\mathscr{H} \equiv \mathscr{H}_{\mathrm{A}} \otimes \mathscr{H}_{\mathrm{B}}$.

Here, $\mathscr{H}_{\mathrm{A}}$ and $\mathscr{H}_{\mathrm{B}}$ could be e.g. the Hilbert spaces of distinguishable particles; or if $ \mathscr{H}_{\mathrm{A}} = L^2(\mathbb{R}^3)$ and $ \mathscr{H}_{\mathrm{B}}= \mathbb{C}^2$, then $\mathscr{H}$ is the Hilbert space of a particle with spin $s=1/2$.

We want to obtain a solution of the time-independent Schrödinger equation: $$H|\psi\rangle = E |\psi\rangle \quad , $$ where $H$ denotes the Hamiltonian. If it is of the form

$$H = H_{\mathrm{A}} \otimes \mathbb{I}_{\mathrm{B}} + \mathbb{I}_{\mathrm{A}} \otimes H_{\mathrm{B}} \quad ,$$

then an eigenstate of $H$ is given by the tensor product of eigenstates of $H_{\mathrm{A}}$ and $H_{\mathrm{B}}$, as we can easily verify: Let $|\varphi\rangle \in \mathscr{H}_{\mathrm{A}}$ and $|\sigma\rangle \in \mathscr{H}_{\mathrm{B}}$ denote some eigenstates of $H_{\mathrm{A}}$ and $H_{\mathrm{B}}$, respectively. We compute

$$H \left(|\varphi\rangle \otimes |\sigma\rangle \right) = H_{\mathrm{A}} |\varphi\rangle \otimes \mathbb{I}_{\mathrm{B}} |\sigma\rangle + \mathbb{I}_{\mathrm{A}} |\varphi\rangle \otimes H_{\mathrm{B}} |\sigma\rangle \quad , $$

which, using that $H_{\mathrm{A}} |\varphi\rangle = E_{\varphi} |\varphi\rangle$ and $H_{\mathrm{B}}|\sigma\rangle = E_{\sigma} |\sigma\rangle $, reduces to $$H \left(|\varphi\rangle \otimes |\sigma\rangle \right) = \left(E_{\varphi} + E_{\sigma} \right)|\varphi\rangle \otimes |\sigma\rangle \quad. $$

Finally, if $H$ is time-independent, then $$ \mathscr{H}\ni|\psi (t)\rangle \equiv e^{-i(E_{\varphi} + E_{\sigma})t} |\varphi\rangle \otimes |\sigma\rangle$$ solves the time-dependent Schrödinger equation, i.e. we find that

$$ i \, \partial_t |\psi(t) \rangle = H |\psi(t)\rangle $$

with $|\psi(0)\rangle =|\varphi\rangle \otimes |\sigma\rangle $.

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    $\begingroup$ Thank you, but in my specific case, why can we assume that $\mathscr{H}$ is separable in the first place? $\endgroup$
    – giofrida
    Commented Apr 13, 2021 at 21:26
  • $\begingroup$ @giofrida Do you mean why we can write $\mathscr{H} \equiv \mathscr{H}_{\mathrm{A}} \otimes \mathscr{H}_{\mathrm{B}}$? Then first of all note that separable Hilbert space means something different. Second, the answer to your question is yes! There are many posts here on stackexchange regarding this. $\endgroup$ Commented Apr 14, 2021 at 6:25
  • $\begingroup$ @giofrida I've edited the answer. Again, there are many questions and answers here discussing these things. Moreover, these topics should be covered in any quantum mechanics textbook. Have a look for example in section 2.3 of these lecture notes. $\endgroup$ Commented Apr 14, 2021 at 6:49

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