# Solution to the time-independent Schrödinger's equation for systems with spin

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.

• 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! Apr 14, 2021 at 15:39

## 1 Answer

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

• Thank you, but in my specific case, why can we assume that $\mathscr{H}$ is separable in the first place? Apr 13, 2021 at 21:26
• @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. Apr 14, 2021 at 6:25
• @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. Apr 14, 2021 at 6:49