I think this might be a dumb question so I apologize in advance. I have a model for a particle subject to a radial potential. Specifically the spatial part of the wavefunction takes the form $$\psi_{\ell,m}(r,\theta,\phi)=R_{\ell}(r)Y_{\ell,m}(\theta,\phi)$$ Here $Y$ denote the Spherical Harmonics, and $R$ is some arbitrary function describing the radial dependence. I'd like to be able to compute this particle's total angular momentum. (In particular the end goal is to compute the eigenstates of the operators $\widehat{\mathbf J_z}$ and $\widehat{J^2}$). From what I've seen on Wikipedia, this is done by using the total angular momentum operator, $$\hat{\mathbf J}=\hat{\mathbf L}+\hat{\mathbf S}$$ Here the L is orbital angular momentum, and it has the form $$\hat{\mathbf L}=-i\hbar (\hat{\mathbf r}\times \nabla)$$ Where r is the position operator. This makes sense - I can apply this operator to a scalar valued wavefunction $\psi$ and get a vector out on the other side. However, I can't see any similar forms for the spin. I've seen the spin operator written in other places as $$\hat{\mathbf S}=\frac{\hbar}{2}(\boldsymbol{\sigma}_x,\boldsymbol{\sigma}_y,\boldsymbol{\sigma}_z)$$ Where the sigmas are the Pauli spin matrices. To be frank I'm not really sure what the tuple of matrices is supposed to mean, and I don't know how to apply it to my wavefunction $\psi$, especially since it is scalar valued.

Can someone help me figure this out? I am an applied mathematician, not a physicist, so I'm having a hard time comprehending some of the literature on quantum mechanics.

  • $\begingroup$ It is not a dumb question, the books you have read do not use proper mathematics. You have tensor products of spaces and of operators. Spin acts trivially on the Hilbert space of coordinates, while L acts trivially on the space of spin. $\endgroup$
    – DanielC
    May 5, 2021 at 18:26
  • $\begingroup$ @Mauricio Comments should only be used to request clarification or suggest edits. You should put any information you find important into an answer. $\endgroup$ May 5, 2021 at 18:27
  • $\begingroup$ @DanielC Comments should only be used to request clarification or suggest edits. You should put any information you find important into an answer. $\endgroup$ May 5, 2021 at 18:28

2 Answers 2


The wavefunction you wrote down is, as you say, the spatial part of the vector. The full Hilbert space is $L^2(\mathbb R^3)\otimes \mathbb C^2$, an element of which has the form

$$|\psi\rangle = \sum_{i=1}^2\int \mathrm d^3 x \ \psi_i(\mathbf x) \bigg(|\mathbf x\rangle \otimes |\alpha_i\rangle\bigg)$$

where $\{|\alpha_i\rangle\}$ is a basis for $\mathbb C^2$.

If the Hamiltonian doesn't feature any spin-splitting, then it can be expressed in the form $H = H_\mathrm{space} \otimes \mathbb I$, which means that its (at least twice-degenerate) eigenvectors can be expressed as eigenvectors of $H_\mathrm{space}$ tensored with arbitrary elements of $\mathbb C^2$.

In contrast, the spin operators to which you refer are of the form $S_i = \mathbb I \otimes \bigg(\frac{\hbar}{2} \sigma_i\bigg)$, with $\sigma_i$ the Pauli matrices, so they leave the spatial part of the vectors alone while acting on the spin part. The orbital angular momentum operators are of the form $L_i \otimes \mathbb I$, as they act exclusively on the spatial part of the vectors while leaving the spin alone. Finally, the total angular momentum operators are $J_i = L_i \otimes \mathbb I + \mathbb I \otimes S_i$, which is colloquially written as $J_i = L_i + S_i$ with the tensor product structure swept under the rug.

As an explicit example, the vector corresponding to a spin-up electron with spatial wavefunction $\psi_{n\ell m}(\mathbf x)$ is

$$ |\psi \rangle = \int \mathbf d^3 x \ \psi_{n\ell m}(\mathbf x ) |\mathbf x\rangle \otimes |\alpha_\uparrow\rangle $$

whose wavefunction can be written $$\langle \mathbf x, \hat z|\psi \rangle = \psi_{n\ell m}(\mathbf x) \pmatrix{1 \\ 0} = \pmatrix{\psi_{n\ell m}(\mathbf x) \\ 0 }$$

The action of the spin operator $\mathbb I \otimes S_x$ on this vector is

$$\mathbb I \otimes S_x |\psi \rangle = \int \mathrm d^3 x \ \psi_{n\ell m}(x) \mathbb I |\mathbf x\rangle \otimes S_x |\alpha_\uparrow\rangle$$ whose wavefunction is $$\langle \mathbf x,\hat z|\mathbb I \otimes S_x|\psi\rangle = \psi_{n\ell m}(\mathbf x) \pmatrix{0\\ \frac{\hbar}{2}}$$

  • $\begingroup$ So from what I gather from your work, any arbitrary operator can be written as a tensor product of a spatial and temporal part, $\hat{\mathbf{A}}=\hat{\mathbf{A}}_{\text{space}}\otimes \hat{\mathbf{A}}_{\text{time}}$. But still, how to I apply that to my wavefunction, even if I write the fully time dependent form $\Psi(\mathbf{r},t)=\psi(\mathbf{r})\exp(-iEt/\hbar)$? I should be expecting a vector-valued output for spin, right? But based on what you have written, I'm going to get a tensor of order $(2,2)$, since you have $\hat{\mathbf S_i}=\mathbf{I}\otimes \frac{\hbar}{2}\boldsymbol\sigma_i$ $\endgroup$
    – K.defaoite
    May 5, 2021 at 19:24
  • $\begingroup$ The tensor product of two matrices, or $(1,1)$ tensors, will yield a $(1+1,1+1)$ tensor $\endgroup$
    – K.defaoite
    May 5, 2021 at 19:24
  • 1
    $\begingroup$ @K.defaoite 1. No, an operator cannot be written as a tensor product of a spatial and temporal part - I don't know what that means. An operator might be time-dependent (in which case it's really a family of operators indexed by a continuous variable $t$), but I don't see how that's relevant here 2. I have added an example to show how a spin operator acts on a vector. $\endgroup$
    – J. Murray
    May 5, 2021 at 19:53
  • $\begingroup$ @K.defaoite More colloquially, the wavefunction of an electron is really a 2-component spinor. $\endgroup$
    – J. Murray
    May 5, 2021 at 19:54

The expression $$\psi_{\ell,m}(r,\theta,\phi)=R_{\ell}(r)Y_{\ell,m}(\theta,\phi)$$ is incomplete because it contains only the spatial part, but neglects the spin part.

Actually the wave function $\psi$ is not just a single position-dependent function, but it is a "vector" of two such functions, together called a spinor wave function. $$\psi(r,\theta,\phi) =\begin{pmatrix}\psi_+(r,\theta,\phi)\\\psi_-(r,\theta,\phi)\end{pmatrix}$$

Now, the orbital angular momentum operator $\hat{\mathbf L}$ (and also other spatial operators like $\hat{\mathbf r}$ or $\nabla$) act on the spatial dependency only. And they act equally on the 2 spinor components.

For example $L_z=-i\hbar\frac{\partial}{\partial\phi}$, the $z$-component of the orbital angular momentum, acts on the spinor wave function like this:

$$\begin{align} L_z\psi(r,\theta,\phi) &=-i\hbar\frac{\partial}{\partial\phi} \begin{pmatrix} \psi_+(r,\theta,\phi) \\ \psi_-(r,\theta,\phi) \end{pmatrix} \\ &=-i\hbar\begin{pmatrix} \frac{\partial\psi_+(r,\theta,\phi)}{\partial\phi} \\ \frac{\partial\psi_-(r,\theta,\phi)}{\partial\phi} \end{pmatrix} \end{align}$$

On the other hand, the spin angular momentum operator $\mathbf S$ (and also its components $S_x$, $S_y$, $S_z$ and the Pauli matrices $\sigma_x$, $\sigma_y$, $\sigma_z$) act differently on the two spinor components. And they act independent of the position ($r,$, $\theta$, $\phi$). So these operators are constant $2\times 2$ matrices, i.e. when applied to a $2$-component spinor they produce another $2$-component spinor.

For example $S_z=\frac{\hbar}{2}\sigma_z$, the $z$-component of the spin angular momentum operator, acts like this:

$$\begin{align} S_z \psi(r,\theta,\phi) &=\frac{\hbar}{2}\begin{pmatrix}1&0\\0&-1\end{pmatrix} \begin{pmatrix}\psi_+(r,\theta,\phi)\\\psi_-(r,\theta,\phi)\end{pmatrix} \\ &=\frac{\hbar}{2} \begin{pmatrix}\psi_+(r,\theta,\phi)\\-\psi_-(r,\theta,\phi)\end{pmatrix} \end{align}$$

From the two examples above you can derive how $J_z$, the $z$-component of the total angular momentum, acts on the spinor function.

$$\begin{align} J_z\psi(r,\theta,\phi) &=(L_z+S_z)\psi(r,\theta,\phi) \\ &=\left(-i\hbar\frac{\partial}{\partial\phi} +\frac{\hbar}{2}\begin{pmatrix}1&0\\0&-1\end{pmatrix}\right) \begin{pmatrix}\psi_+(r,\theta,\phi)\\ \psi_-(r,\theta,\phi)\end{pmatrix} \\ &=\hbar\begin{pmatrix} -i\frac{\partial\psi_+(r,\theta,\phi)}{\partial\phi}+\frac{1}{2}\psi_+(r,\theta,\phi) \\ -i\frac{\partial\psi_-(r,\theta,\phi)}{\partial\phi}-\frac{1}{2}\psi_-(r,\theta,\phi) \end{pmatrix} \end{align}$$

  • $\begingroup$ This is starting to make more sense. Are there explicit expressions for $\psi_+, \psi_-$? What is their physical significance? $\endgroup$
    – K.defaoite
    May 5, 2021 at 23:48
  • $\begingroup$ The terms $\psi_+$ and $\psi_-$ represent the spin-up and spin-down part of the wavefunction, respectivelly. They are the probability amplitude of finding the state with that spin. $\endgroup$ May 6, 2021 at 0:01
  • 1
    $\begingroup$ There is some confusion here due to non consistent notation. This "3 argument" function is actually the vector of functions (also called a spinor) @Thomas Fritsch presented in his answer. The middle argument ($\pm$) represents which entry of the vector the function is being evaluated. Moreover, we should clarify what wavefunction do you want. In principle any normalizable function can be a valid wavefunction. What you seem to want though, are the eigenfunctions, the wavefunctions which are eigenstates of the Hamiltonian. $\endgroup$ May 6, 2021 at 18:22
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    $\begingroup$ The first thing you have to notice is that we will have double the amount of solutions you had before because we have two possibilities for the spin. Let us denote the solutions by $\Psi_{s,l,m}(r,\theta,\phi)$, where s in the quantum number associated with the spin. If your model has no spin interaction (which it does not seem to have), then the eigenfunctions for your Hamiltonian are going to be proportional to the wavefunctions you described in your question. This means this part of the wavefunction can factor out of the vector: $\Psi_{s,l,m}(r,\theta,\phi)= \Phi_s\psi_{l,m}(r,\theta,\phi)$ $\endgroup$ May 6, 2021 at 18:31
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    $\begingroup$ The $\Phi_s$ above is the spin part of the wavefunction, a vector. It is $\Phi_\uparrow = (1,0)$ for spin up and $\Phi_\downarrow = (0,1)$ for spin down. $\endgroup$ May 6, 2021 at 18:36

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