Sometimes in quantum mechanics we come across notation like $Y_{l m_{l}}(\theta, \phi)\chi_{sm_s}$ where $Y_{lm_l}$ is a spherical harmonic representing the spatial part of some particle wavefunction and $\chi_{sm_s}\in \mathcal{H}_S$ is a spinor or vector representing the spin of a particle. Alternatively, in classicaly electromagnetism we may have $Y_{l, m_{l}}(\theta, \phi)$ represents the spatial distribution of an electric/magnetic field and $\chi_{sm_s}$ represents the local vector part of the field.

There is a sense in which these two "objects" are angular momentum representations with indices $l$ and $s$ and it is known that it is possible to express them in a $j, m_j$ basis using $J= L + S$ and Clebsch-Gordan coefficients.*

I am comfortable with this in the case that we have two vectors like $|l, m_l\rangle \in \mathcal{H}_L$ and $|s, m_s\rangle \in \mathcal{H}_S$ and we are interested in describing bases of the tensor Hilbert space $\mathcal{H}_J = \mathcal{H}_L \otimes \mathcal{H}_S$. In that case I know from the theory of angular momentum addition that there are two related bases for $\mathcal{H}_J$. One is expressed as $|l, m_l\rangle \otimes |s, m_s\rangle$ and one is expressed as $|j, m_j;l, s\rangle$ and the two bases are related by Clebsch Gordan coefficients:

\begin{align} |j, m_j; l, s\rangle =& |l, m_l\rangle \otimes |s, m_s\rangle \langle l, m_l, s, m_s |j, m_j; l, s\rangle\\ =& |l, m_l\rangle \otimes |s, m_s\rangle C_{l, m_l, s, m_s}^{j, m} \end{align}

However, something feels like an abuse of notation (or at least a shortcut) when we say $Y_{lm_l}(\theta, \phi)\chi_{sm_s}$ is a tensor product of this sort. I feel like this is being a bit notationally pedantic, but I guess my issue is that I would grant that $Y_{lm_l}$ (as a function in $L_2(\mathbb{S}^2,\mathbb{C})$) is a vector in a Hilbert space, but it feels like $Y_{lm_l}(\theta, \phi)$ is a scalar because the spherical harmonic has been evaluated. It feels like a shortcut to replace the tensor product by simple scalar-vector multiplication.

What is a more "proper" way to write the tensor product in this case? I could see something like $Y_{lm_l}\otimes \chi_{sm_s}$ making sense but then how do we "evaluate it" at a certain point $(\theta, \phi)$? Would we write something like $(Y_{lm_l}\otimes \chi_{sm_s})(\theta, \phi)$ and understand that the output of this function evaluation is a vector in $\mathcal{H}_S$?

This questions is related to the question/answer/comments at Vector Spherical Harmonics and total angular momentum.

*Sometimes $|l, m_l\rangle \otimes |s, m_s\rangle$ is written in a shorter notation as $|l, m_l, s, m_s\rangle$. Perhaps this is a similar notation abbreviation as the one involving $Y_{lm_l}(\theta, \phi)$ and $\chi_{sm_s}$?

  • 2
    $\begingroup$ Hint: $Y\mapsto Y(\theta,\phi)$ is a mapping from Hilbert space to the complex numbers which corresponds to (actually: is) the bra $⟨\theta,\phi|$. For the related case of $\mathbb C^3$, the mapping $\mathbf v\mapsto v_i$ that extracts a given component is, similarly, a linear functional $\mathbb C^3 \to \mathbb C$; in regular QM we rarely denote it as such, but the structure is there. Your attempted notation, $(Y_{lm_l}\otimes \chi_{sm_s})(\theta, \phi)$, is treating the $L^2$ and the spinor halves of the tensor product asymmetrically. Fix that asymmetry and you'll essentially be done. $\endgroup$ Commented Mar 27, 2022 at 20:18
  • 2
    $\begingroup$ To expand @EmilioPisanty 's comment: You can define an operator: $O_{\theta \phi :}\mathcal H_J \longrightarrow \mathcal H_S$ with $O_{\theta \phi} (Y\otimes\chi) = Y(\theta,\phi)\, \chi$. In bra-ket notation we could write e.g. $O_{\theta,\phi}=\langle\theta \phi| \otimes \mathbb I_s$. Other than that, I think you're right regarding the abuse of notation, which, however, is not different from saying that $f(x)$ is a function... etc. $\endgroup$ Commented Mar 27, 2022 at 20:21
  • $\begingroup$ Ah, thanks both for the comments. Yeah so $Y$ is a function on $\mathbb{S}^2$ which means the dual space to the space $Y$ is in is just $\mathbb{S}^2$ so elements of that space can act on $Y$. Just like $\psi(x) = \langle x|\psi \rangle$. If no one does it before I may expand out these comments into an answer. $\endgroup$
    – Jagerber48
    Commented Mar 27, 2022 at 21:38
  • 1
    $\begingroup$ "the dual space to the space $Y$ is in is just $\mathbb S^2$ - I get what you mean, but this is wrong. $Y$ is in $L_2(\mathbb S^2,\mathbb C)$. The (continuous) dual to that is $L_2(\mathbb S^2,\mathbb C)$ itself. If you ignore the requirement of continuity (which you need to do in order to talk about $⟨\theta,\phi|$), then the dual of $L_2(\mathbb S^2,\mathbb C)$ does contain a copy of $\mathbb S^2$ (i.e. $\{⟨\theta,\phi|\}$). But the dual also contains all the linear combinations of them, which are many more linear functionals than just those. $\endgroup$ Commented Mar 28, 2022 at 14:46
  • 1
    $\begingroup$ @JasonFunderberker I didn't say it doesn't make sense ;-). I said that it treats the two tensor halves asymmetrically (which is unambiguously true). Whether that's a problem or not is a subjective matter. $\endgroup$ Commented Mar 28, 2022 at 14:47

1 Answer 1


It's been stipulated that $\chi_{s, m_s} = |s, m_s\rangle\in\mathcal{H}_s$ with $\mathcal{H}_s$ a Hilbert space. Let's generalize and just let $|\chi\rangle \in \mathcal{H}_s$.

Now consider the space $L^2(\mathbb{S}^2, \mathbb{C})$ of square integrable complex functions on the 2-sphere. $L^2(\mathbb{S}^2, \mathbb{C})$ is a Hilbert space. Suppose $f\in L^2(\mathbb{S}^2, \mathbb{C})$. In Dirac notation we would write $f = |f\rangle$. The $L^2(\mathbb{S}^2, \mathbb{C})$ Hilbert space has a basis denoted by $|\theta, \phi\rangle$. which decomposes any $|f\rangle$ as

$$ \langle \theta, \phi|f\rangle = f(\theta, \phi) $$

Consider the spherical harmonics $Y_{l, m_l}(\theta, \phi)$ with $l$ fixed and $-l \le m_l \le l$. The set $$ \{Y_{l, m_l} | -l \le m_l \le l\} = \{ | Y_{l, m_l}\rangle | -l \le m_l \le l\} $$ constites a vector subspace of $L^2(\mathbb{S}^2, \mathbb{C})$ which we denote by $\mathcal{H}_l$. Because $|Y_{l, m_l}\rangle \in L^2(\mathbb{S}^2, \mathbb{C})$ we have $$ \langle \theta, \phi | Y_{l, m_l} \rangle = Y_{l, m_l}(\theta, \phi) $$

We can take the tensor product space $\mathcal{H}_j = \mathcal{H}_l \otimes \mathcal{H}_s$. Let $|Y\rangle \in \mathcal{H}_l$. Then an element of $\mathcal{H}_j = \mathcal{H}_l\otimes \mathcal{H}_s$ could be written as

$$ |Y\rangle \otimes |\chi\rangle $$

One natural thing to do would be do calculate the components of this tensor product element in the original basis. We would do this by

$$ \left(\langle Y_{l, m}|\otimes \langle s, m_s|\right)\left(|Y\rangle \otimes |\chi\rangle \right) = \langle Y_{l, m}|Y\rangle \langle s, m_s|\chi\rangle \in \mathbb{C} $$

This tells us how much overlap the vector has with the $|Y_{l, m}\rangle$ spherical harmonic and $|s, m_s\rangle$ spin state. Note that $\langle Y_{l, m}| \in \mathcal{H}_l^*$ and $\langle s, m_s| \in \mathcal{H}_s^*$ and I guess it follows that $\langle Y_{l, m}|\otimes \langle s, m_s| \in \mathcal{H}_j^*$ so this whole "component finding" formalism is pretty natural.

Another way we could have done this is by using $\langle \theta, \phi|$ instead of $\langle Y_{l, m}|$. If we had had $|Y\rangle = |Y_{l, m}\rangle$ then the result would have been

$$ Y_{l, m}(\theta, \phi)\langle s, m_s|\chi\rangle $$

But what I asked for is a little different. I asked for sense-making of the expression $$ Y_{l, m}(\theta, \phi)|\chi\rangle $$

To get this the operator we need to act on $|Y_{l, m}\rangle \otimes |\chi\rangle$ is

$$ \langle \theta, \phi| \otimes \mathbb{I}_s $$

This is a little odd because $\langle \theta, \phi \in \mathcal{H}_l^*$ but $\mathbb{I}_s \in \mathcal{L}(\mathcal{H}_s)$ (the linear operators from $\mathcal{H}_s\rightarrow \mathcal{H}_s$), not $\mathcal{H}_s^*$ (the linear operators from $\mathcal{H}_s \rightarrow \mathbb{C}$). So this is a little bit of an abnormal construction.

Nonetheless, I think it's valid and does give a rigorous definition of what was requested. @JasonFunderberker points out in the comments that these manipulations give rise to an isomorphism between the spaces of the different objecs I have been manipulating here. This isomorphism likely justifies the abuse of notation which identifies $Y_{l, m}(\theta, \phi)$ with $|Y_{l, m}\rangle$.

  • $\begingroup$ Are you sure $\vert\theta,\phi\rangle \in \mathcal{H}_L$? Intuitively, if you were to write the wavefunction of this state as $\langle\theta',\phi'\vert\theta,\phi\rangle$, it seems to me you'd get to a delta distribution centered at $(\theta,\phi)$, which does not seem to be square-integrable $\endgroup$ Commented Mar 29, 2022 at 1:44
  • 2
    $\begingroup$ @NíckolasAlves I'm hitting a sweet spot of a little more rigorous that the sloppiest physicists, but not perfectly mathematically rigorous. Yes, there are very well known problems with "position basis" representations of $L^2$ functions and these need to be solved with like, rigged Hilbert spaces or something and great care in thinking about domains of things. I'm not aiming for that full rigor here and am comfortable enough with usual physicist sloppiness with $\delta$ functions. So maybe I'm being a little hypocritical in my desire for rigor here. Not sure what more to sya. $\endgroup$
    – Jagerber48
    Commented Mar 29, 2022 at 1:52
  • $\begingroup$ Why do you say the operator $\langle \theta, \phi| \otimes \mathbb{I}_s$ is an 'abnormal construction'? This notation is pretty much used in e.g. quantum information. Note that a bra in essence is just a (linear) operator, too. Anyway, what I was trying to say in the comments was that if you want to choose a particular realization of the tensor product (I think all realizations are unitarily equivalent), then you can choose $\otimes$ as defined in the comments above. Note, however, that it does not really justify the abuse of notation you mentioned at the end. An element of the tensor product $\endgroup$ Commented Mar 29, 2022 at 7:03
  • $\begingroup$ ...space is (e.g.) a function $(\theta,\phi) \mapsto Y(\theta,\phi)\, \chi$. So, if you evaluate the function, then $Y(\theta,\phi)\, \chi \in \mathcal H_S$. But okay, if you're fine with calling something like $f(x)$ a function (which is pretty much used everywhere), then you can, I guess, also see $Y(\theta,\phi)\, \chi$ as a function in the tensor product space. $\endgroup$ Commented Mar 29, 2022 at 7:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.