Imagine you have a triplet spin state, which, in general, can be written as $$|\psi \rangle = \alpha | \uparrow \uparrow \rangle + \beta ( | \downarrow \uparrow \rangle+ | \uparrow \downarrow \rangle) + \gamma | \downarrow \downarrow \rangle,$$ with some complex numbers $\alpha, \beta, \gamma$.

Here, I've written the given state as a superposition of the $m_z=0,-1$ and $+1$ triplet states in the $z$-basis; so, for example, $|\uparrow \uparrow \rangle$ is an eigenstate of $S^{z} =\tfrac{1}{2}( \sigma_1^{z} + \sigma_2^{z}) $ with eigenvalue $m_z = 1$.

My question is: Can you find a total spin operator $S^{\hat{n}} =\tfrac{1}{2}( \sigma_1^{\hat{n}} + \sigma_2^{\hat{n}}) $, defined by some quantization axis $\hat{n}$, such that the above state $|\psi \rangle$ is an eigenstate of $S^{\hat{n}}$?

  • $\begingroup$ We need to translate this into mathematical language and migrate it back to math.SE . Would "representation-theory" be an appropriate tag? $\endgroup$ Mar 29, 2019 at 15:17
  • $\begingroup$ A possible translation is: Can the vector $(\alpha,\sqrt{2}\beta,\gamma)^T$ be an eigenvector of a 3x3 Hermitian traceless matrix with eigenvalue $\pm 1$ or 0? $\endgroup$ Apr 26, 2019 at 15:03
  • $\begingroup$ @CosmasZachos I'm not sure that's quite the same question - the algebra generators, and their linear combinations, form a smaller space than the traceless hermitian matrices. (The former is a two-dimensional manifold isomorphic to $\mathbb S^2$, once normalized; the latter is a six-dimensional manifold once normalized, if I'm counting correctly.) Still, I would appreciate your opinion about the correctness of the arguments in my answer below. $\endgroup$ Apr 26, 2019 at 16:22
  • $\begingroup$ If he is asking for some basis in which his $\psi$ is an eigenvector of similarity-transformed (triplet) $S^z$, then it is easy to find a Unitary $U$ in SU(3) for such a similarity transform. But he is (physically: "quantization axis") asking for a (basically) Real orthogonal matrix, so R in SO(3), instead, corresponding to a physical space 3-rotation. Aware I am sounding like a math-pedant.... $\endgroup$ Apr 29, 2019 at 15:04

1 Answer 1


No, this isn't possible for generic states.

To see the underlying structure in more detail, let's start off by using a correct normalization for your state, $$|\psi \rangle = \alpha | {\uparrow \uparrow} \rangle + \beta \frac{ | {\downarrow \uparrow} \rangle+ | {\uparrow \downarrow} \rangle}{\sqrt{2}} + \gamma | {\downarrow \downarrow} \rangle,$$ and then, as Cosmas Zachos noted (in a now-deleted answer), noting that there's nothing special about the tensor-product structure that underlies this state, and that it can be equally well thought of as a general state $$ |\psi \rangle = \alpha | 1 \rangle + \beta |0 \rangle + \gamma | {-1} \rangle $$ within an $\ell=1$ dipolar representation of $\rm SO(3)$.

Moreover, since all of the (complex) irreducible representations of $\rm SO(3)$ with equal total angular momentum quantum number $\ell$ are equivalent, your question can equally well be phrased in the language of spherical harmonics, which carry one such representation, i.e. anything that you can say about the state $|\psi\rangle$ as you've phrased it can also be made to apply to the linear combination $$ f(\theta,\phi) = \alpha \, Y_{11}(\theta,\phi) + \beta \, Y_{10}(\theta,\phi) + \gamma \, Y_{1,-1}(\theta,\phi) $$ of dipolar spherical harmonics. And in yet another transformation, I will multiply each of these $Y_{\ell m}(\theta, \phi)$ with $r$ to get solid harmonics, which are homogeneous linear combinations of $x$, $y$ and $z$: in other words, anything you can say about the state $|\psi\rangle$ as you've phrased it can also be made to apply to the linear combination \begin{align} g(x,y,z) & = -\alpha \frac{x+iy}{\sqrt{2}} + \beta z + \gamma \frac{x-iy}{\sqrt{2}} \\ & = \frac{-\alpha + \gamma}{\sqrt{2}} x + \frac{\alpha + \gamma}{\sqrt{2}\, i} x + \beta z \\ & = \left( \frac{-\alpha + \gamma}{\sqrt{2}}, \frac{\alpha + \gamma}{\sqrt{2}\, i}, \beta \right) \cdot (x,y,z) \\ & = \mathbf n \cdot \mathbf r, \end{align} i.e. to a complex-valued linear functional $g:\mathbb R^3 \to \mathbb C$ which is fully characterized by the (complex) gradient vector $\mathbf n$.

In these terms, then your question boils down to the following:

  1. Does $\mathbf n$ look like $\mathbf n = (0,0,1)$? (which would imply that $|\psi\rangle$ is an eigenstate of some $S_z$ with eigenvalue $0$)
  2. Or does $\mathbf n$ look like $\mathbf n = \frac{1}{\sqrt{2}}(1,i,0)$? (which would imply that $|\psi\rangle$ is an eigenstate of some $S_z$ with eigenvalue $1$)
  3. Or does $\mathbf n$ look like neither of those? (which would imply that $|\psi\rangle$ is not an eigenstate of any $S_z$)

A key component of these questions is the words "looks like", which here means "equal up to a rigid rotation in $\mathbb C^3$ and up to a global phase", with that rotation drawn from $\rm SO(3)$.

And here the answers are relatively easy to provide:

  1. If $\mathbf n$ looks like $\mathbf n = (0,0,1)$, then $\mathrm{Re}(\mathbf n)$ and $\mathrm{Im}(\mathbf n)$ are parallel, in which case $|\psi\rangle$ is an eigenstate of $\mathbf n \cdot S_z$ with eigenvalue $0$.
  2. If $\mathbf n$ looks like $\mathbf n = \frac{1}{\sqrt{2}}(1,i,0)$, then $\mathrm{Re}(\mathbf n)$ and $\mathrm{Im}(\mathbf n)$ are exactly orthogonal, in which case $|\psi\rangle$ is an eigenstate of $(\mathrm{Re}(\mathbf n)\times \mathrm{Im}(\mathbf n))\cdot S_z = \frac{1}{2}(\mathbf n \times \mathbf n^*)\cdot S_z$ with eigenvalue $1$.
  3. And, conversely, if $\mathrm{Re}(\mathbf n)$ and $\mathrm{Im}(\mathbf n)$ are neither parallel nor orthogonal, then neither of those conclusions is possible.

Examples from point 3 are trivial to construct, and they form almost all of the states in the space (by which I mean: the set of complex numbers $(\alpha,\beta,\gamma)$ which yield an eigenstate of some $S_\hat{n}$ forms a set of measure zero under the regular Lebesgue measure of the space $\mathbb C^3$ that encases them).

I should also note that this conclusion changes if you force $\beta\in \mathbb R$, and $\alpha = -\gamma^*$ (which is a not-unreasonable restriction), in which case $\mathbf n$ is also real-valued and everything collapses so that all states are zero-$S_z$ eigenstates along some direction. The fact that this is possible on the (real-valued) $\ell = 1$ forms the premise of my question How many truly different multipolar charge distributions are there?, which goes on to ask what happens at $\ell = 2$ and higher ─ with the conclusion that at higher angular momentum, even if you restrict to real-valued representations, there are nontrivial states that cannot be reduced to the canonical form of a rotated spin eigenvector.


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