# Is any given triplet spin state an eigenstate of some $j^z$ in the suitable basis?

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 $$S^{\hat{n}}$$?

• We need to translate this into mathematical language and migrate it back to math.SE . Would "representation-theory" be an appropriate tag? – Keith McClary Mar 29 '19 at 15:17
• 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? – Cosmas Zachos Apr 26 '19 at 15:03
• @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. – Emilio Pisanty Apr 26 '19 at 16:22
• 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.... – Cosmas Zachos Apr 29 '19 at 15:04

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.