I am considering the action of $SU(2)$ on two spins. More precisely, I want to determine what states can be reached by acting with an element $\hat R \in SU(2)$ on $|\psi \rangle \in \text{span}\{|{\uparrow \uparrow}\rangle, \enspace |{ \uparrow\downarrow}\rangle, \enspace|{\downarrow \uparrow}\rangle, \enspace |{ \downarrow \downarrow}\rangle\}$. For certain states this question is easy to answer. For instance, starting with any triplet state: $|\psi_1\rangle = |{\uparrow \uparrow }\rangle$, $|\psi_2\rangle = |{\downarrow \downarrow }\rangle$, $|\psi_3 \rangle = \frac{|{\uparrow \downarrow}\rangle + |{\downarrow \uparrow}\rangle}{\sqrt 2}$ we can reach any other triplet state when acting with $\hat R$.

$$(\hat R\otimes\hat R) |\psi_j\rangle \in \text{span} \{| {\uparrow \uparrow} \rangle, \enspace |{\downarrow \downarrow} \rangle, \enspace \frac{|{ \uparrow \downarrow}\rangle + |{ \downarrow \uparrow}\rangle}{\sqrt 2}\} \quad \text{for } j = 1, 2, 3.$$

Likewise, when starting from the singlet state $|\phi \rangle = \frac{|{\uparrow \downarrow}\rangle - |{\downarrow \uparrow} \rangle}{\sqrt 2}$ it is not possible to reach any other state.

$$(\hat R \otimes \hat R) |\phi\rangle \in \text{span} \{ \frac{|{\uparrow \downarrow} \rangle - |{\downarrow \uparrow}\rangle}{\sqrt 2} \}.$$

My question is: when starting from $|{\uparrow \downarrow} \rangle$ what states can be reached by acting with elements from SU(2)? And how do I find a basis for this subspace?

My attempt so far: A generic operator from $SU(2)$ can be written as a matrix,

\begin{align} SU(2) = \{ \begin{pmatrix} \alpha & -\beta^* \\ \beta & \alpha^* \end{pmatrix} \enspace | \enspace \alpha, \beta \in \mathbb C, \enspace |\alpha|^2 + |\beta|^2 = 1 \} \end{align} In this way we can simply choose a generic operator from SU(2) and act on $|{\uparrow\downarrow}\rangle = \begin{pmatrix}1 \\ 0 \end{pmatrix} \otimes \begin{pmatrix}0 \\ 1 \end{pmatrix}$. This results in the following expression:

\begin{align} (\hat R \otimes \hat R)|{\uparrow \downarrow} \rangle = |\alpha|^2 |{\uparrow \downarrow} \rangle - |\beta|^2 |{\downarrow \uparrow} \rangle + \alpha^* \beta |{\downarrow \downarrow}\rangle - \alpha\beta^* |{\uparrow \uparrow}\rangle \end{align} This can be rewritten more clearly in polar form: $\alpha = r_\alpha e^{i\theta_\alpha}$, $\beta = r_\beta e^{i\theta_\beta}$

\begin{align} (\hat R \otimes \hat R)|{\uparrow \downarrow }\rangle = r_\alpha^2 |{\uparrow \downarrow }\rangle - r_\beta^2 |{\downarrow \uparrow} \rangle + r_\alpha r_\beta(e^{i(\theta_\beta - \theta_\alpha)} |{ \downarrow \downarrow}\rangle - e^{-i(\theta_\beta - \theta_\alpha)}|{\uparrow \uparrow}\rangle). \end{align} Thus it seems we can only reach $|{\uparrow \downarrow}\rangle$ and $|{\downarrow \uparrow } \rangle$ with real coefficients and there is some kind of phase relation between the coefficients of $|{\downarrow \downarrow}\rangle$ and $|{\uparrow \uparrow} \rangle$. How can I rewrite these observation as a span of some basis?

  • 1
    $\begingroup$ The spacings in $|\uparrow\downarrow\rangle$ ($|\uparrow\downarrow\rangle$) that you were trying to fix using \! are caused because the arrows are interpreted as binary operators and spaced accordingly (i.e. LaTeX/MathJax is, roughly, trying to apply $\uparrow$ to the pair $|$ and $\downarrow$). The correct solution is to isolate the arrows: $|{\uparrow\downarrow}\rangle$ ($|{\uparrow\downarrow}\rangle$). $\endgroup$ Aug 30, 2021 at 10:14
  • $\begingroup$ Related : What is the symmetry of the pion triplet ( π−,π0,π+ )?. $\endgroup$
    – Frobenius
    Aug 30, 2021 at 11:09
  • 1
    $\begingroup$ The product 4-dimensional Hilbert space is the direct sum of the 3-dimensional Hilbert subspace of the triplets and the 1-dimensional Hilbert subspace of the singlet. Each subspace is invariant under SU(2) that is under the product SU(4) which is the irreducible sum of SU(3) acting on the subspace of the triplets and SU(1) acting on the subspace of the singlet. So under SU(2) triplets are transformed to triplets and the singlet on itself. $\endgroup$
    – Frobenius
    Aug 30, 2021 at 11:26
  • $\begingroup$ The state $\:|{\uparrow \downarrow} \rangle\:$ is the sum of the triplet state $\frac{|{ \uparrow \downarrow}\rangle + |{ \downarrow \uparrow}\rangle}{\sqrt 2} $and the singlet state $\frac{|{\uparrow \downarrow} \rangle - |{\downarrow \uparrow} \rangle}{\sqrt 2}$ except a constant factor. So any $SU(4)= SU(2)\boldsymbol\otimes SU(2)=SU(3)\boldsymbol\oplus SU(1)$ will transform the triplet to another triplet and the singlet to itself. It's concluded that the subspace of states which could be reached is the whole 4-dimensional product Hilbert space. $\endgroup$
    – Frobenius
    Aug 30, 2021 at 13:23

1 Answer 1


Thus it seems we can only reach $|{\uparrow \downarrow}\rangle$ and $|\! \downarrow \uparrow \rangle$ with real coefficients [...].

How can I rewrite these observation as a span of some basis?

You can't. The span of a basis is a vector subspace $V$, which means that if $|{\uparrow \downarrow}\rangle \in V$ then you must also have $i|{\uparrow \downarrow}\rangle \in V$.

If you want a full proof, suppose that the orbit of $|{\uparrow \downarrow}\rangle$ under $\rm SU(2)$ were a vector subspace $V$. Then, as you've shown,

  • $|{\uparrow \downarrow}\rangle \in V$ trivially, and
  • $|{\downarrow \uparrow }\rangle \in V$ (setting $\alpha=0$, $\beta=1$).

Since $V$ is a subspace, this means that their linear combination $|{\uparrow \downarrow}\rangle - |{\downarrow \uparrow }\rangle \in V$ is also in the subspace, which is clearly not true.

  • $\begingroup$ Good point! Thank you for the clear explanation :) $\endgroup$ Aug 30, 2021 at 10:10

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.