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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?

Emilio Pisanty
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Michael Iversen
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1 Answers1

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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.

Emilio Pisanty
  • 137,480