Similarity transformations in $SU(2)$ and rotations

Question

The problem: Show that an arbitrary element of $SU(2)$: $U(\alpha, \hat{n})$ is related to $U(\alpha, \hat{n_3})$, a rotation about the 3rd axis, by a similarity transformation $U(\alpha, \hat{n}) = VU(\alpha, \hat{n_3})V^{-1}$. Construct an appropriate $SU(2)$ transformation $V$.

I am not sure how to approach this entirely. I've been looking at $U(\alpha, \hat{n}) = \cos(\alpha/2)+i\hat{n}\cdot\vec{\sigma}\sin(\alpha/2)$ as two parts, with the cosine part being the result of a dot product between $\hat{n}$ and $\hat{n_{3}}$ and the second part being the result of a cross product that gets multiplied by the vector of Pauli matrices. Is this the correct approach or is there another way to go about solving this problem?

Answer 1

Would the following work?

If $U(\alpha,\hat n)$ is any $SU(2)$ transformation, and $V$ is also unitary and $2\times 2$, then clearly $V^{-1}\cdot U(\alpha,\hat n).V$ is also in $SU(2)$. Choose for $V$ the matrix that diagonalizes $U(\alpha,\hat n)$. Since the eigenvalues of $U(\alpha,\hat n)$ are $e^{\pm i\alpha/2}$, this brings $U$ to the form $U(\alpha,\hat n_3)=e^{-i\alpha\sigma_3}$.

Since $V$ diagonalizes $U(\alpha,\hat n)$, it is the matrix constructed from the eigenvectors of $U(\alpha,\hat n)$ arranged in columns. These eigenvectors are orthogonal and have the form $$ \left(\begin{array}{c} \cos\beta/2 \\ e^{i\gamma} \sin\beta/2 \end{array}\right)\, ,\qquad \left(\begin{array}{c} -e^{i\gamma}\sin\beta/2 \\ \cos\beta/2 \end{array}\right) $$ with $V$ as above, depending on two parameters, so the matrix $V$ that does the trick is $$ V=\left( \begin{array}{cc} \cos \left(\frac{\beta }{2}\right) & -e^{-i \gamma } \sin \left(\frac{\beta }{2}\right) \\ e^{i \gamma } \sin \left(\frac{\beta }{2}\right) & \cos \left(\frac{\beta }{2}\right) \\ \end{array} \right)=R_z(\gamma)R_y(\beta)R_z(-\gamma)\, . $$ Summarizing: \begin{align} V^{-1}\cdot U(\alpha,\hat n)\cdot V&=U(\alpha,\hat n_3)\, ,\\ U(\alpha,\hat n)&=V\cdot U(\alpha,\hat n_3)\cdot V^{-1}\, . \end{align}

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Edit:

The transformation $V$ is of the form $R_z(\gamma)R_y(\beta)R_z(-\gamma)$, i.e. a conjugation of $R_y(\beta)$, meaning $V$ is a rotation about the axis $R_z(\gamma)\hat y$ by $\beta$, i.e. it's a rotation in the $xy$ plane, with the axis $\hat y$ rotated by $\gamma$ about $\hat z$. Calling $\hat y'=R_z(\gamma)\hat y$ for simplicity, the transformation $V$ is a rotation about $\hat y'$ by an angle $\beta$. In terms of Pauli matrices, we would have $$ V= e^{-i\gamma \sigma_z}e^{-i\beta \sigma_y}e^{i\gamma \sigma_z} =\exp \left(-i\beta\, [e^{-i\gamma \sigma_z} \sigma_y e^{i\gamma \sigma_z}]\right)\, . $$

Answer 2

Think about the correspondence between $SU(2)$ and $SO(3)$. As you know that $S^\dagger\sigma_iS$ is a linear combination of the $\sigma^i$ ($S$ is an $SU(2)$ matrix), you can define the matrix $M_{ij}(S)$ as $$ M_{ij}(S)\sigma_j=S^\dagger\sigma_iS\implies M_{ij}(S)=\frac12\text{tr}(\sigma_i S\sigma_j S^\dagger). $$ This law gives you an $SO(3)$ matrix from an $SU(2)$ matrix. It is two to one, as $M(S)=M(-S)$. If you write $$ S=\cos\frac\alpha2\mathbf 1_2+i\sigma_i\hat n_i\sin\frac\alpha2, $$ then you have that, using your notations $$ M_{ij}(S)=U_{ij}(\alpha,\hat n). $$ Tip: you can easily check this result by using axis angle representation for $U$, $$ U_{ij}(\alpha,\hat n)=\delta_{ij}\cos\alpha+n_in_j(1-\cos\alpha)+\epsilon_{ijk}n_k\sin\alpha. $$ Let us approach your problem. The multiplication by $V$ and $V^\dagger$ only affects the $\sigma_i$ part of $U(\alpha,\hat n3)$, as $$ VU(\alpha,\hat n_3)V^\dagger=V\left(\cos\frac\alpha2\mathbf 1_2+i\sigma_3\sin\frac\alpha2\right)V^\dagger=\cos\frac\alpha2\mathbf 1_2+iM_{j3}\sigma_j, $$ so you have that $V$ must be such as $M_{i3}(V)=\hat n_i$, where $\hat n_i$ is the versor on the $LHS$ of your assignment. This means that you must take a rotation matrix that rotates the third axis in the vector $\hat n$. Take any (there is a whole subspace of them), but be sure to note the rotation axis $\hat n'$ and the angle $\alpha'$ of this rotation.

The matrix $V$ is finally constructed as $$ V=\cos\frac{\alpha'}2\mathbf 1_2+i\hat n'_i\sigma_i\sin\frac{\alpha'}2, $$ and you can also say that $-V$ does the trick too, as many other matrices depending on the particular $M$ that you choose to rotate the 3-axis.