Tensors of rotations about an arbitrary vector in C^2

Question

I'm trying to solve the following equation:

$$e^{-i\theta/2 \sigma_{\vec{i}}^A} \otimes e^{-i\theta/2 \sigma_{\vec{i}}^B} |\Psi\rangle_{AB} = e^{i\phi} |\Psi\rangle_{AB} $$

where $e^{i\phi}$ should be expressed in terms of $\theta, \vec{i}$. I've expanded this, factored this, exploited every darn trig identity I can think of, but I don't think I understand tensors enough to actually solve this problem. I don't know, for instance, whether in the expanded form of the above:

$$e^{-i\theta/2 \sigma_{\vec{i}}^A} = cos(\theta/2)\mathbb{I}^A - isin(\theta/2)\sigma_{\vec{i}}^A$$

I should be taking the tensor of $\sigma_{\vec{i}}^A \otimes \sigma_{\vec{i}}^B$, which, given that their both unitary matrices of the same vector should yield the 4x4 identity matrix. That gets me somewhere, but the meaning of the tensor appears to be lost in the process...

I think the point of this exercise is to show that rotating A and B the same amount only changes the global phase of the system... Is this right or am I totally off?

Answer 1

A linear operator is fully determined by its action on a set of basis-vectors. Say $\mathcal{H}=\mathbb{C}^2 $ is the Hilbertspace of a single spin. The Hilbertspace of two spins is given by the tensor-product of $\mathcal{H}$ with itself: $\mathcal{H}\otimes\mathcal{H}=\mathbb{C}^4$. The tensor-product is like the cartesian product with the additional property that one may shuffle scalars around $$ (\alpha v)\otimes w = v\otimes (\alpha w) \equiv \alpha\; v\otimes w$$

Given two arbitrary vector spaces $V$ und $W$ with bases $\{e_k\}$ and $\{f_k\}$, a basis of the product space is immediately given as $\{e_j\otimes f_k\}$.

Operators that act on a tensor-product are also written with a $\otimes$. One may define a linear operator on a tensor-product space by specifying how it acts on each factor, e.g $$ (A\otimes B)(\psi\otimes \phi) \equiv A\psi\otimes B\phi$$ Also, the tensor-product is distributive in both factors $$ (A+B)\otimes C = A\otimes C + B\otimes C$$ $$ A\otimes (B+C) = A\otimes B + A\otimes C $$

I assume your notation $\sigma_{\vec{i}}$ means $\vec{n}\cdot \vec{\sigma}$ with $\vec{n}$ a unit vector and $\vec{\sigma} = (\sigma^1,\sigma^2,\sigma^3)$

You may be aware, that you can express the Pauli matrices in a different basis: $$ (\sigma^1,\sigma^2,\sigma^3) \rightarrow (\sigma^+,\sigma^-,\sigma^3)$$ with $$ \sigma^+ = \sigma^1 + i\sigma^2 $$ and $$ \sigma^- = \sigma^1 - i\sigma^2 $$ which act as raising and lowering operators in the basis where $\sigma^3$ is diagonal.

Since you've already written out the exponential, you can now explicitly calculate the tensor operator by distributivity and observe how it acts on the basis of the product-space.

I have not followed the calculation to the end and truth be told, don't know if the conjecture is correct. But the reasoning you bring forward sounds convincing enough.

Answer 2

$\sigma_{\vec{\imath}}$ is the rotation matrix about the vector $\vec{\imath}$:

$$\sigma_{\vec{\imath}} = \imath_x\sigma_{x}+\imath_y\sigma_{y}+\imath_z\sigma_{z}$$

After expanding it out completely and after some shuffling around, the solution is as follows:

$$e^{-i\theta/2 \sigma_{\vec{\imath}}^A} \otimes e^{-i\theta/2 \sigma_{\vec{\imath}}^B} |\Psi\rangle_{AB} = e^{i\phi/2\sigma_{\vec{\imath}}^A \otimes \sigma_{\vec{\imath}}^B} |\Psi\rangle_{AB}$$