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There is a well-known formula for rotations of Pauli vectors $e^{i\theta \vec{n}\cdot\vec{\sigma}}=\cos{\theta}+i\sin{\theta} \vec{n}\cdot\vec{\sigma}$ with $\vec{\sigma}=(\sigma_x,\sigma_y,\sigma_z)$.

Now I am dealing with a similar formula with an added difficulty: two spin vectors, $\vec{\sigma_1}$ and $\vec{\sigma_2}$, such that my rotation is given by $e^{i\theta \vec{\sigma_1}\cdot\vec{\sigma_2}}$. It is not possible to, a priori, writte the same expression as for one spin because the dot product will produce non-commuting terms, i.e. $[\sigma_x^{1}\sigma_x^{2},\sigma_y^{1}\sigma_y^{2}]\neq 0$ which prevents one from using the BKH formula.

Is there any known expression for such a two-spin rotation?

Thank you!

Edit: My goal is to apply this transformation to a state and see how it transforms. Specifically, I want to evaluate $e^{i\theta \vec{\sigma_1}\cdot\vec{\sigma_2}}|\phi\rangle$, where $|\phi\rangle$ is the bell state $\frac{1}{\sqrt{2}}(|00\rangle+|11\rangle)$

J.Agusti
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  • Are you sure that you want $\sigma_1\cdot \sigma_2$? It is far more like that a qm situation (combining two angular momenta for example) would give you $\sigma_1 \otimes {\mathbb I}+ {\mathbb I}\otimes \sigma_2$ in the exponent. – mike stone Jul 11 '22 at 16:53
  • @mikestone yes, that's what I need. I have a unitary transformation such that $U=e^{i\theta/4(\sigma_1+\sigma_2)^{2}}\simeq e^{i\theta/2(\sigma_1\cdot\sigma_2)}$ and I need to act on a state with this. – J.Agusti Jul 11 '22 at 16:59
  • @mike stone Peculiar as it might be, the reducible matrix in the exponent is still well-defined: $(\vec \sigma_1+\vec \sigma_2)^2/2 -3{\mathbb I}$... – Cosmas Zachos Jul 11 '22 at 17:00
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    So this is $J^2$ for the sum of two spin 1/2s? In that case it can be diagonalized as $1 {\mathbb I}_3 + 0 {\mathbb I}_1$ for the three spin one states and the one spin zero state. Once this is done application to state is easy. – mike stone Jul 11 '22 at 17:07
  • @mikestone I have to think about it, but for me, it is just a two-qubit rotation. Still, I will try to apply what you said. – J.Agusti Jul 11 '22 at 17:24
  • @CosmasZachos I don't really understand your result... When I apply exponential series on the vector product, I always get it to power $(\sigma_1\cdot\sigma_2)^{k}$, i.e. I am not able to reduce it to the unitary for even numbers or the product for odd numbers, as with the single-qubit rotation. – J.Agusti Jul 11 '22 at 17:28
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    It really helps to use the tensor product notation when you have more than one qbit. – mike stone Jul 11 '22 at 17:32

2 Answers2

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Review spin addition for two spin doublets. Your formula is actually simpler, in a sense! No expansion of the exponential is warranted. Expanding it is a hallmark of some misconception.

Anything that happens to the exponent happens to the exponential. You are liable to get confused by the complacent notation you express your exponential in. The exponent of it is proportional to a 4×4 matrix, $$ (\vec\sigma_1\otimes {\mathbb I}+ {\mathbb I} \otimes \vec \sigma_2 )^2/2 -3 {\mathbb I}_4 \\ = \vec \sigma_1\otimes \cdot \vec \sigma_2 +(\vec \sigma_1^2\otimes {\mathbb I}+ {\mathbb I}\otimes \vec \sigma_2^2 )/2 -3 {\mathbb I}_4\\ = \vec \sigma_1\otimes \cdot \vec \sigma_2 ~, $$ where the dimensionality of the unlabelled identity matrices, 2, is obvious from context and omitted for simplicity.

Now, you know from the linked answer that the above 4×4 matrix may be reduced by a 4×4 Clebsch similarity transform change of basis to a diagonal matrix, $$ O ~\vec \sigma_1\otimes \cdot \vec \sigma_2 ~ O^{-1}= 5{\mathbb I}_3 \oplus (-3), $$ with evident 3×3 and 1×1 blocks.

The very same similarity transformation will transform the exponential of your matrix then, the trivial exponential of a diagonal matrix, $$ O~ \exp (i\theta ~ \vec \sigma_1\otimes \cdot \vec \sigma_2 ) ~O^ {-1}= e^{i5\theta}{\mathbb I}_3 \oplus e^{-3i\theta}\\ = \operatorname{diag} (e^{i5\theta},e^{i5\theta},e^{i5\theta} , e^{-3i\theta} ). $$ If you insisted on returning to your original, 1-tensor-2, dysfunctional basis, you trivially reverse the above orthogonal Clebsch equivalence transformation.

Cosmas Zachos
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  • Thank you for the detailed answer and the links, it was been really helpful to understand my unitary transformation! One question: Originally, this unitary acted on the entangled state $U|psi\rangle$, where $|\psi\rangle=(|0,0\rangle+|1,1\rangle)/\sqrt{2}$ and I know how $\sigma_{1}\otimes\sigma_{2}$ acts upon it. Under this transformation, I need to find how the new operators act on the old state or define a new state that lives in the triplet state, right? – J.Agusti Jul 12 '22 at 09:25
  • Right. The Clebsch matrix O is given in the linked answer footnote. – Cosmas Zachos Jul 12 '22 at 09:27
  • ...but your state is 1-2 symmetric, so it's already in the triplet ! So it looks like it is just multiplied by $e^{i5\theta}$... – Cosmas Zachos Jul 12 '22 at 13:36
  • Exactly! I did the maths for the 1-2 symmetric state and it works. Another different story is when I apply this transformation on $|1/2,1/2\rangle\otimes|1/2,-1/2\rangle=(|1,0\rangle+|0,0\rangle)/\sqrt{2}$ which mixes the states. – J.Agusti Jul 12 '22 at 13:42
  • Right. You then have to use the matrix $O^T \operatorname{diagonal} (...) O$ acting on $(1,0,1,0)^T/\sqrt{2}$. Do you see the $(e^{i5\theta}\pm e^{-i3\theta})/2$ s? – Cosmas Zachos Jul 12 '22 at 13:50
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Perhaps a naïve approach: the total spin is $$\vec{S}=\vec{\sigma}_1+\vec{\sigma}_2,$$ and hence the rotation is given by $$ e^{i\theta\vec{n}\vec{S}}=e^{i\theta\vec{n}(\vec{\sigma}_1+\vec{\sigma}_2)}= e^{i\theta\vec{n}\vec{\sigma}_1}e^{i\theta\vec{n}\vec{\sigma}_2}, $$ where the last equality follows from the fact that the operators for the two spins commute.

Roger V.
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    This is true, but my unitary transformation is different: $e^{i\theta(\vec{\sigma_1}+\vec{\sigma_2})^2}\propto e^{i\theta\vec{\sigma_1}\vec{\sigma_2}}$, which is different and does not allow me to use the decomposition you use. – J.Agusti Jul 12 '22 at 09:59
  • @J.Agusti your post is not very clear on whether you are looking for a rotation or for applying the given unitary transformation. – Roger V. Jul 12 '22 at 10:03
  • you are right, this was probable confusing... I edited the original post mentioning what my goal is. – J.Agusti Jul 12 '22 at 10:09
  • @J.Agusti anyhow, the general method is to diagonalize the matrix $\vec{\sigma}_1\vec{\sigma}_2$, exponentiate it, return to the original representation and write it in terms of the two spins. It is a very simple 4-by-4 matrix (exchange interaction). – Roger V. Jul 12 '22 at 10:10
  • I did the maths but I'm not sure about it: $U|\psi\rangle=e^{i\theta/2\vec{\sigma_1}\vec{\sigma_2}}\frac{|1/2,-1/2\rangle\otimes|1/2,-1/2\rangle+|1/2,1/2\rangle\otimes|1/2,1/2\rangle}{\sqrt{2}}=e^{i\theta/2\vec{\sigma_1}\vec{\sigma_2}}\frac{|1,1\rangle+|1,-1\rangle}{\sqrt{2}}=e^{i\theta/2}\frac{|1,1\rangle+|1,-1\rangle}{\sqrt{2}}$, where I expressed the old basis in the triplet basis $|S,M_{s}\rangle$ and used $\vec{\sigma_1}\vec{\sigma_2}|1,M_{s}\rangle=+|1,M_{s}\rangle$. Then, going to the old basis we have $e^{i\theta/2}|\phi\rangle$, i.e. the transformation only adds a phase. – J.Agusti Jul 12 '22 at 11:40