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Let $\vec a$ be a given vector, and $\vec J=[\hat J_x,\hat J_y,\hat J_z]$ the total angular momentum operator. $\vec \beta$ is an unknown vector, to be solved for, which satisfies $$ \vec a=\mathrm{Tr}[\vec J\exp(\vec \beta\cdot\vec J)]. $$

How to solve this equation? The above nonlinear equation has three components, and three unkowns, $\vec \beta$. The existence and uniqueness of the solution are not guaranteed.

I asked several related questions like this, this and this, but I'm still confused about this problem.

The title and tags of this post are not clear. Thanks for any improvement.

John Rennie
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Luessiaw
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  • Just to be sure: You want to find a vector $\beta$ such that this equation holds for a given $a$? And do you mean that the equation holds for each component? I.e. $a_x=\mathrm{Tr} J_x \exp(\beta_x Jx + \beta_y J_y + \beta_z J_z)$ and similarly for $a_y$, $a_z$ - is this the meaning of your notation? – Tobias Fünke Jan 23 '23 at 14:11
  • @TobiasFünke Yes. – Luessiaw Jan 23 '23 at 14:15
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    The system is not necessarily of spin-1/2. It can be arbitrary spin. – Luessiaw Jan 23 '23 at 14:31
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    You might start by selecting cases and utilizing this ref, which you probably have used. The spin 1/2 is tractable, and you must show your work on it. – Cosmas Zachos Jan 23 '23 at 15:29
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    @ CosmasZachos Thanks! I'll work on it. – Luessiaw Jan 23 '23 at 16:18

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Your stumbling block might be starting from seeking a general solution for something belonging to the universal enveloping algebra, something essentially outside Lie Theory (rotations). You are probably aware of the expansion of your exponential operator in terms of finite polynomials of increasing order with the dimensionality of the representation, so your equation complicates substantially with increasing this dimensionality.

Instead of rapid-fire XY-problem questions, you might explore your equations for tractable small dimensionalities.

For instance, for spin 1/2, and $\vec \beta\equiv \beta ~\hat n $, you simply have $$ e^{\vec \beta\cdot \vec \sigma }= I~\cosh\beta + \hat n \cdot \vec \sigma ~ \sinh \beta \implies \\ 2\vec a= \operatorname{Tr} (\vec \sigma ~~ \hat n \cdot \vec \sigma \sinh \beta)= 2 \hat n ~ \sinh \beta. $$

For spin 1, you use the Rodrigues rotation formula, $$ e^{\vec \beta\cdot \vec \sigma }= I+ \sinh\! \beta~\hat n\cdot J +(\cosh\! \beta -1) ~(\hat n\cdot J), \implies \\ \vec a = \operatorname{Tr} (\vec J ~~(\sinh\! \beta~\hat n\cdot J +(\cosh\! \beta -1) ~(\hat n\cdot J)^2)), $$ where you may use the Cayley-Hamilton theorem for 3×3 matrices, etc.

For instance, you may rotate inside the trace $\hat n\cdot\vec J\to J_3$, $$\vec\beta\cdot \vec a= \beta \sinh\beta \operatorname{Tr} ( J_3^2)= 2 \beta \sinh\beta , $$ since odd powers of $J_3$ are traceless. This component, $\hat n ~~\hat n\cdot \vec a =\hat n~\vec \beta \cdot \vec a/\beta $, expands effortlessly to all dimensionalities. Work them out.

But the other two components, $\vec a -\hat n ~ \hat n\cdot \vec a$, are a messy problem requiring (lacking) knowledge of anticommutators. The problem quickly grows in complication with increasing dimensionality, but there may be an elegant solution shortcut lurking there...

Cosmas Zachos
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