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I have the following set of equations: $${\bf M_0}\Phi_0(u_1)|A_0\rangle = {\bf M_1}\Phi_1(u_1)|A_1\rangle $$ $${\bf M_1}\Phi_1(u_2)|A_1\rangle = {\bf M_2}\Phi_2(u_2)|A_2\rangle $$

I want to find $\bf B$ in the following equation$$|A_0\rangle={\bf B}|A_2\rangle$$ I have that it is equal to $${\bf B}=\Phi_0(-u_1){\bf M_0M_1}\Phi_1(u_1-u_2){\bf M_1M_2}\Phi_2(u_2)$$ All of the $M$ matrices are square, complex. $\Phi$ are diagonal matrices such that $\Phi^{-1}(u)=\Phi(-u)$.

I do not understand how to get from the two equations to the third.

Also, $\Phi(u)$ is built very similar to $e^u$.

EDIT::

I think I got it. But it only works if I assume $\bf M\cdot M=1$ M is constructed out of eigenvectors of a different matrix, so does this make sense?

yankeefan11
  • 1,449

1 Answers1

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You can't determine a matrix from what it does to one vector. However, assuming the $M$ matrices are invertible.

$$ \eqalign{| A_1 \rangle &= (M_1 \Phi_1(u_2))^{-1} M_2 \Phi_2(u_2) |A_2>\cr &= \Phi_1(-u_2) M_1^{-1} M_2 \Phi_2(u_2) |A_2>\cr | A_0 \rangle &= (M_0 \Phi_0(u_1))^{-1} M_1 \Phi_1(u_1) |A_1 \rangle\cr &= \Phi_0(-u_1) M_0^{-1} M_1 \Phi_1(u_1) \Phi_1(-u_2) M_1^{-1} M_2 \Phi_2(u_2) |A_2>\cr}$$

so you could take $$ B = \Phi_0(-u_1) M_0^{-1} M_1 \Phi_1(u_1) \Phi_1(-u_2) M_1^{-1} M_2 \Phi_2(u_2)$$

Robert Israel
  • 448,999