How to find the 2-by-2 complex matrix $H$ (if there is one) for which $$H\left[\begin{matrix} a & b \\ b^* & c\end{matrix}\right]H^\dagger=\left[\begin{matrix} a & be^{-i\lambda} \\ b^*e^{i\lambda} & c\end{matrix}\right]$$
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$\begingroup$ What does $b^*$ mean? Conjugate of $b$? And $H^\dagger$? Tranpose, adjoint... of $H$? $\endgroup$– AugSBNov 21, 2016 at 9:19
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$\begingroup$ @AugSB yes, $b^*$ is conjugate and $H^\dagger$ is conjugate transpose $\endgroup$– AdamNov 21, 2016 at 9:22
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1$\begingroup$ Just multiplying and solving the resulting equation system may help? $\endgroup$– vidyarthiNov 21, 2016 at 9:25
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Hint: $$ \begin{pmatrix}a & be^{-i\lambda} \\ b^*e^{i\lambda}&c\end{pmatrix}= \begin{pmatrix}e^{-i\lambda} &\\ &1\end{pmatrix} \begin{pmatrix}a & b\\b^* &c\end{pmatrix} \begin{pmatrix}e^{i\lambda} &\\ &1\end{pmatrix}. $$
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$\begingroup$ how did you come to know about this product? $\endgroup$ Nov 21, 2016 at 9:26
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1$\begingroup$ @vidyarthi Just feeling that it should be something simple and I tried diagonal matrices. $\endgroup$– A.Γ.Nov 21, 2016 at 9:28
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