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I have 2 bases A and B with the following kets:

Base A: $|a_1\rangle$ and $|a_2\rangle$

Base B: $|b_1\rangle = \frac{1}{\sqrt2} \cdot(|a_1\rangle + i\cdot|a_2\rangle)$ $|b_2\rangle = \frac{1}{\sqrt2} \cdot(|a_1\rangle - i\cdot|a_2\rangle)$

Both bases are orthonormal in the 2D Hilbert-space. The linear operator $\hat{T}$ is given in base A:

$\hat{T}= \begin{pmatrix} 1& 0\\ 0& -1 \end{pmatrix}$

Now I should find out how the linear operator $\hat{T}$ looks in the base B image? Should be easy, but I cant wrap my head around it.

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closed as off-topic by Valter Moretti, DavePhD, Brandon Enright, Chris White, BMS May 13 '14 at 6:11

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First you need to compute the change-of-basis-matrix. – Antonio Ragagnin May 12 '14 at 12:10
you can calculate this by knowing that hermitian operator acting on ket vector produces number (eigenvalue) and eigen-vector. $\hat H|\Psi\rangle = \lambda|\Psi\rangle$, and for $|b_1\rangle$ $\lambda = 1$ and for $|b_2\rangle$ $\lambda = -1$ so for base B this operator will look like this: $\begin{pmatrix} 0 & -i\\ i & 0 \\ \end{pmatrix}$ – Gigi Butbaia May 12 '14 at 13:15
You find the unitary operator U that maps $|a_1>$ to $|b_1>$ and $|a_2>$ to $|b_2>$. Then you find T in the b basis by doing $U^\dagger T U$. – alanf May 12 '14 at 13:15
Have a look at the extended edit to this post: - maybe this will already be enough to apply it to your problem! – Martin May 12 '14 at 14:33