# How to make a base tranformation for a linear operator in QM? [closed]

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.

-

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

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "Homework-like questions should ask about a specific physics concept and show some effort to work through the problem. We want our questions to be useful to the broader community, and to future users. See our meta site for more guidance on how to edit your question to make it better" – Valter Moretti, DavePhD, Brandon Enright, Chris White, BMS
If this question can be reworded to fit the rules in the help center, please edit the question.

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: physics.stackexchange.com/q/81400 - maybe this will already be enough to apply it to your problem! –  Martin May 12 '14 at 14:33