How to find matrix representation of an operator in new basis I have recently begun to learn QM and I cannot solve this task:
Let's say I have operator $\hat H = \begin{bmatrix}\epsilon & \upsilon \\ \upsilon & \epsilon \end{bmatrix} ,(\upsilon \in \mathbb{R} \backslash \{0\}) $ in a orthornormal basis defined by $ \lvert\phi_1\rangle $ and $ \lvert\phi_2\rangle $.
Then make new basis defined by $\lvert\phi_1'\rangle = \frac{1}{\sqrt2} (\lvert\phi_1\rangle + \lvert\phi_2\rangle)$ and $\lvert\phi_1'\rangle = \frac{1}{\sqrt2} (\lvert\phi_1\rangle - \lvert\phi_2\rangle)$. Is there an easy way to find matrix representation of $\hat H$ in second basis?
 A: Yes, indeed, you simply need to calculate the matrix elements of the Hamiltonian in the new basis:
\begin{array}
\hat{H}_{11}' = \langle \phi_1'|\hat{H}|\phi_1'\rangle = 
\frac{1}{2}(\langle\phi_1| + \langle\phi_2|)\hat{H}(|\phi_1\rangle + |\phi_2\rangle) = \\
\frac{1}{2}(\langle \phi_1|\hat{H}|\phi_1\rangle + \langle \phi_1|\hat{H}|\phi_2\rangle + \langle \phi_2|\hat{H}|\phi_1\rangle + \langle \phi_2|\hat{H}|\phi_2\rangle) = \frac{1}{2}(\epsilon + v + v + \epsilon) = \epsilon + v.
\end{array}
Similarly we obtain:
\begin{array}
\hat{H}_{12}' = \langle \phi_1'|\hat{H}|\phi_2'\rangle = 
\frac{1}{2}(\langle\phi_1| + \langle\phi_2|)\hat{H}(|\phi_1\rangle - |\phi_2\rangle) = \frac{1}{2}(\epsilon + v - v - \epsilon) = 0,\\
\hat{H}_{21}' = \langle \phi_2'|\hat{H}|\phi_1'\rangle = 
\frac{1}{2}(\langle\phi_1| - \langle\phi_2|)\hat{H}(|\phi_1\rangle + |\phi_2\rangle) = \frac{1}{2}(\epsilon - v + v - \epsilon) = 0,\\
\hat{H}_{22}' = \langle \phi_2'|\hat{H}|\phi_2'\rangle = 
\frac{1}{2}(\langle\phi_2| - \langle\phi_2|)\hat{H}(|\phi_1\rangle - |\phi_2\rangle) = \frac{1}{2}(\epsilon - v - v + \epsilon) = \epsilon -v.
\end{array}
Thus, the new Hamiltonian in matrix representation is
\begin{equation}
\hat{H}' =
\begin{bmatrix}
\epsilon + v & 0\\
0 & \epsilon -v\\
\end{bmatrix}.
\end{equation}
A: The easy way in which you will always get there is just by diagonalizing the system first. When you "solve" a Hamiltonian you always search for the eigenvectors (=eigenstates) and eigenvalues. Let's try and do this for this Hamiltonian now first. We find the eigenvalues through the following characteristic equation:
$\begin{equation} \det(\lambda I - A ) \textbf{v} = 0 \end{equation} $ 
If we do this we get $\det\hat H = \det\begin{bmatrix}\lambda - \epsilon & \upsilon \\ \upsilon & \lambda -\epsilon \end{bmatrix} = ( \lambda - \epsilon)^2 - v^2 = 0 \\ \Rightarrow\lambda = \epsilon \pm v$
Now we solve the system of equations with the eigenvalues for the eigenvectors:
$\hat H = \begin{bmatrix}\lambda - \epsilon & \upsilon \\ \upsilon & \lambda -\epsilon \end{bmatrix} \begin{bmatrix} x\\  y\end{bmatrix} = 0$
I won't do this completely for you, but you will notice that you get exactly the vectors you mentioned $ \lvert\phi_1'\rangle $ and $ \lvert\phi_2'\rangle $. So trying to get the Hamiltonian in that specific base amounts to diagonalizing it here. It will look like 
$\hat H = \begin{bmatrix} \epsilon + v & 0 \\ 0 & \epsilon - v \end{bmatrix} $
Edit: But this is not necessarily always so, as I had to clarify. For very complex Hamiltonians it is easier to transform it to the new basis from this form though. 
