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?


2 Answers 2


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}

  • $\begingroup$ Thank you really much, it's so simple yet I couldn't see that. $\endgroup$
    – k-duda
    Mar 24, 2020 at 15:36

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.

  • $\begingroup$ What you suggest is finding the Hamiltonian in the basis of its eigenvectors, not transforming it to the basis defined by the new basis vectors. These new vectors didn't have to be the same as its eigenbasis (although in this case they are indeed the same). $\endgroup$
    – Roger V.
    Mar 24, 2020 at 12:55
  • $\begingroup$ In this case they are, and it's very easy starting from the eigenvalues and eigenvectors to construct the new Hamiltonian in any new basis, certainly when you have a very complex Hamiltonian. I never said you would always magically have your eigenvectors as the asked basis. $\endgroup$
    – CFRedDemon
    Mar 24, 2020 at 13:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.