Finding matrix representation of Hamiltonian operator

Question

Let the quantum system composed by an orthonormal base with the states $|1\rangle, |2\rangle$ and $|3\rangle$ with all being degenereted states of the observable D with eingenvalue $\delta$. So, being the action of the Hamiltonian operator $H$ given by

$$ H|1\rangle = \Omega|1\rangle + \Omega|3\rangle $$ $$ H|2\rangle = \Omega|2\rangle + \Delta|3\rangle $$ $$ H|3\rangle = \Omega|1\rangle + \Delta|2\rangle + \Omega|3\rangle $$

with $\Omega, \Delta$ real constants, how can I represented this Hamiltonian in matrix form ? Actually, I don't asking for a solution, just for tips that I can start to work in.

Answer 1

Remember, the matrix elements of the Hamiltonian $H_{ij}$ are given by the element $\langle i | H | j \rangle$, where $|i\rangle$ and $|j\rangle$ are orthonormal eigenstates of $H$. So for example we can find the matrix element $H_{23} = H_{32}$ as follows:

$$ H_{23} = \langle 2 | H | 3 \rangle = \Omega \langle 2 | 1 \rangle + \Delta \langle 2 | 2 \rangle + \Omega \langle 2 | 3 \rangle = \Delta \\ H_{32} = \langle 3 | H | 2 \rangle = \Omega \langle 3 | 2 \rangle + \Delta \langle 3 | 3 \rangle = \Delta \\ H_{23} = H_{32} $$

Since the Hamiltonian is always Hermitian and equal to its own adjoint, it is symmetrical, and we recover that here for this pair of matrix elements. Hope that helps!

Answer 2

To represent $H$ in a matrix form, $H_{ij}$, you need basis states that you can represent in matrix form:

$$ e_1 = \left[\begin{array}{c} 1 \\0\\0 \end{array}\right]$$

You have that in your kets: $|j\rangle$ for $j \in (1,2,3)$.

Thus, the state:

$$ \psi = a_1|1\rangle + a_2|2\rangle + a_3|3\rangle $$

is represented as:

$$ \psi_i = \left[\begin{array}{c} a_1 \\a_2\\a_3 \end{array}\right]$$

From here, you write out the nine $H_{ij}$ by inspecting your three equations, e.g. $H_{11} = \Omega$.

Answer 3

The matrix elements $O_{ij}$ of any operator $\hat{O}$ in a basis $|n\rangle$ can easily be shown to be $$O_{ij} = \langle i|\hat{O}|j\rangle.$$

(The convention is to have $i$ represent the rows and $j$ the columns.) This is done in most introductory quantum mechanics courses, and is relatively easy to show if you accept that $|n\rangle$ is an orthonormal basis satisfying the completeness relation $\sum_i |i\rangle\langle i| = 1$.

Using this, your question should be trivial to solve, unless I've misunderstood it...