# Finding matrix representation of Hamiltonian operator

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.

• I'm not familiar with your definition of the action of the Hamiltonian and I don't get what you mean by writing $|H\rangle$. Could you be more explicit or include how you define the action of $H$? Commented Oct 29, 2020 at 17:55
• What is $|H\rangle$? Your notation does not make sense to me: the first and third equations imply that $\Delta = 0$ e.g. and then that gives a linear relation between three orthonormal vectors which is impossible. Commented Oct 29, 2020 at 17:55
• I wrote the expression wrong. Now is correct, sorry about that. Commented Oct 29, 2020 at 18:04

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!

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. $$H_{11} = \Omega$$.

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...