# Matrix of Hamiltonian $H=\frac{ħω}{2}(|1〉〈1|−|2〉〈2|)+\frac{iħχ}{2}(|1〉〈2|−|2〉〈1|)$ [closed]

I have a second order system, with a Hamiltonian $$H=\frac{ħω}{2}(|1〉〈1|−|2〉〈2|)+\frac{iħχ}{2}(|1〉〈2|−|2〉〈1|)$$ where $$|1〉,|2〉$$ form a complete basis for the system. I'm trying to get the matrix that represents H in this system, but am unsure how to go about it. Am I supposed to treat the basis vectors generally? ie $$|1〉=(a_1+ib_1, a_2+ib_2)^T$$, and similarly for $$|2〉$$. Where I then expand out and simplify?

Cheers for any help.

## closed as off-topic by ZeroTheHero, Aaron Stevens, GiorgioP, stafusa, user191954 Mar 10 at 5:28

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• Choose an orthogonal basis of dimension 2. – InertialObserver Mar 2 at 0:40
• Each matrix element is just given by $H_{m,n}=\langle m|H|n\rangle$. Just check any of your intro QM or linear algebra texts :) – Aaron Stevens Mar 2 at 0:51
• With basis you mean orthonormal basis? The basically you can take (1,0) and (0,1) as basis vector representation. – lalala Mar 2 at 2:36

## 1 Answer

$$\newcommand{\bra}[1]{\langle #1 \rvert} \newcommand{\ket}[1]{\lvert #1 \rangle} \newcommand{\bracket}[2]{\langle #1 \vert #2 \rangle}$$

You said your vectors $$\ket 1$$ and $$\ket 2$$ form a complete basis. I assume they are also meant to be orthonormal, i.e. \begin{align} \bracket 1 1 = 1 \\ \bracket 1 2 = 0 \\ \bracket 2 1 = 0 \\ \bracket 2 2 = 1 \end{align} \tag{1}

Given your Hamiltonian $$H = \frac{\hbar\omega}{2} (\ket{1}\bra{1} - \ket{2}\bra{2}) + \frac{i\hbar\chi}{2} (\ket{1}\bra{2} - \ket{2}\bra{1}) \tag{2}$$

it is straight-forward to get the matrix elements by their definition $$H_{mn} = \bra{m}H\ket{n}$$.

For example \begin{align} H_{11} &= \bra{1}H\ket{1} \\ &= \bra{1}\left( \frac{\hbar\omega}{2} (\ket{1}\bra{1} - \ket{2}\bra{2}) + \frac{i\hbar\chi}{2} (\ket{1}\bra{2} - \ket{2}\bra{1}) \right) \ket{1} \\ &= \frac{\hbar\omega}{2} \end{align}

In the last step above you expand the product, and by observing the orthonormality relations (1) most terms of the resulting sum are zero. Only one term remains.

In a similar manner you get \begin{align} H_{12} = \bra{1}H\ket{2} &= \frac{i\hbar \chi}{2} \\ H_{21} = \bra{2}H\ket{1} &= -\frac{i\hbar \chi}{2} \\ H_{22} = \bra{2}H\ket{2} &= -\frac{\hbar \omega}{2} \end{align}

or written as matrix $$\begin{pmatrix} H_{11} & H_{12} \\ H_{21} & H_{22} \end{pmatrix} = \frac{\hbar}{2} \begin{pmatrix} \omega & i\chi \\ -i\chi & -\omega \end{pmatrix}$$

Let me add some clarification about representation, since I got the impression that there was some confusion what this actually means. The Hamiltonian $$H$$ is represented by this matrix $$\begin{pmatrix} H_{11} & H_{12} \\ H_{21} & H_{22} \end{pmatrix}$$. The basis state $$\ket 1$$ is simply represented by the column vector $$\begin{pmatrix} 1 \\ 0 \end{pmatrix}$$, and $$\ket 2$$ is represented by $$\begin{pmatrix} 0 \\ 1 \end{pmatrix}$$. A general state $$\ket \psi$$ would be represented by $$\begin{pmatrix} c_1 \\ c_2 \end{pmatrix}$$.