# The Cooper pair box Hamiltonian in the matrix form

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1. Mar 12, 2015

### maximus123

Hello,

In my problem I need to
We are advised to create the Cooper pair box Hamiltonian in a matrix form in the charge basis for charge
states from 0 to 5. Here is the Hamiltonian we are given

$H=E_C(n-n_g)^2 \left|n\right\rangle\left\langle n\right|-\frac{E_J}{2}(\left|n\right\rangle\left\langle n+1\right|+\left|n+1\right\rangle\left\langle n\right|)$
Which in the matrix form looks like
$\begin{pmatrix} \ddots & & & & &\\ & E_C(0-n_g)^2 & -\frac{E_J}{2} & 0 & 0 &\\ &-\frac{E_J}{2} & E_C(1-n_g)^2 & -\frac{E_J}{2} & 0 &\\ &0 & -\frac{E_J}{2} & E_C(2-n_g)^2 & -\frac{E_J}{2} &\\ &0 & 0 & -\frac{E_J}{2} & E_C(3-n_g)^2 &\\ & & & & &\ddots \end{pmatrix}$
Because we are being asked for this matrix from states 0 to 5 I presume this means

$\begin{pmatrix} E_C(0-n_g)^2 & -\frac{E_J}{2} & 0 & 0 & 0 & 0\\ -\frac{E_J}{2} & E_C(1-n_g)^2 & -\frac{E_J}{2} & 0 & 0 & 0\\ 0 & -\frac{E_J}{2} & E_C(2-n_g)^2 & -\frac{E_J}{2} & 0 & 0\\ 0 & 0 & -\frac{E_J}{2} & E_C(3-n_g)^2 & -\frac{E_J}{2} & 0\\ 0 & 0 & 0 & -\frac{E_J}{2} & E_C(4-n_g)^2 & -\frac{E_J}{2} \\ 0 & 0 & 0 & 0 & -\frac{E_J}{2} & E_C(5-n_g)^2 \end{pmatrix}$

It is then suggested we put this into Mathematica and use the Eigenvalues function to return the eigenvalues so we can then plot the energy bands. I have tried using Mathematica with this matrix but am not getting any results I understand. Is there a method for finding the eigenvalues of this matrix by hand? I am quite lost with this question, any help would be greatly appreciated. Thanks

2. Mar 17, 2015