The Cooper pair box Hamiltonian in the matrix form

Click For Summary
SUMMARY

The discussion focuses on calculating and plotting the energy bands (eigenvalues) of the Cooper pair box Hamiltonian using Mathematica. The Hamiltonian is defined in matrix form for charge states 0 to 5, with parameters E_C and E_J specified as (i) E_C = 70, E_J = 10 and (ii) E_C = 20, E_J = 20. Users are advised to utilize the Eigenvalues function in Mathematica to obtain the eigenvalues for plotting, although some participants express confusion regarding its implementation and seek alternative methods for manual calculation.

PREREQUISITES
  • Understanding of quantum mechanics concepts, particularly the Cooper pair box model.
  • Familiarity with matrix representation of Hamiltonians.
  • Proficiency in using Mathematica for mathematical computations.
  • Knowledge of eigenvalue problems in linear algebra.
NEXT STEPS
  • Learn how to implement matrix operations in Mathematica, specifically using the Eigenvalues function.
  • Research the mathematical derivation of eigenvalues for tridiagonal matrices.
  • Explore visualization techniques in Mathematica for plotting energy bands.
  • Study the physical implications of varying E_C and E_J in the Cooper pair box model.
USEFUL FOR

Physicists, quantum computing researchers, and students studying superconductivity or quantum mechanics, particularly those interested in the Cooper pair box model and its applications in quantum circuits.

maximus123
Messages
48
Reaction score
0
Hello,

In my problem I need to
Use Matematica (or any other program) to calculate and plot energy bands
(eigenvalues) of the Cooper pair box with (i)E_C = 70, E_J = 10 and (ii) E_C = 20,
E_J = 20
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}<br /> \ddots &amp; &amp; &amp; &amp; &amp;\\<br /> &amp; E_C(0-n_g)^2 &amp; -\frac{E_J}{2} &amp; 0 &amp; 0 &amp;\\<br /> &amp;-\frac{E_J}{2} &amp; E_C(1-n_g)^2 &amp; -\frac{E_J}{2} &amp; 0 &amp;\\<br /> &amp;0 &amp; -\frac{E_J}{2} &amp; E_C(2-n_g)^2 &amp; -\frac{E_J}{2} &amp;\\<br /> &amp;0 &amp; 0 &amp; -\frac{E_J}{2} &amp; E_C(3-n_g)^2 &amp;\\<br /> &amp; &amp; &amp; &amp; &amp;\ddots<br /> \end{pmatrix}
Because we are being asked for this matrix from states 0 to 5 I presume this means

\begin{pmatrix}<br /> E_C(0-n_g)^2 &amp; -\frac{E_J}{2} &amp; 0 &amp; 0 &amp; 0 &amp; 0\\<br /> -\frac{E_J}{2} &amp; E_C(1-n_g)^2 &amp; -\frac{E_J}{2} &amp; 0 &amp; 0 &amp; 0\\<br /> 0 &amp; -\frac{E_J}{2} &amp; E_C(2-n_g)^2 &amp; -\frac{E_J}{2} &amp; 0 &amp; 0\\<br /> 0 &amp; 0 &amp; -\frac{E_J}{2} &amp; E_C(3-n_g)^2 &amp; -\frac{E_J}{2} &amp; 0\\<br /> 0 &amp; 0 &amp; 0 &amp; -\frac{E_J}{2} &amp; E_C(4-n_g)^2 &amp; -\frac{E_J}{2} \\<br /> 0 &amp; 0 &amp; 0 &amp; 0 &amp; -\frac{E_J}{2} &amp; E_C(5-n_g)^2<br /> \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
 
I don't know if this thread is relevant anymore, since it is already a week old. But what was the problem with using the Eigenvalues[] function? It should be quite straightforward. The Mathematica documentation can assist you in using it.
 

Similar threads

  • · Replies 13 ·
Replies
13
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 14 ·
Replies
14
Views
3K
Replies
4
Views
1K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 0 ·
Replies
0
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
Replies
3
Views
2K