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Many two-level systems can be described by the following Hamiltonian matrix:

  • $\left<i|\hat {H}|i\right>$ $=\alpha, i=1,2$
  • $\left<i|\hat {H}|j\right> = \beta, i\neq j$

a) Find the e eigenvalues and eigenvectors for the Hamiltonian in the basis given, being careful to normalise.

b) Find the similarity transformation from the initial representation to the representation using the eigenvectors of the molecule. Explicitly transform the Hamiltonian to the eigenvector basis, and comment on the resulting form.

I have done part (a) and the eigenvectors are: $\psi_1=\frac{1}{\sqrt{2}} \begin{bmatrix} 1 \\ -1 \\ \end{bmatrix} $

and $\psi_2=\frac{1}{\sqrt{2}} \begin{bmatrix} 1 \\ 1 \\ \end{bmatrix} $

but I don't understand how to do the change of representation.

By this I mean, I understand the principles of it, but I just don't really get what the vectors for the initial representation are, not sure if I'm being really obtuse here....

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closed as off-topic by FGSUZ, Emilio Pisanty, Buzz, Aaron Stevens, ZeroTheHero Jan 23 at 20:49

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The part (b) just means the following:

$$ \langle \psi_i | H | \psi_j \rangle\ =\ ? \quad \ i,j = 1,2 $$

You can certainly calculate that using the eigenvalue problem $H |\psi_i\rangle = E_i |\psi_i\rangle$ (since you said you've got the eigenvalues) and the orthogonalization condition of eigenvectors. And after this, again represent the eigenstates (in terms of vectors) in this basis.

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  • $\begingroup$ thanks! so i can use the intial i and j vectors as \begin{bmatrix} 1 \\ 0 \\ \end{bmatrix} and \begin{bmatrix} 0 \\ 1\\ \end{bmatrix} $\endgroup$ – M. Collyer Jan 23 at 16:00
  • $\begingroup$ correct, the representation of eigenstates using eigenstates as the basis is just like you wrote! You can guess the form of the Hamiltonian matrix this time, right? Btw, the second part of the Hamiltonian definition should be $\langle i | H | j \rangle$ to follow the description $i \neq j$. I can't edit it because the forum requires at least editing 6 characters. $\endgroup$ – rnels12 Jan 23 at 16:08
  • $\begingroup$ yes, I didn't see that! I have now editing it, thanks! $\endgroup$ – M. Collyer Jan 23 at 22:20

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