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Consider the following Hamiltonian, in arbitrary units:

$$ H = \begin{bmatrix} 0 & 0 & g\\ 0 & 0 & g\\ g & g & 1 \end{bmatrix}$$

where $g<<1$. It is relatively straightforward to find the spectrum of this Hamiltonian analytically. Simply solve the characteristic equation and find that the three eigenvalues of this matrix are:

$$\begin{aligned} \epsilon_1 &= \frac{1}{2}\left(1 - \sqrt{1+8g^2}\right) \approx-4g^2 \\ \epsilon_2 &= 0 \\ \epsilon_3 &= \frac{1}{2}\left(1 + \sqrt{1+8g^2}\right) \approx1+4g^2 \end{aligned}$$

I am asked to calculate the spectrum of this Hamiltonian to second order by pertubative methods. So we split the Hamiltonian up into:

$$ H = H_0 + g \tilde H$$

But of course we realise immediately that $H_0$ is degenerate.

How do we perform perturbation theory on this Hamiltonian? References are welcome, but will not be considered answers, as at the moment I cannot access them.

Note, I am at home, and the only reference I have on the subject is Alastair Rae's Quantum Mechanics, 5th edition. Here there is a section on perturbation theory for degenerate levels, but this approach does not work here. That's because it relies on finding first order corrections to the energy levels, which we know don't exist, thanks to our analytical solution.

I have also partially cheated, by recognising that the vector:

$$ v_2 = \frac{1}{\sqrt{2}} \begin{bmatrix} ~~1~ \\ -1~ \\ ~~0~ \end{bmatrix}$$

is both the eigenvector of both $H$ and $H_0$ for the eigenvalue $0$, but it hasn't helped me much.

Finally, I am aware that there is a question about the same Hamiltonian, but there is no answer for it.

Cosmas Zachos
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Andrea
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1 Answers1

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Indeed, the first order firmly establishes the vanishing of the energy corrections but fails to completely specify the wavefunction corrections, and you must keep going, to 2nd and 3rd order to specify your unknowns in underdetermined systems. Courant and Hilbert (cited here ) describe the procedure, but maybe you don't want to go there... Anyway, the other two eigenvectors are $$ v_1 \propto \frac{1}{\sqrt{2}} \begin{bmatrix} ~~1~ \\ 1~ \\ ~~c~ \end{bmatrix} \qquad c=\left (1-\sqrt{1+8g^2}\right )/2g \approx -2g, \\ v_3 \propto \begin{bmatrix} ~~g~ \\ g~ \\ ~~b~ \end{bmatrix} \qquad b=\left (1+\sqrt{1+8g^2}\right )/2 \approx 1+2g^2, \\ \Longrightarrow v_3 \propto \begin{bmatrix} ~~0~ \\ 0~ \\ ~~1~ \end{bmatrix} + \sqrt{2} g v_1 +O(g^2),...$$ All 3 eigenvectors are mutually orthogonal.

The first order specifies the $v_3$ correction, but one knows nothing about 1-2 mixing, at this order, and so you need to go to the next level, to get those, etc.

Instead, your limited cheating was just enough, and all you need is to project out the $v_2$ subspace and end up with a trivial non-degenerate 2×2 system. Consider the "Foldy-Wouhuysen" transformation, $$ U = \begin{bmatrix} 1/\sqrt{2} & 1/\sqrt{2} & 0\\ -1/\sqrt{2} & 1/\sqrt{2} & 0\\ 0 & 0 & 1 \end{bmatrix}$$ yielding $$ U^\dagger H ~U = \begin{bmatrix} 0 & 0 & 0\\ 0 & 0 & \sqrt{2}g\\ 0 &\sqrt{2} g & 1 \end{bmatrix},$$ an equivalent system to the original, but now with the first subspace, your cheating eigenvector, projected out, and non degenerate to boot, so the second order energy shift is $E^{(2)}=\pm 2g^2$ as per the standard dull formula. The pros call this a 3-scale problem: 0, $g^2$, and 1, as you saw.

Cosmas Zachos
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