In introductory superconductivity one often studies the BCS Hamiltonian

$$H= \begin{pmatrix} \xi & -\Delta \\ -\Delta & -\xi \end{pmatrix} $$

I can find the Eigenvalues and Eigenvectors by writing Eigensystem[H] in Mathematica.

I then obtain the eigenvalues \begin{equation} \begin{split} &E_1 = \sqrt{\Delta^2 + \xi^2}\\ &E_2 = -\sqrt{\Delta^2 + \xi^2} \end{split} \end{equation} which I am happy with.

However, the Eigenvectors are given as \begin{equation} \begin{split} &v_1 = \left\{-\frac{\xi + \sqrt{\Delta^2 + \xi^2}}{\Delta},1\right\},\\[1em] &v_2 = \left\{\frac{-\xi + \sqrt{\Delta^2 + \xi^2}}{\Delta},1\right\} \end{split} \end{equation} which I am not happy with.

In the physics literature this is usually given as \begin{equation} \begin{split} &v_1 = \{u,v\}\\ &v_2 = \{v,u\} \end{split} \end{equation} where \begin{equation} \begin{split} &u^2 = \frac{1}{2}\left(1 + \frac{\xi}{\sqrt{\xi^2 + \Delta^2}} \right)\\ &v^2 = \frac{1}{2}\left(1 - \frac{\xi}{\sqrt{\xi^2 + \Delta^2}} \right). \end{split} \end{equation}

My question is how can I get mathematica to give me the 'correct' eigenvectors.

I think the issue is two fold:

  1. Mathematica does not care about the ratios $a/b$ and $c/d$ in $v_1 = \{a,b\}$ and $v_2 = \{c,d\}$
  2. Mathematica does not understand that I'm looking for the eigenvectors that makes the matrix $P = \{v_1, v_2\}$ unitary
  • 1
    $\begingroup$ If I massage the results I find $v_2^{\mathrm{yours}} = \alpha \{v,u\}$ and $v_1^{\mathrm{yours}} = \beta \{v, - u\}$. For some real constants $\alpha$ and $\beta$. There is a minus sign that doesn't match up. Can you double check the result of the literature? I agree with the $v_{1,2}^{\mathrm{yours}}$. $\endgroup$
    – MannyC
    Jun 25, 2019 at 14:19
  • $\begingroup$ As for point 1., Mathematica does care about the ratio, if the ratio was different they won't be eigenvectors. Rather it doesn't care about the ratio $a/c$. $\endgroup$
    – MannyC
    Jun 25, 2019 at 14:21
  • $\begingroup$ How did you massage the expression to get $\alpha\{v,u\}$? $\endgroup$
    – MOOSE
    Jun 25, 2019 at 14:51
  • $\begingroup$ The notes I'm referring to is here on page 12: phy.ntnu.edu.tw/~changmc/Teach/SM/ch04.pdf $\endgroup$
    – MOOSE
    Jun 25, 2019 at 15:40
  • 1
    $\begingroup$ Take $v_1^{\mathrm{yours}} / v_1$ term by term. Call it $\{x,y\}$. Full simplify $x/y$ assuming $\xi \in \mathbb{R}$ and $\Delta \in \mathbb{R}$. It should be something like $\pm \mathrm{sign}\,\Delta$. $\endgroup$
    – MannyC
    Jun 25, 2019 at 18:55

1 Answer 1


This is what I get with my program

$$H=\left[ \begin {array}{cc} \xi&-\Delta\\-\Delta&- \xi\end {array} \right] $$

Eigenvalues are:

$\lambda_1=\sqrt {{\xi}^{2}+{\Delta}^{2}}$

$\lambda_2=-\sqrt {{\xi}^{2}+{\Delta}^{2}}$

Eigenvectors (not normalized)

$$\vec{v}_1=\left[ \begin {array}{c} 1\\ {\frac {\xi-\sqrt {{ \xi}^{2}+{\Delta}^{2}}}{\Delta}}\end {array} \right] $$

$$\vec{v}_2=\left[ \begin {array}{c} 1\\ {\frac {\xi+\sqrt {{ \xi}^{2}+{\Delta}^{2}}}{\Delta}}\end {array} \right] $$

normalized eigenvectors : $$ \vec{\hat{v}}_1=\frac{1}{\sqrt{\vec{v}_1\cdot \vec{v}_1}}\vec{v}_1$$

$$ \vec{\hat{v}}_2=\frac{1}{\sqrt{\vec{v}_2\cdot \vec{v}_2}}\vec{v}_2$$

so $ \vec{\hat{v}}_i\cdot \vec{\hat{v}}_i=1$ and $ \vec{\hat{v}}_i\cdot \vec{\hat{v}}_j=0$

  • $\begingroup$ Okay, but how are these vectors related to the ones in the literature? (See question above) $\endgroup$
    – MOOSE
    Jun 24, 2019 at 17:46
  • $\begingroup$ this I can't tell you . is your $v_1$ and $v_2$ correct ?. they are not normalized ? $\endgroup$
    – Eli
    Jun 24, 2019 at 17:49
  • $\begingroup$ They are normalized because $u^2 + v^2 = 1$. $\endgroup$
    – MOOSE
    Jun 24, 2019 at 18:06
  • $\begingroup$ I think that the calculation of the eigenvectors is not unique? $\endgroup$
    – Eli
    Jun 24, 2019 at 18:14
  • $\begingroup$ For a general matrix, I agree. But I think the fact that the opetators should anti-commute should give unique eigenvectors somehow. That is after all what follows when performing a bogoliubov transformation on the Hamiltonian $\endgroup$
    – MOOSE
    Jun 24, 2019 at 18:19

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