I'm trying to compute the wavefunction for a Majorana state in an nanowire/superconductor hybrid system, like arXiv: Majorana Fermions and a Topological Phase Transition in Semiconductor-Superconductor Heterostructures.

I use the same ansatz for the wavefunction $\Psi\left(x\right) = Ae^{zx}$ and obtain the characteristic polynomial

$$ z^{4} + 4\left(\mu + 1\right)z^{2} + 8\lambda\Delta z + 4\left(\mu^{2} + \Delta^{2} - V^{2}\right) = 0\text{,} $$

where $\mu$ the chemical potential, $V$ the Zeeman field, $\Delta$ the superconducting gap and I know that $u_{\sigma} = \lambda v_{\sigma}$ where $u_{\sigma}$ describes electron states and $v_{\sigma}$ describes hole states with $\lambda = \pm 1$.To solve the above equation I use the fundamental theorem of algebra:

$$ z^{4} + 4\left(\mu + 1\right)z^{2} + 8\lambda\Delta z + 4\left(\mu^{2} + \Delta^{2} - V^{2}\right) = z^4 - \left(z_1 + z_2 + z_3 + z_4\right)z^3 + \left(z_1 z_2 + z_1 z_3 + z_1 z_4 + 2_2 z_3 + z_2 z_4 + z_3 z_4\right)z^2 - \left(z_1 z_2 z_3 + z_2 z_3 z_4 + z_1 z_3 z_4\right)z + z_1 z_2 z_3 z_4 = 0\text{,} $$

where $z_{i}$ are the roots. We see directly that

$$ \sum_{i = 1}^{4} z_i = 0 \text{ and } \prod_{i = 1}^{4} z_i = 4\left(\mu^{2} + \Delta^{2} - V^{2}\right)\text{.} $$

Now we can study different cases: the most interested case is when $z_{1/2}$ are complex and $z_{3/4}$ real. In this case we obtain

$$ z_{1/2} = a \pm ib $$


$$ z_{3/4} = -a \pm \sqrt{a^{2} - \frac{4\left(\mu^{2} + \Delta^{2} - V^{2}\right)}{a^2 + b^2}}\text{.} $$

In the publication they write that they have 4 boundary condition and 1 condition from the normalization.Okay, two from the fact that the wavefunction must localized at ends, so that I can write

$$ \Psi\left(0\right) = \Psi\left(L\right) = 0\text{,} $$

and the same fact for the derivative of the wavefunction

$$ \Psi^\prime\left(0\right) = \Psi^\prime\left(L\right) = 0\text{.} $$


In the paper above they study two cases. In the first $\left(\mu^{2} + \Delta^{2} - V^{2}\right) > 0$ they say that they have 4 boundary condition and 1 condition from normalization but only 4 coefficients. I can explain that they neglect the two complex wavefunctions. However, in the opposite case $\left(\mu^{2} + \Delta^{2} - V^{2}\right) < 0$ they write that they have 6 coefficients and 6 Condition to solve the equation.

But now I'm completly confused! Why now 6 coefficients and 6 conditions?

  • $\begingroup$ Nobody has an idea? $\endgroup$
    – user27964
    Oct 27 '13 at 12:30
  • $\begingroup$ From the paper: "On the other hand, similar analysis for $C_{0} < 0$ always yields three roots with $\Re [z]<0$ either in $\lambda =1$ or $\lambda =−1$ channels resulting in six coefficients to match." Isn't the answer you were looking for ? $\endgroup$
    – FraSchelle
    Mar 5 '14 at 9:45
  • $\begingroup$ Yes, but at that time I didn't understand this sentence. $\endgroup$
    – user27964
    Mar 5 '14 at 11:23
  • $\begingroup$ Well, I'm not sure I understand the link to te presence or absence of Majorana mode either :-( But the problem is clear from the beginning: they have i) left/right waves, ii) spin up/down waves, iii) electron/hole waves, so in total 8 solutions for their differential equation: 8 in the bridge, 8 in the superconductor. There are also 8 continuity equations on the left of the normal bridge, 8 on the right. So the problem is complete I believe. I would ask Maple or Mathematica to resolve it, since the full solution is always ugly. Did you try some of these ? All the best. $\endgroup$
    – FraSchelle
    Mar 5 '14 at 13:08
  • 1
    $\begingroup$ I solved this problem for few months. But thanks for the arXiv articles, there are very helpful! $\endgroup$
    – user27964
    Mar 5 '14 at 13:35

The solution of this problem is in principle very easy. Since

$$ z^{4} + 4\left(\mu + 1\right)z^{2} + 8\lambda\Delta z + 4\left(\mu^{2} + \Delta^{2} - V^{2}\right) = 0\text{,} $$

is a fourth order equation I can solved this equation by the hand or with Mathematica/Maple and get 4 solutions. The interesting case is that with two purely real and two complex solutions. From this I can write down the general solution for the wavefunction:

$$ \Psi\left(x\right) = \sum_{i = 1}^{4}A_{i}e^{z_{i}x} $$

In the next step I must fix my boundary condition (here I consider just the simplest case with a semi-infinte wire, means my system starts at x = 0 and ends in infinity so that the right Majorana is vanishing). In this case I get three conditions:

$$ \Psi\left(0\right) = \Psi^{\prime}\left(0\right) = 0\text{,} $$

and the normalization

$$ \int dx |\Psi\left(x\right)|^{2} = 1\text{,} $$

means I must fullfill three conditions to fix the $A_{i}$. However, what we need is that $Re\left(z_{i}\right) < 0$ due to the fact that we need a decaying of the wavefunction into the wire. Hence, we can neglect all solutions with $Re\left(z_{i}\right) > 0$.

In the case that $\left(\mu^{2} + \Delta^{2} - V^{2}\right) > 0$ I get two real solutions with the same sign and two complex where the sign of the real part is opposite as the real solutions for each $\lambda$. Therefore we can not fullfill the conditions to fix the amplitudes. In the opposite case that $\left(\mu^{2} + \Delta^{2} - V^{2}\right) < 0$ I get for each $\lambda$ two complex and one real solution for the $z_{i}$ with the same sign for the real part. Means here in this case I can fullfill my conditions for the wavefunction and get a very well result.


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