In his work1, Ettore Majorana studied two related problems. First, an oriented atomic beam passes a point with a vanishing magnetic field; the second half of the paper is devoted to a spin-1/2 particle in a linearly time-dependent magnetic field. Majorana considered the problem about a spin orientation in a dynamic magnetic field with components \(H_{x}\sim -\Delta\), \(H_{y}=0\), and \(H_{z}\sim -vt\), where \(\Delta\) and v are constant values.
The time evolution of a linearly driven two-level system is governed by the Schrödinger equation:
$$\begin{aligned} i\hbar \frac{\partial }{\partial t}\left| \psi \right\rangle =-\frac{1}{2 }\left( \Delta \sigma _{x}+vt\sigma _{z}\right) \left| \psi \right\rangle , \end{aligned}$$
(1)
where \(\sigma _{i}\) are the Pauli matrices.
For a wave function of the form:
$$\begin{aligned} \left| \psi \right\rangle = \begin{pmatrix} \alpha \\ \beta \end{pmatrix} \end{aligned}$$
(2)
this can be rewritten as a system of two coupled ordinary differential equations. Introducing the dimensionless time \(\tau\) and the adiabaticity parameter \(\delta\):
$$\begin{aligned} \tau =\sqrt{\frac{v}{2\hbar }}t,\;\;\;\delta =\frac{\Delta ^{2}}{4v\hbar }, \end{aligned}$$
(3)
and making the substitutions
$$\begin{aligned} \alpha =f\exp \!{\left( \frac{i}{2}\tau ^{2}\right) },\text { }\text { }\text { }\beta =g\exp \!{\left( -\frac{i}{2}\tau ^{2}\right) }, \end{aligned}$$
(4)
we obtain
$$\begin{aligned} {\left\{ \begin{array}{ll} \dot{f}\!&{} =i\sqrt{2\delta }g\exp {(-i\tau ^{2}),} \\ \dot{g}\!&{} =i\sqrt{2\delta }f\exp {(i\tau ^{2}).} \end{array}\right. } \end{aligned}$$
(5)
These can be rewritten for f and g separately
$$\begin{aligned}{} & {} \frac{d^{2}f}{d\tau ^{2}}+2i\tau \frac{df}{d\tau }+2\delta f =0, \end{aligned}$$
(6)
$$\begin{aligned}{} & {} \frac{d^{2}g}{d\tau ^{2}}-2i\tau \frac{dg}{d\tau }+2\delta g =0. \end{aligned}$$
(7)
The substitution (4) is used for obtaining these equations in homogeneous form. In this form, the Laplace transform simplifies the equations. (Note that to obtain Majorana’s equations we have to replace \(\tau \rightarrow \sqrt{2}\tau _{\textrm{M}}\) and \(\sqrt{\delta }\rightarrow -\sqrt{k}/2\), where \(\tau _{\textrm{M}}\) and k are Majorana’s notation.) Following Majorana, the equation for \(f(\tau )\), Eq. (6), can be solved by the two-sided Laplace transform,
$$\begin{aligned} \mathscr {L}[f(\tau )]=\int _{-\infty }^{\infty }\!\!\!e^{-s\tau }f(\tau )\;d\tau =F(s). \end{aligned}$$
(8)
Here F(s) is the Laplace transform of the function \(f(\tau )\). Then we substitute this in Eq. (6), use the theorem about differentiation of the original function, \(\mathscr {L}[\tau f(\tau )]=-F^{\prime }(s)\), and obtain
$$\begin{aligned} s^{2}F(s)-2i\left[ F(s)+sF^{\prime }(s)\right] +2\delta F(s)=0. \end{aligned}$$
(9)
The solution of this first-order differential equation gives
$$\begin{aligned} F(s)=C_{\delta }\exp {\left( -\frac{is^{2}}{4}\right) }s^{-1-i\delta }. \end{aligned}$$
(10)
Here the constant of integration \(C_{\delta }\) could be defined from an initial condition. And then we can find \(f(\tau )\) from the inverse Laplace transform:
$$\begin{aligned} f(\tau )=\lim _{T\rightarrow \infty }\int _{\gamma -iT}^{\gamma +iT}e^{s\tau }F(s)\;ds, \end{aligned}$$
(11)
where \(\gamma\) is a real number so that the contour path of integration is in the region of convergence of F(s).
This integral can be calculated by the steepest descent method28. For the following calculations, the contour could be deformed due to the residue theorem. This requires large times and we need to find the solution in two limits: for large positive time, which means that \(\tau \rightarrow +\infty\), and for large negative time \(\tau \ll 0\). Then, the integration contour in Eq. (11) is over either the contour \(L_{1}\) for \(\tau <0\) or \(L_{2}\) for \(\tau >0\), the contours to be defined. How the contours are chosen is described in detail in Appendix A; these steepest-descent contours \(L_{1,2}\) are demonstrated in Fig. 2.
Our integral (11) has two contributions: the first one is from the saddle point and the second one is from the vicinity of zero. Details of the calculations are presented in Appendix A.
First, we can write the contribution from the saddle point for \(f(\tau )\) as follows
$$\begin{aligned} f(\tau )=C_{\delta }\sqrt{{4\pi }}(-2i\tau )^{-1-i\delta }\exp \!{\left( i\frac{3\pi }{4}-i\tau ^{2}\right) }. \end{aligned}$$
(12)
Second, we describe the contribution from the near-zero region \(s\rightarrow 0\), which is the integral Eq. (11) on the contour \(L_{0}\). Here \(L_{0}\) is the near-zero vicinity contour with the branch cut along the negative axis which is a part of the \(L_{2}\) contour. Here we can neglect the term \(s^{2}/4\) next to \(s\tau\), and then we use the Gamma function in the Hankel integral representation [Ref.29 §12.22]:
$$\begin{aligned} \int _{L_{0}}e^{x}x^{-y}dx\approx \frac{2\pi i}{\Gamma (y)}. \end{aligned}$$
(13)
For using this, we define the replacement \(x=s\tau\) and then for the integral in Eq. (11) we have
$$\begin{aligned} \int _{L_{0}}\tau ^{i\delta }\frac{e^{x}}{x^{1+i\delta }}dx=\tau ^{i\delta } \frac{2\pi i}{\Gamma (1+i\delta )}. \end{aligned}$$
(14)
So, we obtain the approximate solution of Eq. (6) for two cases, \(\tau <0\) and \(\tau >0\), in the general form. The second part of the spinor, \(g(\tau )\), is obtained from Eq. (5) neglecting the terms \(\sim \tau ^{-2}\) because this result is asymptotic. Then using the substitutions Eq. (4), we obtain
$$\begin{aligned}&\tau <0: {\left\{ \begin{array}{ll} \alpha (\tau )=C_{\delta }\sqrt{4\pi }(-2i\tau )^{-i\delta -1}\exp \!\left( -i \frac{\tau ^{2}}{2}+i\frac{3\pi }{4}\right) , \\ \\ \beta (\tau )=C_{\delta }\sqrt{\frac{2\pi }{\delta }}(-2i\tau )^{-i\delta }\exp \!\left( -\frac{i\tau ^{2}}{2}+\frac{i\pi }{4}\right) , \end{array}\right. } \end{aligned}$$
(15a)
$$\begin{aligned}&\tau >0: {\left\{ \begin{array}{ll} \alpha (\tau )=C_{\delta }\sqrt{4\pi }(-2i\tau )^{-i\delta -1}\exp \!\left( – \frac{i\tau ^{2}}{2}+i\frac{3\pi }{4}\right) +C_{\delta }\frac{2\pi i}{ \Gamma (i\delta +1)}\tau ^{i\delta }\exp \!\left( \frac{i\tau ^{2}}{2}\right) , \\ \\ \beta (\tau )=C_{\delta }\sqrt{\frac{2\pi }{\delta }}(-2i\tau )^{-i\delta }\exp \!\left( -\frac{i\tau ^{2}}{2}+\frac{i\pi }{4}\right) +C_{\delta }\sqrt{ \frac{\delta }{2}}\frac{2\pi i}{\Gamma (i\delta +1)}\tau ^{i\delta -1}\exp \! \left( \frac{i\tau ^{2}}{2}\right) . \end{array}\right. } \end{aligned}$$
(15b)
We now consider the initial condition far from the avoided-level crossing, at \(\tau \rightarrow -\infty\), and from Eq. (15a) obtain
$$\begin{aligned} {\left\{ \begin{array}{ll} |\alpha |^{2}=0 \\ |\beta |^{2}=1 \end{array}\right. } . \end{aligned}$$
(16)
To fulfil the normalization condition we have taken the constant of integration \(C_{\delta }\) as
$$\begin{aligned} C_{\delta }=\sqrt{\frac{\delta }{2\pi }}\exp {\left( -\frac{\pi \delta }{2}\right) }, \end{aligned}$$
(17)
where we used that \(i^{i\delta }=e^{-\pi \delta /2}\). Note that the initial condition (16) leaves the phase undefined, so that replacing \(C_{\delta }\rightarrow C_{\delta }e^{i\vartheta }\) with any phase \(\vartheta ~\) would result in the same initial condition.