## 3 Wentzel-Kramers-Brillouin Approximation
\right]\]  
is proportional to the probability-current operator in Equation (17), \(\lambda=0\), \(1\), \(2\), \(3\), \(\ldots\), and \(\Psi_{\lambda=-1}(s_{1},s_{2}\,|\,x,k_{y})\equiv 0\). Furthermore, the transport operator \(\hat{\mathbb{O}}_{\,T}(s_{1},s_{2}\,|\,x,p_{y},\Delta_{0})\) introduced in Equation (27), which connects each two consecutive terms of expansion in Equation (26), is as follows:  
\[\hat{\mathbb{O}}_{\,T}(s_{1},s_{2}\,|\,x,p_{y},\Delta_{0})\] \[= \xi(x)\,\hat{\mathcal{T}}_{0}^{(2)}\otimes\hat{\Sigma}_{0}^{(2)}+ \hat{\mathcal{T}}_{0}^{(2)}\otimes\left[\hat{\Sigma}^{(2)}\cdot\left\{\frac{ \partial}{\partial x}\,\mathbb{S}(x),p_{y}\right\}\,\right]+\Delta_{0}\left[ \hat{\mathcal{T}}^{(2)}\cdot\left\{\frac{\partial}{\partial x}\,\mathbb{S}(x ),p_{y}\right\}\,\right]\otimes\hat{\Sigma}_{0}^{(2)}\] \[= \left[ \begin{array}{c|c}\xi(x)&\partial\,\mathbb{S}(x)/\partial x -ip_{y}&\Delta_{0}[\partial\,\mathbb{S}(x)/\partial x-ip_{y}]&0\\ \partial\,\mathbb{S}(x)/\partial x+ip_{y}&\xi(x)&0&\Delta_{0}[\partial\, \mathbb{S}(x)/\partial x-ip_{y}]\\ 0&\Delta_{0}[\partial\,\mathbb{S}(x)/\partial x+ip_{y}]&\partial\,\mathbb{S}( x)/\partial x-ip_{y}&\xi(x)\end{array} \right]\,\]  
where \(\xi(x)=V(x)-\varepsilon(p_{y})\). Specifically, for \(\lambda=-1\), we obtain:  
\[\hat{\mathbb{O}}_{\,T}(s_{1},s_{2}\,|\,x,p_{y},\Delta_{0})\,\Psi_{0}(s_{1},s_ {2}\,|\,x,k_{y})=0. \tag{30}\]  
For the linear homogeneous equation in Equation (30), a non-trivial solution becomes possible only if its determinant is zero, namely:  
\[\left[\left(\frac{\partial\,\mathbb{S}(x)}{\partial x}\right)^{2}+ p_{y}^{2}\right]^{2}-2\Delta_{0}^{2}\left[\left(\frac{\partial\,\mathbb{S}(x)}{ \partial x}\right)^{4}-p_{y}^{4}\right]+\Delta_{0}^{4}\left[\left(\frac{ \partial\,\mathbb{S}(x)}{\partial x}\right)^{2}+p_{y}^{2}\right]^{2} \tag{31}\] \[-2\xi^{2}(x)\Bigg{[}\left(\frac{\partial\,\mathbb{S}(x)}{ \partial x}\right)^{2}+p_{y}^{2}\Bigg{]}\left(1+\Delta_{0}^{2}\right)+\xi^{4} (x)=0\.\]  
As a quartic equation, Equation (31) could be solved analytically, but in general, its solutions are too tedious to write down. Instead, we use the fact that \(\Delta_{0}\ll 1\). Therefore, as \(\Delta_{0}\longrightarrow 0\), Equation (31) turns into:  
\[\left[\left(\frac{\partial\,\mathbb{S}(x)}{\partial x}\right)^{2}+p_{y}^{2} \right]^{2}-2\Bigg{[}\left(\frac{\partial\,\mathbb{S}(x)}{\partial x}\right)^{ 2}+p_{y}^{2}\Bigg{]}\,\xi^{2}(x)+\xi^{4}(x)=0 \tag{32}\]  
with a set of doubly degenerate solutions for incident energy \(\varepsilon_{0}\) of an electron, given by:  
\[p_{x}(x)\equiv\frac{\partial\,\mathbb{S}(x)}{\partial x}=\pm\sqrt{\left[ \varepsilon_{0}-V(x)\right]^{2}-p_{y}^{2}}\, \tag{33}\]which results in:  
\[\mathbb{S}(x)-\mathbb{S}(x_{0})=\int\limits_{x_{0}}^{x}d\zeta\,\sqrt{\left[\epsilon _{0}-V(\zeta)\right]^{2}-p_{y}^{2}}\;. \tag{34}\]  
For a finite but small \(\Delta_{0}\), we are able to solve Equation (31) by using a perturbation method. Assuming:  
\[p_{x}(x)=p_{x}^{(0)}(x)+\Delta_{0}^{2}\,p_{x}^{(1)}(x)+\ldots\;, \tag{35}\]  
where \(p_{x}^{(0)}(x)=\pm\sqrt{\left[\epsilon_{0}-V(x)\right]^{2}-p_{y}^{2}}\) derived from the zeroth-order equation in Equation (32), we obtain the first-order correction \(p_{x}^{(1)}(x)\) (on the order of \(\backsimeq\Delta_{0}^{2}\)) from the following equation:  
\[4\,p_{x}^{(0)}(x)\,p_{x}^{(1)}(x)\bigg{\{}\left[p_{x}^{(0)}(x) \right]^{2}+p_{y}^{2}\bigg{\}}-2\bigg{\{}\left[p_{x}^{(0)}(x)\right]^{4}-p_{y} ^{2}\bigg{\}} \tag{36}\] \[+2\bigg{\{}\left[p_{x}^{(0)}(x)\right]^{2}+2p_{x}^{(0)}(x)\,p_{x }^{(1)}(x)+p_{y}^{2}\bigg{\}}\left[\epsilon_{0}-V(x)\right]^{2}=0\;,\]  
given by:  
\[p_{x}^{(1)}(x)=\frac{\left[p_{x}^{(0)}(x)\right]^{4}-p_{y}^{4}+\left[p_{x}^{( 0)}(x)\right]^{2}\xi^{2}(x)+p_{y}^{2}\,\xi^{2}(x)}{2\,p_{x}^{(0)}(x)\bigg{\{} \left[p_{x}^{(0)}(x)\right]^{2}+p_{y}^{2}-\xi^{2}(x)\bigg{\}}}\;. \tag{37}\]  
Equations (35)-(37) define the solution for the \(x\)-dependent momentum \(p_{x}(x)\), while the semi-classical action for a given momentum \(p_{x}(x)\) can be determined by Equation (34).