I am trying to solve a stationary Schrodinger equation for a quantum reflection problem. This is usually done by solving a one-dimensional Schrodinger equation
$$\dfrac{\partial^{2}\psi(x)}{\partial x^{2}} - 2(E-V(x))\psi(x)=0$$,
where $\hbar = m = 1$, $E$ is the energy, and $V(x)$ the potential that depends on x only.
In my case, this is slightly more complicated, as the potential $V(x,y)$ is a complicated function of $x$ and $y$. Thus, the Schrodinger equation in two dimensions becomes:
$$\nabla^{2}\psi(x,y) - 2(E-V(x,y))\psi(x,y)=0$$
I don't need to know the result of this equation at every point, only at a specific trajectory. Which takes me to my question:
Can I reduce this two-dimensional problem into a single-dimensional one at a specific trajectory?
What we know is that $y$ is a function of $x$,i.e., $y = x \tan(\theta)$, where $\theta$ is any arbitrary angle. I would like then to rewrite the Schrodinger equation in terms of a new variable $x'= x \sec(\theta)$
$$ \mathcal{L}(x') \psi(x') - 2(E-V(x'))\psi(x')=0$$
where $\mathcal{L}(x')$ is some differential operator. I know this is sketchy territory, but I thought that maybe I could find the form of $\mathcal{L}(x') $ by using the power of hand-waving arguments.If I think of the function $\psi(x,y)$ and want to find the rate at which it changes in a specific direction I can use a directional derivative. Say the direction I am interested in is expressed by $\mathbf{n} =(\cos(\theta),\sin(\theta))$, so we can dot it with the gradient of $\psi(x,y)$
$$ (\dfrac{\partial}{\partial x}\psi(x,y),\dfrac{\partial}{\partial y}\psi(x,y))\cdot (\cos(\theta),\sin(\theta))$$
and at this point I thought if we could perform the change of variables $$y = x \tan(\theta)$$ and $$\frac{\partial}{\partial y}= \frac{\partial x}{\partial y} \frac{\partial}{\partial x}$$ so we would get:
$$ (\dfrac{\partial}{\partial x}\psi(x,x \tan(\theta)),\cot(\theta)\dfrac{\partial}{\partial x}\psi(x,x \tan(\theta))\cdot (\cos(\theta),\sin(\theta))$$
which gives us
$$2\partial_{x}\psi(x,x \tan(\theta) $$
getting it in terms of a new varaible $x' = x \sec(\theta)$, we would end up with:
$$ 2 \cos(\theta)\dfrac{\partial}{\partial x'}\psi(x')$$ and to get to the second derivative we would operate with $\partial_{x'}$, so we'd end up with
$$ \mathcal{L}(x')=2\cos(\theta)\dfrac{\partial^{2}}{\partial x'^{2}}$$
I realise this approach might be silly as there is some information lost by not evaluating the full Laplacian in 2d but I wanted to double check whether this was a dead-end and I should fight with the 2 dimensional pde. Just to note, I am preparing those equations for a numerical solver.