# Finding the ground state energy for a non-symmetric linear potential

Given a potential $$V(x) = \begin{cases} \infty \quad\text{ for } x<0 \\ k\cdot x \,\text{ for } x>0 \text{ with } k>0 \end{cases}$$ I want to find the ground state energy using the variational wave functions ($\alpha$ is the variational parameter) \begin{align} \Psi_1 &= Ax \exp{(-\alpha \cdot x)} \\ \Psi_2 &= Bx \exp(-\alpha \cdot x^2/2) \end{align} Now, this potential looks pretty easy to me at first, basically a wedge shifted onto an infinitely high potential wall on the left, with the slope depending on $k$. Intuitively, I'd say $\Psi (0) = 0$ because of continuity and $\Psi(x<0) \overset{!}{=} 0$.

For the continous derivative $\Psi'(x)$, we can postulate that because $\Psi(0)=0 \Rightarrow \Psi'(0)=0$. Since $\Psi'(x)=Ae^{-\alpha x} (1-\alpha x)$, we can deduce $$\Psi'(0)=A \cdot (1- 0) = 0 \Rightarrow A=0$$ This throws me off, because this means the wavefunction is 0 everywhere? How would I continue? Did I even do the approach right? Any help would be appreciated.

Actually the derivative is not continuous at $0$ since the potential has an infinite discontinuity there. This is a generic feature of infinite discontinuities. Another example would be the flat infinite well between $0$ and $1$, with solutions $A\sin(n\pi x)$, for which the derivative is not continuous at the end points.
(Nota: for $\delta$-well the derivative has a finite discontinuity across the $\delta$.)
You know that the exact solution is an Airy function with its first zero at $x=0$. This does not look much like either of your trial wavefunctions, so you have made a bad guess.