Energy values in case of 1D infinite potential well 
As it is depicted in the picture, there are 4 types of potential wells and if $E_1, E_2, E_3, E_4$ represent the ground state energies of the particle confined in those wells, what should be the correct order of energies $E_1, E_2, E_3, E_4$?
How to find the ordering without doing much of mathematics? Is there any logical argument?
 A: In general, I would say it depends on the values of $V_0$ and $a$.
But we can try to formulate some logical argument, starting from Heisenberg's uncertainty relation $\Delta p \cdot \Delta x \geq \frac{\hbar}{2}$ and the energy formula $E = \frac{\hbar^2 p^2}{2m}$.
For the potential with infinite walls in a) we know that the boundary conditions require that the wave function vanishes exactly at the boundary. So, there is not even a little part of the wave function in the region outside $[0,a]$. Contrarily, for b) the wave function will have non-zero amplitude for $x > a$, hence $\Delta x_b > \Delta x_a$. By the same argument, we can argue that in d) the wave function can also enter $x < 0$. Therefore, $\Delta x_d > \Delta x_b > \Delta x_a$.
Therefore, we can (handwavingly) argue that since the uncertainty relation still has to hold, $\Delta p_a > \Delta p_b > \Delta p_d$, hence $E_a > E_b > E_d$.
For setting c) it's getting a bit more tricky. The wave functions for a linear potential are Airy functions, therefore the solution in the middle region looks clearly different than for all the three other cases. I am not sure how that effects the ground state energy.
I would finish my argument by observing that (compared to b) the linear potential constrains $\Delta x$ more than in b), therefore $\Delta x_c < \Delta x_b$, hence $E_c > E_b$, but I am not sure how to relate $\Delta x_a$ and $\Delta x_c$.
I hope that helps!
