*I was struggling with the structure of SchrÃ¶dinger's equation over time when we consider a real physical process. The problem can be stated in two parts as follows,*

*Is it possible to construct a wave function for a system with a new introduced potential from the previously known wave function solution of the same system before the introduction of the potential. Eg: I want to get the solution of an infinite square well where the dirac delta potential is placed at the center from the usual square well problem without potential. We are not talking about entirely two things. Basically, it is a process of introducing an infinite square well and slowly putting the delta potential at the center over time.*

*The second part is where we don't introduce any kind of bizarre potentials, instead we change the boundary conditions over time. It is like expanding the square well or making some regions in space as inaccessible to the wave function.*

*Note:*

*There is a difference between wave function going to zero at some place and the space itself becomes inaccessible to the wave function.*

*One should not to be confused these things with the concept of perturbation even though it looks similar in some places. In perturbation, there is no change in boundary conditions or any introduction of completely arbitrary potentials. It is a very small change in the initial potential with same conditions maintained at*

*all the time.*

*To understand this, I took a series of problems with different levels in ascending order. First, we try to solve the general schrodinger equation for one dimensional infinite square well with a very little change by moving it to length "a" and "2a". Except for the limits, the solution is written in normal convention of "k" and "E" , we have $$ \psi = A e^{ikx} + B e^{-ikx}$$ and applying the left, right limits, $$A e^{ika} + Be^{-ka} = 0 \,\,\,...(1)\\ A e^{i2ka} + B e^{-i2ka}\,\,\,...(2)$$ (1) gives $$ A e^{i2ka}= -B $$ provided $ e^{ika}$ don't become zero. $$(2) \rightarrow \,\,\, B (e^{-i2ka}-1) = 0 \\ B = 0 \,\,or\,\, e^{-2ika} = 1$$ B = 0 gives A=0 i.e. trivial solution where there is no wave function. It leaves us with $$ e^{-i2ka} = 1 \\ A = -B\\ cos2ka = 1 \,\,and\,\, sin 2ka = 0$$both conditions need to be satisfied if, $$ sin 2ka = 0 \,\,and \,\,cos2ka = 1 \\ 2ka = n\pi \,\,and\,\, 2ka = 2n\pi$$ larger condition is the common one i.e. $$ ka = n\pi$$ which gives us the exact solution $$\psi = A e^{ika} - A e^{-ika} = C sin 2ka $$ where "C" is again an overall constant to be determined by normalization and $ ka = n\pi$ Normalization gives again the same constant [ since, $$ sin^2 2kx = \frac{1 - cos2kx}{2} $$ where integration is from "a" to "2a" i.e. sin 2ka = sin ka = 0 ] i.e. $\sqrt{\frac{2}{a}}$ giving us the complete solution,$$\psi (x) = \sqrt{\frac{2}{a}} sinkx \,\,\,\,and\,\, k = \frac{n\pi}{a}$$*

*It was quite a bit surprise for me, that there is no any kind of reflection about the shift of the coordinate system or the shift about the coordinate syystem. Even though the wave function depends on the position, it doesn't reflect anything i.e. independent of coordinate system.*

*But don't think they are exactly the same in every way because now the limits have changed to give negative values for all x values in between "a" and "2a". But, it doesn't matter since the measurable quantities depend on the square of the wave function - saves the day.*