00:01
Hello students, in this equation it is given that v of x is equal to minus v0 delta x where delta x is the dirac delta function and v0 is any positive real number.
00:13
Here we are asked to find out the wave function.
00:15
For that let's use the schrodinger equation that is minus h cut square by 2m into d square psi by dx square plus v of x into psi equal to e psi where v of x is the potential we can substitute there, e is the total energy and psi is the wave function.
00:52
Now let's substitute the v of x plus v0 delta x, psi is equal to e psi which can be written as minus h cut square divided by 2m dou square psi divided by dou x square minus v0 delta x, psi is equal to e psi.
01:31
Now further let's derivate it with respect to derivate with respect to x implies the equation becomes minus h cut square divided by 2m into d cube psi by dx square sorry dx cube minus v0 delta x into d psi by dx is equal to e psi by e d psi by dx.
02:17
Here let's apply the property of delta dirac function which is that delta x is equal to minus infinity set sx equal to 0, 0 such that x not equal to 0.
02:36
Here at x equal to 0 implies the function as having a discontinuity, a discontinuity as it approaches to minus infinity that is the function let's specify the function d psi by dx has an discontinuity at x is equal to 0...