Particle in abox : continuous functions problem

[g.prabhakar]

I was studying particle in a box from shankar and I couldn't get the following point. If V is infinite at for x > L/2 and x < L/2, so is double derivative of psi. Now Shankar mentions that it follows the derivative of psi has a finite jump. I am not able to get this point because according to my understanding if a function f is such that
f ''(x) = k, a constant, x<=a
f ''(x) = infinite , x>a
then its integration (which comes out to be f '(x) ) from say x=0 to any point x>a (say x=2a) becomes infinite and therefore jump is infinite, (which is contradictory to what shankar mentions)
Where am I going wrong?

 

[Bill_K]

If V is infinite at for x > L/2 and x < L/2, so is double derivative of psi.

If Shankar said that, then he's the one who is wrong. If V is infinite, then ψ''/ψ is infinite. In other words, ψ is 0 for |x| > L/2.

 

[g.prabhakar]

Please correct me if I am wrong but I don't think Shankar is wrong. He is discussing a general case(not specifically particle in a box) where V is infinite for x > L/2. Now time independent schrodinger says
ψ'' = [- 2 m(E-V)/hbar^2)]ψ.
So it is perfectly consistent mathematically that ψ'' is infinite. I couldn't get why ψ'' it can't be infinite? Ofcourse, even ψ can be 0, but the point is how do u decide if either ψ=0 or ψ'' is infinite?

 

[strangerep]

He is discussing a general case(not specifically particle in a box) where V is infinite for x > L/2. Now time independent schrodinger says
ψ'' = [- 2 m(E-V)/hbar^2)]ψ.
So it is perfectly consistent mathematically that ψ'' is infinite. I couldn't get why ψ'' it can't be infinite? Ofcourse, even ψ can be 0, but the point is how do u decide if either ψ=0 or ψ'' is infinite?

Well, "infinite" without further explanation means "undefined".
One needs to be more precise. E.g., you could take V to be a finite constant "C" for
|x| > L/2 and solve the Schroedinger equation for that case, under the conditions
that ψ is suitably continuous everywhere (meaning that ψ and its first derivative
must match at the L/2 boundaries). Then normalize the resulting wavefunction,
and look at what happens as $$C \to \infty$$.