Born's interpretation of psi

Question

According to the Born's postulates, psi should be atleast differentiable to the first order then why does we not require psi in this case to be differentiable at X = a and X = 0.Also psi comes out to be a sinosoid and hence the probability density of the particle varies along the x axis and hence the velocity. But why is this so, classically it should be going under to and fro motion with the constant speed. I am novice in this field. Please help me with this.

Answer 1

Though we usually use it as a first potential over which to solve the Schrödinger equation, the infinite square well (or indeed any potential with an infinite and discontinuous change in potential) has it's subtleties.

You've identified the main one: we ignore the usual requirement that $\frac{\mathrm{d}}{\mathrm{d}x}\psi(x)$ be continuous at the boundaries.

We do—eventually—explain what is going on there, but it helps to have done a least one potential with a finite barrier first.

Consider what the solution would be like if the well was deep (many time the ground-state energy), but finite: it had a maximum of $V_0 \gg E$. In that case the wave-function would not be zero outside the well it would take on the form $$ \psi_{ex} = C e^{\pm i \kappa x} \;,$$ where 'ex' indicates exterior, $\kappa = \sqrt{2m(V_0 - E)}/\hbar$, and $C$ is a constant.

Because the wave-function must be normalizable we could only use the $+$ term on the left side and the $-$ term on the right side.

Then we impose the usual continuity condition including continuity of the derivative, and we would get a smooth curve connecting the sinusoidal term in the well asymptoptically with the horizontal axis.

Now allow $V_0$ to grow. The connecting terms will approach the axis ever more closely over an ever shorter range in $x$. As $V_0$ gets to be very large indeed this will more and more closely approach a discontinuous change in slope (at least if we step back and take out glasses off). In the limit as $V_0$ grows without bound the slope becomes discontinuous at the edge of the well.

So this violation of the usual rules is justifiable as long as you take seriously the notion go infinite potential steps. But the argument that this is so is clearer if you have already done a couple of more involved problems before you are given the argument in the first place. So the justification is often skipped on the first pass over the infinite square well problem.

Answer 2

Concerning OP's first sentence (v1): It is possible to formulate the TISE (understood in weak sense) as an integral equation, which reveals that continuity and/or possible smoothness of the wave function are not mandatory first principles/assumptions, but rather consequences, via a standard bootstrap argument, of properties that the potential $V$ may or may not possess, cf. e.g. my Phys.SE answer here.

Answer 3

Also psi comes out to be a sinosoid and hence the probability density of the particle varies along the x axis and hence the velocity. But why is this so, classically it should be going under to and fro motion with the constant speed. I am novice in this field. Please help me with this.

It is what makes quantum mechanics a different theory than classical mechanics. When solving the Shrodinger equation, the psi solution is interpreted by Bohr as the probability of finding the particle at (x,y,z) when a measurement or an interaction happens. It is in an energy level, and that is all. That is why there exist atomic orbitals, and not orbits.

Quantum mechanical solutions fit the data for small dimensions and masses, and not classical solutions.