Quote from https://en.wikipedia.org/wiki/Copenhagen_interpretation#Principles

"...The wavefunction evolves smoothly in time while isolated from other systems."

However, by the studying, my feeling was that all we need was for the wavefunction to be piecewise continuous, where the function itself assure the probability density and momentum to be continuous.

I.e. Finite square well boundary condition: Suppose a particle in finite square well where the function was bounded(no Dirac Delta Condition).

The the function was cut into $\psi_1$ from $(-\infty,-a]$, $\psi_2$ from $[-a,a]$, and $\psi_3$ from $[a,\infty)$.

When looking at the boundary condition at point $a$, traditionally they write $\psi_2(a)=\psi_3(a)$ and $\psi_2^\prime (a)=\psi_3^\prime(a)$.

I felt like function $\psi_2(a)$ and $\psi_3(a)$ didn't actually carry the meaning at $a$. Instead, since they $\psi_2$ and $\psi_3$ was already the eigenfunciton, the true condition required here ought to be $(\psi_2(a))^*\psi_2(a)=(\psi_3(a))^*\psi_3(a)$ and $(\psi_2^\prime(a))^*\psi_2^\prime (a)=(\psi_3^\prime(a))^*\psi_3^\prime(a)$.

Another example of step well we given:

Step potential well solve the other kind of given boundary condition

Thus quote from Billy, in that case:" ontinuity of probability density implies continuity of wavefunction."

My question were:

  1. Do we still need function itself to be continuous once we obtain the probability density and momentum to be continuous? (Notice the latter case was a weaker condition and, once it was satisfied, all operators and operations could be implied onto the function.)

  2. In the specific case of step well momentum and probability density to be continuous meant $\Psi$ and $\partial_x \Psi$ to be continuous (1D case). However, was that true in general? (ask for proof)

  • $\begingroup$ " didn't actually carry the meaning at a." What does this mean? What "meaning" are you talking about? Why don't you ask the same thing about the boundary condition at $-a$? $\endgroup$
    – Bill N
    Mar 21, 2018 at 20:12
  • $\begingroup$ @BillN Because the mathematical convention here require function to be continuous for it assumed the function to be the "information flow". However, physically speaking, the only "meaningful" quantity at $a$ was the probability density instead of the complex function. Thus $\psi_2(a)=\psi_3(a)$ was actually a mathematical convention, but not the correct interpretation from physics. And in physics, it ought to be the probability density flow was continuous. $\endgroup$
    – J C
    Mar 21, 2018 at 20:31
  • $\begingroup$ Why do you say it's not the correct interpretation from physics? What's wrong about it? Maybe those two conditions taken together are isomorphic with the conservation of the probability current. That's what you need to look at. $\endgroup$
    – Bill N
    Mar 21, 2018 at 21:49
  • $\begingroup$ @BillN $\psi_2(a)=\psi_3(a)$ was a stronger case, I just thought it might not be necessary if the probability density flow was continuous. (i.e. the latter was piece wise continuous where the traditionally asked to be continuous. As a physical quantity, piecewise continuous for the eigenfunction was enough provide the probability density flow was continuous.) On the other hand, if the probability density was continuous, the usual condition for eigenfunction, quantum operator etc. to work was already satisfied. Meaning it was not necessary to require the function itself to be continuous. $\endgroup$
    – J C
    Mar 22, 2018 at 0:25
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