I Why probability current = 0 at infinity? Why must wavefunction be continuous?

[Happiness]

Q1. Why is the probability current $j(x,t)=0$ at $x=\pm\infty$? (See first line of last paragraph below.)

My attempt at explaining is as follows:
For square-integrable functions, at $x=\pm\infty$, $\psi=0$ and hence $\psi^*=0$, while $\frac{\partial\psi}{\partial x}$ and hence $\frac{\partial\psi^*}{\partial x}$ must remain finite for $\psi$ to be differentiable, a requirement for it to be a solution of the Schrodinger's equation. Hence by (2-32), $j(x,t)=\frac{\hbar}{2im}(0-0)=0$.

But more rigorously, we should say as $x\to\pm\infty$, $\psi\to0$ and hence $\psi^*\to0$. Since $\frac{\partial\psi}{\partial x}$ and hence $\frac{\partial\psi^*}{\partial x}$ must remain finite for all values of $x$ and $t$, they must be bounded from above, by say $z_1$, and below, by say $z_2$. Hence $\psi^*z_1\leq\psi^*\frac{\partial\psi}{\partial x}\leq\psi^*z_2$ and as $x\to\pm\infty$, $\psi^*\to0$ and hence $\psi^*\frac{\partial\psi}{\partial x}=0$. Hence $j(x,t)=\frac{\hbar}{2im}(0-0)=0$.

Am I right or missing anything out?

Q2. How does discontinuity in $\psi$ lead to (Dirac) delta functions in $j(x,t)$? (Second line of last paragraph in the photo.)

Suppose at some value of $x$ and $t$, $\frac{\partial\psi}{\partial x}=\pm\infty$. Then $\frac{\partial\psi^*}{\partial x}=\pm\infty^*$. Hence $j(x,t)=\pm\infty-\pm\infty^*$, which could be finite. Then there may not necessarily be any delta function in $j(x,t)$. Isn't it?

Q3. How does delta functions in $j(x,t)$ lead to delta functions in $P(x,t)$? (Third line of last paragraph in the photo.)

Suppose at some value of $x$ (say $x_1$) and $t$ (say $t_1$), $j(x_1,t_1)$ is the $\pm\infty$ of a delta function. Then $\frac{\partial j}{\partial x}=\pm\infty$. By (2-33), $\frac{\partial P}{\partial t}=\mp\infty$. But P may not necessary have delta functions. It could just be discontinous with respect to $t$, say P jumps from 0.1 to 0.2 when $t=t_1$. And so shouldn't we then argue that this discontinuity, and not delta functions as claimed by the text, is unacceptable instead?

 

[PeroK]

Q1. Why is the probability current $j(x,t)=0$ at $x=\pm\infty$? (See first line of last paragraph below.)

View attachment 204169

My attempt at explaining is as follows:
For square-integrable functions, at $x=\pm\infty$, $\psi=0$ and hence $\psi^*=0$, while $\frac{\partial\psi}{\partial x}$ and hence $\frac{\partial\psi^*}{\partial x}$ must remain finite for $\psi$ to be differentiable, a requirement for it to be a solution of the Schrodinger's equation. Hence by (2-32), $j(x,t)=\frac{\hbar}{2im}(0-0)=0$.

But more rigorously, we should say as $x\to\pm\infty$, $\psi\to0$ and hence $\psi^*\to0$. Since $\frac{\partial\psi}{\partial x}$ and hence $\frac{\partial\psi^*}{\partial x}$ must remain finite for all values of $x$ and $t$, they must be bounded from above, by say $z_1$, and below, by say $z_2$. Hence $\psi^*z_1\leq\psi^*\frac{\partial\psi}{\partial x}\leq\psi^*z_2$ and as $x\to\pm\infty$, $\psi^*\to0$ and hence $\psi^*\frac{\partial\psi}{\partial x}=0$. Hence $j(x,t)=\frac{\hbar}{2im}(0-0)=0$.

Am I right or missing anything out?

In general, there are square integrable functions that do not meet these criteria, but they are considered invalid in terms of representing a physical system. In particular, a square integrable function need not have a bounded derivative. A similar example is:

https://www.physicsforums.com/threa...f-position-of-a-particle.853025/#post-5349540

 

[PeroK]

If the function represents a physical system, then it can at least be approximated by a function that is identically $0$ beyond some value of $x$. Hence, of course, all the derivatives are identically $0$ beyond this value.

This technically excludes functions like the Gaussian, but the Gaussian can be approximated by a function that drops to $0$ at some large value of $x$.

In all these calculations, therefore, you could assume that eventually the wave-function and all its derivatives are (effectively) identically $0$. That might be a useful rule of thumb.

 

[Happiness]

If the function represents a physical system, then it can at least be approximated by a function that is identically $0$ beyond some value of $x$. Hence, of course, all the derivatives are identically $0$ beyond this value.

Are there some further elaborations on what it means to be physical?

A wave function that is identically zero beyond some point implies a particle with a non-zero probability of being found here (in the vicinity of $x=0$) cannot simultaneously have a non-zero probability of being found somewhere infinitely far away. I reason this is because at a particular instant, the particle must be somewhere; it cannot be somewhere infinitely far away (being infinitely far away means it goes out of existence?). So in other words, a particle cannot have a probability (say 0.6) of being in existence and a probability (say 0.4) of being out of existence at one instant? And if we add in the time dimension, does being physical means a particle starting with probability = 1 of being in existence cannot have a (non-zero) probability of being out of existence at some future instant?

The assumption that the first derivative $\frac{\partial\psi}{\partial x}$ vanishes at $x=\pm\infty$, is it motivated by some principle of locality? A particle (or wave) located here cannot exert an effect on particles sufficiently far away. So the momentum, energy and all physical quantities of the particle (or wave) at $x=\pm\infty$ must be zero. And since all physical quantities are calculated as some functions of $\psi, x$ and $\frac{\partial\psi}{\partial x}$, the first derivative $\frac{\partial\psi}{\partial x}$ must vanish at $x=\pm\infty$?

 

[PeroK]

I'm not sure i follow much of that. For most experiments, the probability a particle is found outside the solar system is negligible. So, you may as well assume the wave function is 0 beyond that!

Normally we talk about the rate that the probability density falls off - must be exponential eventually. But I just thought that approximating it as 0 beyond the bounds of an experiment was a neat alternative.

Either way an assumption beyond square integrability is needed.