Q1. Why is the probability current ##j(x,t)=0## at ##x=\pm\infty##? (See first line of last paragraph below.)(adsbygoogle = window.adsbygoogle || []).push({});

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?

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

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