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

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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?

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PeroK
Homework Helper
Gold Member
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

Happiness
PeroK
Homework Helper
Gold Member
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