# Discontinuous wave function?

• I
I've been studying the basics of the quantum mechanics, and I found the continuity restraints of the wave function quite suspicious.
What if there is a jump discontinuity on a wave function where the first derivative of which is still continuous? What is the problem with such wave function?

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strangerep
What if there is a jump discontinuity on a wave function where the first derivative of which is still continuous?
This is best discussed (initially at least) by example. Do you have a particular specific example in mind?

If your wavefunction is discontinuous, then how can your first derivative be continuous? Further, the second derivative will be discontinuous and so your Schrodinger equation will be ill-defined.

bhobba
Mentor
If your wavefunction is discontinuous, then how can your first derivative be continuous? Further, the second derivative will be discontinuous and so your Schrodinger equation will be ill-defined.
All these issues are fixed in the Rigged Hilbert Space formalism.

Its based on distribution theory which really should be in the armory of any applied mathematician. It makes Fourier transform theory a snap for example, otherwise you become bogged down is difficult issues of convergence etc.

I recommend the following book:
https://www.amazon.com/dp/0521558905/?tag=pfamazon01-20&tag=pfamazon01-20

Thanks
Bill

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Nugatory
Mentor
If your wavefunction is discontinuous, then how can your first derivative be continuous? Further, the second derivative will be discontinuous and so your Schrodinger equation will be ill-defined.
It would be helpful if you could provide a specific example of a problem involving a discontinuous wave function - I'm still not sure what you're thinking here.

Generally we require that the wave function be continuous across its domain, and expect to find discontinuities in the first derivative only where the potential becomes infinite (infinite square well, delta-function potentials, and the like). Discontinuities in the first derivative do mean that the second derivative is undefined so we can't solve Schrodinger's equation across the discontinuity, but that's not the same thing as saying that it is ill-defined. We can solve Schrodinger's equation on each side of the discontinuity and then take the requirement that the wave function be continuous as a boundary condition. But all of this is standard fare in first-year classes, so I presume that you're thinking of something more difficult.

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The wavefunction of a quantum mechanical partice must be continuous, otherwise, the probabilistic interpretation of the wavefunction fails (I mean of thprobability density, that is, the squared magnitude of the wavefunction $$|\psi(\vec{x})|^{2}dx^{3}$$) which tells you the probability of finding the particle in the volume interval dx^(3). Reducing it to one dimension, imagine a wavefunction with a jump at some x0, so that the quantity above passes from 0.4 to 0.1, what is the true interpretation of that? Also, think about the infinite potential well, where both sides of the wall are defined as $$(V(x=\pm L)=+\infty)$$ In this concrete example, the continuity of the wavefunction states that, at the edges of the well, the wavefunction must vanish since the potential is infinite for x>L or x<-L, which means that the probability of finding the particle in that region has to be identically zero (here you see the probabilist interpretation we talked about before). Having we had a finite barrier, this continuity equation will still hold, however, the wavefunction at the edges will no longer be 0, and we would have $$\psi_{C}(x=L)=\psi_{R}(x=L)$$ so that the central part and right part are linked that way, and analog for the left part at x=-L.

The significance of the wavefunction must be intimately related with the probability of finding the particle, and mathematical functions not satisfying this are just pure mathematical issues, but representing no physical situation.

bhobba
Mentor
The wavefunction of a quantum mechanical partice must be continuous,
Hmmmm. Think about the Dirac Delta function.

As I alluded to the answer lies in Rigged Hilbert Spaces.

The test functions are the physically realizable ones ie are continuously differentiable etc etc and have nice mathematical properties. The dual with all sorts of weird stuff like the Dirac Delta function is introduced for mathematical convenience. That includes non continuous functions etc as well as functions that do not fall off at infinity.

Thanks
Bill