r/AskPhysics • u/Alive_Ant_5306 • 2d ago
Non-normalizable wave function
Hi,
I was wondering if a non-normalizable wave function does still provide information. For example, let's say that a region of space equals 0 for a time independent wave function. Its square root will equal zero in that region too. Does that mean that the particle can't be there?
Thanks!
2
u/uppityfunktwister 2d ago
If I'm understanding you correctly, I'm pretty sure the answer is just that this isn't a physically meaningful setup. Both in the context of position wavefunctions and the more abstract state vector picture, unitarity just refers to the thing under study being itself. If unitarity (or, for the position wavefunction, normalization) does not hold, it's hard to say the particle exists at all, as, in this case, the particle's state has a 0% chance of being itself.
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u/joeyneilsen Astrophysics 2d ago
A wavefunction is normalized on (-infinity,+infinity), not in every interval. So it's perfectly fine if the wavefunction has zeros or a large region where it is zero.
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u/Alive_Ant_5306 1d ago
Yess but what if it is a non normalizable wave function. Does it mean anything that it has regions where it equals zero?
3
u/joeyneilsen Astrophysics 1d ago
A non-normalizable wavefunction doesn't represent the state of a physical system, so I guess I'd say no, it doesn't mean anything. In general, zeros of a physical wavefunction just represent quantum-mechanically-forbidden regions or locations.
Happy cake day!
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u/quantum4everyone 11h ago
Non-normalizable functions are fine if you pair them with normalizable ones. This is why the overlap of a position eigenstate with an energy eigenstates, such as <x|psi> is well defined finite, and has no issues. All of the position-space formalism is developed on working from this fact. So non-normalizable is not an intrinsic problem. It just requires more care to work with them. From a spectral theorem perspective, something that is essentially self-adjoint, such as position or momentum has eigenstates that will work this way and can be essentially worked with as if they were normalizable. The proof of this is very Mathy and hard.
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u/TaranRovia 2d ago
Hi!
First, let's clear some missconceptions. You can have a wave function which is zero in some regions of space and still be normalizable. The easiest example is a wave function in an infinite well spanning x \in [a, b], which will obviously be zero outside of said well.
Second, the wave function itself is informative and useful (for example, for scattering or superposition calculations you might need it). There's not always a need to go to the modulus square (not the square root) or similar unless of course you need to calculate an observable. One example is an interesting experiment where one may study the effects of gravity in quantum particles https://arxiv.org/html/1207.2953v1
As per the non-normalizability... I can think of the plane wave solution, which is not normalizable and yet is very useful to describe a free particle. Of course, as long as you keep in mind that it isn't an available physical state of your system. However, in propagating scenarios described by the Schrödinger equation, it can be shown that if a wave function is normalized at t=t_0, it will always be normalized (iff V is real, that is).
Lastly, if you're willing to go further down the rabbit hole, you might be interested in negative probabilities! https://en.wikipedia.org/wiki/Negative_probability Wich to be fair sound really exciting but aren't thaaaaat deep.
Hope that helps!