Some system can be normalized, some can not

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Normalization in quantum mechanics (QM) refers to the process of ensuring that a wavefunction's total probability equals one. Some systems, like a particle in free space, cannot be normalized because they are not bound, while others, such as a potential step, may present challenges. In practical scenarios, particles are often not truly free, as they typically transition between interactions, making normalization less of a concern. For educational purposes, these concepts help beginners grasp the mathematical foundations of QM. Overall, understanding normalization depends on the specific system being analyzed.
Black Integra
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I'm new and have been studying QM myself for a while. I have a little question about normalization.
Some system can be normalized, some can not (such as potential step).
What does it mean? And how can we solve these problems?
 
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What it means depends on the system.

For instance you cannot normalize the wavefunction for a particle in free space because the particle is not bound.

We don't normally need to "solve" them because we normally have more information than that. Like our particle "in free space" is probably traveling from some interaction to another one ... so it is not actually "free". These puzzles are only used for beginning students to get you used to the math.
 


Black Integra said:
I'm new and have been studying QM myself for a while. I have a little question about normalization.
Some system can be normalized, some can not (such as potential step).
What does it mean? And how can we solve these problems?
I'm very new to QM (taking my first introductory course) but do you mean the case of a plane wave incident on a potential step? If so, it might be because plane waves are an idealization and for a closer to reality model, we take a wavefunction Psi equal to a wave packet rather than a plane wave. I think in that case Psi can be normalized. I may have misunderstood your question though. Feel free to give more details.
 
Time reversal invariant Hamiltonians must satisfy ##[H,\Theta]=0## where ##\Theta## is time reversal operator. However, in some texts (for example see Many-body Quantum Theory in Condensed Matter Physics an introduction, HENRIK BRUUS and KARSTEN FLENSBERG, Corrected version: 14 January 2016, section 7.1.4) the time reversal invariant condition is introduced as ##H=H^*##. How these two conditions are identical?

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