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This question is more like a definition-confusion which is causing me to misunderstand several things. So, I am taking the MIT 8.05 Quantum Physics-II course and the instructor while mentioning the regularity conditions for the energy eigenstates said that the Energy Eigenstates obtained on solving the Time-Independent Schrodinger equation need to be continuous and "bounded" and the derivative of the eigenstates also needs to be "bounded".

He later explains that he is not imposing any normalization conditions since many eigenstates like the momentum eigenstates are a really important set of eigenstates that are not normalizable. But later on, in his notes, he defines what it means for a state to be bound-

A localised energy eigenstate $\psi(x)$ is called a bound state if $\psi(x) \rightarrow 0$ as $|x| \rightarrow \infty$.

This directly contradicts what he imposes as conditions for the energy eigenstates. Can someone help me understand this? I've also attached a screenshot from his notes. This can be found on page 6 in these notes here. enter image description here

Tachyon209
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A function $\psi$ is called bounded on some region $R$ if there exists some $C>0$ such that $|\psi(x)|<C$ for all $x$ in $R$. This does not imply that the function is normalizable; for example, the function $\psi(x)=1$ is bounded on the entire real line, but is clearly not normalizable. This is also different from the author's definition of a bound state.

If I'm interpreting your question correctly, the short answer is that "normalizable" $\neq$ "bounded" $\neq$ "bound."

Etymologically, this definition of a bounded function relates to the fact that the function doesn't escape off to infinity anywhere; its range is restricted to some finite interval.

On the other hand, the definition of a bound state relates to the fact that in quantum mechanics, a bound state is (roughly speaking) localized in some finite region of space. A bound state of an electron in a hydrogen atom is "bound" to the nucleus, insofar as the probability of measuring it some distance $r$ from the nucleus goes to zero as $r\rightarrow \infty$.

J. Murray
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    Aah, so in general, the energy eigenstates need to be bounded, like $\psi(x)=\frac{1}{x^2}$ can't be a viable wave function since it does die off at infinities but it isn't "bounded" since it diverges as $x\rightarrow0$. But, a special type of energy eigenstates, that is, localised energy eigenstates are callled "bound states". Am I correct? – Tachyon209 Aug 27 '20 at 19:38
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    @Tachyon209 Yes, that is how the terminology is commonly used. If it helps, one usually makes the distinction between bound states and scattering states. – J. Murray Aug 27 '20 at 19:40
  • So it was just a terminology problem. Hehe. Thanks a lot. – Tachyon209 Aug 27 '20 at 19:45
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    @Tachyon209 note that in practice a viable wavefunction could be unbounded (in particular, S-wave solutions of Dirac equation for hydrogen atom are). The condition in your lecture notes is needlessly restrictive (although it's good enough for a start). See this question for more discussion. – Ruslan Aug 27 '20 at 21:59