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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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1 Answers1

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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 $r\rightarrow \infty$.

Albatross
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