2
$\begingroup$

In the textbook "Molecular Quantum Mechanics" by Atkins and Friedman [1, p. 11], it says in chapter 1:

… a general function can be expanded in terms of all the eigenfunctions of an operator, a so-called complete set of functions.

Can ANY function be written as a linear combination of eigenfunctions of ANY operator? I think not.

What are the criteria here for which functions can be written as linear combinations of which operators?

References

  1. Atkins, P. W.; Friedman, R. Molecular Quantum Mechanics, 5th ed.; Oxford University Press: Oxford ; New York, 2011. ISBN 978-0-19-954142-3.
$\endgroup$
2
  • $\begingroup$ you should probably post this on the math stackexchange. The sentence you quoted reads like the author is defining a function that will be analyzed further along in the text. $\endgroup$ Apr 1, 2020 at 5:24
  • $\begingroup$ In short, any square-integrable function can be written that way. $\endgroup$
    – Wildcat
    Apr 1, 2020 at 10:20

1 Answer 1

2
$\begingroup$

The complete, formal mathematical answer is a bit complex. It might be better to get that discussion from Math Stack Exchange.

An answer to the world of Chemistry is more straightforward: Yes, for any physically relevant operator and any relevant function, you always can. That’s a fundamental property of quantum mechanics: (functional) states are superpositions of eigenstates.

$\endgroup$
2
  • $\begingroup$ Just a short note for OP: in introductory texts on quantum physics & chemistry one can often encounter phrases like "physically relevant" or "well-behaved function" with the actual meaning being "square-integrable function". There are a lot of mathematical nuances and subtleties but to start from square-integrability requirement would be enough. $\endgroup$
    – Wildcat
    Apr 1, 2020 at 10:35
  • $\begingroup$ That's enough for me! Thank you to Nick, Bob, and Wildcat for answers/comments and to andselisk for editing! $\endgroup$ Apr 1, 2020 at 14:58

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.