Undergrad How to find admissible functions for a domain?

[SemM]

Hi, in a text provided by DrDu which I am still reading, it is given that "the momentum operator P is not self-adjoint even if its adjoint $P^{\dagger}=-\hbar D$ has the same formal expression, but it acts on a different space of functions."

Regarding the two main operators, X and D, each has a domain in $\mathscr{H}$ which represent the set of admissible functions, and that satisfy the completeness relation in $\mathscr{H}$ and the inner product.

In practice in quantum chemistry or in quantum physics, one would like to know the form of these functions. In Bohms "Quantum Theory" Bohm shows an explicitely example of such a function $f(x)=e^{-x^2}g(x)$ where g is a polynomial. This is an ideal example of such a function, however in this book, as well as in others, it seems that these functions "are pulled out of thin air", without any method to it. Is this a trial and error process fueled by the experience of the mathematician to find such a function? There are, probably other forms than the above-given, so my question is, is there a method to find a list of functions in the the domain of an operator?
Thanks!

 

[PeroK]

Hi, in a text provided by DrDu which I am still reading, it is given that "the momentum operator P is not self-adjoint even if its adjoint $P^{\dagger}=-\hbar D$ has the same formal expression, but it acts on a different space of functions."

Regarding the two main operators, X and D, each has a domain in $\mathscr{H}$ which represent the set of admissible functions, and that satisfy the completeness relation in $\mathscr{H}$ and the inner product.

In practice in quantum chemistry or in quantum physics, one would like to know the form of these functions. In Bohms "Quantum Theory" Bohm shows an explicitely example of such a function $f(x)=e^{-x^2}g(x)$ where g is a polynomial. This is an ideal example of such a function, however in this book, as well as in others, it seems that these functions "are pulled out of thin air", without any method to it. Is this a trial and error process fueled by the experience of the mathematician to find such a function? There are, probably other forms than the above-given, so my question is, is there a method to find a list of functions in the the domain of an operator?
Thanks!

"Pulled out of thin air" in this case is simply "at a level of mathematics that the reader is supposed to be familiar with". No advanced physics text is likely to spend much time explaining that $f(x)=e^{-x^2}g(x)$ (where g is a polynomial) is square integrable, infinitely differentiable, and all its derivatives are square integrable. You're supposed to be familiar with mathematics at this level.

This is not "A" for advanced mathematics. The result required here is simply that an exponential "dominates" a polynomial.

 

[SemM]

The result required here is simply that an exponential "dominates" a polynomial.

This is the key to my question, if the exponential term dominates it, how come several exponential functions are dismissed and thus $f(x)=e^{-ix}g(x)$ is not allowed? The square integrable nature of a function, accounting for that you mean $\langle |\psi|^2 \rangle$ can in some cases have to be calculated numerically, should the function be more complex. .ie. $f(x)=e^{(-x^2-x)}g(x)$.

Thanks PeroK.

 

[PeroK]

This is the key to my question, if the exponential term dominates it, how come several exponential functions are dismissed and thus $f(x)=e^{-ix}g(x)$ is not allowed? The square integrable nature of a function, accounting for that you mean $\langle |\psi|^2 \rangle$ can in some cases have to be calculated numerically, should the function be more complex. .ie. $f(x)=e^{(-x^2-x)}g(x)$.

Thanks PeroK.

There's a big difference between real exponentials and complex exponentials. Again, no one writing these books is expecting these sorts of questions.

 

[SemM]

There's a big difference between real exponentials and complex exponentials. Again, no one writing these books is expecting these sorts of questions.

OK, can't complain on that!