# A How to find the domain of functions of an operator

#### SemM

Gold Member
Hi, I have a strange nonlinear operator which yields non-Hermitian solutions when treated in a simple ODE, $H\Psi$=0. It appears from a paper by Dr Du in a different posting, that an operator can be non-self-adjoint in one domain, but be self-adjoint in another domain defined by the interval the operator should operate in (and thus yield hermitian solutions). The statement in the brackets is an assumption I have made.

How can I find the type of wavefunctions that belong to the domain of the operator H?

Thanks!

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#### DrDu

Hi, I have a strange nonlinear operator which yields non-Hermitian solutions when treated in a simple ODE, $H\Psi$=0. It appears from a paper by Dr Du in a different posting, that an operator can be non-self-adjoint in one domain, but be self-adjoint in another domain defined by the interval the operator should operate in (and thus yield hermitian solutions).
Just two remarks: 1. Your operator H in the other post was a linear operator. 2. Self-adjoinedness is a stronger property than Hermiticity.

#### SemM

Gold Member
Just two remarks: 1. Your operator H in the other post was a linear operator. 2. Self-adjoinedness is a stronger property than Hermiticity.
Thanks for this Dr. Du.

I noticed that some intermingle hermiticity with self-adjointedness. I have Kreyszig's "Kreyszigs Functional Analysis" which is really interesting. I will look it up there.

About H, it is H =TT'

where T = $(i\hbar d/dx + \gamma)$ and T' $(-i\hbar d/dx + \gamma)$ which give:

$H= \hbar^2d^2/dx^2 - 2i\hbar\gamma d/dx + \gamma^2$

#### DrDu

Yes, so what is nonlinear here? Apparently $H(c\phi+d\psi)=c(H\phi)+d(H\psi)$, so H is linear.

#### SemM

Gold Member
Yes, so what is nonlinear here? Apparently $H(c\phi+d\psi)=c(H\phi)+d(H\psi)$, so H is linear.
Ok. I just received a posting a few days ago on that its components T and T' are nonlinear, so I assumed that their product would be non-linear. Nevertheless, its non-self-adjointedness, and the domain where it can be self-adjoint is still elusive to me. I read in your article suggestion that this domain is critical to make the operator self-adjoint, and thus get a workable solution. But what does one use to define the domain? I take the boundary conditions are the principal part, and I have a suggestion to use the zero point energy in the system, with the L-interval. That interval will give a workable domain. But unfortunately, after doing numerical calculations on this, there seems to be no value of L that makes the integral complex values vanish, and therefore, I get nontrivial values no matter what L is.

#### DrDu

I suppose you basically want this operator to be defined on a domain from the hilbert space of square integrable functions on the real line, i.e. from $-\infty<x<\infty$. For the operator to be defined, the functions in its domain have to be 2 times differentiable and with this domain the operator is already self-adjoint. A 2 time differentiable (and hence also continuous) square integrable function on the real line vanishes at $\pm \infty$ and so does its first derivative, hence the boundary terms from partial integration vanish, and therefore, the operator is symmetric. As it is also densely defined and the range is equal to that of the Hermitian adjoint, is is also self-adjoint.

#### SemM

Gold Member
I suppose you basically want this operator to be defined on a domain from the hilbert space of square integrable functions on the real line, i.e. from $\-\infty<x<\infty$. For the operator to be defined, the functions in its domain have to be 2 times differentiable and with this domain the operator is already self-adjoint. A 2 time differentiable (and hence also continuous) square integrable function on the real line vanishes at $\pm \infinity$ and so does its first derivative, hence the boundary terms from partial integration vanish, and therefore, the operator is symmetric. As it is also densely defined and the range is equal to that of the Hermitian adjoint, is is also self-adjoint.
Thanks. This is very clear. In this context, I have considered the case $H\psi = 0$. And I gave it initial conditions. The boundary conditions are not clear to me really, because in a sense, the initial conditions $\psi(0)=1$ and $\psi'(0)=0$ are boundary conditions. However, the boundary must be defined in the inner product integral, for $L^2[a,b]$, and finding a and b is the mystery to me.

I tried to simply do the calculation on that ODE with the initial conditions, however this case gave a non-hermitian solution, which has no vanishing complex terms. I am therefore wondering, how I can find the "right" domain when the current solution doesn't obey the inner product (Sommerfeld condition)?

#### SemM

Gold Member
I suppose you basically want this operator to be defined on a domain from the hilbert space of square integrable functions on the real line, i.e. from $-\infty<x<\infty$. For the operator to be defined, the functions in its domain have to be 2 times differentiable and with this domain the operator is already self-adjoint. A 2 time differentiable (and hence also continuous) square integrable function on the real line vanishes at $\pm \infty$ and so does its first derivative, hence the boundary terms from partial integration vanish, and therefore, the operator is symmetric. As it is also densely defined and the range is equal to that of the Hermitian adjoint, is is also self-adjoint.

PS: From your article suggestion, I assumed that that operator is non-self adjoint in $L^2[-\infty,+\infty]$ however it is self-adjoint in $L^2[a,b]$, as this is the case for the momentum operator.

Thanks.

#### DrDu

Maybe you could explain your physical problem in detail, so that we can work out what is the right hamiltonian?

#### SemM

Gold Member
Maybe you could explain your physical problem in detail, so that we can work out what is the right hamiltonian?
Thanks for your suggestion. The physical part of this is a free particle that oscillates outside of a well potential (without a V(x) term) and it is under the effect of a magnetic field, contained in the $\gamma$ term. Its intensity is a parameter in the $\gamma$ term, and the one dimensional independent variable is x. The Hamiltonian H, has therefore the kinetic term, the momentum term and the $\gamma$ term which is the potential parameter, and not an oscillator potential term such as V(x). The initial conditions are $\psi(0)=1$ so it is starts by unity from the start, and $\psi'(0)=0$ so that is a maximum point. Finally, the equation $H\psi=0$ is considered to see the behavior of the wavefunction. A further step is to set $H\psi=E_ 0\psi$ where $E_0$ is the zero-point energy, with L in it. I have already calculated the latter solution as well, however they look even worse, and the integral of the born-sommerfeld condition evaluated across boundary 0 to L gives an even worse expression half a page in MATLAB output with complex conjugates.

It is noteworthy to mention, that $E_0$ would be the solution to the boundary problem, in that it has L in it, however, it does not result in vanishing terms , so the observables are all non-trivial. For the former case, $H\psi=0$ , the same accounts unfortunately.
Thanks

#### DrDu

The physical part of this is a free particle that oscillates outside of a well potential (without a V(x) term)
What do you mean with this?

#### SemM

Gold Member
That is is not bound to a potential well, and is a free particle.

#### DrDu

Thanks for your suggestion. The physical part of this is a free particle that oscillates outside of a well potential (without a V(x) term)
I still don't understand what you mean with "outside". Either there is a potential in your hamiltonian or not.
I also wonder on the form of your hamiltonian. If $\gamma$ is basically the vector potential of the magnetic field, your hamiltonian should look most probably like
$H= (-i\hbar \partial /\partial x +\gamma)^2=T'^2.$
In the hilbert space, this operator has no eigenvectors, as it has a purely continuous spectrum ranging from E=0 to infinity.
However, it has some improper eigenvalues which do not lie in the Hilbert space, but at least do not diverge, $\psi_k(x)=\exp(ikx)$ belonging to $E=(k+\gamma)^2$. The minimum energy solution is for E=0 or $k=-\gamma$.

I saw that you started numerous threads on very similar questions. It is recommended in this forum, to rather discuss them in one thread.

#### SemM

Gold Member
I still don't understand what you mean with "outside". Either there is a potential in your hamiltonian or not.
I also wonder on the form of your hamiltonian. If $\gamma$ is basically the vector potential of the magnetic field, your hamiltonian should look most probably like
$H= (-i\hbar \partial /\partial x +\gamma)^2=T'^2.$
In the hilbert space, this operator has no eigenvectors, as it has a purely continuous spectrum ranging from E=0 to infinity.
However, it has some improper eigenvalues which do not lie in the Hilbert space, but at least do not diverge, $\psi_k(x)=\exp(ikx)$ belonging to $E=(k+\gamma)^2$. The minimum energy solution is for E=0 or $k=-\gamma$.

I saw that you started numerous threads on very similar questions. It is recommended in this forum, to rather discuss them in one thread.
Thanks Dr Du. This is very interesting. The second order term you write:

H= (-i\hbar \partial /\partial x +\gamma)^2=T'^2.

is actually considered as :

H= (-i\hbar \partial /\partial x +\gamma)(i\hbar \partial /\partial x +\gamma)=TT^{*}

because the two operators become Hilbert-adjoint, by the change of the sign in front of i. It is indeed as you otherwise write here. I considered the lowest state, E= 0 and then a numerical solution to it, by using y(0)=1 and y'(0)=0, which gave a complex solution, with the exponential term similar to as you wrote it.

However, as I am also now reading "Fundamentals of the Theory of Operator Algebras vol. 1" and Kreyszig Functional Analysis, I have not found one example where one derives some physically sensible answers from a similar Hamiltonian and its solution. My intention is to study the Hilbert-adjoint pair and how they affect this Hamiltonian and its solution. However, it becomes non-trivial. So therefore, a lot of posts and questions have been asked.

Can one conclude that with this Hamiltonian, the solution will ALWAYS become non-trivial?

#### DrDu

As should have become clear from the discussion up to now, both T and T' are self-adjoint on their own and hence are not adjoint to each other. Furthermore $T'^2 \neq T'T$, so the two hamiltonians are not identical and I wonder whether your version is physically sensible. A hamiltonian $H=T'^2$ also describes the periodic Bloch functions in solid state physics and gives rise to the band structure of periodic solids.

#### SemM

Gold Member
Thanks, maybe the reason I seem not to converge to a notion here on their adjoint properties is because I am referring to too many texts? In Kreyszig it says on two similar operators as T and T* the following:

\quote Kreyszig p. 594:

To simplify the formulas in our further consideration , we define:

A = \beta\big(\alpha Q+i/\alpha D\big)

A^{*} = \beta\big(\alpha Q-i/\alpha D\big)

\end quote Kreyszig p. 594

Note there should be a dot over $\alpha$ in the latter operator. However, I was pretty certain that this signified that the equivalent T and T*

T = \big(\frac{i}{\hbar}\frac{d}{dx}+\gamma\big)

T^{*} = \big(-\frac{i}{\hbar}\frac{d}{dx}+\gamma\big)

Thanks!

#### SemM

Gold Member
A hamiltonian $H=T'^2$ also describes the periodic Bloch functions in solid state physics and gives rise to the band structure of periodic solids.
Thanks Dr Du, this is part of a solid state model indeed.

#### DrDu

Thanks, maybe the reason I seem not to converge to a notion here on their adjoint properties is because I am referring to too many texts?
No, the problem seems to be that you don't read them!
Kreyszig defines D to be the momentum operator, i.e. $D=-i\hbar d/dx$.

#### SemM

Gold Member
No, the problem seems to be that you don't read them!
Kreyszig defines D to be the momentum operator, i.e. $D=-i\hbar d/dx$.
I read them , but I do indeed not understand all.

I am looking at the overall structure of the operator, where D is inverted of its sign. This is what is done in the T and $T^{*}$ pair and then solved as the ODE $TT^{*}\psi=0$

It appears this model is similar to the damped harmonic oscillator.

However, $TT^{*}\psi=0$ does not describe an n-level system, and for this, one needs to do:

$TT^{*}\psi=E\psi$

Having done this, the values of L in E and the form of the solution do not yield vanishing terms in the integrals, so the results are horrible. I am therefore pondering on the form of E for this case. - still however, discounting for any aesthetic form of E, the solution is non trivial.

So my final question is, what does one do with non-trivial results?

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