Understanding Complex Operators: Rules, Boundedness, and Positivity

[SemM]

Hi, from the books I have, it appears that some rules for operators, boundedness, positivity and possibly the definition of the spectrum regard real operators, and not complex operators.

From the complex operator $i\hbar d^3/dx^3$ it appears that it can be defined as not bounded (unbounded) by the conventional rule of

\begin{equation}
||H_s\psi|| \leqslant c||\psi||
\end{equation}

where the interval n, defined for which a trial function exists in an imaginary plane. So for this operator to be unbounded and not satisfy the eqn above, the limits of the integral must have an imaginary value. Then, one gets\begin{equation}
||H_s\psi|| \geqslant c||\psi||
\end{equation}

So one cannot say if that operator is bounded at all, using that regular rule.

Is there any particular geometrical meaning in saying that an operator is unbounded only when it acts on a function defined in a complex interval?

Are there particular rules for complex operators, as all operators defined in the literature I have here are real?

 

[Mark44]

Hi, from the books I have, it appears that some rules for operators, boundedness, positivity and possibly the definition of the spectrum regard real operators, and not complex operators.

From the complex operator $i\hbar d^3/dx^3$ it appears that it can be defined as not bounded (unbounded) by the conventional rule of

\begin{equation}
||H_s\psi|| \leqslant c||\psi||
\end{equation}

You're mixing up the terminology. An operator on a particular region of the complex plane is either bounded or it's unbounded. You don't define it to be either bounded or unbounded. And what you're calling "the conventional rule" is the definition of boundedness.

Also, what is s in the operator $H_s$?

where the interval n, defined for which a trial function exists in an imaginary plane. So for this operator to be unbounded and not satisfy the eqn above, the limits of the integral must have an imaginary value.

I'm not sure this makes any sense at all.

Then, one gets
\begin{equation}
||H_s\psi|| \geqslant c||\psi||
\end{equation}

Not quite. If this operator is unbounded, the inequality would be $||H_s\psi|| > c||\psi||$.
As you wrote the inequalities, if $||H_s\psi|| = c||\psi||$, the operator would be both bounded and unbounded.

So one cannot say if that operator is bounded at all, using that regular rule.

Is there any particular geometrical meaning in saying that an operator is unbounded only when it acts on a function defined in a complex interval?

Are there particular rules for complex operators, as all operators defined in the literature I have here are real?

 

[SemM]

You're mixing up the terminology. An operator on a particular region of the complex plane is either bounded or it's unbounded. You don't define it to be either bounded or unbounded. And what you're calling "the conventional rule" is the definition of boundedness.

Also, what is s in the operator $H_s$?

I'm not sure this makes any sense at all.
Not quite. If this operator is unbounded, the inequality would be $||H_s\psi|| > c||\psi||$.
As you wrote the inequalities, if $||H_s\psi|| = c||\psi||$, the operator would be both bounded and unbounded.

Hi Mark, the operator is:

\begin{equation}
H_s= i\hbar^3\frac{d^3}{dx^3} - x^2
\end{equation}

Note that one can split them up into\begin{equation}
H_{s-a}= i\hbar^3\frac{d^3}{dx^3}
\end{equation}

and\begin{equation}
H_{s-b}= -x^2
\end{equation}

and one can treat them individually in the formula for boundedness.

So I am not sure here, as the former gives non-real values, unless n in the interval is purely imaginary, thus the strange point on the "imaginary interval".

 

[George Jones]

In order to begin to make sense of this thread, we have to know the precise definition of "complex operator" that you are using.

 

[SemM]

In order to begin to make sense of this thread, we have to know the precise definition of "complex operator" that you are using.

George, it is given above.

 

[Mark44]

the operator is: $H_s= i\hbar^3\frac{d^3}{dx^3} - x^2$

Again, what's the significance of s in this equation?

 

[SemM]

Again, what's the significance of s in this equation?

It's none, it's just a notation to define it.

I tested the function $cos(n\pi x)$ in the interval $0-2\pi$

 

[George Jones]

George, it is given above.

I don't see it. Can you please quote the relevant portion of a post in which you give the precise definition of what the general term "complex operator" means. Note that I am not asking for what you consider to be specific examples of complex operators, I am asking for the general definition of a complex operator.

 

[SemM]

I don't see it. Can you please quote the relevant portion of a post in which you give the precise definition of what the general term "complex operator" means. Note that I am not asking for what you consider to be specific examples of complex operators, I am asking for the general definition of a complex operator.

George, it's an operation which is allowed to have complex coefficients, like $H_s$ has above, in its first part. I don't know much about this form of operators, however, I am aware of that they are supposed to act on holomorphic functions, which are not necessarily smooth. Otherwise, I am trying to find whether $H_s$ is bounded or not...

 

[Orodruin]

I am sorry, but it is completely unclear what you mean as you do not seem to be using standard nomenclature. Whether or not you can multiply an operator by a complex number or not depends on the vector space it acts on. If the vector space is a vector space over $\mathbb C$, then multiplication by a complex number is well defined. This has nothing to do with whether the vector space is a function space or not.

If you have a norm in order to talk about bounded operators, it is obvious that $i H$ is a bounded operator if $H$ is a bounded operator as $\newcommand{\norm}[1]{\left|\left| #1 \right|\right|} \norm{iH \psi} = \lvert i \rvert \norm{H\psi} = \norm{H\psi}$ for all elements $\psi$ in the vector space.

 

[Mark44]

Thread closed. This thread is a continuation of this existing thread: https://www.physicsforums.com/threads/proving-that-an-operator-is-unbounded.939201/

@SemM, please don't start a new thread for the same question you're asking in an existing thread.