A Operator mapping in Hilbert space

SeM

Hi, I have an operator given by the expression:

L = (d/dx +ia) where a is some constant. Applying this on x, gives a result in the subspace C and R. Can I safely conclude that the operator L can be given as:

\begin{equation}
L: \mathcal{H} \rightarrow \mathcal{H}
\end{equation}

where H is Hilbert space, with subspaces C and R ?
 

PeroK

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I think you have a confusion of ideas here. I'm not sure where the subspaces come into it.
 

SeM

I thought C and R , being complex and real, are both subspace to the entire space H, composed of R and C. Disregarding the term "subspace", does this look reasonable?
The reason I am not sure is because the projection of L on x (which can be any real value) goes from R to R and C. Is this generalized form right?

\begin{equation}
L : \mathcal{H} \rightarrow \mathcal{H}
\end{equation}
 

PeroK

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I thought C and R , being complex and real, are both subspace to the entire space H, composed of R and C. Disregarding the term "subspace", does this look reasonable?
The reason I am not sure is whether the projection of L on x, being from x (which can be any real value) onto R and C, is correctly given by the form:

\begin{equation}
L : \mathcal{H} \rightarrow \mathcal{H}
\end{equation}
What Hilbert space are we talking about here?
 

SeM

Lx, where x are real values.

L is doing something to z, it is transforming it to 1 +i ax. If x is in R, and R is Hilbert space, then Lx is R and C, still Hilbert space?
 
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PeroK

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Lx, where x are real values.

L is doing something to z, it is transforming it to 1 +i ax. If x is in R, and R is Hilbert space, then Lx is R and C, still Hilbert space?
Sorry, none of that makes much sense. If ##L## is a differential operator it must be acting on functions, not on real or complex numbers.
 

SeM

It is acting on a function y(x) = x. It is written above
 

PeroK

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It is acting on a function y(x) = x. It is written above
In that case, as you have defined it, we have:

##Lx = iax +1##

Which is another function.
 

SeM

So, iax+1 and x are both two functions , respectively in C and R and in R. What can one say about this, in a Hilbert space?
 

fresh_42

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It is acting on a function y(x) = x. It is written above
But you've written something with the variable ##x##, then ##z## and now ##y##. Can you just say, which spaces you're talking about? Perhaps with a complete example.
There is nothing exotic about Hilbert spaces and examples are many. So again:
What Hilbert space are we talking about here?
I'm not sure where the subspaces come into it.
Me neither.
 

PeroK

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So, iax+1 and x are both two functions , respectively in C and R and in R. What can one say about this, in a Hilbert space?
No, in this context ##\mathbb{C}## and ##\mathbb{R}## are fields of scalars. They are not subspaces of a Hilbert Space of functions.
 

SeM

I have this operator, L, and I Have to explain what it does in a Hilbert space on two other operators, the position operator x, and the momentum operator d/dx.

In Kreyszig Functional Analaysis it says:

Let T: H -> H be a bounded inear operator on a complex Hilbert space H.

How does this NOT apply on L? Can't I write:

L: H -> H ?
 

PeroK

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I have this operator, L, and I Have to explain what it does in a Hilbert space on two other operators, the position operator x, and the momentum operator d/dx.

In Kreyszig Functional Analaysis it says:

Let T: H -> H be a bounded inear operator on a complex Hilbert space H.

How does this NOT apply on L? Can't I write:

L: H -> H ?
Your grasp of the material is such that it is difficult to follow what you are asking.

Have you jumped into a functional analysis text on a digression from QM?
 

SeM

Your grasp of the material is such that it is difficult to follow what you are asking.

Have you jumped into a functional analysis text on a digression from QM?

I don't think it is a digression indeed. QM and Hilbert spaces go hand in hand. I have to analyze an operator for its properties in a Hilbert space, and its commutation with other operators. The latter is OK. The former, I am not sure.

If the Operator L is acting on say the position operator x, and it is composed of d/dx + ia, what can one say about this operator in a Hilbert space? And please answer at least something instead of asking me more questions, if possible.

Thanks
 

PeroK

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I don't think it is a digression indeed. QM and Hilbert spaces go hand in hand. I have to analyze an operator for its properties in a Hilbert space, and its commutation with other operators. The latter is OK. The former, I am not sure.

If the Operator L is acting on say the position operator x, and it is composed of d/dx + ia, what can one say about this operator in a Hilbert space? And please answer at least something instead of asking me more questions, if possible.

Thanks
A study of pure functional analysis requires considerable mathematical prerequisites. And, although QM relies on Hilbert Spaces, the mathematical treatment of Hilbert Spaces goes much deeper than is required for an initial study of QM.

In this case, you are confused by the concepts of operators acting on functions in your Hilbert space and operators being combined through functional composition.

I believe you have jumped in the mathematical deep end here.
 

PeroK

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Your question might be:

What is the commutator of ##L## and ##\hat{x}##?
 
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George Jones

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I have this operator, L, and I Have to explain what it does in a Hilbert space on two other operators, the position operator x, and the momentum operator d/dx.

In Kreyszig Functional Analaysis it says:

Let T: H -> H be a bounded inear operator on a complex Hilbert space H.

How does this NOT apply on L? Can't I write:

L: H -> H ?
Are you trying do do a particular problem in Kreysyzig? If so, which one?

A study of pure functional analysis requires considerable mathematical prerequisites. And, although QM relies on Hilbert Spaces, the mathematical treatment of Hilbert Spaces goes much deeper than is required for an initial study of QM.

In this case, you are confused by the concepts of operators acting on functions in your Hilbert space and operators being combined through functional composition.

I believe you have jumped in the mathematical deep end here.
Echoing PeroK;

As functional analysis books go, Kreyszig (my text as a student long before my beard turned white) requires less mathematical prerequisites (e.g., no measure theory or topology) and is more pedagogical than most other functional analysis texts, but even it requires a dose of mathematical sophistication that is administered through material like elementary real analysis.

Quantum mechanics can provide motivation to learn a bit of functional analysis for interest's sake, but the functional analysis does not often help with quantum mechanics, and can actually hinder one's progress in quantum mechanics, i.e., the rigour gained can cause rigour mortis to set in.

Personal digression: Kreyszig's son was a fellow student in a numerical analysis course that I took.
 

SeM

Your question might be:

What is the commutator of ##L## and ##\hat{x}##?
Hi Perok, I have investigated the commutation algebra, and it is fine. I found out , as another post I put on the QM section, that this operator has more algebraic combinations with the imaginary position operator, ix, which is what you write here?

This goes a little bit out of the scope of the project. My project is simply to answer the very general request by the tutor:

"The set up of the Hamiltonian requires a proper set up (i.e. Hilbert space etc)". Can you imagine now, why I am am confused?
 

PeroK

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Hi Perok, I have investigated the commutation algebra, and it is fine. I found out , as another post I put on the QM section, that this operator has more algebraic combinations with the imaginary position operator, ix, which is what you write here?

This goes a little bit out of the scope of the project. My project is simply to answer the very general request by the tutor:

"The set up of the Hamiltonian requires a proper set up (i.e. Hilbert space etc)". Can you imagine now, why I am am confused?
I'm confused! In any case, I don't understand what question you are asking.
 
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SeM

I'm confused! In any case, I don't understand what question you are asking.
Me neither, because I don't understand the tutor's request, and he doesn't want to specify more...
 

SeM

Specifically he says " "The set up of the Hamitonian requires a proper set up (Hilbert space, etc). Since the student does not motivate a concrete system in which the newly proposed Hamiltonian could be explored, this point is crucial."
 

SeM

Are you trying do do a particular problem in Kreysyzig? If so, which one?


No, I am using Kreyszig Operator Algebra to list up the properties of an own operator. I have gone through the commutation algebra. However, when I try to show the operator's properties in a Hilbert space, I am not sure my proof shows that the operator is applicable (or acting) in a Hilbert space. How does one prove that an operator can transform a function, say f= e^ipx in a Hilbert space, even though it is evident from doing a simple operation with it on f?
 

SeM

Your question might be:

What is the commutator of ##L## and ##\hat{x}##?
Dear Perok, why did you bring in the hat operator? Is it because of the mapping of L: H-> H ? Sorry, but much of this is new to me.
 

PeroK

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Specifically he says " "The set up of the Hamitonian requires a proper set up (Hilbert space, etc). Since the student does not motivate a concrete system in which the newly proposed Hamiltonian could be explored, this point is crucial."
I'm sure something must be lost in translation here. In any case, this statement is not a question or problem that I can help you with.

Note that I used ##\hat{x}## as the position operator, to distinguish it from a point ##x## or the function ##f(x) = x##.

You need to talk to your tutor and straighten out what exactly it is that he wants you to learn.
 

SeM

Note that I used ##\hat{x}## as the position operator, to distinguish it from a point ##x## or the function ##f(x) = x##.

Thanks for the clarification. Yes, this and the momentum operators have been clarified already for commutation. All well!
 

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