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]

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]

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?

 

[PeroK]

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]

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]

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]

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]

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]

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]

Your question might be:

What is the commutator of $L$ and $\hat{x}$?

 

[George Jones]

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]

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.

 

[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]

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!

 

[George Jones]

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?

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?

Well, if $x$ lives in the set of real numbers, then $e^{ipx}$ does not live in the standard Hilbert space of (equivalence classes of) square-integrable functions. Restricting $x$ (i.e., the domain of the functions) to the bounded interval $\left[a , b\right]$ seems to help, but, in this case, the momentum operator is unbounded (see page 569). Then, by page 525, a self-adjoint moment momentum operator can't have the Hilbert space all the square-integrable functions on $\left[a , b\right]$ as its domain. Done "properly", this stuff is quite subtle.

 

[SeM]

Well, if $x$ lives in the set of real numbers, then $e^{ipx}$ does not live in the standard Hilbert space of (equivalence classes of) square-integrable functions. Restricting $x$ (i.e., the domain of the functions) to the bounded interval $\left[a , b\right]$ seems to help, but, in this case, the momentum operator is unbounded (see page 569). Then, by page 525, a self-adjoint moment momentum operator can't have the Hilbert space all the square-integrable functions on $\left[a , b\right]$ as its domain. Done "properly", this stuff is quite subtle.

This is what I was hoping for! Thanks! I will give this a thought for a few days.

PS: The restriction to L^2[a,b] works for the operator, and that makes it Hilbert space. If the wavefunction is completely different from the given, and is hermitian, and it normalizes to 1 in the interval a b, then both the Operator and the hermitian solution are all in the Hilbert space, right?

 

[SeM]

Well, if $x$ lives in the set of real numbers, then $e^{ipx}$ does not live in the standard Hilbert space of (equivalence classes of) square-integrable functions.
.

PS: George, does this mean that any solution to a Hamiltonian in a Hilbert space must be continuous and square-integrable and must satisfy ALL the fundamental prescriptions of quantum mechanics, or just the two mentioned here?

A second thing: A Hamitlonian transforms a wavefunction to give a result, whether it is the energy in the Schrödinger eqn, or some other solution. That function is transformed by a i.e derivation. If the Hamiltonian (L) is deriving a wavefunction, and that wavefunction is hermitian, does it automatically mean that L is projecting the wavefunction into its derivative (in the same space)? So L: H-> H , where the wfn is in H before and after the operation? Is the plain and simple confirmation of the action of an operator in Hilbert space?

 

[SeM]

PS: Being entirely frank, I need to clarify a thing. What does this really mean?

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

I thought it means that an operator L is doing some operation on some function in Hilbert space H, and its product is still in Hilbert space. For instance, a square integrable function is operated on, and it yields a second square-integrable function, where both, before and after the arrow, are in Hilbert space?

Thanks!

 

[PeroK]

PS: Being entirely frank, I need to clarify a thing. What does this really mean?

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

I thought it means that an operator L is doing some operation on some function in Hilbert space H, and its product is still in Hilbert space. For instance, a square integrable function is operated on, and it yields a second square-integrable function, where both, before and after the arrow, are in Hilbert space?

Thanks!

Yes, that's what it means. It's really just the definition of a function, $L$, from a set $\mathcal{H}$ to itself (*).

Note that $\mathcal{H}$ is the Hilbert space (in this case of square integrable functions).

But, you know, if you need to ask these questions, you need a crash course in pure mathematics before you dive into functional analysis.

(*) Note that mathematically unless you specify more, then that is all it means. However, in this case there are probably other statements like:

Let $L$ be a linear operator. Or let $L$ be defined by ... etc.

 

[SeM]

Yes, that's what it means. It's really just the definition of a function, $L$, from a set $\mathcal{H}$ to itself (*).

But, you know, if you need to ask these questions, you need a crash course in pure mathematics before you dive into functional analysis.

Its inverse, I had an Advanced Mathematics course in my PhD last year, and I am MORE interested in this and reading on my own, while doing other courses which are completely irrelated. So I take I am doing all I can at the moment. I remember stupid questions are the foundation for many things I have learned, and I ask them places where no one can see me, like here.

(*) Note that mathematically unless you specify more, then that is all it means. However, in this case there are probably other statements like:

Let $L$ be a linear operator. Or let $L$ be defined by ... etc.

Thanks for this Perok. This L, is actually a linear operator, and I take that an example where H is not H anymore would be:

\begin{equation}
P: \mathcal{H} \rightarrow \mathcal{O}
\end{equation}

would be where that operator P, the integral, is sending the function to a different space, i.e. an integration such as this:

\begin{equation}
\int x dxdy = 1/2x^2 + xy
\end{equation}

which is an example of an operation on a one dimensional function , x, making it a two dimensional function when integrated with respected to dx and dy?

Is this an example of

\begin{equation}
P: \mathcal{H} \rightarrow \mathcal{O}
\end{equation}

?

Thanks!

 

[PeroK]

A simpler example of a function that is not a linear operator is the norm. In any Hilbert space you have an inner product and if you take the inner product of a vector with itself you get the norm (squared). So, you could have:

$N: \mathcal{H} \rightarrow \mathbb{R}$

Defined by

$N(f) = ||f|| \equiv \langle f, f \rangle ^{1/2}$

Another, very important example, is the idea of a Linear Functional. Take any function $f \in H$. Define:

$L_f : H \rightarrow \mathbb{C}$

where $L_f(g) = \langle f, g \rangle$

In this case, therefore, $L_f$ maps each function to a complex number. Note that for every $f \in H$ you have a linear functional $L_f$. The set of all linear functionals is called the Dual Space ...

Another example, which I think you are getting at is where the $L$ is an operator, but it produces functions that lie outside the Hilbert Space. For example, you take $H$ to be the space of square integrable functions on some interval $[a, b]$, where the functions vanish at $a$ and $b$. That is:

If $f \in H$, then $f(a) = f(b) = 0$

Now, the differential operator $L = \frac{d}{dx}$ maps functions to functions that may not vanish at the end-points. In this case, $L$ maps functions in the original Hilbert space into a different space, that involves a wider class of functions.

But, you know, I don't really have time to teach anyone functional analysis!

 

[SeM]

A simpler example of a function that is not a linear operator is the norm. In any Hilbert space you have an inner product and if you take the inner product of a vector with itself you get the norm (squared). So, you could have:

$N: \mathcal{H} \rightarrow \mathbb{R}$

Defined by

$N(f) = ||f|| \equiv \langle f, f \rangle ^{1/2}$

Another, very important example, is the idea of a Linear Functional. Take any function $f \in H$. Define:

$L_f : H \rightarrow \mathbb{C}$

where $L_f(g) = \langle f, g \rangle$

In this case, therefore, $L_f$ maps each function to a complex number. Note that for every $f \in H$ you have a linear functional $L_f$. The set of all linear functionals is called the Dual Space ...

Another example, which I think you are getting at is where the $L$ is an operator, but it produces functions that lie outside the Hilbert Space. For example, you take $H$ to be the space of square integrable functions on some interval $[a, b]$, where the functions vanish at $a$ and $b$. That is:

If $f \in H$, then $f(a) = f(b) = 0$Now, the differential operator $L = \frac{d}{dx}$ maps functions to functions that may not vanish at the end-points. In this case, $L$ maps functions in the original Hilbert space into a different space, that involves a wider class of functions.

But, you know, I don't really have time to teach anyone functional analysis!

You pretty much just did! And thanks for the confirmation. I noticed in the book by Kreyszig, that he showed an operator which did like you wrote here, mapping into another spce, where he drawed a sketch of a plane (where one function was before operation) and a perpendicular plane over it, where the function "was sent to" after operation. That sketch illustrated a lot, however, one who is not experienced asks:

Why write a strangely simple formula as this:

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

where one describes that a function from space H is operated to another function in the same space, or as you wrote, a function "from a space H on itself H".

I understand George's example, where one confirms that an operator : is producing an output in H of an input in H, which is square integrable in H before and after operation. This is directly confirming that one or even two of the fundamental pescriptions of quantum mechanics are preserved with THAT operator, by that simple statement:\begin{equation}
L : \mathcal{H} \rightarrow \mathcal{H}
\end{equation}

But why not use this alternative to confirm square integrable of a function?

\begin{equation}
L : \mathbb{R}, \mathbb{C} \rightarrow \mathbb{R}, \mathbb{C}
\end{equation}

?

Is it because one ALWAYS assumes that H is both R and C?

Thanks! Have a good weekend!

 

[PeroK]

But why not use this alternative to confirm square integrable of a function?

\begin{equation}
L : \mathbb{R}, \mathbb{C} \rightarrow \mathbb{R}, \mathbb{C}
\end{equation}

?

Is it because one ALWAYS assumes that H is both R and C?

Thanks! Have a good weekend!

That notation makes no sense, as $H$ is a set/space/Hilbert Space of functions. It's not a set of numbers. An "operator" is just a function that maps functions to functions (or vectors to vectors).

In finite-dimensional vector spaces, Linear Operators are represented by matrices. But, the set of square integrable functions is actually an infinite dimensional vector space. And, this makes this more complicated than the finite-dimensional case.

Note that finite dimensional Hilbert Spaces turn up in QM when you consider spin, for example. And,in these cases, the linear operators are indeed represented by matrices: e.g. the Pauli spin matrices.

You may be confused because both $\mathbb{R}$ and $\mathbb{C}$ are, in fact, Hilbert spaces in their own right. But, in terms of square-integrable functions and QM this is irrelevant.

In terms of Hilbert spaces, very roughly, you have:

$\mathbb{C}$ - not relevant to QM as a Hilbert space, but as a field of scalars

$\mathbb{C}^n$ - Hilbert spaces in which spin states reside

$\mathcal{H} \equiv L^2$ - Hilbert space in which wavefunctions reside.

 

[WWGD]

A simpler example of a function that is not a linear operator is the norm. In any Hilbert space you have an inner product and if you take the inner product of a vector with itself you get the norm (squared). So, you could have:Another, very important example, is the idea of a Linear Functional. Take any function $f \in H$. Define:

$L_f : H \rightarrow \mathbb{C}$

where $L_f(g) = \langle f, g \rangle$

In this case, therefore, $L_f$ maps each function to a complex number. Note that for every $f \in H$ you have a linear functional $L_f$. The set of all linear functionals is called the Dual Space ...

!

Just to add that Riesz representation theorem states that every element of the continuous dual can be written in this way for some fixed $f$, in case that may help the conversation .

 

[SeM]

That notation makes no sense, as $H$ is a set/space/Hilbert Space of functions. It's not a set of numbers. An "operator" is just a function that maps functions to functions (or vectors to vectors).

In finite-dimensional vector spaces, Linear Operators are represented by matrices. But, the set of square integrable functions is actually an infinite dimensional vector space. And, this makes this more complicated than the finite-dimensional case.

Note that finite dimensional Hilbert Spaces turn up in QM when you consider spin, for example. And,in these cases, the linear operators are indeed represented by matrices: e.g. the Pauli spin matrices.

You may be confused because both $\mathbb{R}$ and $\mathbb{C}$ are, in fact, Hilbert spaces in their own right. But, in terms of square-integrable functions and QM this is irrelevant.

In terms of Hilbert spaces, very roughly, you have:

$\mathbb{C}$ - not relevant to QM as a Hilbert space, but as a field of scalars

$\mathbb{C}^n$ - Hilbert spaces in which spin states reside

$\mathcal{H} \equiv L^2$ - Hilbert space in which wavefunctions reside.

Thanks again Perok, this was indeed very useful. I thought C and R where for wavefunctions, but indeed, they are for coordinates. So when one works with wavefunction and their transformation by operators, they are given in H, because H is not for coordinates, but for the infinity and continuity of wavefunctions and their derivatives, integrals and so one. This actually gave me a complete comprehension of the Hilbert space. It has therefore nothing to do with coordinates.Thanks, this was one of the best lectures I have had in a long time.

Cheers