# A Operator mapping in Hilbert space

#### George Jones

Staff Emeritus
Gold Member
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

#### 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?

L: \mathcal{H} \rightarrow \mathcal{H}

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

Homework Helper
Gold Member
2018 Award
PS: Being entirely frank, I need to clarify a thing. What does this really mean?

L: \mathcal{H} \rightarrow \mathcal{H}

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 noone 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:

P: \mathcal{H} \rightarrow \mathcal{O}

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

\int x dxdy = 1/2x^2 + xy

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

P: \mathcal{H} \rightarrow \mathcal{O}

?

Thanks!

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#### PeroK

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Gold Member
2018 Award
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

#### 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:

L : \mathcal{H} \rightarrow \mathcal{H}

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:

L : \mathcal{H} \rightarrow \mathcal{H}

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

L : \mathbb{R}, \mathbb{C} \rightarrow \mathbb{R}, \mathbb{C}

?

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

Thanks! Have a good weekend!

#### PeroK

Homework Helper
Gold Member
2018 Award
But why not use this alternative to confirm square integrable of a function?

L : \mathbb{R}, \mathbb{C} \rightarrow \mathbb{R}, \mathbb{C}

?

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.

SeM

#### WWGD

Gold Member
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 .

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SeM

#### 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

"Operator mapping in Hilbert space"

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