Creation Operator is not a densely defined operator....

In summary, the conversation discusses the creation operator in local quantum field theory and how it is defined as the adjoint of the annihilation operator. There is a mention of the operator not being densely defined, but it is possible to define it as a quadratic form. The use of delta functions and integrals is also brought up in relation to the creation operator. The conversation ends with a request for help and a response offering assistance.
  • #1
FreeBiscuits
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0
Hi everyone,

I am currently preparing myself for my Bachelor thesis in local quantum field theory. I was encouraged by my advisor to read the books of M. Reed and Simon because of my lag of functional analysis experience but I have quite often problems understand the “obvious” conclusions.

For example:

Why is the creation operator not a densely defined operator? And how do I proof that formally correctly? I am not asking for a step by step solution but I have absolutely no idea how to start.

Reed, Simon define the creation operator as:

$$(a^\dagger(p)(\psi))^{(n)}(k_1,...k_n) = \frac{1}{\sqrt{n}} \sum_{l=1}^n \delta(p-k_l) \psi^{(n-1)}(k_1,...,k_{l-1},k_k,...,k_n)$$

as the adjoint of the annihilation operator :

$$(a(p)\psi)^{(n)}(k_1,...k_n) = \sqrt{n+1} \psi^{(n+1)}(p,k_1,...k_n)$$

with $$\psi$$ a Schwartz function.

I really appreciate any hints.
 
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  • #2
Who told you the operator was not densely defined?
 
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  • #3
The book of Simon and Reed. (Fourier Analysis chapter X.7)
"The adjoint of the operator a(p) is not a densely defined operator since it is given formally by" the expression I posted before. There is no more explanation given. They merely mention that it is possible to define $$a^\dagger$$ as a quadratic form.
 
  • #4
I see (books google) that the chapter is about self adjointness.
Relativist fields need creation and annihilation operators. Not only creator operations.
 
  • #5
According to the formulas given by you, the annihilation operator is densely defiened, but the creation operator moves the states in the alleged domain out of the Hilbert space because of the delta function, hence is not defined on this domain. (This is probably what is really claimed in your quoted piece of text.) However, smeared creation operators are densely defined since smearing removes the delta function.
 
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  • #6
In other words, a creation "operator" is actually an operator-valued distribution. See Chapter 10 "Creation and Annihilation Operators" in the book "Quantum Field Theory II" by Zeidler.
 
  • #7
As Arnold remarked, the „trick” comes from the delta-Dirac which „explodes” for p equal to any of the k-s, therefore the equality means something only in terms of distributions, i.e. putting integrals and smearing test functions. Actually, the true operators are defined simply by omitting one of the arguments. These operators are defined on pages 208-209 of R-S, Vol.II. For those of you who do not have R-S and understand German, the notes attached by Prof. Keyl are gold. For those who cannot understand German, use the other notes by Prof. Luecke.
 

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  • #8
I thank all of you.
The attached notes are a great help. I really appreciate it.
 
  • #9
  • #10
I think i see what you are stuck on. I am not sure if my response will help you or complicate the issue so i am rather reluctant to post.

δ(p−kl)
δ is not a function of (p−kl) according to the usual mathematical definition of a function which requires a function to have a definite value for each point in its domain.
as above, the range of the integration should contain the origin
 

1. What is a creation operator?

A creation operator is a mathematical operator that creates a new element or object in a given system. It is often used in quantum mechanics to describe the creation of a particle or excitation in a quantum field.

2. Why is the creation operator not a densely defined operator?

A densely defined operator is one that has a domain that is dense in its range, meaning that for any element in the range, there is a sequence of elements in the domain that converges to it. The creation operator is not densely defined because it cannot create all possible elements in a given system, and therefore, its domain is not dense in its range.

3. What are the implications of the creation operator not being densely defined?

The lack of density in the domain of the creation operator means that it cannot create all possible states or elements in a given system. This can limit its usefulness in certain physical systems and can also affect the completeness and accuracy of mathematical models that use creation operators.

4. How is the creation operator used in quantum mechanics?

In quantum mechanics, the creation operator is used to describe the creation of a particle or excitation in a quantum field. It is often combined with the annihilation operator to create a complete description of the dynamics of a quantum system.

5. What other operators are related to the creation operator?

The creation operator is closely related to the annihilation operator, which removes an element from a given system. These two operators, along with other related operators, form a set known as the ladder operators, which are commonly used in quantum mechanics to describe the dynamics of quantum systems.

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