Field operators - how do they work?

In summary, the quantization of a classical field (such as electric field) involves taking the Fourier transform, multiplying by annihilation and creation operators for a given wavelength, and taking the IFT to return to time space. These operators act upon an oscillator wavefunction to raise or lower the energy state.
  • #1
erkokite
39
0
It seems to me that in the quantization of a classical field, one first takes the Fourier transform of the field to put it in frequency space:
[tex]F \left(X, \omega \right) = \int f(X,t)e^\left(-i \omega t\right)[/tex]

then multiply by the annihilation and creation operators for a given wavelength:

[tex] F \left(X, \omega \right) \sqrt{m \omega / \left(2 \hbar \right)} \left( x + i / \left(m \omega \right) p \right) [/tex]
[tex] F \left(X, \omega \right) \sqrt{m \omega / \left(2 \hbar \right)} \left( x - i / \left(m \omega \right) p \right) [/tex]

Then take the IFT of these to return to time space, which would yield a creation field operator and an annihilation field operator respectively. Note that I used X as a vector and x as an operator.

I understand that these objects increase and decrease the number of particles in the system respectively. But what do they act upon and what do they return when applied?
In single particle QM, these operators act upon an oscillator wavefunction to raise or lower the energy state.

But in QFT, I am guessing that they act upon the Hilbert space ( + time) and return the creation/annihilation operators of a particle of a field (ex. a photon in a Maxwell field), which can be used to define the Hamiltonian of the particle, and the resultant Schrodinger equation, which of course can be used to define the wavefunction of the particle. Ex. when passed a vector value for X, and a scalar value for t, they will return the ladder operators for a single particle state about that point on the Hilbert space.

Is this understanding correct? Please correct any of my misunderstandings (I am sure that there are many). I am an engineer, not a physicist. Please understand and take pity ;).

Many thanks.
 
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  • #2
Alright- it seems that I found one of my answers, thus nullifying some of my earlier understanding, and creating more questions. Creation and annihilation operators act upon a Fock space, which I take it is some abstract mathematical object that engineers don't learn about out. My earlier understanding of a Hilbert space is also, as it seems, incorrect (it is apparently an n-dimensional real euclidean vector space, which is easy enough). To make things even more confusing to me, it seems that a Fock space is a type of Hilbert space (which does make mathematical sense, but confuses me on its physical meaning).
 
  • #3
erkokite said:
My earlier understanding of a Hilbert space is also, as it seems, incorrect (it is apparently an n-dimensional real euclidean vector space, which is easy enough).

A Hilbert space is an infinite dimensional vector space, where an inner product is defined, and the space is complete (i.e. if an infinite sequence approaches a limit, then the limit is in the space). As an example, think of the space of functions where an inner product is defined (e.g. as an integral over an interval)
 
  • #4
A Hilbert space is simply a (real or complex) vector space that has an inner product and is complete w.r.t. the inner product. It can be finite dimensional, countably infinite dimensional, or even uncountably infinite dimensional.

Every n-dimensional vector space, when given an inner product is a Hilbert space.

The physically interesting Hilbert spaces are usually countably-infinite-dimensional.
 
  • #5
Hurkyl said:
It can be finite dimensional, countably infinite dimensional, or even uncountably infinite dimensional.

I always thought it had to be infinite dimensional - Wolfram Mathworld says it can be finite dimensional. Looks like you're right.
 
  • #6
Hurkyl said:
A Hilbert space is simply a (real or complex) vector space that has an inner product and is complete w.r.t. the inner product. It can be finite dimensional, countably infinite dimensional, or even uncountably infinite dimensional.

Every n-dimensional vector space, when given an inner product is a Hilbert space.

The physically interesting Hilbert spaces are usually countably-infinite-dimensional.

Ah... ok. That makes sense to me. Thanks for the clear up. I suppose I should spend more time reading before posting. I'm still lost on my QFT question though. Thoughts?
 
  • #7
In standard QM, you have one Hilbert space for 1-particle wave functions, another Hilbert space for 2-particle wave functions, yet another Hilbert space for 3-particle wave functions, etc. ... In QFT, the Hilbert space is a direct sum of all these n-particle Hilbert spaces, including a trivial space of 0-particle wave functions (the vacuum). Such an enlarged Hilbert space allows mixtures of states with different numbers of particles, which, in turn, allows you to describe processes in which the number of particles changes.
Does it help?
 

1. How do field operators work?

Field operators are mathematical tools used in quantum field theory to describe the behavior of physical fields. They work by assigning operators to each point in space and time, which act on the field at that point. These operators are then used to calculate the field's behavior and interactions with other fields.

2. What is the purpose of field operators?

The main purpose of field operators is to provide a mathematical framework for studying and understanding the behavior of quantum fields. They allow scientists to make predictions and calculations about the properties and interactions of these fields, which play a fundamental role in our understanding of the universe.

3. How do field operators differ from other mathematical tools used in physics?

Unlike other mathematical tools, such as differential equations or matrices, field operators are specifically designed for use in quantum field theory. They take into account the principles of quantum mechanics and special relativity, making them better suited for describing the behavior of subatomic particles and their interactions.

4. Are field operators used in other branches of physics?

Yes, field operators are not limited to quantum field theory and are also used in other branches of physics, such as condensed matter physics and statistical mechanics. In these fields, they are used to describe the behavior of complex systems and materials, rather than subatomic particles.

5. How do scientists use field operators in their research?

Scientists use field operators to make predictions and calculations about the behavior and interactions of quantum fields. These calculations are then compared to experimental data to test the validity of theories and to gain a deeper understanding of the fundamental laws of nature.

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