# QFT general properties operators

1. Jan 15, 2010

### matt8282

Hi all,

I have quite basic questions about the general properties of operators in quantum field theory. When quantizing the free scalar field, for instance, you promote the classical fields to operators and impose suitable commutation relations (canonical quantization). In momentum space the harmonic oscillators decouple and you can find the spectrum by introducing a/a(dagger) operators. All of this is done without specifying the explicit form of these operators. Concering this I have some questions:

1) Without knowing the explicit form of these operators, which restrictions (physical and mathematical) do they underlie? How would mathematicians define these operators in a strict language?

2) Is it just a definition that operators like the field operator phi or a, a(dagger) only act on states of the fock space and noting else? For example when doing fourier transformations you always commute phases exp(ipx) or general functions of x and p with these operators. How do I know that the field operators do not act on these terms? As far as I know it's just definition. What is the argument that the field operators cannot contain derivatives (like in QM the momentum space representation of p) d/dx and this way would act on exp(ipx)?

3) How does the nabla operator act on field operators?

As you can see I have some problems with the mathematical construct of these operators. At the moment I think that it's just a definition that my operators only act on fock space states and nothing else so I only have to take care when commuting operators. But I am not really sure.

Matt

2. Jan 15, 2010

### blechman

the operators $a,a^\dagger$ are perfectly well defined and have a specified form in terms of the field and it's conjugate momentum. I don't understand why you say their form is "unspecified" They are in exact analogy to the SHO with (x,p) replaced by $\phi,\pi$.

In terms of the Fock space language, these are operators that when acting on the Fock space vacuum generate plane wave excitations. Again, they're perfectly well defined by specifying their action on states, just like any other linear operator (a matrix is specified by specifying all if its components).

3. Jan 16, 2010

### matt8282

You are absolutely right, they are well-definied in terms of $$\phi$$ and the momentum conjugate $$\pi$$. What I meant with "there form is unspecified" was that they are functions of $$\phi , \pi$$ and then again depend on the properties of the field operators.

What about my questions 2) and 3)? And what is the argument that my operators do not act on c-numbers like exp(ipx)? Is it just a definition or is it clear (if so, why?)?

Matt

4. Jan 16, 2010

### blechman

I'm not entirely sure I understand your confusion. $\phi(x)$ is a linear operator by the axioms of "second quantized quantum mechanics", and we chose to expand this operator in terms of plane waves, which we can always do. We do this because it provides a nice solution to the Klein-Gordon equation. But since the $\phi$ are operators, their "Fourier components" are also operators. These are the $a,a^\dagger$. Nothing wrong with that.

Operators act on the states. That is, if you like, the definition of an "operator". The fact that they are LINEAR operators means that they don't do anything to c-number coefficients. Again, that's what the "linear" means:

$$T(a\vec{x}+b\vec{y})=aT(\vec{x})+bT(\vec{y})$$

where a,b are c-numbers, and $\vec{x},\vec{y}$ are the "vectors" (in the case of QFT, Fock space elements).

Again, that they are "linear operators" is, if you wish, an axiom of quantum mechanics.

That the derivative on the field now corresponds to a factor of momentum multiplying the creation/annihilation operators is a standard trick of Fourier-transforming:

$$\nabla\phi(x)=\int\frac{d^3k}{(2\pi)^3}\frac{1}{\sqrt{2E_{\vec{k}}}} (\nabla e^{-ikx}a_{\vec{k}})~+~{\rm c.c}\quad=\quad\int\frac{d^3k}{(2\pi)^3}\frac{1}{\sqrt{2E_{\vec{k}}}}(-i\vec{k} e^{-ikx}a_{\vec{k}})~+~{\rm c.c}$$

The fact that these are operators does not change anything.

Hope that helps!

5. Jan 16, 2010

### matt8282

Thank you very much! You really helped me solving my uncertainty.