Is the delta in the commutation relations of QFT a dirac delta or a kronecker?

[silence11]

If it's a dirac delta doesn't it mean it's infinite when x=y? Or is it a sort of kronecker where it's equal to one but the indices x and y are continuous? I'm confused.

 

[strangerep]

Both the Dirac and Kronecker deltas can occur in QFT.

Post a specific example of a CR you don't understand...

 

[haushofer]

If it's a dirac delta doesn't it mean it's infinite when x=y? Or is it a sort of kronecker where it's equal to one but the indices x and y are continuous? I'm confused.

In QFT you can have commutation relations like

$$[\phi^i (x), \pi_j (y)] = \delta^_i \delta^{(D)}(x-y)$$

where phi and pi are conjugate fields, of which you can have, say, N, and D is the dimension of spacetime.

The first delta states that these fields don't commutate if they are different components. That's a Kronecker delta. The second is a distribution, and is there because the fields are really distributions. These kind of relations only make sense if you integrate them with a test function. Otherwise you would naively say that the commutator blows up if x=y.

You can compare it with the commutators in QM; those only make sense if you apply them to a wave function.

 

[geoduck]

The second is a distribution, and is there because the fields are really distributions. These kind of relations only make sense if you integrate them with a test function. Otherwise you would naively say that the commutator blows up if x=y.

You can compare it with the commutators in QM; those only make sense if you apply them to a wave function.

Are all operators distributions then, and not just commutators? In QM, X and P are distributions. In QFT, creation/annihilation operators and fields are distributions?

 

[lugita15]

In QM, X and P are distributions.

Could you elaborate on this?

 

[geoduck]

Could you elaborate on this?

I was just asking haushofer a question. He said fields in QFT are distributions, and I was wondering why. He said you integrate them over test functions, so I was wondering if he meant that any linear operator is a distribution in QM, e.g.,

<x|P|x'> takes ψ(x') and maps it to:

<x|P|ψ>=-i d/dx[ψ(x)]

I have no idea what he meant, so I just wanted to learn :)

 

[tom.stoer]

In QFT fields are operator-valued distributions where (in position space) x is a "continuous index". The delta-distribution is the identity-operator.

In QM the operators are X (and P, ...) - no fields - and there is only a discrete index (i=1,2,3, ... for dimensions etc.) and therefore you get a Kronecker delta as identity-operator. The delta-function in QM is not an operator but a matrix element.