Do commutators hold in interacting theory?

• geoduck
In summary, the Fourier coefficients of an interacting field obey the commutation relations if the Fourier coefficients are taken at equal times.
geoduck
Do the Fourier coefficients of an interacting field obey the commutation relations? I think I was able to show once that they do if the Fourier coefficients are taken at equal times (the coefficients are time-dependent in the interacting theory), but my proof felt shaky.

In any case, does anyone know the answer for equal times, or unequal times?

I do not believe they hold in general because, for example, in QCD the interactions introduce:

\langle \bar{\psi}_i \psi_i \rangle = c

where c is some constant and we sum over flavour indeces. I don't see how standard commutation relations can hold with a non-zero VEV of the product of these fields.

The commutators for elementary field operators must hold in interacting theories otherwise the construction would be ill-defined. But for more complicated operator products which require regularization this need not be the case; it can happen that a quantization anomaly is introduced which spoils the classical Poisson brackets or trivial commutators.

A well-known example is the axial anomaly where the classically conserved charge Q5 fails to commute with the Hamiltonian after quantization & regularization, i.e. [H,Q5] ≠ 0

tom.stoer said:
The commutators for elementary field operators must hold in interacting theories otherwise the construction would be ill-defined. But for more complicated operator products which require regularization this need not be the case; it can happen that a quantization anomaly is introduced which spoils the classical Poisson brackets or trivial commutators.

A well-known example is the axial anomaly where the classically conserved charge Q5 fails to commute with the Hamiltonian after quantization & regularization, i.e. [H,Q5] ≠ 0

Are you saying something like [f(X),P]=i (df/dx) might not hold, where f(X) is a complicated operator product of X's, if f(X) requires regularization?

In QM I am not aware of any anomaly. I am not talking about QM but about QFT.

In QFT you don't have X and P as operators but field operators like $\phi(x)$, $\pi(y)$ and commutators like

$$[\phi(x),\pi(y)] = i\delta(x-y)$$

Let's make an example: the axial charge is constructed from fermionic field operators as

$$Q_5 = \int_{\mathbf{R}^3}d^3r\,\bar{\psi}(x)\,\gamma^0\, \gamma_5 \,\psi(x)$$

Applying the naive commutators before regularization one finds

$$[H,Q_5] = 0$$

which would correspond to a conserved axial current

$$\partial_\mu j^\mu_5 = 0$$

But due to gauge-invariant regularization preserving

$$\partial_\mu j^\mu = 0$$

one is forced to introduce a regulator violating the axial symmetry; therefore one finds

$$[H,Q_5] = \mathcal{A}$$

and therefore

$$\partial_\mu j^\mu_5 = \mathcal{a}$$

where A represents the anomaly (which is a term depending on the spatial dimension and topology).

The commutators of the fermionic Fourier components for $\psi(x)$ remain valid, but the commutators of the Fourier coefficients of $j^\mu_5(x)$ are modified. So what happens is that due to quantization and regularization the structure of a composite operator may change. The conclusion is that at least for composite operators the naive replacement of Poisson brackets with commutators and the naive application of Fourier transformation of field operators need not always be valid.

Don't the commutators e.g. in QED vanish during renormalization due to Z becoming infinite?

Which commutators? For the axial charge?

Starting with the anomaly equation

$$\partial_\mu j^\mu_5 = \mathcal{a}$$

one finds

$$i[H,Q_5] = \partial_0 Q_5 = \int_{\mathcal{M}^D} d^{D-1}x\,a_D(x)$$

for D-dimensional space-time and spatial manifold D (assuming reasonable boundary conditions or compact M). This expression does not vanish in general

Last edited:
tom.stoer said:
Which commutators? For the axial charge?

I think what is meant is that if the canonical commutation relations hold for the bare fields, then what about for the renormalized fields? The commutator should be 1/Z less, where Z is the field strength normalization, which goes to infinity.

This is worrisome because if you form bilinears out of the renormalized fields, then their commutators vanish, which ought to mean that there can be no microcausality?

(assuming reasonable boundary conditions or compact M).

I am always confused by this. If you have:

∫∫∫(div f)d3x

then by Stoke's theorem this is the flux of f over the surface, and if you mandate that fields vanish at the surface, then this term is zero.

But isn't it true for a scalar function g=g(x,y,z):

$$\int\int\int \partial_y g(x,y,z) dxdydz$$

that the above expression also vanishes? So requiring the integrand to be a 3-divergence is a little too strong: if it's merely a partial derivative then wouldn't it vanish?

So not only does ∫∫∫(div f)d3x vanish, but each term in the sum of div f vanishes:

$$\int\int\int \partial_x f_1(x,y,z) dxdydz=\int\int\int \partial_y f_2(x,y,z) dxdydz=\int\int\int \partial_z f_3(x,y,z) dxdydz=0$$

1. What is a commutator in the context of interacting theory?

A commutator is a mathematical operation that measures the difference between two quantities. In interacting theory, commutators are used to describe the relationship between different observables, such as position and momentum, and determine their compatibility.

2. How do commutators behave in interacting theory?

In general, commutators do not hold in interacting theory. This means that the order in which operations are performed can affect the final result, unlike in non-interacting theories where commutators always equal zero.

3. Why do commutators not hold in interacting theory?

In interacting theory, particles can interact and exchange energy, which leads to a breakdown of the commutator relationship. This is due to the uncertainty principle, which states that the more precisely we know one observable, the less precisely we can know another.

4. Are there any exceptions where commutators do hold in interacting theory?

Yes, there are certain special cases where commutators can hold in interacting theory. For example, in certain types of conformal field theories, commutators can still hold even in the presence of interactions.

5. How do we deal with the breakdown of commutators in interacting theory?

To deal with the breakdown of commutators in interacting theory, we use alternative mathematical approaches, such as Feynman diagrams and perturbation theory, to calculate the behavior of particles and their interactions. These methods take into account the non-commuting nature of observables and provide a framework for studying interacting systems.

Similar threads

• Quantum Physics
Replies
3
Views
853
• Quantum Physics
Replies
5
Views
756
• Quantum Physics
Replies
3
Views
1K
• Quantum Physics
Replies
18
Views
3K
• Quantum Physics
Replies
4
Views
1K
• Quantum Physics
Replies
6
Views
420
• Quantum Physics
Replies
24
Views
1K
• Quantum Physics
Replies
3
Views
3K
• Quantum Physics
Replies
13
Views
5K
• Quantum Physics
Replies
93
Views
4K