uncertainty of fields with many harmonic modes

In most basic level introduction to the quantum harmonic oscillator formulation of fields, it is assumed that the commuting variables for the fields $p_m$, $q_m$ are

$$ \lbrack p_m , q_n \rbrack = \delta_{m n} i \hbar $$

which seem to imply that each individual mode holds an uncertainty relation like $ \Delta p_m \Delta q_m \ge \hbar $

now, uncertainties of field values with many modes must be expressed like (assuming the vacuum state, where $\langle E \rangle = \langle E_k \rangle = 0$):

$$ \langle E^2 \rangle = \langle \psi | ( \sum_k{ E_k } )^2 | \psi \rangle = \sum_k{ \langle \psi | E_k^2 | \psi \rangle } $$

but since each mode has some uncertainty in vacuum, it seems to imply that the uncertainty of the net field is infinite, which clearly does not make any sense

Any idea where my assumptions are going wrong?

Answer

The fluctuation in a field at a point is infinite in any field theory, this is because of the reason you state. This is why you need to smear the field over a region with a test function for it to have finite fluctuation, and the reason that the fields are characterized as operator valued distributions.

If you look at the expected value of the square of the field at a point, you consider the point split regulated version:

$$ \langle \phi(x)\phi(0)\rangle = G(x)$$

and take the limit $x\rightarrow 0 $. This is clearly infinite, since G(x) goes as $1\over x^{d-2}$ or as a log in 2d. It is only finite in 0+1 dimensions (quantum mechanics). If you smear the field and look at the square of the smeared operator, you get

$$ \langle \int f(x) \phi(x) \int f(y)\phi(y)\rangle = \int f(x)f(y) G(x-y) d^dx d^dy $$

This is completely finite, since the G(x-y) singularity is always softer than the volume. So free fields always produce well defined after smearing by test functions.

This was analyzed by Bohr and Rosenfeld (for the electromagnetic field) in the early 1930s, at the beginning of quantum field theory.