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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?

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3 Answers 3

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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.

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what is the physical interpretation of the smearing functions? –  diffeomorphism Jul 25 '12 at 3:50
    
@diffeomorphism: The smearing functions aren't physical--- they just tell you what the operators with finite fluctuations are. The smeared fields have finite square expected value. If you use a lattice, the lattice will smear the fields for you, so if you prefer, just put the fields on a lattice so that you don't get divergence in square fluctuations. Smearing allows you to define free fields with no problems. In interacting theories, you need to take the limit again, with more difficulties, so the lattice (or some other regularization) is necessary once again, for renormalization. –  Ron Maimon Jul 25 '12 at 4:25
    
+1. is it not "distribution-valued operators", though? –  Emilio Pisanty Jul 25 '12 at 17:25
    
@EmilioPisanty: "operator valued distribution" is the right term--- when you smear the fields, they become operators, so the smearing takes the singular distributions to nonsingular operators (hence "operator valued", the smearing takes test functions on R^n to normal ordinary nonsingular operators in a linear way). The operators don't have values, but I understand your intuition--- the field values which are typical in a path integral are distributional in nature. This is true, but it is not the term the rigorous QFT people use. –  Ron Maimon Jul 25 '12 at 18:17
    
Hmmm. My original intuition is that the "function" $\langle \hat{\phi}(x)\rangle$ is a distribution. Compared with what you say it depends if I'm seeing it right on the order of "smearing" and "expectation-valuing" - and you seem to be doing both orders. If the operations commute, surely the terminology ought to ;). –  Emilio Pisanty Jul 25 '12 at 21:50

EDITED: @ronmaimon comment is right, this answer does not apply to this question. The below applies if you were asking a different question, how does a single mode of the field have finite noise. But the actual question actually refers to a situation in which an infinite number of modes are present and the fluctuations in a measurement on that field is infinite.

EDIT CONTINUED: the hand-waving resolution to the infinite fluctuation issue is ultimately resolved by roll-off of everything at high frequencies. The modes in the original question must go to infinite energy, if we bound the energy we only have a finite number of modes in a finite region. But higher energy modes have temporal variation faster and faster $f=E/h$, and a spatial variation which gets faster and faster. Any realistic measurement of field strength involves averaging the field over some finite extent. Any such measuring tool then has an effective cut-off of frequency/energy modes that it couples to. So the total fluctuation measured is finite because of this cutoff. It does stand to reason that the smaller we make our measurement the higher frequency modes we include, and the higher the total fluctuations included in our measurement. But until we get to an infinitely small measurement probe (much much MUCH smaller than a quark), we will always have finite fluctuations from any real measuring device.

OLD ANSWER TO THE WRONG QUESTION: When you write a state as a sum over a bunch of other states, you must normalize the sum. So if $\langle \psi_k \rangle$ are the normalized states you are creating your sum from, the state you create is $$| \phi \rangle = \sum_k \alpha_k | \psi_k\rangle$$ where $$1 = \sum_k \alpha_k^*\alpha_k$$ is the normalization condition.

But this means when we are gathering the uncertainty of about about $\hbar$ from each mode, the sum winds up being finite because we are only adding about $\alpha_k^*\alpha_k\hbar$ from each mode.

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sounds almost like a reasonable answer. However, i am assuming we are already in the vacuum, so my state is already normalized. There is nothing inside the vacuum state that can be split in sums like you propose –  diffeomorphism Jul 25 '12 at 3:16
1  
No, this is not so. The fluctuation is infinite. –  Ron Maimon Jul 25 '12 at 3:41
    
@RonMaimon I agree my original answer is not the answer to this question. (I was answering the question of why are the fluctuations in a single mode of the radiation field always finite). –  mwengler Jul 25 '12 at 14:13

I will give a preliminar take on the problem, by no means intended as a complete answer

the uncertainty principle relationship is true for the function space of square-integrable functions where fourier analysis holds, see this for a reference. So, the principle was from the beginning, meant for the net field

Extrapolating the principle to the individual modes is an ill-defined operation that has become nonetheless the standard for quantum fields, but it looks pretty unconvincing, and unmotivated, at least on the surface

individual modes of a free non-interacting fields (for which we accept a fourier transform) are all plane waves, hence they are all state of well-defined momenta and infinite uncertainty in position. Hence it is of no surprise that a naive attempt to express the combined uncertainties lead to a crash. Of course, second quantization is just an ansatz (and one that appears to work well, or so goes the saying) and it does not need any justification except their end results

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