# Vacuum fluctuations of quantum scalar field

Consider a free scalar quantum field

$$H = \int d^3 x \left( \, \Pi(x)^2+(\nabla\phi(x))^2 \right).$$

Introducing the creation and annihilation operators we find the "vacuum catastrophe"

$$H = \int d^3p \, \big(\omega_p a^\dagger(p)a(p) +\frac{1}{2}[a(p),a^\dagger(p)] \big),$$

a diverging energy of the vacuum ground state. No problem, this is just an unobservable constant energy shift, as is argued for example in Peskin Schröder and claimed all over the place when discussing this issue.

However, what is glossed over in this explanation is that the variance of the field at any given point is also infinite:

$$\Delta\phi^2=\langle0|\phi^2|0\rangle=\langle0| \int d^3p \int d^3q \frac{1}{2\sqrt{\omega_q \omega_p}}(a(p)e^{ipx}+a^\dagger(p)e^{-ipx})(a(q)e^{iqx}+a(q)e^{-iqx})|0\rangle \\ =\int d^3p \frac{1}{2\omega_p}=\infty$$

Since the quantum field countains all possible modes, and every mode has groundstate fluctuations which contribute to the field at a given point, the total field fluctuations diverge.

Clearly, the variance of the field is observable by just measuring the field. We can ignore an infinite energy shift but not infinite values of the field can we?

• Dec 11 '20 at 13:58
• also relevant: physics.stackexchange.com/q/192403/84967 Dec 11 '20 at 14:10
• I feel like the fact that $\phi$ is a distribution does not answer the question. The same reasoning should apply to a more regular model, like a linear chain in the thermodynamic limit. The issue is that we have an infinite number of modes and all of them have ground state fluctuations. This might be a different question though. Dec 11 '20 at 14:20

Quantum fields are not operator-valued functions over spacetime, but operator-valued distributions (aka generalized functions). These are linear maps from test functions $$f(x)$$ over spacetime to operators acting on the Hilbert space. Naively, you can say $$\phi(f) = \int d^4 x \; f(x) \phi(x)$$ for $$\phi(x)$$ the quantum field at $$x$$, but really $$\phi(x)$$ doesn't exist mathematically: only $$\phi(f)$$ does.

Instead of computing a quantity that doesn't exist in the QFT $$\left< 0 \right| \phi(x)^2 \left| 0 \right>$$ you could compute the quantity that does exist $$\left< 0 \right| \phi(f)^2 \left| 0 \right>.$$

That will give you a finite answer for any test function $$f$$. Now you can start changing $$f$$ such that it approaches the delta function concentrated at $$x$$. In this limit the value of $$\left< 0 \right| \phi(f)^2 \left| 0 \right>$$ will diverge, which is expected.

To compute this quantity note that $$\left< 0 \right| \phi(f)^2 \left| 0 \right> = \int d^4 x \int d^4 y f(x) f(y) \Delta(x - y).$$

The singularity in the propagator at $$|x - y| = 0$$ is integrable, so the integral above exists and is finite.

• I accept this answer because it gives more details about the procedure. Dec 13 '20 at 19:48

Actually, in Quantum Field Theory one should only calculate something that makes sense. Citing Anthony Zee (QFT in a nutshell): Asking wrong questions give wrong answers.

So what is wrong with the expectation value of the variance of the scalar quantum field? We will not be able to measure it. In order to measure $$\langle 0|\phi^2|0\rangle$$ at a single point we would need an infinite amount of energy, so it is impossible. However, if the field is smeared out we would be able to measure it since in that case the answer of the expectation value would be finite and the amount of energy needed to measure it would be finite.

The underlying problem, however, is the assumption that QFT is valid up to extremely small distances (below Planck scale). QFT is very probably an emerging theory from a more fundamental theory at extreme small scale which could be string theory or some other still not known theory. Therefore it has limited range of validity. QFT is not valid to arbitrary small distances. QFT is actually like the Navier-Stokes equations that describe hydrodynamics very well on a macroscopic scale by a set of differential equations, a mathematics that is supposed to work at any scale, but we all know it will not be longer true on a molecular scale. Another very good example is condensed matter physics where the techniques of QFT are used and give very reasonable results as long as its application does not go below the atomic scale.

However, a frequent approach in QFT that is applied is just to forget about this limited range of validity. Therefore some calculations neglecting the limited range of validity give infinite results. The amazing thing in QFT is, however, that almost all of these problems can be cured. The formalism used for this is called renormalization.

• Could you please elaborate on the smeared out field operator? I tried this approach and still got infinity. Dec 11 '20 at 15:10
• @curio take a test function over spacetime $f(x)$ that decays rapidly away from a certain point $x = x_0$. Then compute $\left< 0 \right| \phi(f)^2 \left| 0 \right>$. Here $\phi(f) = \int d^4 x f(x) \phi(x)$. You should obtain a finite result that depends on the form of $f$. In the limit where $f$ approaches the delta function, the value diverges. That is ok, because the delta function is a generalized function (aka distribution), not a test function, so we shouldn't expect the limit to exist. Dec 13 '20 at 6:44