[Physics] Is zero-point energy real

casimir-effectcosmological-constantpath-integralquantum mechanicsquantum-field-theory

In the context of canonical quantization, the ground state/zero-point energy of a harmonic oscillator is
$$
E_0 = \frac{1}{2}\hbar \omega.
$$
The vacuum is alleged to be permeated with this zero-point energy of quantum fields, hence causing the exorbitantly large cosmological constant issue.

However, the zero-point energy can be normal ordered away (Wick ordering). And path integral formulation automatically takes care of normal ordering. In other words, path integral formulation (and normal ordering canonical quantization) doesn't suffer from the zero-point energy issue.

The Casimir effect is usually cited as an evidence of zero-point energy. Nevertheless, a 2005 paper (https://arxiv.org/abs/hep-th/0503158v1) states that "Casimir forces can be computed without reference to zero-point energies…The Casimir force is simply the (relativistic, retarded) van der Waals force between the metal plates".

So is zero-point energy a spurious artifact of canonical quantization?

Note that in the absence of zero-point energy, the cosmological constant problem still stands because of the vacuum energy shift resulted from spontaneous electroweak symmetry breaking.

Best Answer

This is the data that I use to convince my students. It is the result of a measurement of the velocity of argon atoms. The method is inelastic neutron diffraction, no need to go into details, but one can read those in Fradkin et al, 1994.

In this figure, $mv^2/(2k_B)$ is plotted as a function of temperature. At higher temperature this approaches $3/2\ T$ according to classical physics and equipartition. At low temperature the derivative decreases, in agreement with the third law of thermodynamics that $c_p$ must go to zero.

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[Physics] Green’s function in path integral approach (QFT)

Good question; I remember spending hours trying to understand this when I first learned QFT. Let's address your two main points in turn. First, you say

I don't understand how rhyme these two different pictures.

Let's outline how to connect the two pictures in steps. It's a good exercise to try and work through all of the gory details yourself, so I encourage you to try!

For each admissible classical field configuration $\varphi:\mathbb R^3\to \mathbb R$, let $|\varphi,t\rangle$ denote a field configuration eigenstate at time $t$. Namely, \begin{align} \hat \phi(t,\mathbf x)|\varphi,t\rangle = \varphi( \mathbf x)|\varphi,t\rangle. \end{align} Take special note of the fact that $\hat\phi$ and $\varphi$ are different. The former is a operator valued distribution defined on spacetime, while the latter is a classical field configuration defined on space only.
Show that given admissible classical field configurations $\varphi_a,\varphi_b:\mathbb R^3\to \mathbb R$, there is a simple functional integral expression for the time-ordered expectation value from $|\varphi_a, t_a\rangle$ to $|\varphi_b, t_b\rangle$ of the product of a finite sequence of field operators: \begin{align} \langle \varphi_b, t_b| T\big[\hat \phi(x_1)\cdots \hat \phi(x_n)\big]&|\varphi_a, t_a\rangle \\ &= \int\limits_{\phi(t_a,\mathbf x) = \varphi_a(\mathbf x)}^{\phi(t_b,\mathbf x) = \varphi_b(\mathbf x)} \mathscr D\phi \,\phi(x_1)\cdots \phi(x_n)\,e^{iS_{t_a,t_b}[\phi]} \tag{$\star$} \end{align} where we have defined \begin{align} S_{t_a,t_b}[\phi] = \int_{t_a}^{t_b}dt\int d^3\mathbf x \,\mathscr L_\phi(t) \end{align} and $\mathscr L_\phi$ is the Lagrangian density of the theory.
Show that the expectation value on the left hand side of $(\star)$ can be used to compute a corresponding vacuum expectation value (vev); \begin{align} \lim_{t\to(1-i\epsilon)\infty} \frac{\langle \varphi_b, t| T\big[\hat \phi(x_1)\cdots \hat \phi(x_n)\big]|\varphi_a, -t\rangle}{\langle \varphi_b, t |\varphi_a, -t\rangle} &= \langle 0|T\big[\hat \phi(x_1)\cdots \hat \phi(x_n)\big]|0\rangle \end{align} where $\epsilon$ is a "positive infinitesimal" (namely you take the $\epsilon\to 0$ limit at the end). This is often called the $i\epsilon$ prescription; notice it's basically a clever trick for projecting out the ground state from a general expectation value.
Notice that the functional integration on the right hand side of $(\star)$ can be written as \begin{align} \int\limits_{\phi(t_a,\mathbf x) = \varphi_a(\mathbf x)}^{\phi(t_b,\mathbf x) = \varphi_b(\mathbf x)} \mathscr D\phi \,\phi(x_1)\cdots \phi(x_n)\,e^{iS_{t_a,t_b}[\phi] }=\left(\frac{1}{i}\right)^n \frac{\delta^n Z_{t_a, \varphi_a, t_b, \varphi_b}[J]}{\delta J(x_1)\cdots \delta J(x_n)}\bigg|_{J=0} \end{align} where we have defined \begin{align} Z_{t_a, \varphi_a, t_b, \varphi_b}[J] = \int\limits_{\phi(t_a,\mathbf x) = \varphi_a(\mathbf x)}^{\phi(t_b,\mathbf x) = \varphi_b(\mathbf x)} \mathscr D\phi \,e^{iS_{t_a,t_b}[\phi]+ i\int_{t_a}^{t_b}\int d^3\mathbf x\,J(x)\phi(x)} \end{align}
Combine steps 2-4 to show that if we define \begin{align} W[J] = \lim_{t\to(1-i\epsilon)\infty}\frac{Z_{t_a, \varphi_a, t_b, \varphi_b}[J]}{Z_{t_a, \varphi_a, t_b, \varphi_b}[0]}, \end{align} then we obtain our desired expression which gives vacuum expectation values in terms of path integrals: \begin{align} \langle 0|T\big[\hat \phi(x_1)\cdots \hat \phi(x_n)\big]|0\rangle &= \left(\frac{1}{i}\right)^n \frac{\delta^n W[J]}{\delta J(x_1)\cdots \delta J(x_n)}\bigg|_{J=0} \end{align}

Notice that \begin{align} G^{(n)}(x_1, \dots, x_n) = \langle 0|T\big[\hat \phi(x_1)\cdots \hat \phi(x_n)\big]|0\rangle \end{align} is just a suggestive definition which makes us think of Green's functions. It's suggestive because, for example, $G^{(2)}(x_1, x_2)$, the so called "two-point function," is the Green's function for the corresponding classical field theory.

Second, you ask

Do these Feynman diagram for the two different approaches somehow represent the same scattering amplitude?

The LSZ reduction formula is the answer to the question of how vevs, or equivalently Green's functions, are related to the $S$-matrix and scattering amplitudes, and above we have argued how the canonical formalism (which is formulated in terms of vevs) is related to the functional integral formalism, so we have found how the functional integral formalism allows us to compute the $S$-matrix. In practice, it's true, that you don't see people explicitly using the LSZ reduction formula, but that's because although it conceptually underlies the connection between Green's functions and the $S$-matrix, in practice people have already used LSZ to justify codified rules, namely Feynman rules, that allow one to go directly from Feynman diagrams (which simply represent terms in the perturbative expansions of Feynman integrals) to scattering amplitudes.

Vacuum Catastrophe – Why Isn’t a Large Length (1mm) Cutoff Plausible?

There is a strong argument that zero point energy in field (at least in free EM field) is really just an artifact of dubious glorification of a particular Hamiltonian (the usual one carrying zero point energy terms $\hbar \omega_m/2$) into EM field energy.

Most(all?) measurable things (frequencies of transitions...) are predicted by the theory the same if instead of the problematic Hamiltonian

$$ \sum_{m} \hbar\omega_m(a_m^+a_m + \frac{1}{2}) \tag{*} $$ with the zero point terms, the normally ordered Hamiltonian $$ \sum_{m} \hbar \omega_m a_m^+a_m\tag{**} $$ is used. In fact if we want the quantum energy to correspond to classical Poynting energy in some limit of strong coherent states (a quantum EM field described well by classical light theory) , we can't define quantum energy by $(*)$, because there is no zero-point energy in e.g. classical electrostatic field or plane wave.

Already in classical mechanics we encounter many Hamiltonians that differ in form and value, but are equivalent in their equations of motion. But despite that infinite plethora of Hamiltonians for any given system, we use only one expression to define standard energy, often sum of kinetic and some standard form of potential energy. It seems natural to do similar in QT of field, i.e. we may work with many Hamiltonian operators with different constant shifts, but we should settle on some unique definition of energy in relation to rest of physics. Using (**) seems much better than (*) for this.

Trying to explain the value of the cosmological term via zero point energy of field is like the old idea of explaining mass of electron in terms of electromagnetic mass of some interacting system of charged elements - originally, a cool new idea, but after many trials it didn't really work out, the calculated value was either way too big (charge concentrated to a ball of size limited by experiments to be less than 1e-18 m) or zero (if electron has no parts), and we got back to regarding the electron mass as independent constant of nature, having nothing to do with EM mass.

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[Physics] Green’s function in path integral approach (QFT)

Vacuum Catastrophe – Why Isn’t a Large Length (1mm) Cutoff Plausible?

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