Editing Zero-point energy (section)

== Quantum field theory ==
{{Quantum field theory}}
{{Main|Vacuum expectation value|Vacuum energy|Vacuum state}}

In quantum field theory (QFT), the fabric of "empty" space is visualized as consisting of [[field (physics)|fields]], with the field at every point in space and time being a quantum harmonic oscillator, with neighboring oscillators interacting with each other. According to QFT the universe is made up of matter fields whose [[Quantum|quanta]] are fermions (e.g. [[electron]]s and quarks), force fields whose quanta are bosons (i.e. photons and gluons) and a Higgs field whose quantum is the [[Higgs boson]]. The matter and force fields have zero-point energy.{{sfnp|Milonni|1994|p=35}} A related term is ''zero-point field'' (ZPF), which is the lowest energy state of a particular field.<ref>
{{cite book
 |last=Gribbin |first=J. R.
 |editor-last=Gribbin |editor-first=M.
 |year=1998
 |title=Q is for Quantum: An Encyclopedia of Particle Physics
 |publisher=[[Touchstone Books]]
 |isbn=978-0-684-86315-3|oclc= 869069919
 |bibcode=1999qqep.book.....G
}}</ref> The vacuum can be viewed not as empty space, but as the combination of all zero-point fields.

In QFT the zero-point energy of the vacuum state is called the vacuum energy and the average expectation value of the Hamiltonian is called the vacuum expectation value (also called condensate or simply VEV). The QED vacuum is a part of the vacuum state which specifically deals with quantum electrodynamics (e.g. electromagnetic interactions between photons, electrons and the vacuum) and the QCD vacuum deals with [[quantum chromodynamics]] (e.g. [[color charge]] interactions between quarks, gluons and the vacuum). Recent experiments advocate the idea that particles themselves can be thought of as excited states of the underlying [[quantum vacuum]], and that all properties of matter are merely vacuum fluctuations arising from interactions with the zero-point field.{{sfnp|Battersby|2008}}

Each point in space makes a contribution of {{math|''E'' {{=}} {{sfrac|''ħω''|2}}}}, resulting in a calculation of infinite zero-point energy in any finite volume; this is one reason [[renormalization]] is needed to make sense of quantum field theories. In [[physical cosmology|cosmology]], the vacuum energy is one possible explanation for the cosmological constant{{sfnp|Rugh|Zinkernagel|2002}} and the source of dark energy.<ref name="Dark Energy May Be Vacuum"/>{{sfnp|Wall|2014}}

Scientists are not in agreement about how much energy is contained in the vacuum. Quantum mechanics requires the energy to be large as [[Paul Dirac]] claimed it is, like a [[Dirac sea|sea of energy]]. Other scientists specializing in [[General Relativity]] require the energy to be small enough for [[curvature of space]] to agree with observed [[astronomy]]. The Heisenberg uncertainty principle allows the energy to be as large as needed to promote quantum actions for a brief moment of time, even if the average energy is small enough to satisfy relativity and flat space. To cope with disagreements, the vacuum energy is described as a [[virtual energy]] [[potential]] of positive and negative energy.{{sfnp|Peskin|Schroeder|1995|pp=786–791}}

In quantum [[perturbation theory]], it is sometimes said that the contribution of [[one-loop]] and multi-loop [[Feynman diagram]]s to [[elementary particle]] [[propagator]]s are the contribution of [[quantum fluctuation|vacuum fluctuation]]s, or the zero-point energy to the particle [[mass]]es.

=== Quantum electrodynamic vacuum ===
{{Main|QED vacuum}}

The oldest and best known quantized force field is the [[electromagnetic field]]. [[Maxwell's equations]] have been superseded by quantum electrodynamics (QED). By considering the zero-point energy that arises from QED it is possible to gain a characteristic understanding of zero-point energy that arises not just through electromagnetic interactions but in all [[quantum field theories]].

==== Redefining the zero of energy ====
In the quantum theory of the electromagnetic field, classical wave amplitudes {{mvar|α}} and {{math|''α''*}} are replaced by operators {{mvar|a}} and {{math|''a''<sup>†</sup>}} that satisfy:
<math display="block">\left[a,a^\dagger\right] = 1</math>

The classical quantity {{math|{{abs|''α''}}<sup>2</sup>}} appearing in the classical expression for the energy of a field mode is replaced in quantum theory by the photon number operator {{math|''a''<sup>†</sup>''a''}}. The fact that:
<math display="block">\left[a,a^\dagger a\right] \ne 1</math>
implies that quantum theory does not allow states of the radiation field for which the photon number and a field amplitude can be precisely defined, i.e., we cannot have simultaneous eigenstates for {{math|''a''<sup>†</sup>''a''}} and {{mvar|a}}. The reconciliation of wave and particle attributes of the field is accomplished via the association of a probability amplitude with a classical mode pattern. The calculation of field modes is entirely classical problem, while the quantum properties of the field are carried by the mode "amplitudes" {{math|''a''<sup>†</sup>}} and {{mvar|a}} associated with these classical modes.

The zero-point energy of the field arises formally from the non-commutativity of {{mvar|a}} and {{math|''a''<sup>†</sup>}}. This is true for any harmonic oscillator: the zero-point energy {{math|{{sfrac|''ħω''|2}}}} appears when we write the Hamiltonian:
<math display="block">\begin{align}
H_{cl} &= \frac{p^2}{2m} + \tfrac{1}{2} m \omega^2 {q}^2 \\
&= \tfrac{1}{2} \hbar \omega \left(a a^\dagger + a^\dagger a\right) \\
&=\hbar \omega \left(a^\dagger a +\tfrac{1}{2}\right)
\end{align}</math>

It is often argued that the entire universe is completely bathed in the zero-point electromagnetic field, and as such it can add only some constant amount to expectation values. Physical measurements will therefore reveal only deviations from the vacuum state. Thus the zero-point energy can be dropped from the Hamiltonian by redefining the zero of energy, or by arguing that it is a constant and therefore has no effect on Heisenberg equations of motion. Thus we can choose to declare by fiat that the ground state has zero energy and a field Hamiltonian, for example, can be replaced by:{{sfnp|Itzykson|Zuber|1980|p=111}}
<math display="block">\begin{align}
H_F - \left\langle 0|H_F|0\right\rangle &=\tfrac{1}{2} \hbar \omega \left(a a^\dagger + a^\dagger a\right)-\tfrac{1}{2}\hbar \omega \\
&= \hbar \omega \left(a^\dagger a + \tfrac{1}{2} \right)-\tfrac{1}{2}\hbar \omega \\
&= \hbar \omega a^\dagger a
\end{align}</math>
without affecting any physical predictions of the theory. The new Hamiltonian is said to be [[Normal order|normally ordered]] (or Wick ordered) and is denoted by a double-dot symbol. The normally ordered Hamiltonian is denoted {{math|:''H<sub>F</sub>''}}, i.e.:
<math display="block">:H_F : \equiv \hbar \omega \left(a a^\dagger + a^\dagger a\right) : \equiv \hbar \omega a^\dagger a</math>

In other words, within the normal ordering symbol we can commute {{mvar|a}} and {{math|''a''<sup>†</sup>}}. Since zero-point energy is intimately connected to the non-commutativity of {{mvar|a}} and {{math|''a''<sup>†</sup>}}, the normal ordering procedure eliminates any contribution from the zero-point field. This is especially reasonable in the case of the field Hamiltonian, since the zero-point term merely adds a constant energy which can be eliminated by a simple redefinition for the zero of energy. Moreover, this constant energy in the Hamiltonian obviously commutes with {{mvar|a}} and {{math|''a''<sup>†</sup>}} and so cannot have any effect on the quantum dynamics described by the Heisenberg equations of motion.

However, things are not quite that simple. The zero-point energy cannot be eliminated by dropping its energy from the Hamiltonian: When we do this and solve the Heisenberg equation for a field operator, we must include the vacuum field, which is the homogeneous part of the solution for the field operator. In fact we can show that the vacuum field is essential for the preservation of the commutators and the formal consistency of QED. When we calculate the field energy we obtain not only a contribution from particles and forces that may be present but also a contribution from the vacuum field itself i.e. the zero-point field energy. In other words, the zero-point energy reappears even though we may have deleted it from the Hamiltonian.{{sfnp|Milonni|1994|pp=73–74}}

==== Electromagnetic field in free space ====
From Maxwell's equations, the electromagnetic energy of a "free" field i.e. one with no sources, is described by:
<math display="block">\begin{align}
H_F &= \frac{1}{8\pi}\int d^3r \left(\mathbf{E}^2 +\mathbf{B}^2\right) \\
&=\frac{k^2}{2\pi}|\alpha (t)|^2
\end{align}</math>

We introduce the "mode function" {{math|'''A'''<sub>0</sub>('''r''')}} that satisfies the [[Helmholtz equation]]:
<math display="block"> \left( \nabla^2 + k^2 \right) \mathbf{A}_0(\mathbf{r}) = 0 </math>
where {{math|''k'' {{=}} {{sfrac|''ω''|''c''}}}} and assume it is normalized such that:
<math display="block">\int d^3r \left|\mathbf{A}_0(\mathbf{r})\right|^2 = 1</math>

We wish to "quantize" the electromagnetic energy of free space for a multimode field. The field intensity of free space should be independent of position such that {{math|{{abs|'''A'''<sub>0</sub>('''r''')}}<sup>2</sup>}} should be independent of {{math|'''r'''}} for each mode of the field. The mode function satisfying these conditions is:
<math display="block"> \mathbf{A}_0(\mathbf{r}) = e_{\mathbf{k}}e^{i\mathbf{k}\cdot\mathbf{r}} </math>
where {{math|'''k''' · '''e'''<sub>'''k'''</sub> {{=}} 0}} in order to have the transversality condition {{math|'''∇''' · '''A'''('''r''',''t'')}} satisfied for the Coulomb gauge{{dubious|date=May 2018}} in which we are working.

To achieve the desired normalization we pretend space is divided into cubes of volume {{math|''V'' {{=}} ''L''<sup>3</sup>}} and impose on the field the periodic boundary condition:
<math display="block">\mathbf{A}(x+L,y+L,z+L,t)=\mathbf{A}(x,y,z,t)</math>
or equivalently
<math display="block"> \left(k_x,k_y,k_z\right)=\frac{2\pi}{L}\left(n_x,n_y,n_z\right)</math>
where {{mvar|n}} can assume any integer value. This allows us to consider the field in any one of the imaginary cubes and to define the mode function:
<math display="block">\mathbf{A}_\mathbf{k}(\mathbf{r})= \frac{1}\sqrt{V} e_{\mathbf{k}}e^{i\mathbf{k}\cdot\mathbf{r}}</math>
which satisfies the Helmholtz equation, transversality, and the "box normalization":
<math display="block">\int_V d^3r \left|\mathbf{A}_\mathbf{k}(\mathbf{r})\right|^2 = 1</math>
where {{math|''e''<sub>'''k'''</sub>}} is chosen to be a unit vector which specifies the polarization of the field mode. The condition {{math|'''k''' · ''e''<sub>'''k'''</sub> {{=}} 0}} means that there are two independent choices of {{math|''e''<sub>'''k'''</sub>}}, which we call {{math|''e''<sub>'''k'''1</sub>}} and {{math|''e''<sub>'''k'''2</sub>}} where {{math|''e''<sub>'''k'''1</sub> · ''e''<sub>'''k'''2</sub> {{=}} 0}} and {{math|''e''{{su|b='''k'''1|p=2}} {{=}} ''e''{{su|b='''k'''2|p=2}} {{=}} 1}}. Thus we define the mode functions:
<math display="block">\mathbf{A}_{\mathbf{k}\lambda}(\mathbf{r})=\frac{1}\sqrt{V}e_{\mathbf{k}\lambda}e^{i\mathbf{k}\cdot\mathbf{r}} \, , \quad \lambda = \begin{cases} 1\\2 \end{cases}</math>
in terms of which the vector potential becomes{{clarify|date=May 2018}}:
<math display="block">\mathbf{A}_{\mathbf{k}\lambda}(\mathbf{r},t)=\sqrt{\frac{2\pi\hbar c^2}{\omega_k V}}\left[a_{\mathbf{k}\lambda}(0)e^{i\mathbf{k}\cdot\mathbf{r}}+a_{\mathbf{k}\lambda}^\dagger(0)e^{-i\mathbf{k}\cdot\mathbf{r}}\right]e_{\mathbf{k}\lambda}</math>
or:
<math display="block">\mathbf{A}_{\mathbf{k}\lambda}(\mathbf{r},t)=\sqrt{\frac{2\pi\hbar c^2}{\omega_k V}}\left[a_{\mathbf{k}\lambda}(0)e^{-i(\omega_k t-\mathbf{k}\cdot\mathbf{r})}+a_{\mathbf{k}\lambda}^\dagger(0)e^{i(\omega_k t-\mathbf{k}\cdot\mathbf{r})}\right]
</math>
where {{math|''ω<sub>k</sub>'' {{=}} ''kc''}} and {{math|''a''<sub>'''k'''''λ''</sub>}}, {{math|''a''{{su|b='''k'''''λ''|p=†}}}} are photon annihilation and creation operators for the mode with wave vector {{mvar|k}} and polarization {{mvar|λ}}. This gives the vector potential for a plane wave mode of the field. The condition for {{math|(''k<sub>x</sub>'', ''k<sub>y</sub>'', ''k<sub>z</sub>'')}} shows that there are infinitely many such modes. The linearity of Maxwell's equations allows us to write:
<math display="block">\mathbf{A}(\mathbf{r}t)=\sum_{\mathbf{k}\lambda}\sqrt{\frac{2\pi\hbar c^2}{\omega_k V}}\left[a_{\mathbf{k}\lambda}(0)e^{i\mathbf{k}\cdot\mathbf{r}}+a_{\mathbf{k}\lambda}^\dagger(0)e^{-i\mathbf{k}\cdot\mathbf{r}}\right]e_{\mathbf{k}\lambda}</math>
for the total vector potential in free space. Using the fact that:
<math display="block">\int_V d^3r \mathbf{A}_{\mathbf{k}\lambda}(\mathbf{r})\cdot \mathbf{A}_{\mathbf{k}'\lambda'}^\ast(\mathbf{r})=\delta_{\mathbf{k},\mathbf{k}'}^3\delta_{\lambda,\lambda'}</math>
we find the field Hamiltonian is:
<math display="block">H_F=\sum_{\mathbf{k}\lambda}\hbar\omega_k\left(a_{\mathbf{k}\lambda}^\dagger a_{\mathbf{k}\lambda} + \tfrac{1}{2} \right) </math>

This is the Hamiltonian for an infinite number of uncoupled harmonic oscillators. Thus different modes of the field are independent and satisfy the commutation relations:
<math display="block">\begin{align}
\left[a_{\mathbf{k}\lambda}(t),a_{\mathbf{k}'\lambda'}^\dagger(t)\right]&=\delta_{\mathbf{k},\mathbf{k}'}^3\delta_{\lambda,\lambda'} \\[10px]
\left[a_{\mathbf{k}\lambda}(t),a_{\mathbf{k}'\lambda'}(t)\right]&=\left[a_{\mathbf{k}\lambda}^\dagger(t),a_{\mathbf{k}'\lambda'}^\dagger(t)\right]=0
\end{align}</math>

Clearly the least eigenvalue for {{math|''H<sub>F</sub>''}} is:
<math display="block">\sum_{\mathbf{k}\lambda}\tfrac{1}{2}\hbar\omega_k</math>

This state describes the zero-point energy of the vacuum. It appears that this sum is divergent – in fact highly divergent, as putting in the density factor
<math display="block">\frac{8\pi v^2 dv}{c^3}V</math>
shows. The summation becomes approximately the integral:
<math display="block">\frac{4\pi h V}{c^3}\int v^3 \, dv</math>
for high values of {{mvar|v}}. It diverges proportional to {{math|''v''<sup>4</sup>}} for large {{mvar|v}}.

There are two separate questions to consider. First, is the divergence a real one such that the zero-point energy really is infinite? If we consider the volume {{mvar|V}} is contained by perfectly conducting walls, very high frequencies can only be contained by taking more and more perfect conduction. No actual method of containing the high frequencies is possible. Such modes will not be stationary in our box and thus not countable in the stationary energy content. So from this physical point of view the above sum should only extend to those frequencies which are countable; a cut-off energy is thus eminently reasonable. However, on the scale of a "universe" questions of general relativity must be included. Suppose even the boxes could be reproduced, fit together and closed nicely by curving spacetime. Then exact conditions for running waves may be possible. However the very high frequency quanta will still not be contained. As per John Wheeler's "geons"<ref>{{cite journal|last1=Wheeler|first1=John Archibald|title=Geons|journal=Physical Review|date=1955|volume=97|issue=2|page=511|doi=10.1103/PhysRev.97.511|bibcode=1955PhRv...97..511W}}</ref> these will leak out of the system. So again a cut-off is permissible, almost necessary. The question here becomes one of consistency since the very high energy quanta will act as a mass source and start curving the geometry.

This leads to the second question. Divergent or not, finite or infinite, is the zero-point energy of any physical significance? The ignoring of the whole zero-point energy is often encouraged for all practical calculations. The reason for this is that energies are not typically defined by an arbitrary data point, but rather changes in data points, so adding or subtracting a constant (even if infinite) should be allowed. However this is not the whole story, in reality energy is not so arbitrarily defined: in general relativity the seat of the curvature of spacetime is the energy content and there the absolute amount of energy has real physical meaning. There is no such thing as an arbitrary additive constant with density of field energy. Energy density curves space, and an increase in energy density produces an increase of curvature. Furthermore, the zero-point energy density has other physical consequences e.g. the Casimir effect, contribution to the Lamb shift, or anomalous magnetic moment of the electron, it is clear it is not just a mathematical constant or artifact that can be cancelled out.{{sfnp|Power|1964|pp=31–33}}

==== Necessity of the vacuum field in QED ====
The vacuum state of the "free" electromagnetic field (that with no sources) is defined as the ground state in which {{math|''n''<sub>'''k'''''λ''</sub> {{=}} 0}} for all modes {{math|('''k''', ''λ'')}}. The vacuum state, like all stationary states of the field, is an eigenstate of the Hamiltonian but not the electric and magnetic field operators. In the vacuum state, therefore, the electric and magnetic fields do not have definite values. We can imagine them to be fluctuating about their mean value of zero.{{citation needed| reason=entire derivation with out source|date=May 2024}}

In a process in which a photon is annihilated (absorbed), we can think of the photon as making a transition into the vacuum state. Similarly, when a photon is created (emitted), it is occasionally useful to imagine that the photon has made a transition out of the vacuum state.{{sfnp|Dirac|1927}} An atom, for instance, can be considered to be "dressed" by emission and reabsorption of "virtual photons" from the vacuum. The vacuum state energy described by {{math|Σ<sub>'''k'''''λ''</sub> {{sfrac|''ħω<sub>k</sub>''|2}}}} is infinite. We can make the replacement:
<math display="block">\sum_{\mathbf{k}\lambda}\longrightarrow\sum_{\lambda}\left (\frac{1}{2\pi} \right )^3 \int d^3 k = \frac{V}{8\pi^3} \sum_\lambda \int d^3 k</math>
the zero-point energy density is:
<math display="block">\begin{align}
\frac{1}{V}\sum_{\mathbf{k}\lambda}\tfrac{1}{2}\hbar\omega_k &=\frac{2}{8\pi^3}\int d^3 k \tfrac{1}{2}\hbar\omega_k \\
&= \frac{4\pi}{4\pi^3} \int dk\,k^2 \left(\tfrac{1}{2}\hbar\omega_k\right) \\
&=\frac{\hbar}{2\pi^2 c^3} \int d\omega\,\omega^3
\end{align}</math>
or in other words the spectral energy density of the vacuum field:
<math display="block">\rho_0(\omega)=\frac{\hbar\omega^3}{2\pi^2c^3}</math>

The zero-point energy density in the frequency range from {{math|''ω''<sub>1</sub>}} to {{math|''ω''<sub>2</sub>}} is therefore:
<math display="block">\int_{\omega_1}^{\omega_2} d\omega\rho_0(\omega) = \frac{\hbar}{8\pi^2c^3}\left(\omega_2^4-\omega_1^4\right)</math>

This can be large even in relatively narrow "low frequency" regions of the spectrum. In the optical region from 400 to 700&nbsp;nm, for instance, the above equation yields around 220 [[erg]]/cm<sup>3</sup>.

We showed in the above section that the zero-point energy can be eliminated from the Hamiltonian by the normal ordering prescription. However, this elimination does not mean that the vacuum field has been rendered unimportant or without physical consequences. To illustrate this point we consider a linear dipole oscillator in the vacuum. The Hamiltonian for the oscillator plus the field with which it interacts is:
<math display="block">H=\frac{1}{2m}\left(\mathbf{p}-\frac{e}{c}\mathbf{A}\right)^2 + \tfrac{1}{2}m\omega_0^2\mathbf{x}^2 + H_F</math>

This has the same form as the corresponding classical Hamiltonian and the Heisenberg equations of motion for the oscillator and the field are formally the same as their classical counterparts. For instance the Heisenberg equations for the coordinate {{math|'''x'''}} and the canonical momentum {{math|'''p''' {{=}} ''m'''''ẋ''' +{{sfrac|''e'''''A'''|''c''}}}} of the oscillator are:
<math display="block">\begin{align}
\mathbf{\dot{x}}&=(i\hbar)^{-1}[\mathbf{x}.H] = \frac{1}{m}\left(\mathbf{p}-\frac{e}{c}\mathbf{A}\right) \\
\mathbf{\dot{p}}&=(i\hbar)^{-1}[\mathbf{p}.H]
\begin{align}&=\tfrac{1}{2}\nabla\left(\mathbf{p}-\frac{e}{c}\mathbf{A}\right)^2-m\omega_0^2\mathbf{\dot{x}} \\
&=-\frac{1}{m} \left[\left(\mathbf{p}-\frac{e}{c}\mathbf{A}\right) \cdot \nabla\right] \left[-\frac{e}{c}\mathbf{A}\right] - \frac{1}{m} \left(\mathbf{p}-\frac{e}{c}\mathbf{A}\right) \times \nabla \times \left[-\frac{e}{c}\mathbf{A}\right] -m\omega_0^2 \mathbf{\dot{x}} \\
&= \frac{e}{c}(\mathbf{\dot{x}}\cdot\nabla)\mathbf{A} + \frac{e}{c}\mathbf{\dot{x}} \times \mathbf{B} -m\omega_0^2 \mathbf{\dot{x}}
\end{align}\end{align}</math>
or:
<math display="block">\begin{align}
m \mathbf{\ddot{x}} &= \mathbf{\dot{p}} - \frac{e}{c} \mathbf{\dot{A}} \\
&= -\frac{e}{c} \left[\mathbf{\dot{A}} - \left(\mathbf{\dot{x}} \cdot \nabla\right) \mathbf{A}\right] + \frac{e}{c} \mathbf{\dot{x}} \times \mathbf{B} - m\omega_0^2\mathbf{x} \\
&= e\mathbf{E} + \frac{e}{c} \mathbf{\dot{x}} \times \mathbf{B} - m\omega_0^2\mathbf{x}
\end{align}</math>
since the rate of change of the vector potential in the frame of the moving charge is given by the convective derivative
<math display="block">\mathbf{\dot{A}}=\frac{\partial\mathbf{A}}{\partial t} + (\mathbf{\dot{x}} \cdot \nabla) \mathbf{A}^3 \,.</math>

For nonrelativistic motion we may neglect the magnetic force and replace the expression for {{math|''m'''''ẍ'''}} by:
<math display="block">\begin{align}
\mathbf{\ddot{x}}+\omega_0^2\mathbf{x}
&\approx \frac{e}{m}\mathbf{E} \\
&\approx \sum_{\mathbf{k}\lambda}
\sqrt{\frac{2\pi\hbar\omega_k}{V}} \left[a_{\mathbf{k}\lambda}(t) + a_{\mathbf{k}\lambda}^\dagger(t)\right] e_{\mathbf{k}\lambda}
\end{align}</math>

Above we have made the electric dipole approximation in which the spatial dependence of the field is neglected. The Heisenberg equation for {{math|''a''<sub>'''k'''''λ''</sub>}} is found similarly from the Hamiltonian to be:
<math display="block">\dot{a}_{\mathbf{k}\lambda} = i \omega_k a_{\mathbf{k}\lambda} + ie \sqrt\frac{2\pi}{\hbar \omega_k V} \mathbf{\dot{x}} \cdot e_{\mathbf{k}\lambda}</math>
in the electric dipole approximation.

In deriving these equations for {{math|'''x'''}}, {{math|'''p'''}}, and {{math|''a''<sub>'''k'''''λ''</sub>}} we have used the fact that equal-time particle and field operators commute. This follows from the assumption that particle and field operators commute at some time (say, {{math|''t'' {{=}} 0}}) when the matter-field interpretation is presumed to begin, together with the fact that a Heisenberg-picture operator {{math|''A''(''t'')}} evolves in time as {{math|''A''(''t'') {{=}} ''U''<sup>†</sup>(''t'')''A''(0)''U''(''t'')}}, where {{math|''U''(''t'')}} is the time evolution operator satisfying
<math display="block">i\hbar\dot{U} = HU \,,\quad U^\dagger(t) = U^{-1}(t) \,,\quad U(0) = 1 \,.</math>

Alternatively, we can argue that these operators must commute if we are to obtain the correct equations of motion from the Hamiltonian, just as the corresponding Poisson brackets in classical theory must vanish in order to generate the correct Hamilton equations. The formal solution of the field equation is:
<math display="block">a_{\mathbf{k}\lambda}(t)=a_{\mathbf{k}\lambda}(0)e^{-i\omega_{k}t}+ie \sqrt{\frac{2\pi}{\hbar \omega_k V}} \int^t_0dt'\,e_{\mathbf{k}\lambda}\cdot\mathbf{\dot{x}}(t')e^{i\omega_k\left(t'-t\right)}</math>
and therefore the equation for {{math|''ȧ''<sub>'''k'''''λ''</sub>}} may be written:
<math display="block">\mathbf{\ddot{x}}+\omega^2_0\mathbf{x}=\frac{e}{m}\mathbf{E}_0(t)+\frac{e}{m}\mathbf{E}_{RR}(t)</math>
where
<math display="block">\mathbf{E}_0(t)=i\sum_{\mathbf{k}\lambda} \sqrt{\frac{2\pi\hbar \omega_k}{V}}\left[a_{\mathbf{k}\lambda}(0)e^{-i\omega_kt}-a^\dagger_{\mathbf{k}\lambda}(0)e^{i\omega_kt}\right]e_{\mathbf{k}\lambda}</math>
and
<math display="block">\mathbf{E}_{RR}(t)=-\frac{4\pi e}{V} \sum_{\mathbf{k}\lambda} \int^t_0dt'\left[e_{\mathbf{k}\lambda}\cdot\mathbf{\dot{x}}\left(t'\right)\right]\cos\omega_k\left(t'-t\right)</math>

It can be shown that in the [[Abraham–Lorentz force|radiation reaction]] field, if the mass {{mvar|m}} is regarded as the "observed" mass then we can take
<math display="block">\mathbf{E}_{RR}(t)=\frac{2e}{3c^3}\mathbf{\ddot{x}}</math>

The total field acting on the dipole has two parts, {{math|'''E'''<sub>0</sub>(''t'')}} and {{math|'''E'''<sub>''RR''</sub>(''t'')}}. {{math|'''E'''<sub>0</sub>(''t'')}} is the free or zero-point field acting on the dipole. It is the homogeneous solution of the Maxwell equation for the field acting on the dipole, i.e., the solution, at the position of the dipole, of the wave equation
<math display="block">\left[\nabla^2-\frac{1}{c^2}\frac{\partial^2}{\partial t^2}\right]\mathbf{E}=0</math>
satisfied by the field in the (source free) vacuum. For this reason {{math|'''E'''<sub>0</sub>(''t'')}} is often referred to as the "vacuum field", although it is of course a Heisenberg-picture operator acting on whatever state of the field happens to be appropriate at {{math|''t'' {{=}} 0}}. {{math|'''E'''<sub>''RR''</sub>(''t'')}} is the source field, the field generated by the dipole and acting on the dipole.

Using the above equation for {{math|'''E'''<sub>''RR''</sub>(''t'')}} we obtain an equation for the Heisenberg-picture operator <math>\mathbf{x}(t)</math> that is formally the same as the classical equation for a linear dipole oscillator:
<math display="block">
\mathbf{\ddot{x}} + \omega^2_0\mathbf{x}-\tau \mathbf{\overset{...}{x}}=\frac{e}{m}\mathbf{E}_0(t)
</math>
where {{math|''τ'' {{=}} {{sfrac|2''e''<sup>2</sup>|3''mc''<sup>3</sup>}}}}. in this instance we have considered a dipole in the vacuum, without any "external" field acting on it. the role of the external field in the above equation is played by the vacuum electric field acting on the dipole.

Classically, a dipole in the vacuum is not acted upon by any "external" field: if there are no sources other than the dipole itself, then the only field acting on the dipole is its own radiation reaction field. In quantum theory however there is always an "external" field, namely the source-free or vacuum field {{math|'''E'''<sub>0</sub>(''t'')}}.

According to our earlier equation for {{math|''a''<sub>'''k'''''λ''</sub>(''t'')}} the free field is the only field in existence at {{math|''t'' {{=}} 0}} as the time at which the interaction between the dipole and the field is "switched on". The state vector of the dipole-field system at {{math|''t'' {{=}} 0}} is therefore of the form
<math display="block">|\Psi\rangle=|\text{vac}\rangle|\psi_D\rangle \,,</math>
where {{math|{{!}}vac⟩}} is the vacuum state of the field and {{math|{{!}}''ψ<sub>D</sub>''⟩}} is the initial state of the dipole oscillator. The expectation value of the free field is therefore at all times equal to zero:
<math display="block">\langle\mathbf{E}_0(t)\rangle=\langle\Psi|\mathbf{E}_0(t)|\Psi\rangle=0</math>
since {{math|''a''<sub>'''k'''''λ''</sub>(0){{!}}vac⟩ {{=}} 0}}. however, the energy density associated with the free field is infinite:
<math display="block">\begin{align}
\frac{1}{4\pi} \left\langle \mathbf{E}^2_0(t) \right\rangle &= \frac{1}{4\pi} \sum_{\mathbf{k}\lambda} \sum_{\mathbf{k'}\lambda'} \sqrt{\frac{2\pi\hbar \omega_k}{V}} \sqrt{\frac{2\pi\hbar \omega_{k'}}{V}} \times \left\langle a_{\mathbf{k}\lambda}(0)a^\dagger_{\mathbf{k'}\lambda'}(0)\right\rangle \\
&= \frac{1}{4\pi}\sum_{\mathbf{k}\lambda}\left (\frac{2\pi\hbar \omega_k}{V} \right )\\
&= \int^\infin_0dw\,\rho_0(\omega)
\end{align}</math>

The important point of this is that the zero-point field energy {{math|''H<sub>F</sub>''}} does not affect the Heisenberg equation for {{math|''a''<sub>'''k'''''λ''</sub>}} since it is a c-number or constant (i.e. an ordinary number rather than an operator) and commutes with {{math|''a''<sub>'''k'''''λ''</sub>}}. We can therefore drop the zero-point field energy from the Hamiltonian, as is usually done. But the zero-point field re-emerges as the homogeneous solution for the field equation. A charged particle in the vacuum will therefore always see a zero-point field of infinite density. This is the origin of one of the infinities of quantum electrodynamics, and it cannot be eliminated by the trivial expedient dropping of the term {{math|Σ<sub>'''k'''''λ''</sub> {{sfrac|''ħω<sub>k</sub>''|2}}}} in the field Hamiltonian.

The free field is in fact necessary for the formal consistency of the theory. In particular, it is necessary for the preservation of the commutation relations, which is required by the unitary of time evolution in quantum theory:
<math display="block">\begin{align}
\left[z(t),p_z(t)\right]&=\left[U^\dagger(t)z(0)U(t),U^\dagger(t)p_z(0)U(t)\right]\\
&=U^\dagger(t)\left[z(0),p_z(0)\right]U(t)\\
&=i\hbar U^\dagger(t)U(t)\\
&=i\hbar
\end{align}</math>

We can calculate {{math|[''z''(''t''),''p<sub>z</sub>''(''t'')]}} from the formal solution of the operator equation of motion
<math display="block">\mathbf{\ddot{x}} + \omega^2_0\mathbf{x}-\tau \mathbf{\overset{...}{x}}=\frac{e}{m}\mathbf{E}_0(t)</math>

Using the fact that
<math display="block">\left[a_{\mathbf{k}\lambda}(0),a^\dagger_{\mathbf{k'}\lambda'}(0)\right]=\delta^3_\mathbf{kk'},\delta_{\lambda\lambda'}</math>
and that equal-time particle and field operators commute, we obtain:
<math display="block">\begin{align}
[z(t),p_z(t)]&=\left[z(t),m\dot{z}(t)\right]+\left[z(t),\frac{e}{c}A_z(t)\right] \\
&=\left[z(t),m\dot{z}(t)\right] \\
&= \left (\frac{i\hbar e^2}{2\pi^2mc^3} \right ) \left (\frac{8\pi}{3} \right ) \int^\infin_0\frac{d\omega\,\omega^4}{\left(\omega^2-\omega^2_0\right)^2+\tau^2\omega^6}
\end{align}</math>

For the dipole oscillator under consideration it can be assumed that the radiative damping rate is small compared with the natural oscillation frequency, i.e., {{math|''τω''<sub>0</sub> ≪ 1}}. Then the integrand above is sharply peaked at {{math|''ω'' {{=}} ''ω''<sub>0</sub>}} and:
<math display="block">\begin{align}
\left[z(t),p_z(t)\right]&\approx \frac{2i\hbar e^2}{3\pi mc^3}\omega^3_0 \int^\infin_{-\infin} \frac{dx}{x^2 + \tau^2\omega^6_0} \\
&= \left (\frac{2i\hbar e^2 \omega^3_0}{3\pi mc^3} \right )\left (\frac{\pi}{\tau\omega^3_0} \right ) \\
&=i\hbar
\end{align}</math>
the necessity of the vacuum field can also be appreciated by making the small damping approximation in
<math display="block">\begin{align}
&\mathbf{\ddot{x}} + \omega^2_0\mathbf{x}-\tau \mathbf{\overset{...}{x}}=\frac{e}{m}\mathbf{E}_0(t) \\
&\mathbf{\ddot{x}}\approx-\omega^2_0\mathbf{x}(t) && \mathbf{\overset{...}{x}}\approx-\omega^2_0\mathbf{\dot{x}}
\end{align}</math>
and
<math display="block">\mathbf{\ddot{x}}+\tau\omega^2_0\mathbf{\dot{x}}+\omega^2_0\mathbf{x}\approx\frac{e}{m}\mathbf{E}_0(t)</math>

Without the free field {{math|'''E'''<sub>0</sub>(''t'')}} in this equation the operator {{math|'''x'''(''t'')}} would be exponentially dampened, and commutators like {{math|[''z''(''t''),''p<sub>z</sub>''(''t'')]}} would approach zero for {{math|''t'' ≫ {{sfrac|1|''τω''{{su|p=2|b=0}}}}}}. With the vacuum field included, however, the commutator is {{math|''iħ''}} at all times, as required by unitarity, and as we have just shown. A similar result is easily worked out for the case of a free particle instead of a dipole oscillator.{{sfnp|Milonni|1981}}

What we have here is an example of a "fluctuation-dissipation elation". Generally speaking if a system is coupled to a bath that can take energy from the system in an effectively irreversible way, then the bath must also cause fluctuations. The fluctuations and the dissipation go hand in hand we cannot have one without the other. In the current example the coupling of a dipole oscillator to the electromagnetic field has a dissipative component, in the form of the zero-point (vacuum) field; given the existence of radiation reaction, the vacuum field must also exist in order to preserve the canonical commutation rule and all it entails.

The spectral density of the vacuum field is fixed by the form of the radiation reaction field, or vice versa: because the radiation reaction field varies with the third derivative of {{math|'''x'''}}, the spectral energy density of the vacuum field must be proportional to the third power of {{mvar|ω}} in order for {{math|[''z''(''t''),''p<sub>z</sub>''(''t'')]}} to hold. In the case of a dissipative force proportional to {{math|'''ẋ'''}}, by contrast, the fluctuation force must be proportional to <math>\omega</math> in order to maintain the canonical commutation relation.{{sfnp|Milonni|1981}} This relation between the form of the dissipation and the spectral density of the fluctuation is the essence of the fluctuation-dissipation theorem.<ref name="ReferenceB"/>

The fact that the canonical commutation relation for a harmonic oscillator coupled to the vacuum field is preserved implies that the zero-point energy of the oscillator is preserved. it is easy to show that after a few damping times the zero-point motion of the oscillator is in fact sustained by the driving zero-point field.<ref name="ReferenceG">{{cite journal|last1=Senitzky|first1=I. R.|title=Dissipation in Quantum Mechanics. The Harmonic Oscillator|journal=Physical Review|date=1960|volume=119|issue=2|page=670|doi=10.1103/PhysRev.119.670|bibcode=1960PhRv..119..670S}}</ref>

=== Quantum chromodynamic vacuum ===
{{Main|QCD vacuum}}

The QCD vacuum is the vacuum state of quantum chromodynamics (QCD). It is an example of a ''[[non-perturbative]]'' vacuum state, characterized by a non-vanishing [[condensate (quantum field theory)|condensate]]s such as the [[gluon condensate]] and the [[quark condensate]] in the complete theory which includes quarks. The presence of these condensates characterizes the confined phase of [[quark matter]]. In technical terms, gluons are [[Vector boson|vector]] [[gauge boson]]s that mediate [[strong interaction]]s of quarks in quantum chromodynamics (QCD). Gluons themselves carry the color charge of the strong interaction. This is unlike the photon, which mediates the [[Electromagnetic force|electromagnetic interaction]] but lacks an electric charge. Gluons therefore participate in the strong interaction in addition to mediating it, making QCD significantly harder to analyze than QED (quantum electrodynamics) as it deals with [[nonlinear equation]]s to characterize such interactions.

=== Higgs field ===
{{Main|Higgs mechanism}}

[[File:Mecanismo de Higgs PH.png|thumb|The potential for the Higgs field, plotted as function of {{math|''ϕ''<sup>0</sup>}} and {{math|''ϕ''<sup>3</sup>}}. It has a ''Mexican-hat'' or ''champagne-bottle profile'' at the ground.]]
The Standard Model hypothesises a field called the Higgs field (symbol: {{mvar|ϕ}}), which has the unusual property of a non-zero amplitude in its ground state (zero-point) energy after renormalization; i.e., a non-zero vacuum expectation value. It can have this effect because of its unusual "Mexican hat" shaped potential whose lowest "point" is not at its "centre". Below a certain extremely high energy level the existence of this non-zero vacuum expectation [[Spontaneous symmetry breaking|spontaneously breaks]] [[electroweak]] [[gauge symmetry]] which in turn gives rise to the Higgs mechanism and triggers the acquisition of mass by those particles interacting with the field. The Higgs mechanism occurs whenever a charged field has a vacuum expectation value. This effect occurs because scalar field components of the Higgs field are "absorbed" by the massive bosons as degrees of freedom, and couple to the fermions via Yukawa coupling, thereby producing the expected mass terms. The expectation value of {{math|''ϕ''<sup>0</sup>}} in the ground state (the vacuum expectation value or VEV) is then {{math|⟨''ϕ''<sup>0</sup>⟩ {{=}} {{sfrac|''v''|{{sqrt|2}}}}}}, where {{math|''v'' {{=}} {{sfrac|{{abs|''μ''}}|{{sqrt|''λ''}}}}}}. The measured value of this parameter is approximately {{val|246|u=GeV/c2}}.<ref name="PDGreview2012">
{{cite web
 |title=Higgs bosons: theory and searches
 |url=http://pdg.lbl.gov/2012/reviews/rpp2012-rev-higgs-boson.pdf
 |website=PDGLive
 |publisher=Particle Data Group
 |date=12 July 2012
 |access-date=15 August 2012
}}</ref> It has units of mass, and is the only free parameter of the Standard Model that is not a dimensionless number.

The Higgs mechanism is a type of [[superconductivity]] which occurs in the vacuum. It occurs when all of space is filled with a sea of particles which are charged and thus the field has a nonzero vacuum expectation value. Interaction with the vacuum energy filling the space prevents certain forces from propagating over long distances (as it does in a superconducting medium; e.g., in the [[Ginzburg–Landau theory]]).