Once the muonic molecular ion state is formed, the shielding by the muon of the positive charges of the proton of the triton and the proton of the deuteron from each other allows the triton and the deuteron to tunnel through the Coulomb barrier in time span of order of a nanosecond[13] The muon survives the d-t muon-catalyzed nuclear fusion reaction and remains available (usually) to catalyze further d-t muon-catalyzed nuclear fusions. Each exothermic d-t nuclear fusion releases about 17.6 MeV of energy in the form of a "very fast" neutron having a kinetic energy of about 14.1 MeV and an alpha particle α (a helium-4 nucleus) with a kinetic energy of about 3.5 MeV.[6] An additional 4.8 MeV can be gleaned by having the fast neutrons moderated in a suitable "blanket" surrounding the reaction chamber, with the blanket containing lithium-6, whose nuclei, known by some as "lithions," readily and exothermically absorb thermal neutrons, the lithium-6 being transmuted thereby into an alpha particle and a triton.[note 6]
https://en.wikipedia.org/wiki/Magnetohydrodynamics
https://en.wikipedia.org/wiki/Fusion_energy_gain_factor
Necessity of the vacuum field in QED[edit]
The vacuum state of the "free" electromagnetic field (that with no sources) is defined as the ground state in which nkλ = 0 for all modes (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.
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.[55] 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 ∑kλ ħωk2is infinite. We can make the replacement:
the zero-point energy density is:
or in other words the spectral energy density of the vacuum field:
The zero-point energy density in the frequency range from ω1 to ω2 is therefore:
This can be large even in relatively narrow "low frequency" regions of the spectrum. In the optical region from 400 to 700 nm, for instance, the above equation yields around 220 erg/cm3.
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:
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 x and the canonical momentum p = mẋ +eAc of the oscillator are:
![{\displaystyle {\begin{aligned}\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{aligned}&={\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{aligned}}\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2fc7bd73dc1f0fc5519f1427d30efec3b7806a54)
or:
![{\displaystyle {\begin{aligned}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{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b33f34c188c9b38ac5935b2e8777dedbc487dfbd)
since the rate of change of the vector potential in the frame of the moving charge is given by the convective derivative
For nonrelativistic motion we may neglect the magnetic force and replace the expression for mẍ by:
![{\displaystyle {\begin{aligned}\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{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7d599af4e0a15961249f4381fd95ca0b826637f4)
Above we have made the electric dipole approximation in which the spatial dependence of the field is neglected. The Heisenberg equation for akλ is found similarly from the Hamiltonian to be:
In the electric dipole approximation.
In deriving these equations for x, p, and akλ 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, t = 0) when the matter-field interpretation is presumed to begin, together with the fact that a Heisenberg-picture operator A(t) evolves in time as A(t) = U†(t)A(0)U(t), where U(t) is the time evolution operator satisfying
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:
and therefore the equation for ȧkλ may be written:
where:
![{\displaystyle \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 _{k}t}-a_{\mathbf {k} \lambda }^{\dagger }(0)e^{i\omega _{k}t}\right]e_{\mathbf {k} \lambda }}](https://wikimedia.org/api/rest_v1/media/math/render/svg/79c7762e3d4255c5c4f4d6d0407e36584e3ed596)
and:
![{\displaystyle \mathbf {E} _{RR}(t)=-{\frac {4\pi e}{V}}\sum _{\mathbf {k} \lambda }\int _{0}^{t}dt'\left[e_{\mathbf {k} \lambda }\cdot \mathbf {\dot {x}} \left(t'\right)\right]\cos \omega _{k}\left(t'-t\right)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/66305499878dc5dcb445e9e54a8a3508aab1d37e)
It can be shown that in the radiation reaction field, if the mass m is regarded as the "observed" mass then we can take:
The total field acting on the dipole has two parts, E0(t) and ERR(t). E0(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
![{\displaystyle \left[\nabla ^{2}-{\frac {1}{c^{2}}}{\frac {\partial ^{2}}{\partial t^{2}}}\right]\mathbf {E} =0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0eb3f09893b6543462f2067d166cefebbecc7811)
satisfied by the field in the (source free) vacuum. For this reason E0(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 t = 0. ERR(t) is the source field, the field generated by the dipole and acting on the dipole.
Using the above equation for ERR(t) we obtain an equation for the Heisenberg-picture operator that is formally the same as the classical equation for a linear dipole oscillator:
where τ = 2e23mc3. 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 E0(t).
According to our earlier equation for akλ(t) the free field is the only field in existence at 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 t = 0 is therefore of the form
where |vac⟩ is the vacuum state of the field and |ψD⟩ is the initial state of the dipole oscillator. The expectation value of the free field is therefore at all times equal to zero:
since akλ(0)|vac⟩ = 0. however, the energy density associated with the free field is infinite:
The important point of this is that the zero-point field energy HF does not affect the Heisenberg equation for akλ since it is a c-number or constant (i.e. an ordinary number rather than an operator) and commutes with akλ. 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 ∑kλ ħωk2 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:
![{\displaystyle {\begin{aligned}\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{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7bd879d669ee3c5c9fae50a1f314e765c58a3635)
We can calculate [z(t),pz(t)] from the formal solution of the operator equation of motion
Using the fact that
![{\displaystyle \left[a_{\mathbf {k} \lambda }(0),a_{\mathbf {k'} \lambda '}^{\dagger }(0)\right]=\delta _{\mathbf {kk'} }^{3},\delta _{\lambda \lambda '}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5b319f53134810ccc6a5524bc4cc97a6cd854405)
and that equal-time particle and field operators commute, we obtain:
![{\displaystyle {\begin{aligned}[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 ^{2}mc^{3}}}\right)\left({\frac {8\pi }{3}}\right)\int _{0}^{\infty }{\frac {d\omega \,\omega ^{4}}{\left(\omega ^{2}-\omega _{0}^{2}\right)^{2}+\tau ^{2}\omega ^{6}}}\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/41cf948249a6ae51282a1225d60b45d737e07c12)
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., τω0 ≪ 1. Then the integrand above is sharply peaked at ω = ω0 and:
![{\displaystyle {\begin{aligned}\left[z(t),p_{z}(t)\right]&\approx {\frac {2i\hbar e^{2}}{3\pi mc^{3}}}\omega _{0}^{3}\int _{-\infty }^{\infty }{\frac {dx}{x^{2}+\tau ^{2}\omega _{0}^{6}}}\\&=\left({\frac {2i\hbar e^{2}\omega _{0}^{3}}{3\pi mc^{3}}}\right)\left({\frac {\pi }{\tau \omega _{0}^{3}}}\right)\\&=i\hbar \end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/871074648f797bba79ddb93f1c0a4d7bdafb5378)
the necessity of the vacuum field can also be appreciated by making the small damping approximation in
and
Without the free field E0(t) in this equation the operator x(t) would be exponentially dampened, and commutators like [z(t),pz(t)] would approach zero for t ≫ 1τω2
0. With the vacuum field included, however, the commutator is 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.[98]
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 x, the spectral energy density of the vacuum field must be proportional to the third power of ω in order for [z(t),pz(t)] to hold. In the case of a dissipative force proportional to ẋ, by contrast, the fluctuation force must be proportional to in order to maintain the canonical commutation relation.[98] This relation between the form of the dissipation and the spectral density of the fluctuation is the essence of the fluctuation-dissipation theorem.[77]
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.[99]
https://en.wikipedia.org/wiki/Zero-point_energy#Necessity_of_the_vacuum_field_in_QED
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