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The standard heuristic argument for why quantum effects of gravity become important at the Planck scale is to consider the length scales at which both quantum field theory (QFT) and general relativity (GR) become crucial in order to explain physical phenomena.

For QFT this is when the length scale is of order the Compton wavelength of a particle, $$l_{c}=\frac{h}{mc},$$ since if one attempts to confine a particle within this length then it is possible for pair creation to occur and so the concept of a particle breaks down and QFT is required.

For GR, this is when the length scale is of order the Schwarzschild radius of a particle, $$l_{s}=\frac{2Gm}{c^{2}},$$ since compressing the mass of a particle to within this radius results in the formation of a black hole which requires GR to understand its behaviour.

As such, one expects that when these to lengths are of the same order, i.e. $l_{c}\sim l_{s}$, that the quantum effects of gravity become important. This occurs when $$\frac{h}{mc}\sim\frac{2Gm}{c^{2}}\Rightarrow m^{2}\sim\frac{hc}{2G}\sim m_{P}^{2}$$ That is when the mass of the particle is of the same order as the Planck mass.

What I'm unsure about, is why the Compton wavelength? Why not the de Broglie wavelength as this is the length scale at which the quantum nature of an object becomes evident?

Is it simply because QFT is consistent with special relativity and standard quantum mechanics is not and so it is the scale at which QFT becomes crucial that sets the scale for quantum gravity?

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  • $\begingroup$ The answers here may help physics.stackexchange.com/q/112572 $\endgroup$ – anna v Nov 2 '17 at 16:23
  • $\begingroup$ @annav Thanks for the link. So is the point that QFT is the "language" that one must use in relativistic theories and so it is at the length scales that QFT becomes crucial to explain physical phenomena that one must consider? $\endgroup$ – user35305 Nov 2 '17 at 16:32
  • $\begingroup$ @annav Ok. So can one say that because QFT is essential for describing relativistic quantum mechanics it is the scale that QFT becomes important that sets the “quantum scale”? $\endgroup$ – user35305 Nov 2 '17 at 19:51
  • $\begingroup$ I will try to write an answer because I may be confusing you. $\endgroup$ – anna v Nov 3 '17 at 6:15
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From the QFT of prof.Tong

We learn that particle-anti-particle pairs are expected to be important when a particle of mass m is localized within a distance of order

λ=h/mc

At distances shorter than this, there is a high probability that we will detect particle-anti-particle pairs swarming around the original particle that we put in. The distance λ is called the Compton wavelength . It is always smaller than the de Broglie wavelength

One has to remember that c is the maximum velocity due to special relativity and thus the smallest wavelength of a specific particle of mass m.

He continues:

If you like, the de Broglie wavelength is the distance at which the wavelike nature of particles becomes apparent; the Compton wavelength is the distance at which the concept of a single pointlike particle breaks down completely

Quantum field theory is based on relativistic quantum mechanics: creation and annihilation operators operate on free particle wavefunctions, of the Dirac field for fermions, the Klein Gordon for bosons and the quantized Maxwell equations for photons. These free particles are considered point particles, and the QFT expansions of the solutions depend on the fact that higher order corrections are of smaller value than the first order. there is a high probability that we will detect particle-anti-particle pairs swarming around the original particle that we put in. means that the series expansion breaks down and the QFT calculations dependent on point like particles is no longer reliable.

At the same time , classical general relativity calculations break down, ending in singularities, above the Schwarzschild radius of a particle,

What I'm unsure about, is why the Compton wavelength? Why not the de Broglie wavelength as this is the length scale at which the quantum nature of an object becomes evident?

The maximum velocity is c for a particle of rest mass m, and thus describes the maximum possible momentum of a deBroglie wave

Is it simply because QFT is consistent with special relativity and standard quantum mechanics is not and so it is the scale at which QFT becomes crucial that sets the scale for quantum gravity?

QFT is based on standard quantum mechanics. Its creation and annihilation operators act on relativistic wavefunctions of point particles in order to describe the Feynman diagrams used in the calculations. It is really the scale where QFT breaks down as a perturbative expansion, and the underlying basis of point particles .

On the other hand it is expected that quantization of gravity will get rid of the singularities of classical general relativity. Already in the Big Bang model a fuzzy region is introduced where effective quantum mechanics is used for the calculations. At Planck mass energies there are no longer the standard model point particles, and in various models a Planck mass is used as the particle in expansions of quantized gravity, ( see note 96 in link) but the whole thing is not rigorous . A definitive quantization of gravity will define what sort of perturbative expansions will be valid at those energy scales.

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  • $\begingroup$ Thanks for the detailed answer. So can one say that the Compton wavelength is the limiting case of the de Broglie wavelength? Is the point that we are considering (hypothetical) measurements of a particle. To localise a particle we “bounce” a photon at it and the higher the energy of the photon the more accurate the measurement is. If the energy of the photon is of order the mass energy of the particle then we are attempting to confine the particle within its Compton wavelength. At this level the concept of a particle breaks down altogether and QFT is essential to describe it?! $\endgroup$ – user35305 Nov 3 '17 at 11:13
  • $\begingroup$ well QFT as used in particle physics is inadequete to describe a particle with velocity c, remembert in special relativity this means infinite relativistic mass ( en.wikipedia.org/wiki/… ) . QFT describes particles as wavepackets, using point particles as an underlying level , from low energies up the the planck energies. At planck energies QFT as known for particle physics breaks down because the underlyin fields of point particles is not valid. $\endgroup$ – anna v Nov 3 '17 at 13:32
  • $\begingroup$ A new quantum field theory is needed , as with the inflation model , because at those energies the normal particles are irrelevant. This needs quantization of gravity. $\endgroup$ – anna v Nov 3 '17 at 13:32
  • $\begingroup$ please keep in mind that there is not one only QFT. It is a tool that can be used on other underlying fields with creation and annihilation operators describing a perturbative expansion nucleartalent.github.io/Course2ManyBodyMethods/doc/pub/fci/pdf/… $\endgroup$ – anna v Nov 3 '17 at 13:46
  • $\begingroup$ So is it the case that the current perturbative QFT description of a particle breaks down at the length scale of its Compton wavelength, and the GR description breaks down at the scale of its Schwarzschild radius. When these length scales are of the same order this indicates that we need a new description of the situation, a quantum theory of gravity? $\endgroup$ – user35305 Nov 3 '17 at 14:09
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User anna v has already given a correct answer. In this answer we try to summarize.

In a nutshell, the Planck scale of quantum gravity is determined by the 3 physical constants $G$, $c$ and $h$.

  1. When the wavelength $\lambda$ becomes of the order of the Schwarzschild radius, the rest energy $mc^2$ becomes of the order of the gravitational energy $Gm^2/\lambda$.

  2. When the wavelength $\lambda$ becomes of the order of the Compton wavelength, the rest energy $mc^2$ becomes comparable with the energy $hc/\lambda$ of a quantum.

In contrast, the de Broglie wavelength $h/|{\bf p}|$ lacks information about relativity theory, and fail to identify the pertinent characteristic scale.

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  • $\begingroup$ Thanks for the clarification. So is the point that the Compton wavelength takes into account relativity (whereas, as you said, the de Broglie wavelength) and hence is the relevant length scale to be considering? Furthermore, as anna v has pointed out, is it the case that the current perturbative QFT description of a particle breaks down at the length scale of its Compton wavelength, and the GR description breaks down at the scale of its Schwarzschild radius.... $\endgroup$ – user35305 Nov 3 '17 at 14:09
  • $\begingroup$ When these length scales are of the same order this indicates that we need a new description of the situation, a quantum theory of gravity? $\endgroup$ – user35305 Nov 3 '17 at 14:09

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