Review of the Quantum Tetrahedron – part 1

Over the last few posts the level of mathematics as been rather high  so I’ve decided to review our basic understanding of the Quantum Tetrahedron in this post. This review is based on the work of Carlo Rovelli and Francesca Vidotto. Other parts of this review will look at the graviton propagator and also coherent states.

If we pick a simple geometrical object ,an elementary portion of space, such as a small tetrahedron t, not necessarily regular.


The geometry of a tetrahedron is characterized by the length of its sides, the area of its faces, its volume, the dihedral angles at its edges, the angles at the vertices of its faces, and so on. These are all local functions of the gravitational field, because geometry is the same thing as the gravitational field. These geometrical quantities are related to one another. A set of independent quantities is provided for instance by the six lengths of the sides, but these are not appropriate for studying quantization, because they are constrained by inequalities. The length of the three sides of a triangle, for instance, cannot be chosen arbitrarily: they must satisfy the triangle inequalities. Non-trivial inequalities between dynamical variables, like all global features of phase space, are generally difficult to implement in quantum theory.

Instead, we choose the four vectors


defined for each triangle a as 1/2 of the outward oriented vector product of two edges bounding the triangle.


These four vectors have several nice properties. Elementary geometry shows that they can be equivalently defined in one of the two following ways:

  •  The vectors La are outgoing normals to the faces of the tetrahedron and their norm is equal to the area of the face.
  • The matrix of the components


where M is the matrix formed by the components of three edges of the tetrahedron that emanate from a common vertex.

The vectors La have the following properties:

  •  They satisfy the closure relation:


  • The quantities La determine all other geometrical quantities such as areas, volume, angles between edges and dihedral angles between faces.
  • All these quantities, that is, the geometry of the tetrahedron, are invariant under a common SO(3) rotation of the four La. Therefore a tetrahedron is determined by an equivalence class under rotations of a quadruplet of vectors La’s satisfying


  •  The area Aa of the face a is |La|.
  • The volume V is determined by the properly oriented triple product of any three faces:


We can describe the gravitational field in terms of triads and tetrads. If the tetrahedron is small compared to the scale of the local curvature, so that the metric can be assumed to be locally flat andLa can be identified with the flux of the triad field


across the face a, then


Since the triad is the gravitational field, this gives the explicit relation between La and the gravitational field. Here the triad is defined in the 3d hyperplane determined by the tetrahedron.

The above give all the ingredients for jumping to quantum gravity. The geometry of a real physical tetrahedron is determined by the gravitational field, which is a quantum field. Therefore the normals La can be described by quantum operators, if we take the quantum nature of gravity into account. These will obey commutation relations. The commutation relation can be obtained from the hamiltonian analysis of GR, by promoting Poisson brackets to operators,  but ultimately they are quantization postulates. The simplest possibility -see post Quantum tetrahedra and simplicial spin networks, is:


where lo is a constant proportional h and with the dimension of an area. These commutation relations are realizations of the algebra of SU(2), reflecting again the rotational symmetry in the description of the tetrahedron. This is useful, for instance we see that C as defined  is precisely the generator of common rotations and therefore the closure condition  is an immediate condition of rotational invariance, which is what we want: the geometry is determined by the La up to rotations, which here are gauge.

The constant lo must be related to the Planck scale , which is the only dimensional constant in quantum gravity. Setting,

reviewpart1equ1.13where γ is a dimensionless parameter of the order of unity that fixes the precise scale of the theory.

One consequence is immediate -see post Discreteness of area and volume in quantum gravity, the quantity Aa = |La| behaves as total angular momentum. As this quantity is the area, it follows immediately that the area of the triangles bounding any tetrahedron is quantized with eigenvalues:


This is the gist of loop quantum gravity. The result extends to any surface, not just the area of the triangles bounding a tetrahedron.

Say that the quantum geometry is in a state with area eigenvalues    j1, …, j4. The four vector operators La act on the tensor product H of four representations of SU(2), with respective spins j1, …, j4. That is, the Hilbert space of the quantum states of the geometry of the tetrahedron at fixed values of the area of its faces is


Have to take into account the closure equation, which is a condition the states must satisfy, if they are to describe a tetrahedron. But C is nothing else than the generator of the global diagonal action of SU(2) on the four representation spaces. The states that solve the closure equation, namely


are the states that are invariant under this action, namely the states in the subspace:


The volume operator V is well defined in K because it commutes with C, namely it is rotationally invariant. Therefore we have a well-posed eigenvalue problem for the self-adjoint volume operator on the Hilbert space K. As this space is finite dimensional, it follows that its eigenvalues are discrete. Therefore we have the result that the volume has discrete eigenvalues as well. In other words, there are ‘quanta of volume’ or ‘quanta of space’: the volume of our tetrahedron can grow only in discrete steps.

Eigenvalues of the volume

Computing the volume eigenvalues for a quantum of space whose sides have minimal non vanishing area. Recall that the volume operator V is determined by:


where the operators La satisfy the commutation relations:


If the faces of the quantum of space have minimal area, the Casimir of the corresponding representations have minimal non-vanishing value. Therefore the four operators La act on the fundamental representations j1 = j2 = j3 = j4 = 1/2 . Therefore they are proportionalto the self-adjoint generators of SU(2), which in the fundamental representation are Pauli matrices. That is:


The proportionality constant has the dimension of length square, is of Planck scale and is fixed by comparing with the commutation relations of the Pauli matrices. This gives


The Hilbert space on which these operators act is therefore

H = H½ x H½xH½xH½

This is the space of objects with 4 spinor indices A, B = 0, 1, each being in the ½- representation of SU(2).


The operator La acts on the a-th index. Therefore the volume operator acts as:


Now implementing the closure condition . Let


We only have to look  for subspaces that are invariant under a common rotation for each space Hji , namely we should look for a quantity with four spinor indices that are invariant under rotations. What is the dimension of this space? Remembering that for SU(2) representations:

implies that:


Since the trivial representation appears twice, the dimension of reviewpart1equ1.57kkkkis two. Therefore there must be two independent invariant tensors with four indices. These are easy to guess, because the only invariant objects available are ε(AB) and σ(AB), obtained raising the indices of the Pauli matrices σi:


Therefore two states that spanreviewpart1equ1.57kkkkare:


These form a non orthogonal basis inreviewpart1equ1.57kkkk. These two states span the physical SU(2)-invariant part of the Hilbert space, that gives all the shapes of our quantum of space with a given area. To find the eigenvalues of the volume it suffices to diagonalize the 2×2 matrix V²nm:


An straightforward calculation with Pauli matrices gives:


so that,


and the diagonalization gives the eigenvalues


The sign depends on the fact that this is the oriented volume square, which depends on the relative orientation of the triad of normal chosen. Inserting the value for α, we have finally the eigenvalue of the non oriented volume reviewpart1equ1.65

About 10¹°º quanta of volume of this size fit into a cm³.

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