Attempts at relativistic QM based on S-1 A proper description of particle physics should incorporate both quantum mechanics and special relativity. However historically combining quantum mechanics and relativity was highly non-trivial. Today we review some of these attempts. The result of this effort is relativistic quantum field theory - consistent description of particle physics.
Quantum mechanics Time evolution of the state of the system is described by Schrödinger equation: where H is the hamiltonian operator representing the total energy. For a free, spinless, nonrelativistic particle we have: in the position basis, P = ih H = 1 2m P2 and S.E. is: where ψ(x, t) = x ψ, t is the position-space wave function.
Relativistic generalization? Obvious guess is to use relativistic energy-momentum relation: Schrödinger equation becomes: Not symmetric in time and space derivatives
Relativistic generalization? Obvious guess is to use relativistic energy-momentum relation: Schrödinger equation becomes: Applying i t on both sides and using S.E. we get Klein-Gordon equation looks symmetric
Special relativity (physics is the same in all inertial frames) Space-time coordinate system: x µ = (ct, x) define: x µ = ( ct, x) or where is the Minkowski metric tensor. Its inverse is: that allows us to write:
Interval between two points in space-time can be written as: ds 2 = (x x ) 2 c 2 (t t ) 2 = g µν (x x ) µ (x x ) ν = (x x ) µ (x x ) µ General rules for indices: repeated indices, one superscript and one subscript are summed; these indices are said to be contracted any unrepeated indices (not summed) must match in both name and height on left and right side of any valid equation
Two coordinate systems (representing inertial frames) are related by Lorentz transformation matrix translation vector Interval between two different space-time points is the same in all inertial frames: which requires:
Notation for space-time derivatives: matching-index-height rule works: For two coordinate systems related by derivatives transform as: which follows from not the same as will prove later!
Is K-G equation consistent with relativity? Physics is the same in all inertial frames: the value of the wave function at a particular space-time point measured in two inertial frames is the same: This should be true for any point in the space-time and thus a consistent equation of motion should have the same form in any inertial frame. Is that the case for Klein-Gordon equation?
Klein-Gordon equation: in 4-vector notation: in 4-vector notation: = d Alambertian operator Is it equivalent to: Since? K-G eq. is manifestly consistent with relativity!
Is K-G consistent with quantum mechanics? Schrödinger equation (first order in time derivative) leaves the norm of a state time independent. Probability is conserved: = d 3 xρ(x) ψ, t ψ, t = t d 3 x ρ t = d 3 x. j = S j.ds = 0 ρ t = ψ ψ t + ψ ψ t = i 2m ψ 2 ψ i 2m ψ 2 ψ = i 2m.(ψ ψ ψ ψ ). j Gauss s law j(x) = 0 at infinity
Is K-G consistent with quantum mechanics? Schrödinger equation (first order in time derivative) leaves the norm of a state time independent. Probability is conserved: = d 3 xρ(x) ψ, t ψ, t = t d 3 x ρ t = d 3 x. j = S j.ds = 0 Klein-Gordon equation is second order in time derivative and the norm of a state is NOT in general time independent. Probability is not conserved. Klein-Gordon equation is consistent with relativity but not with quantum mechanics.
Dirac attempt Dirac suggested the following equation for spin-one-half particles (a state carries a spin label, a = 1,2): ψ, a, t Consistent with ψ(x, Schrödinger t) = x ψ, equation t for the Hamiltonian: Dirac equation is linear in both time and space derivatives and so it might be consistent with both QM and relativity. Squaring the Hamiltonian yields:
can be written as anticommutator and also can be written as H 2 Eigenvalues of ψ(x, t) should = x ψ, satisfy t the correct relativistic energymomentum relation: and so we choose matrices that satisfy following conditions: it can be proved (later) that the Dirac equation is fully consistent with relativity. we have a relativistic quantum mechanical theory!
Discussion of Dirac equation to account for the spin of electron, the matrices should be 2x2 but the minimum size satisfying above conditions is 4x4 - two extra spin states H is traceless, and so 4 eigenvalues are: E(p), E(p), -E(p), -E(p) negative energy states no ground state also a problem for K.-G. equation Dirac s interpretation: due to Pauli exclusion principle each quantum state can be occupied by one electron and we simply live in a universe with all negative energy states already occupied. Negative energy electrons can be excited into a positive energy state (by a photon) leaving behind a hole in the sea of negative energy electrons. The hole has a positive charge and positive energy antiparticle (the same mass, opposite charge) called positron (1927)
Quantum mechanics as a quantum field theory Consider Schrödinger equation for n particles with mass m, moving in an external potential U(x), with interparticle potential It is equivalent to: is the position-space wave function. if and only if satisfies S.E. for wave function!
and is a quantum field and its hermitian conjugate; they satisfy commutation relations: is the vacuum state state with one particle at state with one particles at and another at Total number of particles is counted by the operator: commutes with H!
creation operators commute with each other, and so without loss of generality we can consider only completely symmetric functions: we have a theory of bosons (obey Bose-Einstein statistics). For a theory of fermions (obey Fermi-Dirac statistics) we impose and we can restrict our attention to completely antisymmetric functions: We have nonrelativistic quantum field theory for spin zero particles that can be either bosons or fermions. This will change for relativistic QFT.
Lorentz invariance based on S-2 Lorentz transformation (linear, homogeneous change of coordinates): that preserves the interval : All Lorentz transformations form a group: product of 2 LT is another LT identity transformation: inverse: for inverse can be used to prove:
Infinitesimal Lorentz transformation: thus there are 6 independent ILTs: 3 rotations and 3 boosts not all LT can be obtained by compounding ILTs! +1 proper -1 improper proper LTs form a subgroup of Lorentz group; ILTs are proper! Another subgroup - orthochronous LTs, ILTs are orthochronous!
When we say theory is Lorentz invariant we mean it is invariant under proper orthochronous subgroup only (those that can be obtained by compounding ILTs) Transformations that take us out of proper orthochronous subgroup are parity and time reversal: orthochronous but improper nonorthochronous and improper A quantum field theory doesn t have to be invariant under P or T.
How do operators and quantum fields transform? Lorentz transformation (proper, orthochronous) is represented by a unitary operator that must obey the composition rule: infinitesimal transformation can be written as: are hermitian operators = generators of the Lorentz group from using and expanding both sides, keeping only linear terms in we get: since are arbitrary general rule: each vector index undergoes its own Lorentz transformation!
using and expanding to linear order in we get: These comm. relations specify the Lie algebra of the Lorentz group. We can identify components of the angular momentum and boost operators: and find:
in a similar way for the energy-momentum four vector we find: P µ = (H/c, P i ) using and expanding to linear order in we get: or in components: in addition: Comm. relations for J, K, P, H form the Lie algebra of the Poincare group.
Finally, let s look at transformation of a quantum scalar field: Recall time evolution in Heisenberg picture: this is generalized to: P x = P µ x µ = P x Ht x is just a label we can write the same formula for x-a: e +ip a/ e ip x/ φ(0)e +ip x/ e ip a/ = φ(x a) we define space-time translation operator: and obtain:
Similarly: Derivatives carry vector indices: is Lorentz invariant
Canonical quantization of scalar fields Hamiltonian for free nonrelativistic particles: based on S-3 Fourier transform: we get: a(x) = d 3 p (2π) 3/2 eip x ã(p) d 3 x (2π) 3 eip x = δ 3 (p) can go back to x using: d 3 p (2π) 3 eip x = δ 3 (x)
Canonical quantization of scalar fields (Anti)commutation relations: Vacuum is annihilated by : [A, B] = AB BA is a state of momentum, eigenstate of with is eigenstate of with energy eigenvalue:
Relativistic generalization Hamiltonian for free relativistic particles: Is this theory Lorentz invariant? spin zero, but can be either bosons or fermions Let s prove it from a different direction, direction that we will use for any quantum field theory from now: start from a Lorentz invariant lagrangian or action derive equation of motion (for scalar fields it is K.-G. equation) find solutions of equation of motion show the Hamiltonian is the same as the one above
A theory is described by an action: where is the lagrangian. Equations of motion should be local, and so where is the lagrangian density. Thus: is Lorentz invariant: For the action to be invariant we need: the lagrangian density must be a Lorentz scalar!
Any polynomial of a scalar field is a Lorentz scalar and so are products of derivatives with all indices contracted. Let s consider: arbitrary constant = 1, c = 1 and let s find the equation of motion, Euler-Lagrange equation: (we find eq. of motion from variation of an action: making an infinitesimal variation in and requiring the variation of the action to vanish) integration by parts, and δφ(x) = 0 at infinity in any direction (including time) is arbitrary function of x and so the equation of motion is Klein-Gordon equation
Solutions of the Klein-Gordon equation: one classical solution is a plane wave: is arbitrary real wave vector and The general classical solution of K-G equation: where and are arbitrary functions of, and is a function of k (introduced for later convenience) if we tried to interpret as a quantum wave function, the second term would represent contributions with negative energy to the wave function!
real solutions: k k thus we get: (such a k µ is said to be on the mass shell)
Finally let s choose so that is Lorentz invariant: manifestly invariant under orthochronous Lorentz transformations on the other hand sum over zeros of g, in our case the only zero is k 0 = ω for any the differential is Lorentz invariant it is convenient to take for which the Lorentz invariant differential is:
Finally we have a real classical solution of the K.-G. equation: where again:,, For later use we can express in terms of : where and we will call. Note, is time independent.
Constructing the hamiltonian: Recall, in classical mechanics, starting with lagrangian as a function of coordinates and their time derivatives we define conjugate momenta and the hamiltonian is then given as: In field theory: and the hamiltonian is given as: hamiltonian density
In our case: Inserting we get:
d 3 x (2π) 3 eip x = δ 3 (p)
From classical to quantum (canonical quantization): coordinates and momenta are promoted to operators satisfying canonical commutation relations: operators are taken at equal times in the Heisenberg picture
We have derived the classical hamiltonian: We kept ordering of a s unchanged, so that we can easily generalize it to quantum theory where classical functions will become operators that may not commute. The hamiltonian of the quantum theory: (2π) 3 δ 3 (0) = V see the formula for delta function is the total zero point energy per unit volume we are free to choose: the ground state has zero energy eigenvalue.
Summary: is equivalent to: for: We have rederived the hamiltonian of free relativistic bosons by quantization of a scalar field whose equation of motion is the Klein- Gordon equation (starting with manifestly Lorentz invariant lagrangian). does not work for fermions, anticommutators lead to trivial hamiltonian!
The spin-statistics theorem Hamiltonian of free spin zero particles: based on S-4 How to add Lorentz invariant interactions?
Let s split the hermitian free field into: where: time evolved with : For Lorentz transformation (proper, orthochronous) we have found: Thus and are Lorentz scalars. We will have local, Lorentz invariant interactions if we take the interaction lagrangian density to be a hermitian function of and.
Time dependent perturbation theory in QM: the transition amplitude to start with an initial state at time and end with a final state at time is: where is the perturbing hamiltonian in the interaction picture, where is the perturbing hamiltonian in the Schrödinger picture. T is the time ordering symbol a product of operators to its right is to be ordered, not as written, but with operators at later times to the left of those at earlier times. Specifying : H 1 (x, 0) = F(ϕ + (x, 0), ϕ (x, 0)) H I (x, t) = F(ϕ + (x, t), ϕ (x, t))
Time evolution operator: satisfies: α, t = U I (t, t 0 ) α, t 0 i t U I(t, t 0 ) = H I (t)u I (t, t 0 ) with initial condition: U I (t 0, t 0 ) = 1 It is equivalent to the integral equation: U I (t, t 0 ) = 1 i t t 0 H I (t )U I (t, t 0 )dt which we can solve iteratively: U I (t, t 0 ) = 1 i t t 0 H I (t ) 1 i t t 0 H I (t )U I (t, t 0 )dt dt
U I (t, t 0 ) = 1 i H I (t ) 1 i t 0 t t t 0 H I (t )U I (t, t 0 )dt dt more iterations: using: U I (t, t 0 ) = 1 + ( i) t t +( i) 2 dt t 0 t 0 dt H I (t ) + ( i) 2 t t 0 t dt t t t t t t dt dt H I (t )H I (t ) t 0 t 0 t 0 dt H I (t )H I (t )H I (t ) + dt dt H I (t )H I (t ) = 1 dt dt T {H I (t )H I (t )} t 0 t 0 2 t 0 t 0 and with n! for higher order terms, we get: U I (t, t 0 ) = 1 + ( i) T exp t dt H I (t ) + ( i)2 t 0 2! t i dt H I (t ) t 0 t t dt dt T {H I (t )H I (t )} + t 0 t 0
In order for the transition amplitude ordering must be frame independent! to be Lorentz invariant, the time Time ordering of two space-time points is: frame independent if their separation is timelike: frame dependent if their separation is spacelike: Thus for we require: But not in general satisfied! modified Bessel function nonzero for any r > 0 even for m = 0!
Let s try using only particular linear combinations: we then have: and both can be 0 only if we choose commutators and Consistent describing interacting spin-zero particles can be obtained only with commutators. Spin-zero particles are bosons.
In addition: is hermitian! we can replace which doesn t modify commutation relations and leads to: Thus we have found that is the fundamental object we have to use to build a consistent lagrangian. And we have to use commutators. If we choose anticommutators, we find, which leads to a trivial lagrangian that is at most linear in fields.