44.2 Poincar´e symmetry and scalar fields

44.2 Poincar´e symmetry and scalar fields

Concept links · terms present in this machine draft; source roles are unverified: complex inner product · Lie algebra · Lie algebra representation · unitary representation · group action

Returning to the case of a single real relativistic field, the dual phase space $\mathcal{M}$ carries an action of the Poincar´e group $\mathcal{P},$ , and the quantum field theory will come with a unitary representation of this group, in much the same way that the non-relativistic case came with a representation of the Euclidean group $E(3)$ (see section 38.3). The Poincar´e group acts on the space of solutions to the Klein-Gordon equation since its action on functions on space-time commutes with the Casimir operator

$$P^2=\frac{{\partial }^2}{\partial t^2}-\frac{{\partial }^2}{\partial x_1^2}-\frac{{\partial }^2}{\partial x_2^2}-\frac{{\partial }^2}{\partial x_3^2}$$

This Poincar´e group action on Klein-Gordon solutions is by the usual action on functions

$$\varphi \to u(a,\Lambda )\varphi =\varphi ({\Lambda }^{-1}(x-a))\text{\hspace{1em}(44.5)}$$

induced from the group action on Minkowski space. On fields $\Phi (\mathbf{x})$ the action is

$$\Phi \to u(a,\Lambda )\Phi =\Phi (\Lambda x+a)$$

Quantization should give unitary operators $U(a,\Lambda )$ , which act on field operators by

$$\hat{\Phi }\to U(a,\Lambda )\hat{\Phi }(x)U(a,\Lambda {)}^{-1}=\hat{\Phi }(\Lambda x+a)\text{\hspace{1em}(44.6)}$$

The $U(a,\Lambda )$ will provide a unitary representation of the Poincar´e group on the quantum field theory state space, acting by intertwining operators of the sort discussed in the finite dimensional context in chapter 20. We would like to construct these operators by the usual method: using the moment map to get a quadratic polynomial on phase space, quantizing to get Lie algebra representation operators, and then exponentiating to get the $U(a,\Lambda )$

This will require that the symplectic structure on the phase space ${\mathcal{H}}_1$ be Poincar´e invariant. The Poisson bracket relations on the position space fields

$$\{\Phi (\mathbf{x}),\Pi ({\mathbf{x}}^{\prime })\}={\delta }^3(\mathbf{x}-{\mathbf{x}}^{\prime }),\{\Phi (\mathbf{x}),\Phi ({\mathbf{x}}^{\prime })\}=\{\Pi (\mathbf{x}),\Pi ({\mathbf{x}}^{\prime })\}=0$$

are easily seen to be invariant under the action of the Euclidean group of spatial translations and rotations by

$$\Phi (\mathbf{x})\to \Phi (R\mathbf{x}+\mathbf{a}),\Pi (\mathbf{x})\to \Pi (R\mathbf{x}+\mathbf{a})$$

(since the delta-function is). Things are not so simple for the rest of the Poincar´e group, since the definition of the $\Phi (\mathbf{x}),\Pi (\mathbf{x})$ is based on a choice of the distinguished $t=0$ hyperplane. In addition, the complicated form of the relativistic complex structure $J_r$ in these coordinates (see equation 43.15) makes it dificult to see if this is invariant under Poincar´e transformations.

Taking Fourier transforms, recall that solutions to the Klein-Gordon equation can be written as (see 43.3)

$$\varphi (t,\mathbf{x})=\frac{1}{(2\pi {)}^{3/2}}{\int }_{{\mathbf{R}}^4}\delta (p_0^2-{\omega }_{\mathbf{p}}^2)f(p)e^{i(-p_{0}t+\mathbf{p}\cdot \mathbf{x})}{d}^{4}p$$

so these are given by functions (actually distributions) $f(\mathbf{p})$ on the positive and negative energy hyperboloids. The complex structure $J_r$ is +$i$ on functions on the negative energy hyperboloid, −$i$ on functions on the positive energy hyperboloid. The action of the Poincar´e group preserves $J_r,$ since it acts separately on the negative and positive energy hyperboloids. It also preserves the Hermitian inner product (see equations 43.17 and 43.18), and thus gives a unitary action on ${\mathcal{M}}_{J_r}^{+}={\mathcal{H}}_1$

Just as for the finite dimensional case in chapter 25 and the non-relativistic quantum field theory case in section 38.3, we can find for each element $L$ of the Lie algebra of the group acting (here the Poincar´e group $\mathcal{P})$ a quadratic expression in the $A(\mathbf{p}),\overline{A}(\mathbf{p})$ (this is the moment map ${\mu }_L)$ . Quantization then gives a corresponding normal ordered quadratic operator in terms of the operators $a^{†}(\mathbf{p}),a(\mathbf{p})$

44.2.1 Translations

For time translations, we have already found the Hamiltonian operator $\hat{H}$ , which gives the infinitesimal translation action on fields by

$$\frac{\partial \hat{\Phi }}{\partial t}=[\hat{\Phi },-i\hat{H}]$$

The behavior of the field operator under time translation is given by the standard Heisenberg picture relation for operators

$$\hat{\Phi }(t+a_0)=e^{i{a}_0\hat{H}}\hat{\Phi }(t)e^{-i{a}_0\hat{H}}$$

For the infinitesimal action of spatial translations on ${\mathcal{H}}_1$ , the momentum operator is the usual

$$\mathbf{P}=-i\nabla$$

(the convention for the Hamiltonian is the opposite sign $\begin{array}{r}H=i\frac{\partial }{\partial t})\end{array}$ . On fields the infinitesimal action will be given by an operator $\hat{\mathbf{P}}$ satisfying the commutation relations

$$[-i\hat{\mathbf{P}},\hat{\mathbf{\Phi }}]=\nabla \hat{\mathbf{\Phi }}$$

(see the discussion for the non-relativistic case in section 38.3 and equation 38.10). Finite spatial translations by a will act by

$$\hat{\Phi }(\mathbf{x})\to e^{-i\mathbf{a}\cdot \hat{\mathbf{P}}}\hat{\Phi }(\mathbf{x})e^{i\mathbf{a}\cdot \hat{\mathbf{P}}}=\hat{\Phi }(\mathbf{x}+\mathbf{a})$$

The operator needed is the quadratic operator

$$\hat{\mathbf{P}}={\int }_{{\mathbf{R}}^3}\mathbf{p}{a}^{†}(\mathbf{p})a(\mathbf{p})d^3\mathbf{p}\text{\hspace{1em}(44.7)}$$

which in terms of fields is given by

$$\hat{\mathbf{P}}=-{\int }_{{\mathbf{R}}^3}:\hat{\Pi }(\mathbf{x})\nabla \hat{\Phi }(\mathbf{x}):d^3\mathbf{x}$$

One can see that this is the correct operator by showing that it satisfies the commutation relation 38.10 with $\hat{\Phi }.$ , using the canonical commutation relations for $\hat{\Phi }$ and $\pi$.

Note that here again moment map methods could have been used to find the expression for the momentum operator. This is a similar calculation to that of section 38.3 although one needs to keep track of a factor of −$i$ caused by the fact that the basic Poisson bracket relations are

$$\{\Phi (\mathbf{x}),\Pi (\mathbf{x})\}={\delta }^3(\mathbf{x}-{\mathbf{x}}^{\prime })\mathrm{versus}\{\Psi (\mathbf{x}),\overline{\Psi }(\mathbf{x})\}=i{\delta }^3(\mathbf{x}-{\mathbf{x}}^{\prime })$$

44.2.2 Rotations

Concept links · terms present in this machine draft; source roles are unverified: Lie algebra · Lie algebra representation · unitary representation · irreducible representation

We can use the same method as for translations to find the quadratic combinations of coordinates on $M$ corresponding to the Lie algebra of the rotation group, which after quantization will provide the angular momentum operators. The action on Klein-Gordon solutions will be given by the operators

$$\begin{array}{rl}\mathbf{L}=\mathbf{X}\times \mathbf{P}=\mathbf{x}\times -i\nabla & \text{position space}\\ =-i{\nabla }_p\times \mathbf{p}& \text{momentum space}\end{array}$$

The corresponding quadratic operators will be

$$\begin{array}{rl}\hat{\mathbf{L}}& =-{\int }_{{\mathbf{R}}^3}:\hat{\pi }(\mathbf{x})(\mathbf{x}\times \nabla )\hat{\varphi }(\mathbf{x}):d^3\mathbf{x}\\ & ={\int }_{{\mathbf{R}}^3}{a}^{†}(\mathbf{p})(\mathbf{p}\times i{\nabla }_{\mathbf{p}})a(\mathbf{p})d^3\mathbf{p}\end{array}\text{\hspace{1em}(44.8)}$$

which, again, could be found using the moment map method, although we will not work that out here. One can check using the canonical commutation relations that the components of this operator satisfy the so(3) commutation relations

$$[-i{\hat{L}}_j,-i{\hat{L}}_k]={ϵ}_{jkl}(-i{\hat{L}}_l)\text{\hspace{1em}(44.9)}$$

and that, together with the momentum operators $\hat{P},$ they $\mathrm{give}$ a Lie algebra representation of the Euclidean group $E(3)$ on the multi-particle state space.

Note that the operators $\hat{\mathbf{L}}$ commute with the Hamiltonian $\hat{H}$ , and so will act on the energy eigenstates of the state space, providing unitary representations of the group SO(3) on these energy eigenspaces. Energy eigenstates will be characterized by the irreducible representation of $SO(3)$ they are in, so as spin $s=0,s=1,..$ . states.

44.2.3 Boosts

Concept links · terms present in this machine draft; source roles are unverified: Lie algebra · Lie algebra representation · group action · Lie bracket

From the point of view that a symmetry of a physical theory corresponds to a group action on the theory that commutes with time translation, Lorentz boosts are not symmetries because they do not commute with time translations (see the commutators in equation 42.2). From the Lagrangian point of view though, boosts are symmetries because the Lagrangian is invariant under them. From our Hamiltonian point of view, they act on phase space, preserving the symplectic structure. They thus have a moment map, and quantization will give a quadratic expression in the field operators which, when exponentiated, will give a unitary action on the multi-particle state space.

Note that boosts preserve not only the symplectic structure, but also the relativistic complex structure $J_r,$ since they preserve the decomposition of momentum space coordinates into separate coordinates on the positive and negative energy hyperboloids. As a result, when expressed in terms of creation and annihilation operators, the quadratic boost operators $\hat{\mathbf{K}}$ will have the same form as the operators $\hat{\mathbf{P}}$ and $\hat{\mathbf{L}},$ an integral of a product involving one creation and one annihilation operator. The boost operators will be given in momentum space by

$$\hat{\mathbf{K}}=i{\int }_{{\mathbf{R}}^3}{\omega }_{\mathbf{p}}{a}^{†}(\mathbf{p}){\nabla }_{\mathbf{p}}a(\mathbf{p})d^3\mathbf{p}\text{\hspace{1em}(44.10)}$$

One can check that this gives a Poincar´e Lie algebra representation on the multi-particle state space, by evaluating first the commutators for the Lorentz group Lie algebra, which, together with 44.9, are (recall the Lie bracket relations 40.1 and 40.2)

$$[-i{\hat{L}}_j,-i{\hat{K}}_k]={ϵ}_{jkl}(-i{\hat{K}}_l),[-i{\hat{K}}_j,-i{\hat{K}}_k]=-{ϵ}_{jkl}(-i{\hat{L}}_l)$$

The commutators with the momentum and Hamiltonian operators

$$[-i{\hat{K}}_j,-i{\hat{P}}_j]=-i\hat{H},[-i{\hat{K}}_j,-i\hat{H}]=-i{\hat{P}}_j$$

show that the rest of the non-zero Poincar´e Lie algebra bracket relations (equations 42.2) are satisfied. All of these calculations are easily performed using the expressions 43.22, 44.7, 44.8, and 44.10, for $\hat{H},\hat{\mathbf{P}},\hat{\mathbf{L}},\hat{\mathbf{K}}$ and theorem 25.2 (generalized from a sum to an integral), which reduces the calculation to that of the commutators of

$${\omega }_{\mathbf{p}},\mathbf{p},\mathbf{p}\times i{\nabla }_{\mathbf{p}},i{\omega }_{\mathbf{p}}{\nabla }_{\mathbf{p}}$$

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