38.3 Spatial symmetries

38.3 Spatial symmetries

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

We saw in chapter 19 that the action of the group $E(3)$ on physical space ${\mathbf{R}}^3$ induces a unitary action on the space ${\mathcal{H}}_1$ of solutions to the free particle Schr¨odinger equation. Quantization of this phase space with this group action produces a multi-particle state space carrying a unitary representation of the group $E(3)$ . There are several diferent actions of the group $E(3)$ that one needs to keep track of here. Given an element $(\mathbf{a},R)\in E(3)$ one has:

• An action on ${\mathbf{R}}^3$ , by

$$\mathbf{x}\to R\mathbf{x}+\mathbf{a}$$

• A unitary action on ${\mathcal{H}}_1$ induced by the action on ${\mathbf{R}}^3$ , given by

$$\psi (\mathbf{x})\to u(\mathbf{a},R)\psi (\mathbf{x})=\psi (R^{-1}(\mathbf{x}-\mathbf{a}))$$

on wavefunctions, or, on Fourier transforms by

$$\tilde{\psi }(\mathbf{p})\to \tilde{u}(\mathbf{a},R)\tilde{\psi }(\mathbf{p})=e^{-i\mathbf{a}\cdot R^{-1}\mathbf{p}}\tilde{\psi }(R^{-1}\mathbf{p})$$

Recall from chapter 19 that this is not an irreducible representation of $E(3)$ , but an irreducible representation can be constructed by taking the space of solutions that are energy eigenfunctions with fixed eigenvalue $\begin{array}{r}\dot{E}=\frac{|\mathbf{p}{|}^2}{2m}\end{array}$

$E(3)$ will act on distributional fields $\Psi (\mathbf{x})$ by

$$\Psi (\mathbf{x})\to (\mathbf{a},R)\cdot \Psi (\mathbf{x})=\Psi (R\mathbf{x}+\mathbf{a})\text{\hspace{1em}(38.8)}$$

This is because elements of ${\mathcal{H}}_1$ can be written in terms of these distributional fields as

$$\Psi (\psi )={\int }_{{\mathbf{R}}^3}\Psi (\mathbf{x})\psi (\mathbf{x})d^3\mathbf{x}$$

and $E(3)$ will act on $\Psi (\psi )$ by

$$\begin{array}{rl}\Psi (\psi )\to (\mathbf{a},R)\cdot \Psi (\psi )& ={\int }_{{\mathbf{R}}^3}\Psi (\mathbf{x})\psi (R^{-1}(\mathbf{x}-\mathbf{a}))d^3\mathbf{x}\\ & ={\int }_{{\mathbf{R}}^3}\Psi (R\mathbf{x}+\mathbf{a})\psi (\mathbf{x})d^3\mathbf{x}\end{array}$$

(using invariance of the integration measure under $E(3)$ transformations). More generally, if elements of ${\mathcal{H}}_1$ are multi-component functions ${\psi }_j$ (for instance in the case of spin $\frac{1}{2}$ wavefunctions), the (double cover of) the $E(3)$ group may act by

$${\psi }_j(\mathbf{x})\to \sum _k{\Omega }_{jk}{\psi }_k(R^{-1}(\mathbf{x}-\mathbf{a}))$$

on wavefunctions, and

$${\Psi }_j(\mathbf{x})\to \sum _k({\Omega }^{-1}{)}_{jk}{\Psi }_k(R\mathbf{x}+\mathbf{a})$$

on distributional fields (see section 38.3.3).

• The action of $E(3)$ on ${\mathcal{H}}_1$ is a linear map preserving the symplectic structure. We thus expect by the general method of section 20.2 to be able to construct intertwining operators, by taking the quadratic functions given by the moment map, quantizing to get a Lie algebra representation, and exponentiating to get a unitary representation of $E(3)$ . More specifically, we will use Bargmann-Fock quantization, and the method carried out for a finite dimensional phase space in section 25.3. We end up with a representation of $E(3)$ on the quantum field theory state space $\mathcal{H},$ given by unitary operators $U(\mathbf{a},R)$

It is the last of these that we want to examine here, and as usual for quantum field theory, we don’t want to try and explicitly construct the multi-particle state space $H$ and see the $E(3)$ action on that construction, but instead want to use the analog of the Heisenberg picture in the time-translation case, taking the group to act on operators. For each $(\mathbf{a},R)\in E(3)$ we want to find operators $U(\mathbf{a},R)$ that will be built out of the field operators, and act on the field operators as

$$\hat{\Psi }(\mathbf{x})\to U(\mathbf{a},R)\hat{\Psi }(\mathbf{x})U(\mathbf{a},R{)}^{-1}=\hat{\Psi }(R\mathbf{x}+\mathbf{a})\text{\hspace{1em}(38.9)}$$

38.3.1 Spatial translations

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

For spatial translations, we want to construct momentum operators $\hat{\mathbf{P}}$ such that the $-i\hat{\mathbf{P}}$ give a unitary Lie algebra representation of the translation group. Exponentiation will then give the unitary representation

$$U(\mathbf{a},1)=e^{-i\mathbf{a}\cdot \hat{\mathbf{P}}}$$

Note that these are not the momentum operators $P$ that act on ${\mathcal{H}}_1$ , but are operators in the quantum field theory that will be built out of quadratic combinations of the field operators. By equation 38.9 we want

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

or the derivative of this equation

$$[-i\hat{\mathbf{P}},\hat{\Psi }(\mathbf{x})]=\nabla \hat{\Psi }(\mathbf{x})\text{\hspace{1em}(38.10)}$$

Such an operator $\hat{\mathbf{P}}$ can be constructed in terms of quadratic combinations of the field operators by our moment map methods. We find (generalizing theorem 25.1) that the quadratic expression

$${\mu }_{-}\nabla =i{\int }_{{\mathbf{R}}^3}\Psi (\mathbf{x})(-\nabla )\overline{\Psi }(\mathbf{x})d^3\mathbf{x}$$

is real (since ∇ is skew-adjoint) and satisfies

$$\{{\mu }_{-}\nabla ,\Psi (\mathbf{x})\}=\nabla \Psi (\mathbf{x}),\{{\mu }_{-}\nabla ,\overline{\Psi }(\mathbf{x})\}=\nabla \overline{\Psi }(\mathbf{x})$$

Using the Poisson bracket relations, this can be checked by computing for instance (we’ll do this just for $d=1)$

$$\begin{array}{rl}\{{\mu }_{-\frac{d}{dx}},\overline{\Psi }(x)\}& =i\left\{\int \Psi (y)(-\frac{d}{dy})\overline{\Psi }(y)dy,\overline{\Psi }(x)\right\}\\ & =-i\int \{\Psi (y),\overline{\Psi }(x)\}\frac{d}{dy}\overline{\Psi }(y)dy\\ & =\int \delta (x-y)\frac{d}{dy}\overline{\Psi }(y)dy=\frac{d}{dx}\overline{\Psi }(x)\end{array}$$

Quantization replaces $\Psi ,\overline{\Psi }$ by ${\hat{\Psi }}^{†},\hat{\Psi }$ and gives the self-adjoint expression

$$\hat{\mathbf{P}}={\int }_{{\mathbf{R}}^3}{\hat{\Psi }}^{†}(\mathbf{x})(-i\nabla )\hat{\Psi }(\mathbf{x})d^3\mathbf{x}\text{\hspace{1em}(38.11)}$$

for the momentum operator. In chapter 37 we saw that, in terms of momentum space annihilation and creation operators, this operator is

$$\hat{\mathbf{P}}={\int }_{{\mathbf{R}}^3}\mathbf{p}{a}^{†}(\mathbf{p})a(\mathbf{p})d^3\mathbf{p}$$

which is the integral over momentum space of the momentum times the numberdensity operator in momentum space.

38.3.2 Spatial rotations

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

For spatial rotations, we found in chapter 19 that these had as generators the angular momentum operators

$$\mathbf{L}=\mathbf{X}\times \mathbf{P}=\mathbf{X}\times (-i\nabla )$$

acting on ${\mathcal{H}}_1$ . Just as for energy and momentum, we can construct angular momentum operators in the quantum field theory as quadratic field operators, in this case getting

$$\hat{\mathbf{L}}={\int }_{{\mathbf{R}}^3}{\hat{\Psi }}^{†}(\mathbf{x})(\mathbf{x}\times (-i\nabla ))\hat{\Psi }(\mathbf{x})d^3\mathbf{x}\text{\hspace{1em}(38.12)}$$

These will generate the action of rotations on the field operators. For instance, if $R(\theta )$ is a rotation about the $x_3$ axis by angle ${\theta }_{;}$ we will have

$$\hat{\Psi }(R(\theta )\mathbf{x})=e^{-i\theta {\hat{L}}_3}\hat{\Psi }(\mathbf{x})e^{i\theta {\hat{L}}_3}$$

The operators $\hat{\mathbf{P}}$ and $\hat{\mathbf{L}}$ together give a representation of the Lie algebra of $E(3)$ on the multi-particle state space, satisfying the $E(3)$ Lie algebra commutation relations

$$[-i{\hat{P}}_j,-i{\hat{P}}_k]=0,[-i{\hat{L}}_j,-i{\hat{P}}_k]={ϵ}_{jkl}(-i{\hat{P}}_l),[-i{\hat{L}}_j,-i{\hat{L}}_k]={ϵ}_{jkl}(-i{\hat{L}}_l)\text{\hspace{1em}(38.13)}$$

$\hat{\mathbf{L}}$ could also have been found by the moment map method. Recall from section $8.3$ that, for the $SO(3)$ representation on functions on ${\mathbf{R}}^3$ induced from the $SO(3)$ action on ${\mathbf{R}}^3$ , the Lie algebra representation is $\left(\mathrm{for}1\in \mathfrak{so}(3)\right)$

$${\rho }^{\prime }(\mathbf{l})=-\mathbf{x}\times \nabla$$

The action on distributions will difer by a minus sign, so we are looking for a moment map $\mu$ such that

$$\{\mu ,\Psi (\mathbf{x})\}=\mathbf{x}\times \nabla \Psi (\mathbf{x})$$

and this will be given by

$${\mu }_{-\mathbf{x}\times \nabla }=i{\int }_{{\mathbf{R}}^3}\Psi (\mathbf{x})(-\mathbf{x}\times \nabla )\overline{\Psi }(\mathbf{x})d^3\mathbf{x}$$

After quantization, this gives equation 38.12 for the angular momentum operator $\hat{\mathbf{L}}$

38.3.3 Spin 1/2 fields

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

For the case of two-component wavefunctions describing spin $\begin{array}{l}\frac{1}{2}\end{array}$ particles satisfying the Pauli-Schr¨odinger equation (see chapter 34 and section $37.5)$ , the groups $U(1),U(n)$ (for multiple kinds of spin $\frac{1}{2}$ particles) and the ${\mathbf{R}}^3$ of translations act independently on the two spinor components, and the formulas for $\hat{N},\hat{X}$ and $\hat{\mathbf{P}}$ are just the sum of two copies of the single component equations. As discussed in section 34.2, the action of the rotation group on solutions in this case requires the use of the double cover $SU(2)$ of $SO(3)$ , with $SU(2)$ group elements Ω acting on two-component solutions $\psi$ by

$$\psi (\mathbf{x})\to \Omega \psi (R^{-1}\mathbf{x})$$

(here $R$ is the $SO(3)$ rotation corresponding to $\Omega )$ . This action can be thought of as an action on a tensor product of ${\mathbf{C}}^2$ and a space of functions on ${\mathbf{R}}^3$ , with the matrix Ω acting on the ${\dot{\mathbf{C}}}^2$ factor, and the action on functions the induced action from rotations on ${\mathbf{R}}^3$ . On distributional fields, the action will be by the inverse

$$\Psi (\mathbf{x})\to {\Omega }^{-1}\Psi (R\mathbf{x})$$

(where $\Psi$ has two components).

The $SU(2)$ action on quantum fields will be given by a unitary operator $U(\Omega )$ satisfying

$$U(\Omega )\hat{\Psi }(\mathbf{x})U^{-1}(\Omega )={\Omega }^{-1}\hat{\Psi }(R\mathbf{x})$$

which will give a unitary representation on the multi-particle state space. The Lie algebra representation on this state space will be given by the sum of two terms

$$\hat{\mathbf{J}}=\hat{\mathbf{L}}+\hat{\mathbf{S}}$$

corresponding to the fact that this comes from a representation on a tensor product. Here the operator $\hat{\mathbf{L}}$ is just two copies of the single component version (equation 38.12) and comes from the same source, the induced action on solutions from rotations of ${\mathbf{R}}^3$ . The $\cdot {\mathrm{spin}}^{,5}$ operator $\hat{\mathbf{S}}$ comes from the $SU(2)$ action on the ${\mathbf{C}}^2$ factor in the tensor product description of solutions and is given by

$$\hat{\mathbf{S}}={\int }_{{\mathbf{R}}^3}{\hat{\Psi }}^{†}(\mathbf{x})\left(\frac{1}{2}\sigma \right)\hat{\Psi }(\mathbf{x})d^3\mathbf{x}\text{\hspace{1em}(38.14)}$$

It mixes the two components of the spin $\frac{1}{2}$ field, and is a new feature not seen in the single component $\left({}^{66}\mathrm{spin}{0}^3\right)$ theory.

It is a straightforward exercise using the commutation relations to show that these operators Jb satisfy the su(2) commutation relations and have the expected commutation relations with the two-component field operators. They also commute with the Hamiltonian, providing an action of $SU(2)$ by symmetries on the multi-particle state space.

States of this quantum field theory can be produced by applying products of operators $a_j^{†}(\mathbf{p})$ for various choices of p and $j=1,2$ to the vacuum state. Note that the $\widetilde{E(3)}$ Casimir operator $\hat{\mathbf{J}}\cdot \hat{\mathbf{P}}$ does not commute with the $a_j^{†}(\mathbf{p})$ . If one wants to work with states with a definite helicity (eigenvalue of $\hat{\mathbf{J}}\cdot \hat{\mathbf{P}}$ divided by the square root of the eigenvalue of the operator $|\hat{\mathbf{P}}{|}^2)$ , one could instead write wavefunctions as in equation 34.10, and field operators as

$${\hat{\Psi }}_{\pm }(\mathbf{x})=\frac{1}{(2\pi {)}^{\frac{3}{2}}}{\int }_{{\mathbf{R}}^3}{e}^{i\mathbf{p}\cdot \mathbf{x}}{a}_{\pm }(\mathbf{p})u_{\pm }(\mathbf{p})d^3\mathbf{p}$$ $${\hat{\Psi }}_{\pm }^{†}(\mathbf{x})=\frac{1}{(2\pi {)}^{\frac{3}{2}}}{\int }_{{\mathbf{R}}^3}{e}^{-i\mathbf{p}\cdot \mathbf{x}}{a}_{\pm }^{†}(\mathbf{p})u_{\pm }^{†}(\mathbf{p})d^3\mathbf{p}$$

Here the operators $a_{\pm }(\mathbf{p}),a_{\pm }^{†}(\mathbf{p})$ would be annihilation and creation operators for helicity eigenstates. Such a formalism is not particularly useful in the nonrelativistic case, but we mention it here because its analog in the relativistic case will be more significant.

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