How is the energy-momentum spectral measure defined in relation to the energy-momentum generators?

Question

Consider a strongly continuous unitary representation of the (restricted) Poincaré group $\mathcal P_+^{\uparrow} \ni(a,\Lambda) \mapsto U(a,\Lambda) \in \mathcal U(\mathcal H)$ over a separable Hilbert space $\mathcal H$. By strong continuity of the one-parameter Abelian group $\mathbb R \ni t\mapsto U(ta,\mathbb I) \in \mathcal U(\mathcal H)$, one is allowed the construction of a quadruple of strongly commuting self-adjoint generators $\{p_\mu\}, \mu=0,1,2,3\,,$ via the following definition:

$$ p_\mu |\psi\rangle := -i\frac{d}{dx^\mu}\Bigg|_{x^\mu=0} U(x^\mu,\mathbb I)|\psi\rangle \,, $$

where $|\psi\rangle \in \mathcal H$ is so chosen that the above limit $x^\mu\to0$ exists.

I am trying to understand how this translates to saying that there is a projection-valued measure $E:\mathfrak B(\mathbb R^4) \to\mathcal B(\mathcal H)$, which is a map from the $\sigma$-algebra of Borel subsets of $\mathbb R^4$ to the set of projection operators on $\mathcal H$, such that for any $|\psi\rangle \in \mathcal H, p\in\mathbb R^4$,

$$ \langle \psi, E(d^4p)\psi\rangle \stackrel{\color{red}?}= \langle \psi, ``E(p)" \psi\rangle d^4p \equiv ``E_\psi(p)\ d^4p" \,,$$

where $``E":\mathbb R^4 \to \mathcal B(\mathcal H)$ is a distributional map. Furthermore, $U(x,\mathbb I) = \int_{\mathbb R^4} \exp(ip\cdot x)\ E(d^4p)$ and therefore,

$$ \langle \psi, U(x,\mathbb I)\psi\rangle = \int_{\mathbb R^4} \exp(ip\cdot x)\ E_\psi(p)\ d^4p \,.$$

My problem is that I do not understand why $U$ is described as an integral over the projection-valued measure. Naively, I would think that $U(x,\mathbb I) = \exp(ip\cdot x)$. Why is this expectation wrong?

Moreover, how does that equality highlighted with a red question mark work? Is it a definition?

Kindly help. I am not very familiar with measure theory. I am studying this in the context of non-perturbative foundations of QFT.

Answer 1

The spectral theorem for self-adjoint operators says the following:

If $A: D(A)\to H$ is self-adjoint then there is a projection valued measure $dP$ on $\Bbb R$ so that $A\psi=\int_{\Bbb R} \lambda\, dP(\lambda)\psi$ for all $\psi\in D(A)$.

If you have a bunch of commuting self-adjoint $A_1,...,A_n$ then the corresponding projection valued measures commute. Meaning $\int_{X}dP_i\int_Y dP_j = \int_{Y}dP_j \int_XdP_i$ for all $i,j\in\{1,..,n\}$, $X,Y\subset \Bbb R$.

This allows us to take the cross-product of these measures to recover a projection valued measure on $\Bbb R^n$, $dP_{1,..,n}= dP_1\cdot...\cdot dP_n$. This is the measure that is meant with $E$. Using this definition we recover the equation you are asking about: $$\int_{\lambda\in\Bbb R^4}\exp(i\lambda x) dP_1(\lambda_1)...dP_4(\lambda_4) =\prod_i \int_\Bbb R \exp(i\lambda_i x_i)dP_i(\lambda_i) = \prod_iU(x_ie_i,\Bbb1)=U(x,\Bbb1).$$ Here $x_i,\lambda_i$ are the $i$-th components of $x,\lambda$ and $e_i$ the unit vector in $i$ direction.

With regard to your question:

My problem is that I do not understand why $U$ is described as an integral over the projection-valued measure. Naively, I would think that $U(x,\Bbb 1)=\exp(ip\cdot x)$. Why is this expectation wrong?

In part this is because your notation confuses generators with numbers over which you integrate (ie the double-valuation of the name $p$).

Using the notation from the first yellow box above:

For any self-adjoint operator $A$ you have that the unitary group $U_t=\exp(itA)$ associated to it is given by $$U_t=\int_\Bbb R \exp(it\lambda)dP(\lambda)$$

In a way you can understand $dP(\lambda)$ as the projection onto the $\lambda$-mode (if $\lambda$ is an eigenvalue of $A$, then $dP$ has an atom at $\lambda$). An integral $\int f(\lambda)dP(\lambda)\psi$ says "decompose a vector into its eigenmodes wrt $A$, then multiply the $\lambda$-part with $f(\lambda)$.

You may want to orient yourself to the example on $L^2(\Bbb R)$ given by the operator $\hat x$, "multiplication with the variable".