vanhees71 said:
I can't make sense of Eq. (3) in #4.
Even though Eq(3) is nearly as old as QM, I am not surprised to hear you saying that. This is because of the unfortunate fact that many people are still unfamiliar with the great work of Herman Wyle on the use of group theory in QM:
As early as 1925, Weyl considered the projective unitary representation of the abelian group of translations \mathbb{R}^{2} in phase space: \begin{align*} \pi : & \mathbb{R}^{2} \to \mbox{PU}(\mathcal{H}) \\ & ( \alpha , \beta ) \mapsto e^{\frac{i}{\hbar} (\beta Q - \alpha P)} \end{align*} , where [Q , P] = i \hbar. Indeed, it is easy to show that \pi (\alpha , \beta ) is a projective unitary operator on \mathcal{H}: \pi (\alpha , \beta) \pi (\alpha^{\prime} , \beta^{\prime}) = e^{- \frac{i}{2 \hbar} (\alpha \beta^{\prime} - \beta \alpha^{\prime})} \ \pi (\alpha + \alpha^{\prime} , \beta + \beta^{\prime}) . Then, Weyl extends the projective representation \pi to a representation of the (twisted) group algebra of \mathbb{R}^{2} on the same Hilbert space \mathcal{H} (group algebra is another unfamiliar concept, I suppose!). In English, for a function a ( q , p ) on phase space, Weyl takes the associated operator to be A \equiv \hat{\pi} (a) = \int_{\mathbb{R}^{2}} d \alpha d \beta \ \tilde{a}( \alpha , \beta ) \ \pi (\alpha , \beta ) , \ \ \ \ (1) where \tilde{a} ( \alpha , \beta ) = \frac{1}{(2 \pi \hbar)^{2}} \int_{\mathbb{R}^{2}} dq dp \ e^{- \frac{i}{\hbar} (\beta q - \alpha p)} \ a ( q , p ) , is the Fourier transform of a. The operator \hat{\pi}(a) is, at least formally, self-adjoint if a(q,p) is real-valued, and is well-defined and of trace class if a(q,p) belongs to the Schwartz space \mathcal{S}(\mathbb{R}^{2}), i.e. if a is C^{\infty} and decreases, together with all its derivatives, faster than the reciprocal of any polynomial at infinity. If the function a is C^{\infty} and grows, together with all its derivatives, at most polynomially at infinity, then \hat{\pi}(a) will be (in general) a densely defined unbounded operator. And, of course, if \tilde{a} \in L^{1}(\mathbb{R}^{2}), then \hat{\pi}(a) is a bounded (i.e., continuous) operator: \lVert \hat{\pi}(a) \rVert \leq \int_{\mathbb{R}^{2}} d \alpha d \beta \ \lvert \tilde{a}( \alpha , \beta ) \rvert .
Okay, let us now derive Eq(3) of #4 from the Weyl formula (1) which we now write as H \equiv \hat{\pi}(h) = \int_{\mathbb{R}^{2}} d \alpha d \beta \ \tilde{h}( \alpha , \beta ) \ \pi ( \alpha , \beta ) . \ \ \ \ (2) In the coordinate representation, i.e. when we consider the realization of the space \mathcal{H} in the form L^{2}(\mathbb{R}), an observable H : L^{2}(\mathbb{R}) \to L^{2}(\mathbb{R}) is characterized as an integral operator defined by its kernel (or matrix elements): (H \Psi )(q) = \int_{\mathbb{R}} d \bar{q} \ H(q , \bar{q}) \Psi (\bar{q}) , \ \ \ \forall \Psi \in L^{2} (\mathbb{R}) . Alternatively, as Eq(3) of #4 says, we may define the observable H in terms of its (Weyl) symbol, which is a function h(q,p) on phase space. We will now show that the symbol h(x,p) corresponds to the integral kernel H(x,y) \equiv \langle x | H |y \rangle by means of \langle x | H | y \rangle = \frac{1}{2 \pi \hbar} \int_{\mathbb{R}} dp \ h \left( \frac{x + y}{2} , p \right) \ e^{\frac{i}{\hbar} p (x - y)} . Okay, let us calculate the following matrix element of the operator H of Eq(2): \langle q - \frac{\lambda}{2} | H | q + \frac{\lambda}{2} \rangle = \int d \alpha d \beta \ \tilde{h} ( \alpha , \beta ) \ \langle q - \frac{\lambda}{2} | \pi ( \alpha , \beta ) | q + \frac{\lambda}{2} \rangle . \ \ \ (3) The matrix element of \pi can be calculated easily from the factorized form \pi ( \alpha , \beta ) = e^{- \frac{i}{2 \hbar} \alpha \beta} \ e^{\frac{i}{\hbar} \beta Q} \ e^{- \frac{i}{\hbar} \alpha P} . Using e^{- \frac{i}{\hbar} \alpha P} | q \rangle = |q + \alpha \rangle and e^{\frac{i}{\hbar} \beta Q} |q \rangle = e^{\frac{i}{\hbar} \beta q }|q \rangle, we obtain \langle q - \frac{\lambda}{2} | \pi ( \alpha , \beta ) | q + \frac{\lambda}{2} \rangle = e^{\frac{i}{\hbar} \beta q} \ \delta ( \lambda + \alpha ) . Substituting this in (3) then multiplying the result by e^{\frac{i}{\hbar} \lambda p} and integrating over \lambda, we find \int_{\mathbb{R}} d \lambda \ e^{\frac{i}{\hbar} \lambda p} \langle q - \frac{\lambda}{2} | H | q + \frac{\lambda}{2} \rangle = \int_{\mathbb{R}^{2}} d \alpha d \beta \ \tilde{h} ( \alpha , \beta ) \ e^{\frac{i}{\hbar}( \beta q - \alpha p )} . We recognize the RHS as the inverse Fourier transform h (q , p) of \tilde{h}( \alpha , \beta ): \int_{\mathbb{R}} d \lambda \ e^{\frac{i}{\hbar} \lambda p} \langle q - \frac{\lambda}{2} | H | q + \frac{\lambda}{2} \rangle = h (q , p ) . Thus \int_{\mathbb{R}} dp \int_{\mathbb{R}} d \lambda \ e^{\frac{i}{\hbar} ( \lambda - \lambda^{\prime} ) p} \langle q - \frac{\lambda}{2} | H | q + \frac{\lambda}{2} \rangle = \int_{\mathbb{R}} dp \ h(q , p) \ e^{- \frac{i}{\hbar} \lambda^{\prime} p} . The p-integral on the LHS produces 2 \pi \hbar \delta ( \lambda - \lambda^{\prime} ), then the \lambda-integration gives \langle q - \frac{\alpha}{2} | H |q + \frac{\alpha}{2} \rangle = \frac{1}{2 \pi \hbar} \int dp \ h(q,p) \ e^{- \frac{i}{\hbar} \alpha p} . Now, you have it: define x = q - \frac{\alpha}{2} , \ \ y = q + \frac{\alpha}{2}, and you find \langle x | H | y \rangle = \frac{1}{2 \pi \hbar} \int dp \ h \left( \frac{x + y}{2} , p \right) \ e^{\frac{i}{\hbar} p ( x - y ) } . \ \ \ (4) Conversely, you can start from (4) and arrive at the Weyl formula (2).
Mathematicians generalized the above Weyl’s work on the abelian group \mathbb{R}^{2n} to an arbitrary locally compact Lie group G and developed what we now know as “abstract harmonic analysis” or “functional analysis on locally compact groups”: Given a unitary representation \pi of a locally compact group G on some Hilbert space \mathcal{H}: \pi : G \to \mbox{U}(\mathcal{H}), g \mapsto \pi (g), and a “nice” function a : G \to \mathbb{R} on G, then for all a \in L^{1}(G), the operator \hat{\pi}(a) = \int_{G} d \mu (g) \ a(g) \ \pi (g) , is bounded \lVert \hat{\pi} \rVert \leq \int_{G} d \mu (g) \ \lvert a (g) \rvert . Indeed, it is easy to show that \hat{\pi} : L^{1}(G) \to \mathcal{L}(\mathcal{H}) is a non-degenerate \ast-homomorphism from the group algebra \left( L^{1}(G , d \mu ) , \ast \right) into the algebra of bounded operators on the same Hilbert space. That is, \hat{\pi} (a) \ \hat{\pi} (b) = \hat{\pi} ( a \ast b ),\hat{\pi}^{\ast}(a) = \hat{\pi} ( a^{\ast} ), where \left( a \ast b \right) ( g ) = \int_{G} d \mu (h) \ a (h) \ b (h^{-1}g), is the convolution product on L^{1}(G), and a \mapsto a^{\ast} is an involution on L^{1}(G) defined by a^{\ast} (g) = \Delta (g^{-1}) \bar{a} (g^{-1}), where \Delta : G \to \mathbb{R} is the (continuous) modular function on G, a group homomorphism \Delta (g) \Delta (h) = \Delta (gh). Moreover, we can show that there exists a bijective correspondence between the continuous unitary representations \pi (G) in \mathcal{H} and the non-degenerate \ast-representations \hat{\pi}\left( L^{1}(G) \right) in the same Hilbert space.
So, Eq(3) of #4 is in fact the reason why we now do abstract harmonic analysis on locally compact Lie groups.
