# Spectral Measures: Helffer-Sjöstrand

Given a Hilbert space $\mathcal{H}$.

Consider a Hamiltonian:
$$H:\mathcal{D}(H)\to\mathcal{H}:\quad H=H^*$$

Regard a function:
$$f\in\mathcal{C}^\infty_0(\mathbb{R}):\quad f(\mathbb{R})\subseteq\mathbb{R}$$

And an extension:
$$f_E\in\mathcal{C}^\infty_0(\mathbb{C}):\quad f_E\restriction_\mathbb{R}=f\quad\bar{\partial}f_E\restriction=0$$

Then one has:
$$f(H)=-\frac{1}{\pi}\int_\mathbb{C}\overline{\partial}f_E(z)R(z)\mathrm{d}\lambda_\mathbb{C}(z)$$
How can I check this?

#### Solutions Collecting From Web of "Spectral Measures: Helffer-Sjöstrand"

This is only a check!

The spectrum is neglible:
$$0\leq\lambda_\mathbb{C}(\sigma(H))\leq\lambda_\mathbb{C}(\mathbb{R})=0$$

By functional calculus:
$$\langle\left(-\frac{1}{\pi}\int_{\mathbb{C}\setminus\mathbb{R}}\frac{\overline{\partial}f_E(z)}{z-H}\mathrm{d}\lambda_\mathbb{C}(z)\right)\varphi,\psi\rangle=-\frac{1}{\pi}\int_{\mathbb{C}\setminus\mathbb{R}}\int_{\sigma(H)}\frac{\overline{\partial}f_E(z)}{z-\lambda}\mathrm{d}\mu_{\varphi\psi}(\lambda)\mathrm{d}\lambda_\mathbb{C}(z)$$

By Fubini’s theorem:
$$-\frac{1}{\pi}\int_{\mathbb{C}\setminus\mathbb{R}}\int_{\sigma(H)}\frac{\overline{\partial}f_E(z)}{z-\lambda}\mathrm{d}\mu_{\varphi\psi}(\lambda)\mathrm{d}\lambda_\mathbb{C}(z)=-\frac{1}{\pi}\int_{\sigma(H)}\int_{\mathbb{C}\setminus\mathbb{R}}\frac{\overline{\partial}f_E(z)}{z-\lambda}\mathrm{d}\lambda_\mathbb{C}(z)\mathrm{d}\mu_{\varphi\psi}(\lambda)$$

By dominated convergence:
$$\int_{\mathbb{C}\setminus\mathbb{R}}\frac{\overline{\partial}f_E(z)}{z-\lambda}\mathrm{d}\lambda_\mathbb{C}(z)=\lim_{\varepsilon\to0^+}\int_{|\Im z|\geq\varepsilon}\frac{\overline{\partial}f_E(z)}{z-\lambda}\mathrm{d}\lambda_\mathbb{C}(z)$$

Note the identity:
$$\mathrm{d}\lambda_\mathbb{C}=\mathrm{d}x\mathrm{d}y:\quad\mathrm{d}\overline{z}\wedge\mathrm{d}z=2i\mathrm{d}x\wedge\mathrm{d}y$$

By holomorphy one gets:
$$d\left(\frac{f_E(z)}{z-\lambda}\mathrm{d}z\right)=\frac{\overline{\partial}f_E(z)}{z-\lambda}\mathrm{d}\overline{z}\wedge\mathrm{d}z$$

By Stokes’ theorem:
$$\frac{1}{2i}\int_{|\Im z|\geq\varepsilon}\frac{\overline{\partial}f_E(z)}{z-\lambda}\mathrm{d}\overline{z}\wedge\mathrm{d}z=\frac{1}{2i}\int_{|\Im z|=\varepsilon}\frac{f_E(z)}{z-\lambda}\mathrm{d}z$$

Regard the integrand:
$$h(\varepsilon):=\frac{f_E(x+i\varepsilon)}{x-\lambda+i\varepsilon}-\frac{f_E(x-i\varepsilon)}{x-\lambda-i\varepsilon}$$

By almost analyticity:
$$\overline{\partial}f_E\restriction_\mathbb{R}=0:\quad f_E(x\pm i\varepsilon)\approx f(x)\pm(-i)f'(x)\varepsilon$$

The integrand becomes:
$$h(\varepsilon)\approx\frac{1}{\pi}\frac{\varepsilon}{(x-\lambda)^2+\varepsilon^2}(-2\pi i)\left\{f(x)+f'(x)(x-\lambda)\right\}$$

The nascent delta gives:
$$\lim_{\varepsilon\to0^+}\frac{1}{2i}\int_{-\infty}^\infty h(\varepsilon)\mathrm{d}x=\frac{1}{2i}(-2\pi i)f(\lambda)=-\pi f(\lambda)$$

Inserting the expression:
$$-\frac{1}{\pi}\int_{\mathbb{C}\setminus\mathbb{R}}\frac{\overline{\partial}f_E(z)}{z-H}\mathrm{d}\lambda_\mathbb{C}(z)=-\frac{1}{\pi}\int_{\sigma(H)}(-\pi)f(\lambda)\mathrm{d}\mu_{\varphi\psi}(\lambda)=\langle f(H)\varphi,\psi\rangle$$

Concluding Helffer-Sjöstrand.