I'm puzzling out the way one can determine $exp(iA)$ operator for an unbounded $A$. More precisely I would like to know, how could I deal with the momentum operator $p_x = -i\hbar\partial_x$.

I'm trying to follow the instruction which I learned from Frederic Schullers lectures (https://www.youtube.com/watch?v=GbqA9Xn_iM0 lecture 10 and 11)

The instruction is as following:

1) Construct the positive real-valued measure $\mu_{\psi}$ using Stieltjes inversion formula:

$\mu_{\psi}((-\infty,\ \lambda]) = \lim_{\delta\to 0^{+}}\lim_{\epsilon\to 0^{+}}\frac{1}{\pi} \int\limits_{-\infty}^{\lambda+\delta}dt \ Im<\psi, R_p(t + i\epsilon)\psi> $, where $R_p(z)$ is a resolvent operator.

2)Construct complex-valued measure $\mu_{\psi, \phi}$ by polarization formula:

$\mu_{\psi, \phi}(\Omega) = \frac{1}{4} [\mu_{\psi+\phi}(\Omega) - \mu_{\psi-\phi}(\Omega)+i\mu_{\psi-i\phi}(\Omega)-i\mu_{\psi+i\phi}(\Omega)], \Omega$ is a Borel set in $\mathbb{R}$

3) Construct projection-valued measure $P$ as following:

$<\psi, P(\Omega)\phi> := \int \chi_\Omega d\mu_{\psi, \phi}$

4) Calculate the integral:

$exp(i\hbar\partial_x) := \int_{\mathbb{R}} e^{i\lambda}\ P(d\lambda)$

My achievments are really poor. Actually, I've just calculated the resolvent for the momentum operator: $R_p(z) = \frac{i}{\hbar} \int\limits_{0}^{\infty}dt\ e^{izt\hbar^{-1}}u(t)$, where $u(t)\psi(x) = \psi(x-t)$. But I got in trouble even at the first step while calculating Stieltjes integral. I've achieved such thing to calculate:

$\frac{i}{h}\int\limits_{0}^{\infty}da\ e^{iza\hbar^{-1}}\ \int_{\mathbb{R}} dx\ \psi^{*}(x)u(a)\psi(x)$ and I don't know how to deal with it.

Maybe someone tried this way and can give me some advices? However, I bet it is not the shortest way to rigorously construct exponential of unbounded operator. Are they any other ideas to define exp? If they are, what is the main purpose of the spectral theorem and all of equastions I mentioned before. It seems to be not constructive in this case.


The spectral resolution of $A=-i\partial_x$ is given by $$ E(S)f = \frac{1}{2\pi}\int_{S}e^{is x}\left(\int_{-\infty}^{\infty}f(w)e^{-isw}dw\right)ds. $$ You can determine this using the Stieltjes Inversion Formla. As a simple check, note that $E(\mathbb{R})=(\hat{f})^{\vee}=f$. That is $E(\mathbb{R})=I$ as expected. You can perform a change of variable in $s$ to get a modified version that is suited to your scaled version of $A$. The spectrum is continuous, which rules out discontinuities in the spectral measure, which is consistent with the above form.

For the case at hand (not your scaled case,) the exponential $e^{iAt}f$ is $$ e^{itA}f = \int_{\mathbb{R}}e^{its}d_sE(-\infty,s]f= \frac{1}{2\pi}\int_{-\infty}^{\infty}e^{ist}e^{isx}\left(\int_{-\infty}^{\infty}e^{-isw}dw\right)ds \\ \implies (e^{itA}f)(x) = f(x+t). $$

  • $\begingroup$ Thank you, Mr. DisintegratingByParts! However, after several failed attemptions I couldn't derive spectral resolution from Stieltjes Inversion Formula. Maybe you saved drafts and could share them with, please. $\endgroup$ – Max Borovkov Sep 11 '17 at 17:04

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