E.g., if I have a time independent wavefunction [itex]\psi(x)[/itex] with Fourier transform [itex]\tilde{\psi}(k)[/itex], in computing the expectation of momentum are we calculating the principal value(adsbygoogle = window.adsbygoogle || []).push({});

[tex]

\lim_{R \to \infty} \int_{-R}^{R} dk\,\lvert \tilde{\psi}(k)\lvert^2\, \hbar k

[/tex]

instead of the improper integral

[tex]

\int_{-\infty}^{\infty} dk\,\lvert \tilde{\psi}(k)\lvert^2\, \hbar k = \lim_{R_1, R_2 \to \infty} \int_{-R_1}^{R_2} dk\,\lvert \tilde{\psi(k)}\rvert^2\,\hbar k

[/tex]

I ask because a solution from the 8.04 Quantum Physics course on MIT OCW only makes sense taking the principal value of the improper integral for the momentum operator's expectation, and not taking the mathematical definition of an improper integral. Is it assumed integrals over the real line are principal values in quantum physics? Fourier transforms are usually interpreted as principal values, correct? Or no?

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# Are the improper integrals used in calculating expectations to be interpreted as principle values?

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