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

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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
[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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The correct mathematical setting for QM are what's known as Rigged Hilbert Spaces - which, basically, is Hilbert Spaces with distribution theory stitched on. In distribution theory the principal value is generally used. Why is the subject of a full development of it which really should be part of any applied mathematicians armoury. The following is a really good book for that:
https://www.amazon.com/dp/0521558905/?tag=pfamazon01-20

Thanks
Bill
 
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Thanks bhobba! Maybe I'll take that course on distributions offered by coursera in January (though it's in French). Is distribution theory something I should study by itself, or is it something one would pick up studying grad level QM books?
 
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You should study it on your own. It will enrich not just understanding QM and that damnable Dirac Delta function, but many areas of applied math. Its treatment of Fourier Transforms is simply so easy and elegant you won't want to do it any other way. If you come across other approaches with deep theorems of convergence etc you will scratch your head - why bother.

I highly recommend you get the book I linked to for your library. I have it, but learned distribution theory from other sources - wish I started with that book - many other texts require a good knowledge of functional analysis - which fortunately I had - but isn't really required.

Thanks
Bill
 
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Sounds like something worth checking out. Thanks for the recommendation.
 

strangerep

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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?
It would help if you posted the actual MIT integral you refer to.

Generally, when there are poles on the integration contour, one (tries to) interpret the integral as a Cauchy Principal Value integral, since (if it exists), it's compatible with the notion of a Lebesgue integral.
 
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It would help if you posted the actual MIT integral you refer to.

Generally, when there are poles on the integration contour, one (tries to) interpret the integral as a Cauchy Principal Value integral, since (if it exists), it's compatible with the notion of a Lebesgue integral.
The integral was

[tex]\langle \hat{p}\rangle =
\int_{-\infty}^{\infty} dk\,\Big(
\frac{2}{\pi a}
\frac{\sin^2(ka/2)}
{k^2}
\Big)\,\hbar k =
\frac{2\hbar}{\pi a}
\int_{-\infty}^{\infty} dk\,\Big(
\frac{\sin^2(ka/2)}
{k}
\Big)
[/tex]

which was argued to be zero because it is integrating an odd function from [itex]-\infty[/itex] to [itex]+\infty[/itex], which makes sense if we're taking the limit as [itex]R \to \infty[/itex] of integrating from [itex]-R[/itex] to [itex]+R[/itex] so that we really can say the domain of integration is symmetric about [itex]k = 0[/itex].
 

George Jones

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Gold Member
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Interesting. Simpler examples with odd functions: ##\int_{-\infty}^\infty x ~dx## and ##\int_{-\infty}^\infty \sin x ~dx##. The definition given in the text that I used in first-year: if ##f## is continuous on ##\mathbb{R}##, then ##\int_{-\infty}^\infty f \left(x \right)~dx## is defined as

$$\int_{-\infty}^\infty f \left(x \right)~dx = \lim_{a \rightarrow -\infty} \int_a^c f \left(x \right)~dx + \lim_{b \rightarrow \infty} \int_c^bf \left(x \right)~dx$$

where ##c## is any real number. By this definition, both the integrals that I have given are divergent.
 

George Jones

Staff Emeritus
Science Advisor
Gold Member
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[tex]\int_{-\infty}^{\infty} dk\,\Big(
\frac{\sin^2(ka/2)}
{k}
\Big)
[/tex]
Let's take a closer at the integrand. As ##k## goes to ##\pm \infty##, the integrand goes as ##1/k##; as ##k## goes to zero, the integrand goes as ##k^2/k = k##. Split the range of integration into

$$\int_{-\infty}^\infty = \int_{-\infty}^{-L} + \int_{-L}^{-s} +\int_{-s}^0 + \int_0^s + \int_s^L + \int_L^\infty ,$$

where ##L## is a sufficiently large number and ##s## is a sufficiently small number. Because the integrand is odd, ##\int_{-L}^{-s} + \int_s^L =0##

All the other integrals, including the parts at zero, are improper. To get a feel for what happens, use the standard definitions for improper integrals (take PV at zero) and the the approximations that I gave above.
 

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