Principal Values & Fourier Transforms in Quantum Physics

In summary: The result is$$\int_{-\infty}^\infty \frac{\sin^2 \left(k a/2 \right)}{k}~dk \approx \int_{-\infty}^{\infty} \frac{k - k^3/24}{k}~dk = 2 \pi i$$The ##- k^3/24## comes from the fourth-order term in the Taylor series of ##\sin^2 \left(k a/2 \right)##. All the other terms cancel. The imaginary part of the integral is zero, so the integral is zero.It would help if you posted the actual MIT integral you refer to.Generally, when there are poles on the integration
  • #1
homer
46
0
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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  • #2
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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Likes homer
  • #3
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?
 
  • #4
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
 
  • #5
Sounds like something worth checking out. Thanks for the recommendation.
 
  • #6
homer said:
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.
 
  • #7
strangerep said:
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].
 
  • #8
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.
 
  • #9
homer said:
[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.
 

1. What is the concept of principal values in quantum physics?

In quantum physics, principal values refer to the most probable outcomes of a measurement or observation in a quantum system. These values are determined by the wave function of the system and are used to calculate probabilities of different outcomes.

2. How are principal values calculated in quantum physics?

The calculation of principal values in quantum physics involves using the mathematical tool of Fourier transforms. This allows for the conversion of a wave function in position space to a wave function in momentum space, where principal values can be determined.

3. What is the significance of Fourier transforms in quantum physics?

Fourier transforms play a crucial role in quantum physics as they allow for the analysis and understanding of the wave nature of particles. They also aid in the calculation of principal values and probabilities in quantum systems.

4. Can principal values change in quantum systems?

In most cases, principal values in quantum systems remain constant as they are determined by the wave function. However, in systems with time-dependent potentials, the principal values can change over time as the wave function evolves.

5. How are Fourier transforms used in practical applications of quantum physics?

Fourier transforms are used in a wide range of practical applications in quantum physics, including quantum computing, quantum cryptography, and quantum communication. They also have applications in fields such as signal processing and data compression.

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