Undergrad Question about inverse operators differential operators

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The discussion focuses on the interpretation of the operator A = 1/(d/dx) in quantum mechanics, specifically how it acts on a normalized wave function ψ in the space of square integrable functions. It is suggested that the operator can be understood through Fourier transforms, where the behavior of Aψ can be analyzed by considering the convergence of integrals involving negative or fractional derivatives. Clarification is provided on the notation used in integrals, emphasizing that both forms of writing integrals are valid, with the latter being more common in physics. The conversation highlights the mathematical foundations necessary for understanding differential operators in quantum mechanics. Overall, the thread explores the complexities of defining and interpreting inverse differential operators.
cmcraes
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Hi all, so I'm not sure if what I'm asking is trivial or interesting, but is there any general or canonical way to interpret say, The follwing operator? (Specifically in the study of quantum mechanics):

A = 1/(d/dx) (I do not mean d-1/dx-1, which is the antiderivative operator )

How would Aψ behave and what (if any) eigenvalues would It have? I'm assuming ψ is in the space of square integrable functions and is normalized.

Thanks!
 
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cmcraes said:
Hi all, so I'm not sure if what I'm asking is trivial or interesting, but is there any general or canonical way to interpret say, The follwing operator? (Specifically in the study of quantum mechanics):

A = 1/(d/dx) (I do not mean d-1/dx-1, which is the antiderivative operator )

How would Aψ behave and what (if any) eigenvalues would It have? I'm assuming ψ is in the space of square integrable functions and is normalized.

Thanks!

You can make sense of it in terms of Fourier transforms:

\psi(x) = \frac{1}{2 \pi} \int dk e^{i k x} \tilde{\psi}(k)
(\frac{d}{dx})^n \psi(x) = \frac{1}{2 \pi} \int (ik)^n e^{i k x} \tilde{\psi}(k)

If the integral on the right converges for some value of n, then you can let that be the definition of (\frac{d}{dx})^n \psi(x) for n negative or even fractional.
 
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stevendaryl said:
You can make sense of it in terms of Fourier transforms:

\psi(x) = \frac{1}{2 \pi} \int dk e^{i k x} \tilde{\psi}(k)
(\frac{d}{dx})^n \psi(x) = \frac{1}{2 \pi} \int (ik)^n e^{i k x} \tilde{\psi}(k)

If the integral on the right converges for some value of n, then you can let that be the definition of (\frac{d}{dx})^n \psi(x) for n negative or even fractional.

Hi, thanks for your answer. Where does the d come from in your first formula? I've never seen it there before.
 
That's just the notation for integration \int dk is an integral over k
 
cmcraes said:
Hi, thanks for your answer. Where does the d come from in your first formula? I've never seen it there before.
There are two equivalent ways to write an integral: ##\int f(x)dx## or ##\int dxf(x)##. The latter is often used in physics, because their expressions of ##f(x)## are frequently quite long, such that it is helpful to note at the start which is the integration variable. The first one is, I think, more traditional and the usual one in mathematics.
 
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