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I just started studying quantum mechanics using the textbook Introduction to Quantum Mechanics by Griffith. Under the section of solving the Shrodinger equation for a Dirac delta potential, he mentioned that the first derivative must be ordinarily continuous except when potential is infinite. I have 2 queries regarding this.

  1. Firstly, what is the mathematical meaning of ordinarily continuous? Does it mean continuous?

  2. Secondly, he mentioned that the first derivative of the wavefunction is ordinarily continuous since 'the limit on the right is zero'. But evidently, the integral is non-zero when it involves the Dirac delta function, although the potential is not infinite at $x=0$. Where's the issue here?

Frobenius
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lel
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1 Answers1

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Griffiths is appealing to the semantic meaning: ordinarily=usually.

For OP's other questions, in particular the bootstrap equation (2.127), see also e.g. my related Phys.SE answer here.

Qmechanic
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  • I see. Based on what i read on https://en.wikipedia.org/wiki/Uniform_continuity under (ordinary) continuous, I thought he meant uniform continuous, hence the clarification. Regarding the second question, I am referring specifically at (2.127). Griffith argues that the first derivative is ordinarily continuous since the RHS converges to 0, which is not the case as seen from (2.128) regardless of how small $\varepsilon$ is. – lel May 09 '22 at 12:14
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    Here "ordinarily" is an adverb mdifying "is" rather than an adjective modifying "continuous". – mike stone May 09 '22 at 12:28
  • @mike stone: Good point. – Qmechanic May 09 '22 at 12:32
  • I have read the post you refered to me. My 2nd question still remains unresolved. – lel May 09 '22 at 13:47