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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?

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  • $\begingroup$ Concerning the last sentence (v4): The Dirac delta potential is (minus) infinite at $x=0$. $\endgroup$
    – Qmechanic
    May 9 at 17:37
  • $\begingroup$ Do u mean that 2.124 meant that $ d\phi/dx$ is continuous except at points where the probability of potential function approaches infinity? So that it also holds for the Dirac delta function? $\endgroup$
    – lel
    May 10 at 0:41
  • $\begingroup$ $\uparrow$ Yes. $\endgroup$
    – Qmechanic
    May 10 at 4:34
  • $\begingroup$ Ah I see thanks so much! $\endgroup$
    – lel
    May 10 at 8:58

1 Answer 1

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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.

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  • $\begingroup$ I see. Based on what i read on 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. $\endgroup$
    – lel
    May 9 at 12:14
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    $\begingroup$ Here "ordinarily" is an adverb mdifying "is" rather than an adjective modifying "continuous". $\endgroup$
    – mike stone
    May 9 at 12:28
  • $\begingroup$ @mike stone: Good point. $\endgroup$
    – Qmechanic
    May 9 at 12:32
  • $\begingroup$ I have read the post you refered to me. My 2nd question still remains unresolved. $\endgroup$
    – lel
    May 9 at 13:47

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