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In all such derivations, Griffiths suggests assuming the constants zero (in this case A) to eliminate the $f(x)$ when it goes to infinity, but zero * infinity is undetermined instead of zero, so how are these derivations valid? Is my understanding wrong here? (I have assumed "blows up" also means infinity)

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    $\begingroup$ Honestly, if this is tripping you up, I strongly suggest (re)learning calculus before studying quantum mechanics. There's already too many confusing things in QM, basic calculus should not be one of them. $\endgroup$ – Prahar Mitra Apr 20 at 9:35
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    $\begingroup$ Note that we very strongly discourage the use of images of text and equations on the site. You should instead use a combination of text and Mathjax. Apart from readability, mathjax and text can be searched using the site engine, whereas miages cannot. $\endgroup$ – StephenG Apr 20 at 10:18
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What you don't understand is Griffiths is not evaluating $0 \cdot x$ when $x$ is infinite, but rather $0 \cdot x$ as $x$ approaches infinity. There is a big difference between the two (in fact, this very difference makes Calculus work). As $x$ tends to infinity, the product $0 \cdot x$ tends to $0$.

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  • $\begingroup$ Thanks for the answer, can you elaborate some more on that difference? $\endgroup$ – Mohd. Farhan Hassan Apr 20 at 18:40
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I think the clearest way to explain it is this. You are thinking of the indeterminate form

$$\lim_{x\to \infty}f(x)g(x) $$

where $$ \lim_{x\to \infty}f(x)=0\\\lim_{x\to \infty}g(x)=\infty$$

this is indeed indeterminate.

But in this case you don't only have $\lim_{x\to \infty}f(x)=0$, you just set $f(x)=0$ for all $x$ (in the notation of your question, $f(x)=A$ and $g(x)=e^{kx}). $

It is obvious that $0\cdot e^{kx}=0$ for any $x$, so $\lim_{x\to \infty}0\cdot e^{kx}=\lim_{x\to \infty}0=0$, or in more simple terms, by just setting $A=0$ you merely remove the term from the solution.

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  • $\begingroup$ But what would you say if there was an expression like $0.\infty$ and the expressions do not tend to 0 or infinity but are exactly 0 and infinity. Or even a better example- take the indeterminate form $0/0$ where the numerator and denominator tend to 0. But what about an expression $0\0$ where the numerators and denominators are exactly 0 and do not tend to it. What will you say in that case. Are these 2expressions that I wrote undefined ones. First one is undefined because the reals is open set at infinity and second one is undefined because of division by 0. Or is there anything else to add $\endgroup$ – Shashaank Apr 24 at 5:58
  • $\begingroup$ @Shashaank There is no such thing as "the expression is exactly $\infty$" in standard analysis, $\infty$ is not a number, but just a notation we use to say that a function is unbounded. The expression $0/0$ is not defined, so what you said at the end of your comment is correct. The expression $0\cdot f(x)$ is defined and it simply equals $0$, so of course its limit is $0$. $\endgroup$ – user2723984 Apr 25 at 9:13
  • $\begingroup$ Yeah I was thinking that. So the thing is that expressions like $0/0$ or even $1/0$ are undefined if the numerators and denominators are exactly 1 or 0. But when the numerator and denominator tend to 0 or 1, then they are indeterminate forms ($1/0$ is not indeterminate as it will tend to infty anyways). So what happens when these values tend to 0 or 1 is what makes them indeterminate. If the the values are exact they are undefined. Is that all correct $\endgroup$ – Shashaank Apr 25 at 9:18
  • $\begingroup$ The point I wanted to highlight is that $0\0$ or other indeterminate forms are undefined expressions if the values are exact and not tending to ( this should hold true for all the indeterminate forms). But when value are not exactly 0 or whatever but tending to it the th expression is not undefined but can take any value and that’s is it is said indeterminate form. We can calculate their limits to know what they tend to in the given expression. Is that all ok $\endgroup$ – Shashaank Apr 25 at 9:25
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In the fragment cited it is not claimed that $0*\infty\rightarrow 0$ - rather it is claimed that the corresponding term is not present in the solution, since it would diverge. Saying $A=0$ is an unrigorous "physicist" way os stating this.

In some more complicated problems a need may arise of actually considering non-zero coefficient $A$, e.g., if instead of a square potential, one considers a smoother one, going to zero at infinity. In this case one would have to assure that function $A(x)$ is such that $$\lim_{x\rightarrow -\infty} A(x)e^{-\kappa x}\rightarrow 0.$$

Quantum mechanics book by Landau & Livshits contains solutions for many smooth potentials of inetrest, in the chapter devoted to one-dimensional Schrödinger equation, which are worth working through for deeper understanding (although this involves a lot of tedious math with hypergeometric and other special fucntions.)

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You can define this with more mathematical rigor, but to say $e^{\kappa x}$ goes to infinity for $x\to\infty$ actually asks what the limit of $e^{\kappa x_n}$ is, if you choose $x_n$ to be a series with limit $\infty$, for example $x_n=n$, which itself means that, while all the $x_n$ actually never go to infinity, there is still no real number $y$ large enough to NOT find a special $n$, lets call it $N$, such that $x_n>y$ for all $n>N$.

So if the limit of $x_n$ is $\infty$, the limit of $e^{\kappa x_n}$ is also $\infty$, while again, no actual value ever is. While the single values never actually would be infinity, a problem arises, if you want to integrate over the whole expontential function (with boundary $\infty$), because this integral would diverge, hence would not yield a physical useful wavefunction. But, if you multiply this series by 0, you get a series that is just 0 everywhere, and the integral converges (being 0 in this region).

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