Problem 2 in "Quantum Theory for Mathematicians", Solving for the travel time of a particle in a potential

In summary, @BvU has been trying to solve problem 2 in Hall’s QM book for ages, but is having trouble. He thinks that it might be because the book is titled Quantum Theory for Mathematicians and he’s not a mathematician, so he is looking for help from someone who could provide a formal proof. @BvU is unsure of what to do next.
  • #1
haziq
3
1
Homework Statement
Problem 2 in Chapter 2 of Hall’s QM book. See pictures below
Relevant Equations
See photo below
9DC9B877-BAD4-4B03-943A-F785EF133E35.jpeg
910F8033-9470-490C-9F50-8329C5AAAADC.jpeg
4ADAC378-42EC-4A12-88CA-53E329483451.jpeg

I’ve been trying to solve this for ages. Would really appreciate some hints. Thanks
 
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  • #2
Hello @haziq ,
:welcome: ##\qquad ## !​

haziq said:
Homework Statement:: Problem 2 in Chapter 2 of Hall’s QM book. See pictures below
Relevant Equations:: See photo below

solve this for ages

And you are aware that with 'this' you mean the step from $$\dot x(t)=\sqrt{{2\left (E_0-V\left ( x\left (t \right )\right) \right) \over m}} $$ to ##t= ...## by separation ?

PF guidelines require that you actually post your best attempt before we are allowed to assist ...

PS notice how much sharper it looks with ##\LaTeX## ?
##\ ##
 
  • #3
Hi @BvU, sorry for the confusion. That’s not the problem I was referring to. I just included that for context. That’s actually problem 1 and it was pretty easy to solve. With regards to problem 2, I’m completely lost. Probably because the book is titled Quantum Theory for Mathematicians and I’m not a mathematician. Perhaps I could share my attempt after someone gives me some hints?
 
  • #4
BvU said:
PF guidelines require that you actually post your best attempt before we are allowed to assist .
 
  • #5
@BvU Fair enough :smile:. Here’s what I’ve deduced so far (for part a)
  • Assuming ##V’(x_1) \neq 0##, we need to show that ##t=lim_{h \rightarrow x_1} \int_{x_0}^{h} {\sqrt{\frac {m} {E_0 - V(y)}} dy} \in \mathbb{R}## since the upper bound is a vertical asymptote.
  • I *think* ##V’(x_1) \neq 0## implies that ##lim_{x \rightarrow x_1}{E_0 - V(x) \neq 0}##. Not sure how to prove this formally. I just applied the definition of derivative and did some sketchy algebra and thought this is plausible.
Not sure what to do next...
 
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  • #6
Ad 2a) It's just that the singularity of the integrand at ##y=x_1## is integrable. If ##V'(x_1) \neq 0##, then you have
$$V(y)=V(x_1) + (y-x_1) V'(x_1) + \mathcal{O}[(y-x_1)^2],$$
and thus around the singularity the integral behaves like
$$\Delta t_{\epsilon}=\int_{x_1-\epsilon}^{x_1} \mathrm{d} y \sqrt{\frac{m}{-2V'(x_1)(y-x_1)}}.$$
Since ##V'(x_1)>0## you have
$$\Delta t_{\epsilon}=-\sqrt{2mV'(x_1) (x_1-y)}|_{y=x_1-\epsilon}^{x_1}= \sqrt{\frac{2 m}{V'(x_1)} \epsilon}.$$
The total time is
$$t=\int_{x_0}^{x_1-\epsilon} \mathrm{d} y \sqrt{\frac{m}{2 [E_0-V(y)]}} + \Delta t_{\epsilon}.$$
Since ##t## doesn't depend on ##\epsilon## and ##\Delta t_{\epsilon} \rightarrow 0## for ##\epsilon \rightarrow 0## the total time is finite.

If ##V'(x_1)=0##, the above Taylor expansion starts at best with the quadratic term, i.e.,
$$V(y)=V(x_1) + \frac{1}{2} (y-x_1)^2 V''(x_1) + \mathcal{O}[(y-x_1)^3],$$
and the above analysis shows that ##\Delta t_{\epsilon}## diverges logarithmically, i.e., even in this case the time ##t \rightarrow \infty##. If also ##V''(x_1)=0## the divergence gets worse, i.e., for ##V'(x_1)=0## the time always ##t \rightarrow \infty##.
 
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