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

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

I’ve been trying to solve this for ages. Would really appreciate some hints. Thanks

Last edited:
Hello @haziq ,

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## ?
##\ ##

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?

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

@BvU Fair enough . 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...

berkeman
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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