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

[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

 

[BvU]

Hello @haziq ,

$\qquad$ !​

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 $\LaTeX$ ?
$\$

 

[haziq]

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]

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

 

[haziq]

@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...

 

[vanhees71]

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 $t \rightarrow \infty$.