Calculate Revival Time of a Wave in Griffiths' QM 2nd Ed.

[Abhishek11235]

In Griffiths,Quantum Mechanics 2nd edition,Chapter 2,he gives a problem to calculate the revival time of a wave. Revival time is defined as the time taken by a wave to go from one side(x=0) to other side(x=a). Now let's calculate the revival time with 2 methods.

Method 1:

Now to go from 1 position to other and coming back to same place is 1 oscillation. The time period of Oscillation is:
$$T= 2π/\omega $$
We have $$E= 1/2 ka^2$$ where E is energy and a is width or amplitude. Since $$k=m(\omega)^2$$ we have after substituting in energy equation and then in the Equation for time period we find:
$$T= \sqrt{2m/E}×a×π$$

Method 2:

Now,by defination,the particle covers distance 2a with average velocity v given by:

$$v=\sqrt{2E/m}$$

So $vT=2a$ gives
$$T= a\sqrt{2E/m}$$

The question is why the above 2 methods give different result?

After checking solution manual,I found method 2 answer to be correct.

 

[kuruman]

The problem in Griffiths (2.39) states "Show that the wave function of a particle in the infinite square well returns to its original form after a quantum revival time $T=4ma^2/\pi\hbar$. That is $\Psi(x,T)=\Psi(x,0)$ for any state, (not just a stationary state)." I assume you are interested in part (b) which asks "What is the classical revival time, for a particle of energy E bouncing back and forth between the walls?" Well, your method 2 doesn't have the correct dimensions for $T$. Check your algebra. Your method 1 has the energy expression $E=1/2ka^2$. What is $k$? If it's some kind of spring constant, note that the potential is a box with infinite walls not a harmonic oscillator potential.