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- Homework Statement
- A particle moves in a 1D symmetric infinite square well of length ##2a##. Inside the well, the potential is ##V(x) = \gamma \delta(x)##. For sufficiently large ##\gamma##, calculate the time required for the particle to tunnel from being in the ground state of the well extending from ##x = -a## to ##x = 0## to the ground state of the well extending from ##x=0## to x=a##.

- Relevant Equations
- ##\Delta t \Delta E \geq \hbar/2##

I am guessing time-energy uncertainty relation is the way to solve this. I solved the Schrodinger equation for both the regions and used to continuity at ##x=-a, 0,a## and got ##\psi(-a<x<0) = A\sin(\kappa(x+a))## and ##\psi(0<x<a) = -A\sin(\kappa(x-a))## where ##\kappa^2 = 2mE/\hbar^2##. Looking at the discontuity in the derivative, I got the transcendental equation

$$\tan(\kappa a) = -\frac{\hbar^2 \kappa}{m\gamma}$$

Setting ##x = \kappa a##,

$$\tan(x) = -\frac{\hbar^2 }{m\gamma a}x$$.

Imitating what one does for finite square well case, I plotted both the curves. One root is at ##x=0##. But that would mean ##\kappa = E = 0##. So that cannot represent a bound state. Next root for ##\gamma a = m/\hbar^2## (which means ##\tan(x) = -x##) is ##x=2.029##. This implies

$$\kappa = \frac{2.029}{a} = 2.029\frac{\hbar^2\gamma}{m}$$

This gives

$$E = \frac{2m\kappa^2}{\hbar^2} = \frac{2m}{\hbar^2}\left(\frac{4.117 \hbar^4\gamma^2}{m^2}\right) = \frac{8.234\hbar^2\gamma^2}{m}$$.

Now ##\Delta t \Delta E \geq \hbar/2##. But $$E = p^2/2m \implies \Delta E = p\Delta p/m= p\hbar/(2m \Delta x) = p\hbar/(2m a) = \sqrt{2mE} \hbar/(2ma) = \sqrt{\frac{2E}{ma^2}}\frac{\hbar}{2}$$

Thus $$\Delta E = \sqrt{\frac{2E}{ma^2}}\frac{\hbar}{2}$$

Finally,

$$\Delta t = \frac{\hbar}{2\Delta E} = \sqrt{\frac{ma^2}{2E}}$$

Substituting ##E = \frac{8.234\hbar^2\gamma^2}{m}## I get

$$\Delta t = 0.4964\frac{ma\gamma}{\hbar}$$

Does all of this make sense? Is there a simpler method and I am over-complicating an extremely simple problem?

$$\tan(\kappa a) = -\frac{\hbar^2 \kappa}{m\gamma}$$

Setting ##x = \kappa a##,

$$\tan(x) = -\frac{\hbar^2 }{m\gamma a}x$$.

Imitating what one does for finite square well case, I plotted both the curves. One root is at ##x=0##. But that would mean ##\kappa = E = 0##. So that cannot represent a bound state. Next root for ##\gamma a = m/\hbar^2## (which means ##\tan(x) = -x##) is ##x=2.029##. This implies

$$\kappa = \frac{2.029}{a} = 2.029\frac{\hbar^2\gamma}{m}$$

This gives

$$E = \frac{2m\kappa^2}{\hbar^2} = \frac{2m}{\hbar^2}\left(\frac{4.117 \hbar^4\gamma^2}{m^2}\right) = \frac{8.234\hbar^2\gamma^2}{m}$$.

Now ##\Delta t \Delta E \geq \hbar/2##. But $$E = p^2/2m \implies \Delta E = p\Delta p/m= p\hbar/(2m \Delta x) = p\hbar/(2m a) = \sqrt{2mE} \hbar/(2ma) = \sqrt{\frac{2E}{ma^2}}\frac{\hbar}{2}$$

Thus $$\Delta E = \sqrt{\frac{2E}{ma^2}}\frac{\hbar}{2}$$

Finally,

$$\Delta t = \frac{\hbar}{2\Delta E} = \sqrt{\frac{ma^2}{2E}}$$

Substituting ##E = \frac{8.234\hbar^2\gamma^2}{m}## I get

$$\Delta t = 0.4964\frac{ma\gamma}{\hbar}$$

Does all of this make sense? Is there a simpler method and I am over-complicating an extremely simple problem?