Can the Time-Energy Uncertainty Relation Simplify the Schrodinger Equation?

[Wannabe Physicist]

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?

 

[anuttarasammyak]

I observe physical dimension of your result is
dim[\frac{ma\gamma}{\hbar}]=ML^2T^{-1}
, not T which we expect. Instead I see
dim[\frac{a\hbar}{\gamma}]=T
Is it helpful for checking your result ?

 

[Wannabe Physicist]

Yes. Thank you.