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  1. What is quantum tunneling

  2. How is it possible?

I tried reading up on itbut i couldn't exactly grasp the concept.

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The formal description

To answer, you need to establish an equivalence of tunneling exponent and action of theory given for specific solution in euclidean time.

Suppose that you have system which is described by one-dimensional QM with coordinate $q$ and the potential $V(q)$ with $V(q_{0}) = 0$. Suppose now that such potential has barrier which divides local minimum at point $q_{0}$ and area where $V(q)$ is the same (given by point $q_{1}$). The QM state with $q = q_{0}$ is thus metastale.

As you know, the probability in the unit of time of its decay is given as $$ \Gamma = Ae^{-S_{b}}, \quad S_{b} = 2\int \limits_{q_{0}}^{q_{1}}\sqrt{2MV(q)}dq $$

The action for such system is given by $$ S = \int dt\left( \frac{M}{2}\left(\frac{dq}{dt}\right)^{2} - V(q)\right) $$ Suppose now that you perform formal transformation $t = i\tau$. Now the action takes the form $$ S_{E} = \int d\tau \left(\left(\frac{dq}{dt}\right)^{2} + V(q)\right) $$ Corresponding EOM takes the form $$ M\frac{d^{2}q}{d\tau^{2}} = -\frac{\partial (-V)}{\partial q} $$ Corresponding integral of motion $$ E_{E} = \frac{M}{2}\left( \frac{dq}{d\tau}\right)^{2} - V(q) $$ is called euclidean energy; note tht it doesn't coincide with the usual energy.

Suppose now that you are looking for the solution $q_{b}$, for which $q_{b} = q_{0}$ at $\tau \to -\infty$ and $q_{b} = q_{1}$ at $\tau = \infty$; euclidean energy for such solution is zero. Now, the action for such configuration is $$ S_{E}[q_{b}] = 2 \int \limits_{-\infty}^{0}\sqrt{2MV(q(\tau))}d\tau = \left|d\tau = \sqrt{\frac{M}{2V(q_{b})}}dq_{b}\right| = S_{b} $$ We come to the statement that in euclidean time $\tau$ the solution $q(\tau) = q_{b}(\tau)$ describes tunneling effect.

Why is it possible?

We see that tunneling solutions exist in euclidean time; they doesn't exist in the usual time. What is the reason for this? The reason is the energy conservation. In the usual time $t$, if we define the solution with the given energy, then it will be the same for all times; thus the solution $q_{b}$ given below doesn't exist in usual time. In euclidean time, however, the usual notion of energy isn't conserved, and existence of $q_{b}$ solution is possible.

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