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I asked my question on Quora and i got various conflicting answers. So, I decided that here I could get a definite answer. Unless the question is not too vague.

According to quantum mechanics, is there non-zero probability for an electron in ground state to undergo quantum tunnelling through a given potential barrier even if the electron does not obtain some additional energy?

(So, we got an electron in ground state... No energy is added to the electron in our imaginary case.. Can the electron quantum tunnel?)

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Quantum tunneling through a potential barrier is not a matter of adding energy. Tunneling occurs because a wavefunction is nonzero on both sides of a potential barrier. This is a purely quantum mechanical phenomenon.

For example, if you compute the scattering states for a delta function potential--an infinitely high potential barrier--you find a nonzero transmission coefficient. This means that there is a nonzero probability of the particle tunneling through the barrier, no extra energy required. The transmission coefficient will scale with the energy of the incident particle, but transmission is not dependent on addition of any extra energy.

Your mention of "ground state" is a bit unclear, since for a potential like the positive delta function, there do not exist any bound states. However, my argument would apply to any potential configuration, such as the adjacent finite square well potential, which admits both bound and scattering states. Essentially, tunneling will occur not because the energy of the particle, but if the wavefunction is nonzero on both sides of whatever potential barrier exists in your configuration.

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  • $\begingroup$ I can't get your idea. In order for tunneling to occur, the electron must finish up in an energy state equal to its initial energy state. The ground state is the lowest energy state the electron can get into - anywhere it went would be a higher energy state, and you’d have violated the conservation of energy. So, it seems that the tunnelling probability for an electron in ground state (without adding energy from the environment) is completely zero (not just effectively). $\endgroup$
    – Iv Nik
    Commented Feb 19, 2023 at 10:44
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    $\begingroup$ @IvNik Your problem appears to be that you just don't understand quantum tunnelling very well. $\endgroup$
    – Gert
    Commented Feb 19, 2023 at 11:09
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The confusion here is due to thinking of tunneling as a dynamic process, which is usually ultimately traced to intuition from classical/Newtonian mechanics, from thinking of electrons as particles, as well as from the semantic meaning of term tunneling in everyday life. If thinking of electron as a wave, tunneling is no mystery at all - sound waves tunnel through walls, electromagnetic waves tunnel through dielectric and so on. And this applies to standing waves (like those corresponding to a ground state), as well as to propagating waves.

Classical background
In response to comments, I would like to expand here on why tunneling is impossible in classical mechanics, and why these arguments do not apply in QM.

Let us consider a classical particle moving in one dimension in potential $U(x)$. Its motion is described by equation (aka Newton's second law) $$ m\ddot{x}=-\frac{d}{dx}U(x), \tag{1} $$ where $x(t)$ is a function of time, dependent on the initial conditions, $x(0), \dot{x}(0)$, and called trajectory.

Equation (1) has a first integral, called energy. It is easy to obtain by multiplying the equation above with $\dot{x}(t)$ and integrating over time: $$ \dot{x}\left[m\ddot{x}+\frac{d}{dx}U(x)\right]=\frac{d}{dt}\left[\frac{m\dot{x}^2}{2}+U(x)\right]=0\Rightarrow E\left[x,\dot{x}(t)\right]=\frac{m\dot{x}^2}{2}+U(x)=\text{const}\tag{2} $$ That is, combination $\frac{m\dot{x}^2}{2}+U(x)$ remains constant along any trajectory, although this constant might be different for different trajectories. This is known as energy conservation. Energy conservation is fundamental in the sense that it follows from the homogeneity of time, which restricts the possible forms of physics laws (Noether's theorem). The specific form of the energy conservation (i.e., the specific form of this first integral) is however not fundamental - it is specific to the given form of the laws of motion, which, as we know, do not apply in quantum mechanics.

The behavior of the system can be represented as a phase portrait, which shows the trajectories as functions of both position $x$ and velocity $\dot{x}$. The lines in this portrait correspond to constant energy. Note that to speak about a constant energy along a line, we need to be able to calculate the energy at every point $x,\dot{x}$, i.e., the position and the velocity must be known simultaneously.

As we follow a line in a phase portrait, increase in potential energy $U(x)$ is accompanied by the decrease of kinetic energy $m\dot{x}^2/2$ and vice versa, so that their sum remains constant. Since $x(t)$ is a real function (and so is its derivative $\dot{x}$), we cannot have situations where $\frac{m\dot{x}^2}{2}<0$. Thus, for a specified energy, the regions with $\frac{m\dot{x}^2}{2}<0$ are (classically) forbidden regions.

The math above is what physicists refer to as classical intuition regarding the motion of a particle. It is different from laymen intuition, since it is grounded in solid logical and mathematical arguments. Nevertheless, in this case there is a simple laymen analogy with a ball rolling in a rugged surface, whose shape is described by $U(x)$. A less precise analogy is with a train or a person walking in such a potential - less precise, since we do not have energy conservation, but apparently it is this thinking that gives the name to the phenomenon - tunneling, as if we could pass through a barrier even we do not have enough energy to surmount it; as if there was a hole at the energy of the particle.

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Why does this picture fail in quantum mechanics?
Many elements of the analysis given above are not true in quantum mechanics:

  • The equation of motion (1) does not apply - instead we have to use Schrödinger equation. The closest that we get to this equation is the Ehrenfest theorem, which applies only to average quantities.
  • Position and momentum (and hence velocity) cannot be measured simultaneously with infinite precision (Heisenberg uncertainty principle), which means that calculating a function that is a function of both position and momentum, and demanding that it remains constant does not make sense. At best we could talk about average energy $\langle H\rangle = \langle\frac{p^2}{2m}\rangle+\langle U(x)\rangle$. While this average is conserved, the actual energy is a random function with tails at high energies allowing passing above the barriers.
  • Note that conservation of energy still holds. Indeed, it is encoded in the Schrödinger equation, where evolution of the wave function is governed by the Hamiltonian, i.e., the energy operator. Since the energy operator commutes with the Hamiltonian (i.e., with itself), its average does not change with time, whereas a particle placed in an eigenstate of this operator remains in this eigenstate.
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  • $\begingroup$ Comments have been moved to chat; please do not continue the discussion here. Before posting a comment below this one, please review the purposes of comments. Comments that do not request clarification or suggest improvements usually belong as an answer, on Physics Meta, or in Physics Chat. Comments continuing discussion may be removed. $\endgroup$
    – Buzz
    Commented Feb 20, 2023 at 17:11

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