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Since the energy needed in the hopping process is provided by the electric field, no thermal activation is required anymore and a field induced tunneling current is expected


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In some sense, there is some kind of classical tunneling. For example, a high jumper, if (s)he bends her/his body over the bar, can pass over the bar in such a way that its center of gravity is always below the bar, and his kinetic + potential energy is always less than mgh, where m is her/his mass and h is the height of the bar. See, e.g., A. Cohn, M. ...


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One should always remember that quantum mechanics predicts probabilities and not energy distributions . The energy a particle will have is an eigenvalue of the energy operator operating on the wavefunction, but the probability of finding a particle at (x,y,z) at time t is given by the complex conjugate square of the wavefunction which is the solution of the ...


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As mentioned in the link you provided, it is due to Heisenberg's uncertainty relation $-$ during the short-lasting tunneling, the particle may temporarily borrow some energy from the potential of the barrier, so sometimes it can jump over it. Well, the energy and time may be depicted as a sort of Fourier transform pair (see Fourier transform) because the ...


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The particle doesn't borrow energy. Your idea that it gains energy is a guess about what is happening during the experiment, and it is wrong. You are thinking of a particle in a tunnelling experiment as being like a ball rolling up a hill that is too high for the ball to get over the top with the amount of energy you have given it. The ball rolls part way ...


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On a quantum level, particles don't really have "momentum". They're waves. The way the Schrodinger equation works, they move faster if they have a shorter wavelength. So we defined momentum based on the wavelength. Kinetic energy also is part of the whole conservation of energy thing, so we have a very good reason to define it how we did, but again, it's ...


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You mean quantum tunneling? The particle doesn't really "borrow" energy, actually a particle will have a higher probability of tunneling through a barrier if it has a high kinetic energy. A naive analogy would be that the more energetic the bullet is, the higher the probability it has of piercing through a wall, that is, tunneling. What makes quantum ...


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I don't think it's quite that simple. The resistivity may be related to the barrier potential energy separating the two reservoirs, which directly modulates tunneling in the obvious way (higher energy = less tunneling). But there is one barrier which classically you'd describe with an infinite resistance (a Dirac $\delta$-function potential) which admits ...


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Firstly, the transmission is always between zero and one, and there are some sweet spots where it is one exactly so in between two sweet spots it has to decrease then increase again. But that's a bit uninformative and uneducational. And you asked about why, so let's look at the why of tunneling and quantum dynamics in general. You start with a wave packet ...


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The transmission probability depends on the relation between the barrier width and the de Broglie's wavelength of the electron (within the barrier). The waves reflected from the front and the end of the barrier interfere constructively or destructively depending on this relation. The description of a similar phenomenon for light can be found, e.g., at ...


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I) Recall that the probability current 1D QM is $$\tag{1} J(x)~:=~\frac{i\hbar}{2m}W(\psi,\psi^{\ast}) (x), $$ where $$\tag{2} W(\psi,\psi^{\ast}) (x)~:=~\psi(x) \psi^{\prime}(x)^{\ast}- \psi^{\prime}(x)\psi(x)^{\ast}$$ is the Wronskian. Unitarity of the $S$-matrix is equivalent to the statement that $$ \lim_{x\to -\infty} W(\psi,\psi^{\ast}) ...



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