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I'm slightly confused about the idea of potential barriers and how they affect quantum particles. I understand that when finding the wave function for this particle, the amplitude of the function on the right of the barrier (assuming the particle is moving from the left to the right) is less than on the left simply because there is a lower probability of it transmitting through.

But what I don't understand is why are the wavelengths of the particle on the left and on the right of the barrier the same? Surely on the right it should be longer, since the particle must have lost some energy in order to overcome the potential - and since energy is inversely proportional to the wavelength it means it must have increased?

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The square modulus of the wavefunction $|\psi(x)|^2$ of the particle gives the probability of the particle being at a certain location $x$. This probability, as you say, is lower on the other side of the barrier. However, when you find the energy of the particle, you are solving the Schrodinger equation, and the solution for the energy must be constant - in fact energy must be conserved, so it must be the same before and after the barrier.

The particle doesn't "overcome" the potential as you say, in fact if it did it wouldn't be tunnelling through, and would be following classical rules. However, quantum mechanically, the particle doesn't need to have a certain energy to "spend" to be able to overcome the barrier, because it is possible for it to tunnel through it, without losing any energy. Therefore, the wavelength of the particle would remain constant.

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  • $\begingroup$ So is it like the wavelength is inversely proportional to the total energy of the particle, not just the kinetic energy by itself? $\endgroup$ – BigWig May 19 '18 at 17:09
  • $\begingroup$ The wavelength is indeed proportional to the kinetic energy. If you consider a potential barrier, the sum of kinetic and potential energy on the left and the right is the same, and if the potential is either zero outside the barrier, or is simply the same on the right and the left, then the kinetic energy will also be constant on both sides. So this is why the wavelength would stay constant. $\endgroup$ – neradha May 19 '18 at 17:15

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