I Entropic effects of the Uncertainty Principle?

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The Heisenberg uncertainty principle indicates that a particle's location cannot be precisely defined, raising questions about its role in thermal energy transfer and entropy. Uncertainty is argued to reduce information and, consequently, entropy rather than facilitate thermal energy transfer at the quantum level. The discussion highlights the significance of Hermite functions in visualizing the relationship between spatial and Fourier domains. Quantum phenomena like tunneling and alpha particle emission illustrate how uncertainty allows particles to escape classical limits, contributing to energy transfer. Overall, uncertainty plays a critical role in quantum mechanics, influencing thermal energy dynamics.
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Per the Heisenberg uncertainty principle, a particle does not have a precisely defined location. Does such uncertainty contribute to the transfer of thermal energy (i.e. entropy)? Is uncertainty the primary means for the transfer of thermal energy at the quantum level?
 
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It is rather the opposite, the uncertainty reduces the amount of information, and hence the amount of entropy. One way to get some feeling for this is to look at the Hermite functions. They are eigenfunctions of the Fourier transform, and allow you to nicely visualize how the volume increases (simulateneously in the spatial and Fourier domain) when you include more of them.

Planck's constant sort of gives you the unit for how that amount of information gets counted.
 
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Thanks for your reply. I'm thinking of effects like quantum tunneling. alpha particle emission, current leakage from electronics, etc. in which particles escape their classical range limits. AFAIK, without uncertainty, those effects would not occur, and their corresponding transfer of thermal energy would not happen.
 
For the quantum state ##|l,m\rangle= |2,0\rangle## the z-component of angular momentum is zero and ##|L^2|=6 \hbar^2##. According to uncertainty it is impossible to determine the values of ##L_x, L_y, L_z## simultaneously. However, we know that ##L_x## and ## L_y##, like ##L_z##, get the values ##(-2,-1,0,1,2) \hbar##. In other words, for the state ##|2,0\rangle## we have ##\vec{L}=(L_x, L_y,0)## with ##L_x## and ## L_y## one of the values ##(-2,-1,0,1,2) \hbar##. But none of these...

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