B Wavefunction and Lorentz Invariance

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The wavefunction's lack of Lorentz invariance indicates that non-relativistic quantum mechanics is being utilized, which is typically the focus in undergraduate education. This non-invariance suggests that the framework is an approximation, suitable for scenarios where relativistic effects are minimal. Consequently, it remains a practical and effective tool for understanding quantum systems at low energies. However, for high-energy scenarios, the limitations of this approach become significant. Understanding these implications is crucial for advancing to more complex quantum theories that incorporate relativity.
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What are the implications that the wavefunction is not Lorentz invariant?
What are the implications that the wavefunction is not Lorentz invariant?
 
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cosmanino2050 said:
TL;DR Summary: What are the implications that the wavefunction is not Lorentz invariant?

What are the implications that the wavefunction is not Lorentz invariant?
Mostly it means that we’re doing non-relativistic quantum mechanics, the only kind we’re likely to encounter in undergraduate classes. This is an approximation that is valid, accurate, and very useful as long as we’re dealing with energies small enough that relativistic effects are negligible.
 
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Thread 'Angular Momentum Vector and Its Magnitude'

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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