I Understanding No Energy Degeneracy in Sakurai's Quantum Mechanics

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The discussion focuses on the concept of energy degeneracy in quantum mechanics as presented in Sakurai's "Modern Quantum Mechanics." It clarifies that the statement regarding no degeneracies in the angular momentum quantum number \( l \) refers specifically to the absence of different \( l \) values yielding the same energy eigenvalue, distinguishing it from the degeneracies associated with the magnetic quantum number \( m \). The text contrasts this with the isotropic harmonic oscillator, where multiple states can share the same energy for a given \( l \). The conversation highlights the significance of symmetry in determining degeneracies, such as SO(4) for the hydrogen atom and SU(3) for the 3D harmonic oscillator. Understanding these distinctions is crucial for grasping the underlying principles of quantum mechanics.
euphoricrhino
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Hello, I'm hoping someone can help me understand a statement in Sakurai Modern Quantum Mechanics (3rd edition).
In particular, in the section that describes free particle in infinite spherical well (page 198, section 3.7.2), after the text has shown that for a given ##l## value, the energy eigenvalues correspond to the nodes of spherical Bessel functions, the last paragraph went on and said

"It should be noted that this series of energy levels shows no degeneracies in ##l##. Indeed, such degenerate energy levels are impossible, except for any accidental equality between zeros of spherical Bessel functions of different order."

How can there be no energy degeneracy for a given ##l##? Don't all the ##m## state for the same ##l## count as degeneracies?

If I understand this statement as specifically referring to the radial part of the wave function (hence ##m## degeneracies are not in scope), the next few pages (pp201) discussed degeneracies of isotropic oscillator, where the text stated

"Quite unlike the square well, the three-dimensional isotropic harmonic oscillator has degenerate energy eigenvalues in the ##l## quantum number. There are three states (all ##l=1##) for ##N=1##. For ##N=2## there are five states with ##l=2##, plus one state with ##q=1## and ##l=0##..."
where clearly the ##m## states are counted as degeneracies here.

Any insightful explanations are greatly appreciated!
 
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After some thoughts, I think I misunderstood the statement. Degeneracy in ##l## means two different ##l## values ending up at the same energy eigenvalue. So the two cases are indeed what the text claims to be :)
 
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I think the latter is what Sakurai means. The degeneracy wrt. the "magnetic quantum number" ##m## is valid for all radially symmetric potentials due to rotation invariance of the problem. Degeneracy wrt. to the angular-momentum quantum number ##\ell## is special and indicates additional symmetries. E.g., for the non-relativistic Coulomb problem (hydrogen atom) it's an SO(4) symmetry. For the 3D harmonic symmetric oscillator it's SU(3).
 
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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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