Golden ratios in quantum mechanics?

Loren Booda
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Do golden ratios appear in quantum mechanics - such as with the standard model, string theory or quantum gravity?
 
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I don't think so, Loren--at least, not yet. However, I was very startled to see the question! Your natural tendency to think far out of the box is your talent.

In a world dominated by two fold symmetry, 1.414... to 1 is a common ratio we see occurring often. Nature seems to prefer 2:1 over any other.

0.866... is the square root of 3. We don't see it as often. Quarks seem to the the first known, fundamental 3-fold symmetry.

The golden ratio is about 5-fold symmetries. To my knowledge, there is yet to be discovered any elemental 5-fold symmetries in nature.
 
It appears in relation to the quantum dimension of the (nontrivial) anyon in the Fibonacci anyon model. Whether or not you consider this a physical example is another matter...
 
Is it true that golden ratios do not play as significant a part in the microverse as they play in the macroverse?
 
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. Towards the end of the first lecture for the Qiskit Global Summer School 2025, Foundations of Quantum Mechanics, Olivia Lanes (Global Lead, Content and Education IBM) stated... Source: https://www.physicsforums.com/insights/quantum-entanglement-is-a-kinematic-fact-not-a-dynamical-effect/ by @RUTA
I am reading WHAT IS A QUANTUM FIELD THEORY?" A First Introduction for Mathematicians. The author states (2.4 Finite versus Continuous Models) that the use of continuity causes the infinities in QFT: 'Mathematicians are trained to think of physical space as R3. But our continuous model of physical space as R3 is of course an idealization, both at the scale of the very large and at the scale of the very small. This idealization has proved to be very powerful, but in the case of Quantum...
Thread 'Lesser Green's function'
The lesser Green's function is defined as: $$G^{<}(t,t')=i\langle C_{\nu}^{\dagger}(t')C_{\nu}(t)\rangle=i\bra{n}C_{\nu}^{\dagger}(t')C_{\nu}(t)\ket{n}$$ where ##\ket{n}## is the many particle ground state. $$G^{<}(t,t')=i\bra{n}e^{iHt'}C_{\nu}^{\dagger}(0)e^{-iHt'}e^{iHt}C_{\nu}(0)e^{-iHt}\ket{n}$$ First consider the case t <t' Define, $$\ket{\alpha}=e^{-iH(t'-t)}C_{\nu}(0)e^{-iHt}\ket{n}$$ $$\ket{\beta}=C_{\nu}(0)e^{-iHt'}\ket{n}$$ $$G^{<}(t,t')=i\bra{\beta}\ket{\alpha}$$ ##\ket{\alpha}##...

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