Undergrad Hamiltonian of a particle moving on the surface of a sphere

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The discussion centers on the Hamiltonian of a spin-1 particle constrained to move on the surface of a sphere, expressed as H=ω/ħ L². The original Hamiltonian for such a particle is H=L²/(2mR²), leading to the conclusion that ω must equal ħ/(2mR²) for consistency. The dimensional analysis confirms that ω has the correct units of s⁻¹. The participants seek to establish the validity of this relationship and clarify the role of ω as a parameter. The conclusion affirms that ω is indeed defined as ħ/(2mR²).
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In a quantum mechanical exercise, I found the following Hamiltonian:

Consider a particle of spin 1 constrained to move on the surface of a sphere of radius R with Hamiltonian ##H=\frac{\omega}{\hbar}L^2##. I knew that the Hamiltonian of a particle bound to move on the surface of a sphere was ##H=\frac{L^2}{2mR^2}## and then to get the same Hamiltonian should be ##\omega=\frac{\hbar}{2mR^2}## which dimensionally fits, but is it right? How can this equality be proved?
 
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As you write yourself, it's just a definition for a parameter ##\omega##. So what should be right or wrong with it?
 
@vanhees71 So ##\omega## is just a parameter which should be equal to ##\frac{\hbar}{2mR^2}## and with dimensions ##s^{-1}##
 
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Time reversal invariant Hamiltonians must satisfy ##[H,\Theta]=0## where ##\Theta## is time reversal operator. However, in some texts (for example see Many-body Quantum Theory in Condensed Matter Physics an introduction, HENRIK BRUUS and KARSTEN FLENSBERG, Corrected version: 14 January 2016, section 7.1.4) the time reversal invariant condition is introduced as ##H=H^*##. How these two conditions are identical?

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