I 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).
Salmone
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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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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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