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Adwit
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- Zettili's Quantum Mechanics book Chapter 7 ( Rotation in Classical Physics ). I don't understand the calculation. How do "sinϕ - cosϕ sinϕ" become 0 ?
How do "sinϕ - cosϕ sinϕ" become 0 ?
The first order approximation for ##\cos \phi## is ##1##. From its Taylor series.Adwit said:Summary:: Zettili's Quantum Mechanics book Chapter 7 ( Rotation in Classical Physics ). I don't understand the calculation. How do "sinϕ - cosϕ sinϕ" become 0 ?
View attachment 258773
How do "sinϕ - cosϕ sinϕ" become 0 ?
Off topic, but the expansion shouldn't behilbert2 said:Or, in more detail, if I expand the sine and cosine to first two terms, the result is
##\sin \phi - \cos \phi \sin \phi \approx \left(\phi + \frac{1}{6}\phi^3 \right) - \left(1 - \frac{1}{2}\phi^2 \right)\left(\phi - \frac{1}{6}\phi^3 \right)\\ = \phi + \frac{1}{6}\phi^3 - \phi + \frac{1}{6}\phi^3 + \frac{1}{2}\phi^3 - \frac{1}{12}\phi^5\\ = \frac{5}{6}\phi^3 - \frac{1}{12}\phi^5 = \mathcal{O}(\phi^3 )##
Note that even if the trigonometric expression were just slightly different,
##\frac{101}{100}\cdot\sin\phi - \cos\phi \sin\phi##,
then there would also be a 1st order term in the expansion.
Well, it reads "terms of order ##\phi^3##" and basically means that it's some function of ##\phi## that goes to zero equal or faster than the function ##\phi^3##. In your case, you are substituting ##\phi## that is the variable by a very very small number ##\delta##, then when I say ##\mathcal{O}(\delta^3)## means that the error of the approximation $$\sin\delta - \sin\delta\cos\delta \approx 0$$ is, in the worst case, proportional to ##\delta^3##.Adwit said:View attachment 258793 I understand phi. But what is this symbol that looks like O ?
It's the socalled Landau symbol. The meaning as that with a capital O:Adwit said:View attachment 258793 What does it mean ? I recognize phi. But what is this symbol that looks like O ? I have never seen this in my life.
Adwit said:With all due respect hilbert2, isn't the plus sign of 1st "sinϕ expansion" actually going to be minus sign? I indicate it with red colour.
View attachment 258823
By the way, thanks for your explanation.
The rotation problem in quantum mechanics refers to the difficulty in describing the rotational motion of particles at the quantum level. This is because classical mechanics, which is used to describe the motion of macroscopic objects, does not apply to particles at the quantum level.
The rotation problem is important because it is a fundamental aspect of the behavior of particles at the quantum level. Understanding how particles rotate and interact with each other is crucial in many areas of physics, such as atomic and molecular physics, condensed matter physics, and quantum field theory.
The rotation problem is solved by using the mathematical framework of quantum mechanics, which takes into account the wave-like nature of particles. This allows for the description of rotational motion in terms of wave functions and operators, rather than classical variables such as position and momentum.
The rotation problem has many practical applications, such as in the development of quantum computers, which rely on the manipulation of particles' rotational states to perform calculations. It is also important in the study of molecular and atomic structures, as well as in the development of new materials with specific properties.
Yes, there are still some challenges in fully understanding and solving the rotation problem in quantum mechanics. One of the main challenges is the development of a unified theory that can accurately describe both quantum and classical systems. Additionally, there is ongoing research into the effects of gravity on quantum rotational systems, which may lead to new insights and challenges in the future.