Undergrad Rotation problem in Quantum Mechanics

[Adwit]

TL;DR

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 ?

 

[PeroK]

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 ?

The first order approximation for $\cos \phi$ is $1$. From its Taylor series.

PS There's a better answer below ...

 

[Gaussian97]

Well, of course, it doesn't become 0, simply when you expand in powers of $\delta$ you get that
$$\sin \phi - \cos \phi \sin \phi =0+ \mathcal{O}(\delta^3)$$
Because you only seem to be interested in terms up to order $\delta^2$ you simply can put a zero there.

 

[hilbert2]

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.

 

[Gaussian97]

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.

Off topic, but the expansion shouldn't be
$$\sin \phi - \cos\phi\sin\phi = \frac{1}{2}\phi^3-\frac{1}{8}\phi^5+\mathcal{O}(\phi^6)$$?

 

[hilbert2]

Yes, if you include the 4th and 5th order terms in the cosine and sine, then the expansion should change a bit in the fifth order term. I'm not sure how you got $1/2$ for the coefficient of 3rd order term.

Edit: Ok, I calculated it again and you seem to be correct, sorry.

 

[Adwit]

What does it mean ? I recognize phi. But what is this symbol that looks like O ? I have never seen this in my life.

 

[Gaussian97]

View attachment 258793 I understand phi. But what is this symbol that looks like O ?

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 $\delta^3$.
For example, for $\delta=0.1$ you have $$\sin\delta - \sin\delta\cos\delta = 4,986 \cdot 10^{-4} = 0.4986 \ \delta^3$$
And for $\delta=0.01$ you have
$$\sin\delta - \sin\delta\cos\delta = 4,99987 \cdot 10^{-7} = 0.499987\ \delta^3$$
So, you see that is really proportional to $\delta^3$, and that is actually proportional to $\frac{1}{2}\delta^3$ (this is what I mean in my post #5).

 

[Adwit]

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.

By the way, thanks for your explanation.

 

[vanhees71]

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.

It's the socalled Landau symbol. The meaning as that with a capital O:
$$f(x)=g(x)+\mathcal{O}(x^n),$$
that
$$\lim_{x \rightarrow 0} \frac{f(x)-g(x)}{x^n}=\text{finite}.$$
Roughly speaking it means that $f(x)$ and $g(x)$ for small $x$ differ by a quantity of order $x^n$.

With a little o its:
$$f(x) = g(x) + \mathcal{o}(x^n) \; \Rightarrow \; \lim_{x \rightarrow 0} \frac{f(x)-g(x)}{x^n}=0.$$
Roughly speaking it means that $f(x)$ and $g(x)$ differ by a quantity going faster to 0 than $x^n$.

The above example can be made a bit more rigorous and at the same time saving some notation. All you want to establish is that the expression has no linear term in the expansion around $\phi=0$ with the Landau-symbol notation this reads
$$\sin \phi-\cos \phi \sin \phi = \sin \phi (1-\cos \phi)=[\phi+\mathcal{O}(\phi^3)] \mathcal{O}(\phi^2)=\mathcal{O}(\phi^3).$$
This already tells you that the expression goes like $\phi^3$ and has no linear (nor quadratic) term.

 

[hilbert2]

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.

Yes, that was the reason for the wrong result. Thanks.