I Extending an infinitesimal operator

[davidge]

I notice that in Quantum Mechanics when extending an infinitesimal operation to a finite one, we should end with the exponential. For example: (rf. Sakurai, Modern Quantum Mechanics)

$$
D(\boldsymbol{\hat n}, d\phi) = 1 - \frac{i}{\hbar} ( \boldsymbol {J \cdot \hat n})d\phi
$$

This is the infinitesimal version of the rotation operator, where $\boldsymbol{\hat n}$ is a unit vector and $J$ is the angular-momentum operator. Now the "finite" version of the rotation operator, that generates a rotation by a finite angle (meaning non-infinitesimal angle) is

$$
\lim_{N \to \infty}[1 - \frac{i}{\hbar} (\boldsymbol{J \cdot \hat n})\frac{\phi}{N}]^N = exp[- \frac{i}{\hbar} (\boldsymbol{J \cdot \hat n})d\phi]
$$

My question here is how does one know that the correct procedure for going to infinitesimal to finite operations is to take that limit above?

 

[Fightfish]

The idea behind taking this limit is simply this: to get a finite rotation, we make infinitely many infinitesimal rotations (albeit with the proper weighting to ensure that we do indeed arrive at the desired finite rotation angle in the end).

 

[davidge]

@Fightfish Thanks for replying.

to get a finite rotation, we make infinitely many infinitesimal rotations

But why this procedure takes that form in my example? For example, why don't divide all terms by N and after exponentiate it? Why that particular form of dividing the angle $\phi$ by $N$ and after exponentiating the two terms?

Is this because in the limit as $N$ tends to infinity $(\phi / N)$ tends to $d\phi$, and also we should mutiply the infinitesimal operator $N$ times? But why should we multiply it instead of summing up?

 

[Fightfish]

Hmm...I'm not sure where your confusion is.
Let's consider applying $D(\varphi)$ twice. If $\varphi$ is small enough, then $D(\varphi) D(\varphi) = D(2\varphi)$.

Conversely, the opposite decomposition holds as well: $D(\varphi) = D(\varphi/2) D(\varphi/2)$. In a similar fashion I could also do $D(\varphi) = D(\varphi/3) D(\varphi/3) D(\varphi/3)$ and $D(\varphi) = [D(\varphi/N) ]^N$. This decomposition is only true if $\varphi/N$ is small enough of course - so to ensure that, we go ahead and take the limit $N \to \infty$.

So, you see that what we are doing is chopping up the finite angle $\varphi$ into $N$ parts, and do $N$ small rotations of $\varphi/N$ each in order to achieve the final overall rotation.

You do not divide the whole expression by $N$ because what we are looking for is $D(\varphi/N)$, not $D(\varphi)/N$.

 

[davidge]

Ah, ok. I see now. Thank you !