# I Extending an infinitesimal operator

1. Feb 28, 2017

### 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?

Last edited: Feb 28, 2017
2. Feb 28, 2017

### 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).

3. Feb 28, 2017

### davidge

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?

Last edited: Feb 28, 2017
4. Feb 28, 2017

### 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 $D(\varphi)/N$.

5. Feb 28, 2017

### davidge

Ah, ok. I see now. Thank you !!!