# Confusing step in Sakurai Chapter 1

1. Oct 8, 2011

### gitano

Hi,

At one point in Chapter 1 of Sakurai he is deriving the momentum operator in the position basis - I just don't see how he makes some of the mathematical leaps (at least leaps for me) between the following expressions

$$\int dx' |x' + \Delta x'> < x' | \alpha > = \int dx' | x' > < x' - \Delta x' | \alpha >$$
$$= \int dx' | x' > \left ( < x' | \alpha > - \Delta x' \frac{\partial}{\partial x'} < x' | \alpha > \right )$$

I guess I kind of see that the $$|x' + \Delta x' >$$ acting on the position wavefunction of alpha is the same as translating the position wavefunction to the right by $$\Delta x'$$. Is this the logic here or is there a more mathematical way of showing it.

Now, the transition to the last equality is even less intuitive for me.

2. Oct 8, 2011

### Bill_K

If you think of <x'|α> as a function f(x'), then by Taylor series, f(x' - Δx') ≈ f(x') - Δx'(∂/dx')f(x')

3. Oct 9, 2011

### gitano

So the fact that you are cutting off the taylor expansion at the second term is ok because you are only dealing with an infinitesimal change in $$\langle x' | \alpha \rangle ?$$ So it is still exact and not an approximation?

4. Oct 9, 2011

### Fredrik

Staff Emeritus
In a physics book, there's really no difference between those two options. When an author uses the word "infinitesimal", it's not a reference to some definition of "numbers that are smaller than all positive real numbers but still >0". It's just a word that lets you know that the next equality you see is a Taylor expansion with all but a finite number of terms thrown away. I'm not sure that's what these authors actually mean, but it's definitely the best way to make sense of their statements. I haven't seen an example of where it's wrong to interpret them this way, and I doubt that these authors have even seen a definition of an infinitesimal.

The first equality is a change of variables. Set x''=x'+Δx'. When you're done expressing everything in terms of x'' instead of x', you just drop one of the primes, because the value of an integral doesn't depend on what symbol you use.