I don't understand about infinitesimal translation

[Another]

TL;DR

I don't understand about infinitesimal translation

and

But

The infinitesimal translation denoted by equ 1.6.12 and 1.6.32

And then i understand about equation 1.6.35 but equation 1.6.36

Why they take limit N go to inf ? , multiply 1/ N ? and power N ?

So is the relationship below still true?

$F(Δ x'\hat{x}) = 1 - \frac{i p_x \dot{} Δ x'\hat{x} }{\bar{h}} = lim_{N → ∞} (1 - \frac{i p_x Δ x'}{N\bar{h}})^N$ ?

 

[PeroK]

If you want to displace a system by $\Delta x$, you can do that in $N$ smaller steps of $\frac{\Delta x}{N}$. Each of these steps is carried out by the operator that translates a state by $\frac{\Delta x}{N}$. Carrying out one operation after another is represented by operator multiplication. So, in simple notation:
$$T(\Delta x) = T(\frac{\Delta x}{N})^N$$
You can then imagine taking the limit of this as $N \rightarrow \infty$. And, in this case, looking at the form that $T(\frac{\Delta x}{N})$ takes, the limit is an exponential.

 

[vanhees71]

Another, maybe simpler, way is to realize that
$$\mathcal{T}(x+\mathrm{d} x) =\mathcal{T}(\mathrm{d} x) \mathcal{T}(x) $$
and thus (using natural units with $\hbar=1$)
$$\mathcal{T}(x+\mathrm{d}x)-\mathcal{T}(x) = [\mathcal{T}(\mathrm{d} x)-1]\mathcal{T}(x) =-\mathrm{i}\hat{p} \mathrm{d} x \mathcal{T}(x)$$
or
$$\mathcal{T}'(x)=-\mathrm{i} \hat{p} \mathcal{T}(x) \; \Rightarrow \; \mathcal{T}(x)=\exp(-\mathrm{i} \hat{p} x).$$
It's also realized easily in the position representation of a wave function [EDIT: Corrected in view of #4]:
$$\mathcal{T}(a) \psi(x)=\exp(-\mathrm{i} \hat{p} a) \psi(x) = \exp(-a \mathrm{d}_x) \psi(x) = (1-a \mathrm{d}x + a^2/2 \mathrm{d}_x^j+\cdots) \psi(x) = \sum_{j=0}^{\infty} \frac{1}{j!} (-a)^j \mathrm{d}_x^2 \psi(x)=\psi(x-a).$$

 

[Ishika_96_sparkles]

It's also realized easily in the position representation of a wave function:
$$\mathcal{T}(a) \psi(x)=\exp(-\mathrm{i} \hat{p} a) \psi(x) = \exp(-a \mathrm{d}_x) \psi(x) = (1-a \mathrm{d}x + a^2/2 \mathrm{d}_x^2 +\cdots) \psi(x) = \sum_{j=0}^{\infty} \frac{1}{j!} (-a)^j \mathrm{d}_x^2 \psi(x)=\psi(x-a).$$

Shouldn't the last part read as follows?

$$\mathcal{T}(a) \psi(x)=\exp(-a \mathrm{d}_x) \psi(x) = (1-a \mathrm{d}_x + (a^2/2!) \, \mathrm{d}_x^2 +\cdots) \psi(x) = \sum_{j=0}^{\infty} \frac{1}{j!} (-a)^j \mathrm{d}_x^j \psi(x)=\psi(x-a).$$

 

[vanhees71]

Sure, I'll correct it.