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A derivation of momentum as generator of translation

In this page, part 2 “Momentum as generator of translations”, I don’t understand this step:

$$T(x)=\lim_{N \rightarrow \infty}(T(x/N))^N =\lim_{N \rightarrow \infty}\left(1-\frac{ixp}{N\hbar}\right)^N$$

How is the 2nd equality derived?

My attempt:

$$p_x=i\hbar \lim_{a\rightarrow 0} \frac{T(ax)-\mathbb{I}}{a}=i\hbar \lim_{N \rightarrow \infty} \frac{T(x/N)-\mathbb{I}}{1/N}$$

$$T(x/N)=\lim_{N\rightarrow \infty} \left(\mathbb{I}- \frac{ip }{N\hbar}\right)$$

I don’t know where the x in the numerator comes from.

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  • $\begingroup$ You have a typo in the first equality. $T(x)/N \to T(x/N)$. Does that solve your doubt? $\endgroup$
    – MannyC
    Commented Feb 23, 2020 at 16:50
  • $\begingroup$ @MannyC thanks! I’ve corrected it. But I still have problem in that step... $\endgroup$
    – RicknJerry
    Commented Feb 23, 2020 at 17:10

1 Answer 1

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When you change your limits from $a\to 0$ to $N\to\infty$ you need to take care of your dimensions. What you are physically doing is slicing a length $x$ into $N$ parts such that each infinitesimal segment has length $a$. This also makes it clear that $a$ has dimensions of length. Thus the correct variable substitution is $$a\to\frac{x}{N}$$ This should fix your error.

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  • $\begingroup$ If in $f(x+a)$, $x$ has the dimensions of length then even $a$ has the same dimensions. $\endgroup$ Commented Feb 23, 2020 at 17:26
  • $\begingroup$ For future reference my above comment is a response to the question of why should $a$ have dimensions when it is just a parameter in the derivative. $\endgroup$ Commented Feb 23, 2020 at 17:34

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