Assumption that time evolution operator first order in $dt$

Question

I'm having some trouble with this part of Sakurai's derivation of the time evolution operator in QM (page 70):

Because of continuity, the infinitesimal time-evolution operator must reduce to the identity operator as $dt$ goes to zero, $$lim_{dt to 0} mathscr{U}(t_0+dt,t_0)=1,$$ and as in the translation case, we expect the difference between $mathscr{U}(t_0+dt,t_0)$ and 1 to be of first order in $dt$.

(The same argument is made for the translation operator, but no further explanation given.) How does this limit imply that $mathscr{U}$ must be first order in $dt$? Surely terms of higher order in $dt$ would also reduce to zero in the limit, or are there other considerations at work here?

Roger Vadim · Accepted Answer

The terms of higher order in $dt$ would reduce to zero faster. What the author says effectively is that the operator is continuous at $t_0$, and can be expanded in Taylor series in $dt$ up to the first order. The smaller is $dt$, the more precise is the expansion, the following results become exact in the limit $dtrightarrow 0$.

Correct answer by Roger Vadim on May 27, 2021

Luke Pritchett · Answer

The text doesn't say that $mathcal{U}$ must be first order in $dt$. It says that $mathcal{U}(t_0+dt,t_0)-1$ must be first order in $dt$.
Suppose that $mathcal{U}(t_0+dt,t_0)-1$ has a Taylor series expansion in $dt$ so that
$$mathcal{U}(t_0+dt,t_0)-1 = mathcal{U}_0 + mathcal{U}_1 dt+mathcal{U}_2 (dt)^2+dots $$
If we take the limit as $dtrightarrow 0$ on both sides of this equation we get
$$1-1 = mathcal{U}_0+0+0+dots$$
Therefore $mathcal{U}_0 = 0$ and so
$$mathcal{U}(t_0+dt,t_0)-1 =  mathcal{U}_1 dt+mathcal{U}_2 (dt)^2+dots $$.
This is what it means to say that the difference between $mathcal{U}(t_0+dt,t_0)$ and $1$ is first order in $dt$.

Assumption that time evolution operator first order in $dt$

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