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In section 4.4 and 4.5 of "The Theoretical Minimum: Quantum Mechanics" book, there are a explanation about time evolution operator U(t).

The author explains that:

$U^\dagger (t)U(t) = I$

In sequence, he suposes the idea of an incremental change in time and says:

$U^\dagger (\epsilon )U(\epsilon) = I$

First, he consider the case where $\epsilon$ is zero and says that it should be obvious that in this case the time-evolution operator is merely the unit operator I.

So, he does the folowing aprroximation:

$U(\epsilon) = I - i \epsilon H$

He uses this approximation to introduce the quantum Hamiltonian later in the book.

My doubt is that, I accepted the argument for $U^\dagger (t)U(t) = I$, but, I did not understand why $U(0)$ is obviously $I$, i.e. $U(0) = I$. Why $U(0) = I$ ?

Daniel
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1 Answers1

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The evolution operator is an operator which expresses the wavefunction at time $t$ through the wavefunction at $t=0$: \begin{equation} |\psi(t)\rangle = U(t) |\psi(0)\rangle, \end{equation} this is Eq. (4.1) from "The Theoretical Minimum: Quantum Mechanics". At $t=0$, it should be \begin{equation} |\psi(0)\rangle = U(0) |\psi(0)\rangle \end{equation} for arbitrary initial wavefunction $|\psi(0)\rangle$. This can be true only for $U(0) = I$.

E. Anikin
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