Blake Test

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This general solution depends on the initial state of the wavefunction \(\psi (\mathbf{r}, 0)\), along with a time-dependent complex exponential term which we will call \(\mathbf{U}(t)\).

\[\mathbf{U}(t) = e^{-\frac{i \mathbf{H}}{\hbar} t}\]

This complex exponential term is known as the evolution operator, and it has some important qualities. This operator evolves a system from its initial prepared state \(\psi (\mathbf{r}, 0)\) to the state after time \(t\) \(\psi (\mathbf{r}, 0)\):

\[\mathbf{U}(t) \psi (\mathbf{r}, 0) = \psi (\mathbf{r}, t)\]

Though \(\mathbf{U}(t)\) transforms a state forward in time, it’s important to notice what is preserved with this transformation. Consider two states different initial states \(\ket{\psi (0)}\) and \(\ket{\phi (0)}\) of a system. We can compare these two states by taking an inner product \(\braket{\psi (0)|\phi (0)}\). Now let’s move both forward in time:

\[\ket{\psi (t)} = \mathbf{U}(t) \ket{\psi (t)} \\ \ket{\phi (t)} = \mathbf{U}(t) \ket{\phi (t)}\]

With an inner product:

\[\braket{\psi (t)|\phi (t)} = \bra{\psi(t)} U^{*} (t) U(t) \ket{\phi (t)} = \bra{\psi(t)} e^{\frac{i \mathbf{H}}{\hbar} t} e^{-\frac{i \mathbf{H}}{\hbar} t} \ket{\phi (t)} = \braket{\psi (0)|\phi (0)}\]

The time-evolution operator preserves the inner product. Two states that are different at one point in time, will also be different at any future time.