- Time Evolution Operator

1. Time Symmetry: Schrödinger Picture

The inner product between two states should be preserved under time evolution:

$\langle\phi|\psi\rangle = \langle\phi|U^{\dagger}(\Delta t)U(\Delta t)|\psi\rangle$, where $U(\Delta t)|\psi\rangle = |\psi;\Delta t\rangle$

Therefore, $U^{\dagger}(\Delta t)U(\Delta t) = 1$

This means that $U(\Delta t)$ is a unitary operator.

2. General Form of Unitary Operator $U$

$U=e^{i\Omega}$, where $\Omega^{\dagger} = \Omega$

3. Representation of $U(\Delta t)$

$U(\Delta t) = e^{i\Delta t \Omega}$

4. Verification of Group Property

$U(\Delta t_2)U(\Delta t_1) = U(\Delta t_1 + \Delta t_2)$ holds:

$$U(\Delta t_2)U(\Delta t_1) = e^{i\Delta t_2 \Omega}e^{i\Delta t_1 \Omega} = e^{i(\Delta t_1+\Delta t_2)\Omega}$$

5. Approximation of $U(\Delta t)$

For small $\Delta t$:

$U(\Delta t) \approx 1 + i\Omega \Delta t$

6. Unitary Equivalence Transformation: Heisenberg Picture

$|x’\rangle \to |x’; \Delta t\rangle( = U^{\dagger}(\Delta t)|x’\rangle )$ $\leftrightarrow$ $x|x’\rangle = x U(\Delta t)U^{\dagger}(\Delta t)|x’\rangle = x’|x’\rangle$

From $x U(\Delta t)U^{\dagger}(\Delta t)|x’\rangle = x’|x’\rangle$, multiplying both sides by $U^{\dagger}(\Delta t)$:

$U^{\dagger}(\Delta t)x U(\Delta t)U^{\dagger}(\Delta t)|x’\rangle = x’U^{\dagger}(\Delta t)|x’\rangle$

Since $U^{\dagger}(\Delta t)|x’\rangle = |x’; \Delta t\rangle$:

$U^{\dagger}(\Delta t)x U(\Delta t)|x’; \Delta t\rangle = x(\Delta t)|x’; \Delta t\rangle$

Therefore, $x \to x(\Delta t) = U^{\dagger}(\Delta t)x U(\Delta t)$

7. Approximate Calculation of Unitary Equivalence Transformation

Using $U(\Delta t) = 1 + i\Omega \Delta t$:

$$\begin{align*} &U^{\dagger}(\Delta t)x U(\Delta t) \\ &= (1 - i\Omega \Delta t)x(1 + i\Omega \Delta t)\\ &= x + i \Delta t[x,\Omega] + O((\Delta t)^2) \end{align*}$$$$\begin{align*} &U^{\dagger}(\Delta t)x U(\Delta t)|x'; \Delta t\rangle\\ &\approx (x + i \Delta t[x,\Omega])|x'; \Delta t\rangle = x(\Delta t)|x'; \Delta t\rangle \end{align*}$$

Therefore, $-\Delta x+ i \Delta t[x,\Omega]=0$, which means $i[x,\Omega] = \frac{\Delta x}{\Delta t}$

This gives us $i \hbar \frac{\Delta x}{\Delta t} = [x,-\hbar \Omega]$, so $[x,H] = [x,-\hbar \Omega]$

Therefore, $\Omega = -\frac{H}{\hbar}$

8. Exponential Representation of $U(\Delta t)$

Since $U(\Delta t) = e^{i\Omega \Delta t}$ and $\Omega = -\frac{H}{\hbar}$:

$$U(\Delta t) = e^{-\frac{i}{\hbar}H \Delta t}$$

This is the time evolution operator in quantum mechanics, showing how quantum states evolve over time according to the Hamiltonian $H$.