Time evolution and the Schrödinger equation
Contents:
# Time is a parameter
Time $\to$ just a parameter, not an operator / observable.
Eg. Einstein’s photon box experiment: he tried to challenge $\Delta E\Delta t \geq \frac{\hbar}{2}$ by building a time operator. Pauli also showed that there cannot exist a self-adjoint (Hermitian) time operator $\hat T$ s.t. $[\hat H, \hat T] = i\hbar$.
So $\Delta t$ is not the uncertainty of a time operator. It’s a proxy:
$$ \Delta t \equiv \frac{\Delta A}{\left|\frac{d\langle A\rangle}{dt}\right|} $$
$\to$ by observing how physical dynamical operators change w.r.t. the parameter $t$ (the time it takes $\langle A\rangle$ to move by one $\Delta A$). This is the Mandelstam-Tamm version.
# How does a state ket change in time?
Say at $t = t_0$ it is $|\alpha\rangle$. At $t > t_0$ we don’t expect the state to remain $|\alpha\rangle$, so we denote it as
$$ |\alpha, t_0; t\rangle $$
(we write $\alpha, t_0$ to remind us that it used to be $\alpha$ at $t_0$). We know, by definition,
$$ \boxed{\lim_{t\to t_0}|\alpha, t_0; t\rangle = |\alpha\rangle} $$
or, we’ll use a short hand: $|\alpha, t_0; t_0\rangle = |\alpha, t_0\rangle$.
Now we need a time evolution operator, something to bring in the changes in time from $t_0$ to $t$. We’ll call it $\mathcal U(t, t_0)$. The defining equation:
$$ \boxed{|\alpha, t_0; t\rangle = \mathcal U(t, t_0)|\alpha, t_0\rangle} $$
# The time evolution operator is unitary
We know that
$$ |\alpha, t_0\rangle = \sum_{a'}c_{a'}(t_0)|a'\rangle $$
and at a later time
$$ |\alpha, t_0; t\rangle = \sum_{a'}c_{a'}(t)|a'\rangle $$
In general we expect
$$ |c_{a'}(t)| \neq |c_{a'}(t_0)| $$
(but if $[\hat H, \hat A] = 0$, then they’re equal! We’ll see why in the energy eigenkets notes.)
However, since $\sum_{a'}|c_{a'}(t_0)|^2$ talks of the total prob., the sum must stay equal (and 1):
$$ \implies \sum_{a'}|c_{a'}(t_0)|^2 = \sum_{a'}|c_{a'}(t)|^2 $$
Stated another way, if the state ket is normalized, then it’s forever normalized:
$$ \langle\alpha, t_0|\alpha, t_0\rangle = 1 \implies \langle\alpha, t_0; t|\alpha, t_0; t\rangle = 1 $$
and from $|\alpha, t_0; t\rangle = \mathcal U|\alpha, t_0\rangle$ $\implies \langle\alpha, t_0; t| = \langle\alpha, t_0|\mathcal U^\dagger$
$$ \implies \boxed{\mathcal U^\dagger\mathcal U = 1} $$
$\implies$ the time evolution operator is unitary.
Note how probability conservation in time led to normalization, which led to the unitary condition.
# Composition
We need $\mathcal U$ to have another feature, composition:
$$ \mathcal U(t_2, t_0) = \mathcal U(t_2, t_1)\mathcal U(t_1, t_0), \qquad t_2 > t_1 > t_0 $$
And we also define the infinitesimal $\mathcal U(t_0 + dt, t_0)$:
$$ |\alpha, t_0; t_0 + dt\rangle = \mathcal U(t_0 + dt, t_0)|\alpha, t_0\rangle $$
As $dt \to 0$, the small $\mathcal U(t_0 + dt, t_0)$ must go to the identity.
We assert (with no proof) that all of this is met by
$$ \boxed{\mathcal U(t_0 + dt, t_0) = 1 - i\Omega dt} $$
where $\Omega$ is a Hermitian operator, $\Omega^\dagger = \Omega$. Why? We’ll see. It satisfies the composition property
$$ \mathcal U(t_0 + dt_1 + dt_2, t_0) = \mathcal U(t_0 + dt_1 + dt_2, t_0 + dt_1)\mathcal U(t_0 + dt_1, t_0) $$
since $(1 - i\Omega dt_2)(1 - i\Omega dt_1) = 1 - i\Omega(dt_1 + dt_2)$ to first order. And we can also check that $\mathcal U^\dagger\mathcal U = 1$ if we ignore the $(dt)^2$ terms:
$$ (1 + i\Omega dt)(1 - i\Omega dt) = 1 + \Omega^2dt^2 \approx 1 $$
So, it’s not the perfect estimate… but it’s right to first order, which is all we need.
# The Hamiltonian as the generator of time translations
The operator $\Omega$ has dimensions of $\text{time}^{-1}$. Any observable like that? Recall $E = \hbar\omega$ $\to$ an angular frequency.
Now, we know that the Hamiltonian is the generator of time translations (from classical mechanics), so we’ll represent $\Omega$ with the Hamiltonian:
$$ \implies \hat\Omega = \frac{\hat H}{\hbar} $$
$$ \implies \boxed{\mathcal U(t_0 + dt, t_0) = 1 - \frac{i\hat H dt}{\hbar}} $$
$\hat H$ is assumed to be Hermitian. Compare it with the other two generators we’ve seen:
$$ \mathcal J(d\vec x') = 1 - \frac{i\vec p\cdot d\vec x'}{\hbar}, \qquad \mathcal D(\hat n, d\phi) = 1 - \frac{i\vec J\cdot\hat n}{\hbar}d\phi, \qquad \mathcal U(t_0 + dt, t_0) = 1 - \frac{i\hat H dt}{\hbar} $$
$\to$ momentum generates translations, angular momentum generates rotations, and the Hamiltonian generates time translations.
# The Schrödinger equation
Using composition,
$$ \mathcal U(t + dt, t_0) = \mathcal U(t + dt, t)\mathcal U(t, t_0) = \left(1 - \frac{i\hat H dt}{\hbar}\right)\mathcal U(t, t_0) $$
$$ \implies \mathcal U(t + dt, t_0) - \mathcal U(t, t_0) = -\frac{i\hat H dt}{\hbar}\mathcal U(t, t_0) $$
and we can write
$$ \boxed{i\hbar\frac{\partial\mathcal U(t, t_0)}{\partial t} = \hat H\mathcal U(t, t_0)} $$
$\to$ the Schrödinger equation for the time evolution operator.
Now we can obtain this for state kets too: multiply the equation with $|\alpha, t_0\rangle$ (from the right),
$$ i\hbar\frac{\partial\mathcal U}{\partial t}|\alpha, t_0\rangle = \hat H\mathcal U(t, t_0)|\alpha, t_0\rangle $$
$|\alpha, t_0\rangle$ can go inside the derivative, as it’s independent of $t$, and $\mathcal U|\alpha, t_0\rangle = |\alpha, t_0; t\rangle$
$$ \implies \boxed{i\hbar\frac{\partial}{\partial t}|\alpha, t_0; t\rangle = \hat H|\alpha, t_0; t\rangle} $$
# Solving for the time evolution operator
We need to find solutions to
$$ i\hbar\frac{\partial\mathcal U}{\partial t} = \hat H\mathcal U $$
3 cases:
1. $\hat H$ is independent of time. Then
$$ \boxed{\mathcal U(t, t_0) = \exp\left(-\frac{i\hat H(t - t_0)}{\hbar}\right)} $$
for all $t, t_0$. Differentiate it and you get back $-\frac{i\hat H}{\hbar}\mathcal U$. It’s also what you get by compounding $N$ infinitesimal steps, like we did for finite translations:
$$ \lim_{N\to\infty}\left(1 - \frac{i\hat H(t - t_0)}{N\hbar}\right)^N = \exp\left(-\frac{i\hat H(t - t_0)}{\hbar}\right) $$
2. $\hat H$ is time dependent, but $[\hat H(t_1), \hat H(t_2)] = 0$. Then
$$ \boxed{\mathcal U(t, t_0) = \exp\left[-\frac{i}{\hbar}\int_{t_0}^{t}dt'\hat H(t')\right]} $$
3. The Hamiltonians at different times don’t commute in general. Some crazy thing called a Dyson series.
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