# An Intro to Time Evolution: The Heisenberg and Schrödinger Pictures

A quantum state is just a function that describes the probabilistic nature of a particle (or particles) in terms of measurable quantities. Measurable quantities are represented by hermitian operators that act on the state to give possible values. But what happens to a quantum state over time? How does it change? How do the measurable quantities change? Here I will elaborate on what is called “time evolution”, a method of evolving states and operators.
Evolving a state to a later time, and including time dependence are done in the same way.
$\left| \Psi (x,0) \right> \rightarrow \left| \Psi (x,t) \right>$
This is the Schrödinger picture, which evolves states. In the Heisenberg picture, operators evolve. The pictures are equivalent, but are suited for different purposes. One can’t talk about one without talking about the other.
To evolve a state, we want to construct a linear operator that changes the argument of the function.
$\mathscr{U} \left| \Psi (x,0) \right> = \left| \Psi(x,t) \right>$
The operator should be unitary, to conserve probability.
$P = \left< \Psi(x,0) | \Psi(x,0) \right> = \left< \Psi(x,t) | \Psi(x,t) \right> = \left< \Psi(x,0) \left| \mathscr{U}^{\dagger} \mathscr{U} \right| \Psi(x,0) \right>$
$\therefore \mathscr{U}^{\dagger} \mathscr{U} = 1$

One way to construct this operator is to solve the time-dependent Schrödinger equation, with initial conditions imposed on it. For simplicity, let’s assume that the Hamiltonian operator $H$ has no time dependence itself.
$\frac{\partial}{\partial t} \Psi = \frac{-i}{\hbar} H \Psi$
$ln(\Psi) = \frac{-i}{\hbar} \int H dt$
$\Psi (x,t) = e^{\frac{-i}{\hbar} H t} A$
$\Psi(x,0) = A$
$\Psi (x,t) = e^{\frac{-i}{\hbar} H t} \Psi (x,0)$
$\therefore \mathscr{U} = e^{\frac{-i}{\hbar} H t}$
The resulting operator $\mathscr{U}$ is a unitary time evolution operator.

In this simple case, one can also use the same approach used to construct the translation operator, except it will translate through time instead of space.
$\Psi (x,t) \rightarrow \Psi (x,t + \Delta t)$
$\Psi (x, t + \Delta t) = \Psi(x,t) + \Delta t \frac{d}{dt} \Psi(x,t) + \Delta t^2 \frac{1}{2!} \frac{d^2}{dt^2} \Psi(x,t) + ... = e^{\Delta t \frac{d}{dt}} \Psi(x,t)$
$\therefore \mathscr{U} = e^{\Delta t \frac{d}{dt}} = e^{\frac{-i}{\hbar} H \Delta t}$

The Schrödinger-style time evolution can be transformed into Heisenberg-style by pulling the time evolution operators from the state $\Psi$ and attaching them to the operator $\mathscr{O}$.
$\left< \Psi (x,0) \left| \mathscr{O} \right| \Psi (x,0) \right> \rightarrow \left< \Psi (x,t) | \mathscr{O} | \Psi (x,t) \right>$
$\left< \Psi (x,t) | \mathscr{O} | \Psi (x,t) \right> = \left< \Psi (x,0) \left| e^{\frac{i}{\hbar} H t} \mathscr{O} e^{\frac{-i}{\hbar} H t} \right| \Psi (x,0) \right> = \left< \Psi (x,0) \left| \mathscr{O}(t) \right| \Psi (x,0) \right>$
The resulting time-evolved operator is then:
$\mathscr{O}(t) = e^{\frac{i}{\hbar} H t} \mathscr{O} e^{\frac{-i}{\hbar} H t}$

Taking the time derivative of this general operator will give the Heisenberg equation of motion. Plugging in an operator for $\mathscr{O}$ will give an equation describing how that operator evolves with time.
$\frac{d}{dt} \mathscr{O}(t) = \frac{i}{\hbar} e^{\frac{i}{\hbar} H t} H \mathscr{O} e^{\frac{-i}{\hbar} H t} + e^{\frac{i}{\hbar} H t} \frac{\partial \mathscr{O}}{\partial t} e^{\frac{-i}{\hbar} H t} + \frac{-i}{\hbar} e^{\frac{i}{\hbar} H t} \mathscr{O} H e^{\frac{-i}{\hbar} H t}$
The exponential operator $e^{\frac{\pm i}{\hbar} H t}$ commutes with $H$, since it is made of only $H$ operators. In the first term and last term, the exponential operator can act on the $\mathscr{O}$ to evolve it into $\mathscr{O}(t)$, as shown previously. So the expression can be reduced to:
$\frac{d}{dt} \mathscr{O}(t) = \frac{1}{i \hbar} [\mathscr{O}(t), H] + e^{\frac{i}{\hbar} H t} \frac{\partial \mathscr{O}}{\partial t} e^{\frac{-i}{\hbar} H t}$
This is the Heisenberg equation of motion.

So, let’s try out a specific operator, to see how it will evolve with time.
First, we need to define the Hamiltonian operator:
$H = \frac{p^2}{2m} + V(x)$
The first term is kinetic energy in terms of momentum, and the second term is potential energy in terms of position. A Hamiltonian can be arbitrarily more complicated, but this form is fairly general, and relatively simple.
So let’s see how the position operator evolves over time, so plug in $x$:
$\frac{d}{dt} x = \frac{1}{i \hbar} [x, H] + e^{\frac{i}{\hbar} H t} \frac{\partial x}{\partial t} e^{\frac{-i}{\hbar} H t}$
The operator $x$ has no explicit time dependence, so $\frac{\partial x}{\partial t} = 0$. So we are left with:
$\frac{d}{dt} x = \frac{1}{i \hbar} [x, H]$
$[x,H] = [x, \frac{p^2}{2m} + V(x)] = [x, p^2]/2m$
Since $V(x)$ depends only on $x$, it commutes with $x$.
$[x,V(x)] = 0$
However, $x$ does not commute with $p$. This is what gives the uncertainty principle between position and momentum.
$[x,p] = i \hbar$
$\therefore [x,p^2] = 2 i \hbar p$
Using this result we can show:
$[x,H] = i \hbar \frac{p}{m}$
$\therefore \frac{d}{dt} x = \frac{p}{m}$
The equation of motion that describes the evolution of $x$ is then:
$x(t) = \int \frac{p}{m} dt$

The same can be done for the momentum operator.
$\frac{d}{dt} p = \frac{1}{i \hbar} [p, H]$
$[p,H] = [p, \frac{p^2}{2m} + V(x)] = [p,V(x)] = -i \hbar \frac{\partial}{\partial x} V(x)$
$\frac{d}{dt} p = -\frac{\partial}{\partial x} V(x)$
The equation of motion that describes the evolution of $p$ is then:
$p(t) = -\int \frac{\partial}{\partial x} V(x) dt$
This corresponds to Newton’s law, and shows how momentum is linked to the potential energy $V(x)$. The derivative of potential energy with respect to distance gives force.

To make this more familiar, let’s set $V(x) = 0$, to get the free-particle scenario (no forces acting on the particle).
So now $\frac{d}{dt} p = 0$ and the equations of motion become:
$p(t) = p(0)$
$x(t) = x(0) + \frac{p(0)}{m} t$
which describes a particle moving at constant momentum (constant velocity).