# Time evolution of states

### Postulates

We now consider states that can change over time.

A system can be in some initial state $\ket{\psi(t_0)}$ and then evolve to some other state, $\ket{\psi(t)}$, at some time $t \gt t_0$ in the future.

The following postulates are both assumptions, but they are justified by empirical success. Later on we shall see why they make sense.

#### The linearity postulate

States evolve linearly over time.

That is, if a system is in the state $\ket{\psi(t_0)}$, defined at some initial time $t_0$, then the state will evolve in time to the state $$\ket{\psi(t)} = U(t-t_0)\ket{\psi(t_0)}$$ where $U = U(t)$ is a time-dependent linear operator, such that $$U(0) = \mathbf{I}$$

Nb. We previously defined a particular version of the linearity postulate and used it in considering the two-slit experiment.

It is often convenient to use subscripts and to assume that $t_0 = 0$ (without loss of generality). Then, the postulate becomes $$\ket{\psi_t} = U(t)\ket{\psi_0}$$

#### The unitarity postulate

Inner products between state remain constant over time.

That is, given two states $\ket{\psi(t_0)}, \ket{\phi(t_0)}$, defined at a time $t_0$, then for all $t \gt t_0$ $$\braket{\psi(t)}{\phi(t)} = \braket{\psi(t_0)}{\phi(t_0)}$$

In subscripts, and with $t_0 = 0$, $$\braket{\psi_t}{\phi_t} = \braket{\psi_0}{\phi_0}$$

### The evolution operator

We can find out what properties the operator $U$ has by combining the two postulates.

We say that a linear operator $M$ is unitary if it satisfies $$M^\dagger M = \mathbf{I}$$

Clearly, unitary operators are Hermitian. A unitary operator is analogous to a complex number having unit length, that is, a pure phase number.

The linear operator, $U$, defined in the linearity postulate, is a unitary operator. That is, $$U^\dagger U = \mathbf{I}$$

To show why this is true, first note that the conjugate version of the linearity postulate is $$\bra{\phi(t)} = \bra{\phi(t_0)}U^\dagger(t)$$

Then, given two $\ket{\phi(t)}, \ket{\psi(t)}$ arbitrarily chosen states, \begin{align*}\braket{\phi(t_0)}{\psi(t_0)} &= \braket{\phi(t)}{\psi(t)} & \text{ unitarity}\\&= \bra{\phi(t_0)}U^\dagger(t)U(t)\ket{\psi(t_0)} & \text{ linearity}\end{align*}

Since this holds for any particular time, we can set the two states equal to basis states, $\left\{\ket{e_i}:i=1,2,\cdots,N\right\}$ say, and hence calculate the components of $U^\dagger(t)U(t)$, $$\bra{e_i}U^\dagger(t)U(t)\ket{e_j} = \braket{e_i}{e_j} = \delta_{ij}$$

This is the same as saying $$U^\dagger U = \mathbf{I}$$ and so $U$ is unitary.

We call $U$ an evolution operator.

### The Hamiltonian operator

Given that the evolution operator $U$ satisfies
$$U(0)=\mathbf{I}$$
and is linear, we can suppose that if $\Delta t = t - t_0$ is small then $U$ at time $\Delta t$ is $$U(\Delta t) = \mathbf{I} + W\Delta t$$ where $W$ is a linear operator and we assume $\Delta t^2 \approx 0$.

For convenience, however, we usually write $W$ in terms of another operator $H$, and Planck's constant, $\hbar = \tfrac{h}{2\pi}$, where $$W = -\mathrm{i}\frac{H}{\hbar}$$

Then, the evolution operator at time $\Delta t$ is $$U(\Delta t) = \mathbf{I} -\mathrm{i}\frac{H}{\hbar}\Delta t$$ and the conjugate statement is $$U^\dagger(\Delta t) = \mathbf{I} + \mathrm{i}\frac{H^\dagger}{\hbar}\Delta t$$

Since $U$ is unitary, we can write \begin{align*}\mathbf{I} &= U^\dagger U\\&= U^\dagger(\Delta t) U(\Delta t)\\&= \left(\mathbf{I} + \mathrm{i}\frac{H^\dagger}{\hbar}\Delta t\right) \left(\mathbf{I} -\mathrm{i}\frac{H}{\hbar}\Delta t\right)\\&= \mathbf{I} + \mathrm{i}\frac{H^\dagger - H}{\hbar}\Delta t & \text{ since } \Delta t^2 \approx 0\end{align*}

This implies that $$\mathrm{i}\frac{H^\dagger - H}{\hbar}\Delta t = 0$$ and since we assuming that $\Delta t \gt 0$ we must have $$H^\dagger = H$$ and so $H$ is a Hermitian operator.

$H$ is called the Hamiltonian observable of the system. It corresponds to energy and its eigenvalues are the discrete energy bands associated with the system.

### Schrodinger's equation

Given a state $\ket{\psi}$ of some system, there exists a Hermitian operator, $H$, called the Hamiltonian, associated with the system, such that $$\frac{\partial}{\partial t}\ket{\psi} = -\mathrm{i}\frac{H}{\hbar}\ket{\psi}$$

This relationship is the generalised version of Schrodinger's equation.

To see why it's true, consider some state $\ket{\psi(t_0)}$, at time $t = t_0$. Then, \begin{align*}\left.\frac{\partial}{\partial t}\right|_{t=t_0} \ket{\psi(t_0)} &= \lim_{\Delta t \rightarrow 0}\frac{\ket{\psi(t_0 + \Delta t)} - \ket{\psi(t_0)}}{\Delta t}\\&= \lim_{\Delta t \rightarrow 0}\frac{U(\Delta t)\ket{\psi(t_0)} - \ket{\psi(t_0)}}{\Delta t}\\&= \lim_{\Delta t \rightarrow 0}\frac{U(\Delta t) - \mathbf{I}}{\Delta t}\ket{\psi(t_0)}\\&= \lim_{\Delta t \rightarrow 0}\frac{-\mathrm{i}\frac{H}{\hbar}\Delta t}{\Delta t}\ket{\psi(t_0)}\\&= \lim_{\Delta t \rightarrow 0}-\mathrm{i}\frac{H}{\hbar}\ket{\psi(t_0)}\\&= -\mathrm{i}\frac{H}{\hbar}\ket{\psi(t_0)}\end{align*}

Since the time $t_0$ was chosen arbitrarily, we can write, more generally, $$\frac{\partial}{\partial t}\ket{\psi} = -\mathrm{i}\frac{H}{\hbar}\ket{\psi}$$