Time evolution of states YouTube


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
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, $$\frac{\partial}{\partial t}\ket{\psi} = -\mathrm{i}\frac{H}{\hbar}\ket{\psi}$$