# Time-Dependent Solutions: Propagators and Representations

*Michael Fowler, UVa *

### Introduction

We’ve spent most of the course so far concentrating on the eigenstates of the Hamiltonian, states whose time-dependence is merely a changing phase. We did mention much earlier a superposition of two different energy states in an infinite well, resulting in a wave function sloshing backwards and forwards. It’s now time to cast the analysis of time dependent states into the language of bras, kets and operators. We’ll take a time-independent Hamiltonian $H,$ with a complete set of orthonormalized eigenstates, and as usual

$$i\hslash \frac{\partial \psi \left(x,t\right)}{\partial t}=-\frac{{\hslash}^{2}}{2m}\frac{{\partial}^{2}\psi \left(x,t\right)}{\partial {x}^{2}}+V\left(x\right)\psi \left(x,t\right),$$

Or, as we would now write it

$$i\hslash \frac{\partial}{\partial t}|\psi \left(x,t\right)\rangle =H|\psi \left(x,t\right)\rangle .$$

Since $H$ is itself time independent, this is very easy to integrate!

$|\psi \left(x,t\right)\rangle ={e}^{-iH\left(t-{t}_{0}\right)/\hslash}|\psi \left(x,{t}_{0}\right)\rangle .$

*propagator*, because it describes how the wave propagates from its initial configuration, and is usually denoted by $U:$

$|\psi \left(x,t\right)\rangle =U\left(t-{t}_{0}\right)|\psi \left(x,{t}_{0}\right)\rangle .$

$U\left(t-{t}_{0}\right)={e}^{-iH\left(t-{t}_{0}\right)},\text{so}\text{\hspace{0.17em}}{U}^{\u2020}\left(t-{t}_{0}\right)={e}^{i{H}^{\u2020}\left(t-{t}_{0}\right)}={e}^{iH\left(t-{t}_{0}\right)}={U}^{-1}\left(t-{t}_{0}\right).$

$\begin{array}{c}\langle \psi \left(x,t\right)|\psi \left(x,t\right)\rangle =\langle \psi \left(x,{t}_{0}\right)|{U}^{\u2020}U\left(t-{t}_{0}\right)|\psi \left(x,{t}_{0}\right)\rangle \\ =\langle \psi \left(x,{t}_{0}\right)|\psi \left(x,{t}_{0}\right)\rangle \end{array}$

*same*as the eigenstates of $H$, and if $H|{\psi}_{n}\rangle ={E}_{n}|{\psi}_{n}\rangle ,$ then

${e}^{-iH\left(t-{t}_{0}\right)/\hslash}|{\psi}_{n}\rangle ={e}^{-i{E}_{n}\left(t-{t}_{0}\right)/\hslash}|{\psi}_{n}\rangle .$

$\begin{array}{c}|\psi \left(t\right)\rangle ={e}^{-iH\left(t-{t}_{0}\right)/\hslash}{\displaystyle \sum _{n=1}^{\infty}|{\psi}_{n}\rangle \langle {\psi}_{n}}|\psi \left({t}_{0}\right)\rangle \\ ={\displaystyle \sum _{n=1}^{\infty}{e}^{-i{E}_{n}\left(t-{t}_{0}\right)/\hslash}|{\psi}_{n}\rangle \langle {\psi}_{n}}|\psi \left({t}_{0}\right)\rangle .\end{array}$

$\begin{array}{c}|\psi \left({t}_{0}\right)\rangle ={\displaystyle \sum |{\psi}_{n}\left({t}_{0}\right)\rangle}\langle {\psi}_{n}\left({t}_{0}\right)|\psi \left({t}_{0}\right)\rangle \\ ={\displaystyle \sum {c}_{n}|{\psi}_{n}\left({t}_{0}\right)\rangle}\end{array}$

$|\psi \left(t\right)\rangle ={\displaystyle \sum {c}_{n}{e}^{-i{E}_{n}\left(t-{t}_{0}\right)/\hslash}|{\psi}_{n}\left({t}_{0}\right)\rangle}.$

$\langle E\rangle =\langle \psi |H|\psi \rangle ={\displaystyle \sum {\left|{c}_{n}\right|}^{2}{E}_{n}}$

*x*is

$\langle \psi \left(t\right)|x|\psi \left(t\right)\rangle ={\displaystyle \sum _{n,m}{c}_{n}^{*}{c}_{m}{e}^{i({E}_{n}-{E}_{m})\left(t-{t}_{0}\right)/\hslash}\langle {\psi}_{n}\left({t}_{0}\right)|x|{\psi}_{m}\left({t}_{0}\right)\rangle}$_{}

*not*in general time-independent. (It

*is*real, of course, on adding the $n,m$ term to the $m,n$ term.)

*time-independent*Hamiltonian. The important extension to a system in a time-dependent external field, such as an atom in a light beam, will be given later in the course.

### The Free Particle Propagator

$$\begin{array}{c}U\left(t\right)={e}^{-iHt/\hslash}={\displaystyle \underset{-\infty}{\overset{\infty}{\int}}{e}^{-iH\text{\hspace{0.05em}}t/\hslash}\frac{dk}{2\pi}|k\rangle \langle k|}\\ ={\displaystyle \underset{-\infty}{\overset{\infty}{\int}}{e}^{-i\hslash \text{\hspace{0.05em}}{k}^{2}t/2m}\frac{dk}{2\pi}|k\rangle \langle k|}.\end{array}$$

$\begin{array}{c}\langle x|U\left(t,0\right)|{x}_{0}\rangle \text{\hspace{0.17em}}={\displaystyle \underset{-\infty}{\overset{\infty}{\int}}{e}^{-i\hslash {k}^{2}t/2m}\frac{dk}{2\pi}\langle x|k\rangle \langle k|{x}_{0}\rangle}\\ ={\displaystyle \underset{-\infty}{\overset{\infty}{\int}}{e}^{-i\hslash {k}^{2}t/2m}\frac{dk}{2\pi}{e}^{-ik\left({x}_{0}-x\right)}}\\ =\sqrt{\frac{m}{2\pi \hslash it}}{e}^{im{\left({x}_{0}-x\right)}^{2}/2\hslash t},\end{array}$

*Question*: Where did we go wrong?

*Answer*: Notice first that ${\left|\psi \left(x,t\right)\right|}^{2}$ is constant throughout space. This means that the normalization, $\int {\left|\psi \left(x,t\right)\right|}^{2}}dx=\infty $! And, as we’ve seen above, the normalization stays constant in time$\u2014$the propagator is unitary. Therefore, our initial wave function must have had infinite norm. That’s exactly right$\u2014$we took the initial wave function $\psi (x,t=0)=\delta (x-{x}_{0})=|{x}_{0}\rangle $.

*not*represent wave functions individually normalizable in the usual sense. The only meaningful wave functions are

*integrals over a range*of such kets, such as $\int dx\psi \left(x\right)}|x\rangle $. In an integral like this, notice that states $|x\rangle $ within some tiny $x$ -interval of length $\delta x,$ say, have total weight $\psi \left(x\right)\delta x$, which goes to zero as $\delta x$ is made smaller, but by writing $\psi (x,t=0)=\delta (x-{x}_{0})=|{x}_{0}\rangle $ we took a single such state and gave it a finite weight. This we can’t do.

*near*a point develops. To find out, we must apply the propagator to a legitimate wave function$\u2014$one that is normalizable to begin with. The simplest “localized particle” wave function from a practical point of view is a Gaussian wave packet,

$$\psi \left({x}^{\prime},0\right)={e}^{i{p}_{0}{x}^{\prime}/\hslash}\frac{{e}^{-{{x}^{\prime}}^{2}/2{d}^{2}}}{{\left(\pi {d}^{2}\right)}^{1/4}}.$$

$\begin{array}{c}\psi (x,t)={\displaystyle \int U(x,t;{x}^{\prime},0){e}^{i{p}_{0}{x}^{\prime}/\hslash}\frac{{e}^{-{{x}^{\prime}}^{2}/2{d}^{2}}}{{\left(\pi {d}^{2}\right)}^{1/4}}d{x}^{\prime}}\\ =\sqrt{\frac{m}{2\pi \hslash it}}{\displaystyle \int {e}^{im{(x-{x}^{\prime})}^{2}/2\hslash t}}{e}^{i{p}_{0}{x}^{\prime}/\hslash}\frac{{e}^{-{{x}^{\prime}}^{2}/2{d}^{2}}}{{\left(\pi {d}^{2}\right)}^{1/4}}d{x}^{\prime}.\end{array}$

$\underset{-\infty}{\overset{\infty}{\int}}d{x}^{\prime}{e}^{-a{{x}^{\prime}}^{2}+b{x}^{\prime}}=\sqrt{\frac{\pi}{a}}{e}^{{b}^{2}/4a}$.

$$b=-\frac{im}{\hslash t}\cdot \left(x-\frac{{p}_{0}t}{m}\right)$$, $$a=\frac{1}{2{d}^{2}}-\frac{im}{2\hslash t}$$.

$$\psi \left(x,t\right)=\frac{{\pi}^{-1/4}}{\sqrt{d\left(1+\frac{i\hslash t}{m{d}^{2}}\right)}}\mathrm{exp}\left(\frac{im{x}^{2}}{2\hslash t}\right)\mathrm{exp}\left(-\frac{\frac{im}{\hslash t}{\left(x-\frac{{p}_{0}t}{m}\right)}^{2}}{2\left(1+\frac{i\hslash t}{m{d}^{2}}\right)}\right)$$

where the second exponential is the term ${e}^{{b}^{2}/4a}$. As written, the small $t$ limit is not very apparent, but some algebraic rearrangement yields:

$$\psi \left(x,t\right)=\frac{{\pi}^{-1/4}}{\sqrt{d\left(1+i\hslash t/m{d}^{2}\right)}}\mathrm{exp}\left(-\frac{{\left(x-{p}_{0}t/m\right)}^{2}}{2{d}^{2}\left(1+i\hslash t/m{d}^{2}\right)}\right)\mathrm{exp}\left(\frac{i{p}_{0}}{\hslash}\left(x-{p}_{0}t/2m\right)\right)$$.

*probability density*:

$${\left|\psi \left(x,t\right)\right|}^{2}=\frac{1}{\sqrt{\pi \left({d}^{2}+{\hslash}^{2}{t}^{2}/{m}^{2}{d}^{2}\right)}}\cdot {\mathrm{exp}}^{-\frac{{\left(x-{p}_{0}t/m\right)}^{2}}{\left({d}^{2}+{\hslash}^{2}{t}^{2}/{m}^{2}{d}^{2}\right)}}$$.

*velocities*$\langle \Delta v\rangle $ within the packet, hence the gradual spreading $\langle \Delta v\rangle t$ in $x$ -space.

### Schrödinger and Heisenberg Representations

Assuming a Hamiltonian with no explicit time dependence, the time-dependent Schrödinger equation has the form

$i\hslash \frac{\partial}{\partial t}|\psi \left(x,t\right)\rangle =H|\psi \left(x,t\right)\rangle $

and as discussed above, the formal solution can be expressed as:

$|\psi \left(x,t\right)\rangle ={e}^{-iHt/\hslash}|\psi \left(x,t=0\right)\rangle .$

*physically significant time dependent quantities*are of the form

$\langle \phi \left(t\right)|A|\psi \left(t\right)\rangle =\langle \phi \left(0\right)|{e}^{iHt/\hslash}A{e}^{-iHt/\hslash}|\psi \left(0\right)\rangle $

where $A$ is an operator, which we are assuming has no explicit time dependence.

*not*change in time, it remained at $|\psi \left(0\right)\rangle ,$ but the

*operators*evolved according to:

${A}_{H}(t)={e}^{iHt/\hslash}{A}_{H}(0){e}^{-iHt/\hslash}.$

$i\hslash \frac{d{A}_{H}(t)}{dt}=\left[{A}_{H}(t),H\right].$

The Hamiltonian itself does not change in time$\u2014$energy is conserved, or, to put it another way, $H$ commutes with ${e}^{-iHt/\hslash}.$ But for a nontrivial Hamiltonian, say for a particle in one dimension in a potential,

$H={p}^{2}/2m+V(x)$

*will*have time-dependence, parallel to the classical case: the

*kinetic*energy of a swinging pendulum varies with time. (For a particle in a potential in an energy eigenstate the expectation value of the kinetic energy is constant, but this is not the case for any other state, that is, for a superposition of different eigenstates.) Nevertheless, the

*commutator*of $x,p$ will be time-independent:

$\begin{array}{c}\left[{x}_{H}(t),{p}_{H}(t)\right]={e}^{iHt/\hslash}\left[{x}_{H}(0),{p}_{H}(0)\right]{e}^{-iHt/\hslash}\\ ={e}^{iHt/\hslash}i\hslash {e}^{-iHt/\hslash}\\ =i\hslash .\end{array}$

$$\left[{x}_{H}(t),\text{\hspace{0.17em}}\frac{{p}^{2}{}_{H}(t)}{2m}\right]=\frac{i\hslash {p}_{H}(t)}{m}$$

so

$$\frac{d{x}_{H}(t)}{dt}=\frac{{p}_{H}(t)}{m}$$

$$\frac{d{p}_{H}(t)}{dt}=\text{\hspace{0.17em}}\frac{1}{i\hslash}\left[{p}_{H}(t),V({x}_{H}(t))\right]=-\nabla V({x}_{H}(t)).$$

*Exercise*: check this.

*classical*laws of motion!

*Ehrenfest’s Theorem*, that the

*expectation values*of operators in a quantum state follow the classical laws of motion, follows immediately, by taking the expectation value of both sides of the operator equation of motion in a quantum state.

### Simple Harmonic Oscillator in the Heisenberg Representation

For the simple harmonic oscillator, the equations are easily integrated to give:

$\begin{array}{l}{x}_{H}(t)={x}_{H}(0)\mathrm{cos}\omega t+({p}_{H}(0)/m\omega )\mathrm{sin}\omega t\\ {p}_{H}(t)={p}_{H}(0)\mathrm{cos}\omega t-m\omega {x}_{H}(0)\mathrm{sin}\omega t.\end{array}$

*annihilation operator*$a$ is:

$a\left(t\right)={e}^{iHt/\hslash}a\left(0\right){e}^{-iHt/\hslash}$

with

$H=\hslash \omega \left({a}^{\u2020}\left(t\right)a\left(t\right)+{\scriptscriptstyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\right).$

$\begin{array}{c}i\hslash \frac{d}{dt}a\left(t\right)=\left[a\left(t\right),H\right]\\ =\hslash \omega \left[a\left(t\right),{a}^{\u2020}\left(t\right)a\left(t\right)\right]\\ =\hslash \omega \left[a\left(t\right),{a}^{\u2020}\left(t\right)\right]a\left(t\right)\\ =\hslash \omega \text{\hspace{0.05em}}\text{\hspace{0.05em}}a\left(t\right)\end{array}$

so

$a\left(t\right)=a\left(0\right){e}^{-i\omega t}.$

Actually, we could have seen this as follows: if $|n\rangle $ are the energy eigenstates of the simple harmonic oscillator,

${e}^{-iHt/\hslash}|n\rangle ={e}^{-in\text{\hspace{0.05em}}\hslash \omega t/\hslash}|n\rangle ={e}^{-in\omega t}|n\rangle .$

*only*nonzero matrix elements of the annihilation operator $\widehat{a}$ between energy eigenstates are of the form

$\begin{array}{c}\langle n-1|a\left(t\right)|n\rangle =\langle n-1|{e}^{iHt/\hslash}a\left(0\right){e}^{-iHt/\hslash}|n\rangle \\ ={e}^{i\omega \left(n-1\right)t}\langle n-1|a\left(0\right)|n\rangle {e}^{-i\omega nt}\\ =\langle n-1|a\left(0\right)|n\rangle {e}^{-i\omega t}.\end{array}$

*all*energy matrix elements (trivially so for most of them, since they’re identically zero), and the eigenstates of the Hamiltonian span the space, it is true as an

*operator*equation.