II The Schrödinger equation

For simplicity consider a particle moving in one dimension, with position measured by a coordinate $x$. A freely propagating wave with wave number $k$ and angular frequency $\omega$ is described by

$${\mathrm{\Psi }}_k(x;t)=A\exp (ikx-i\omega t).$$

(7)

For light (photons), $\omega =kc$, and we know that ${\mathrm{\Psi }}_k(x;t)$ solves the wave equation

$$\frac{{\partial }^2\mathrm{\Psi }(x;t)}{\partial t^2}=c^2\frac{{\partial }^2\mathrm{\Psi }(x;t)}{\partial x^2}.$$

(8)

Indeed, if we insert $\mathrm{\Psi }(x;t)={\mathrm{\Psi }}_k(x;t)$ into this equation it reduces to

$$-{\omega }^2{\mathrm{\Psi }}_k(x;t)=-c^2{k}^2{\mathrm{\Psi }}_k(x;t).$$

(9)

Let us multiply the left-hand side of Eq. (9) with ${(i\mathrm{\hslash })}^2=-{\mathrm{\hslash }}^2$, and the right-hand side with ${(-i\mathrm{\hslash })}^2=-{\mathrm{\hslash }}^2$. With the help of the Planck and de Broglie relations, we then find (check that both forms are indeed identical!)

$$E^2{\mathrm{\Psi }}_k(x;t)=c^2{p}^2{\mathrm{\Psi }}_k(x;t).$$

(10)

For arbitrary $\mathrm{\Psi }(x;t)$, we now repeat this manipulation for the wave equation (8) itself:

$${\left(i\mathrm{\hslash }\frac{\partial }{\partial t}\right)}^2\mathrm{\Psi }(x;t)=c^2{\left(-i\mathrm{\hslash }\frac{\partial }{\partial x}\right)}^2\mathrm{\Psi }(x;t).$$

(11)

This reformulation turns out to be very convenient. Notice that the operations

	$i\mathrm{\hslash }{\displaystyle \frac{\partial }{\partial t}}{\mathrm{\Psi }}_k(x;t)=E{\mathrm{\Psi }}_k(x;t),$		(12)
	$-i\mathrm{\hslash }{\displaystyle \frac{\partial }{\partial x}}{\mathrm{\Psi }}_k(x;t)=p{\mathrm{\Psi }}_k(x;t),$		(13)

allow us to directly read off the energy and momentum of the photons with wavefunction (7). If $\mathrm{\Psi }(x;t)$ is not of the specific form ${\mathrm{\Psi }}_k(x;t)$, the result of these operations is not just a multiplication with a number ($E$ or $p$). However, if we define

	$i\mathrm{\hslash }{\displaystyle \frac{\partial }{\partial t}}\mathrm{\Psi }(x;t)\equiv \widehat{E}\mathrm{\Psi }(x;t),$		(14)
	$-i\mathrm{\hslash }{\displaystyle \frac{\partial }{\partial x}}\mathrm{\Psi }(x;t)\equiv \widehat{p}\mathrm{\Psi }(x;t),$		(15)

the result always fulfils the wave equation

$${\widehat{E}}^2\mathrm{\Psi }(x;t)=c^2{\widehat{p}}^2\mathrm{\Psi }(x;t).$$

(16)

Notice the similarity to Eq. (10). We call $\widehat{E}$ the energy operator and $\widehat{p}$ the momentum operator. The mathematical properties of these operators are explored later on in this course.
Points to remember

• •

  Physical observables are associated with operators.

• •

  Momentum operator ${\widehat{p}}_x=-i\mathrm{\hslash }\frac{\partial }{\partial x}$, ${\widehat{p}}_y=-i\mathrm{\hslash }\frac{\partial }{\partial y}$, ${\widehat{p}}_z=-i\mathrm{\hslash }\frac{\partial }{\partial z}$;

• •

  Energy operator $\widehat{E}=i\mathrm{\hslash }\frac{\partial }{\partial t}$.

• •

  Position operator $\widehat{x}=x$, $\widehat{y}=y$, $\widehat{z}=z$.

• •

  E.g.:
  ${\widehat{p}}_x\widehat{x}f(x)=-i\mathrm{\hslash }\frac{d}{dx}(xf(x))$
  $\widehat{x}{\widehat{p}}_{x}f(x)=-xi\mathrm{\hslash }\frac{d}{dx}(f(x))$.

For light, we have ${\widehat{E}}^2\mathrm{\Psi }(x;t)=c^2{\widehat{p}}^2\mathrm{\Psi }(x;t)$ in correspondence to the relation $E^2=c^2{p}^2$ for energy and momentum of the photons. This correspondence shall guide as a heuristic principle from classical mechanics to quantum mechanics.

For particles with finite mass (at non-relativistic speed), the total energy is given by $E=p^2/2m+V(x)$, where $T=p^2/2m=m{v}^2/2$ is the kinetic energy and $V(x)$ is the potential energy (recall that the force acting on the particle is $F(x)=-dV/dx$).

Reading the correspondence principle given above backwards, one finds the Schrödinger equation

$$\widehat{E}\mathrm{\Psi }(x;t)=\frac{{\widehat{p}}^2}{2m}\mathrm{\Psi }(x;t)+\widehat{V}\mathrm{\Psi }(x;t).$$

(17)

Let us decode this equation and express it in usual mathematical notation:

$$i\mathrm{\hslash }\frac{\partial \mathrm{\Psi }(x;t)}{\partial t}=-\frac{{\mathrm{\hslash }}^2}{2m}\frac{{\partial }^2\mathrm{\Psi }(x;t)}{\partial x^2}+V(x)\mathrm{\Psi }(x;t).$$

(18)

In the steps above we associate to the potential energy an operator $\widehat{V}$, which changes $\mathrm{\Psi }(x;t)$ into

$$\widehat{V}\mathrm{\Psi }(x;t)\equiv V(x)\mathrm{\Psi }(x;t).$$

(19)

It is only consequent to also introduce the position operator

$$\widehat{x}\mathrm{\Psi }(x;t)\equiv x\mathrm{\Psi }(x;t)$$

(20)

and the operator of kinetic energy $\widehat{T}\equiv {\widehat{p}}^2/2m$. The combination

$$\widehat{H}\equiv \widehat{T}+\widehat{V}={\widehat{p}}^2/2m+\widehat{V}$$

(21)

is called the Hamilton operator (or Hamiltonian), which is of central importance in quantum mechanics. It corresponds to the total energy in classical mechanics.

We now can restate the Schrödinger equation in the compact form

$$i\mathrm{\hslash }\frac{\partial }{\partial t}\mathrm{\Psi }(x;t)=\widehat{H}\mathrm{\Psi }(x;t).$$

(22)

Even more compactly we can write

$$\widehat{E}\mathrm{\Psi }=\widehat{H}\mathrm{\Psi },$$

(23)

where we suppressed the arguments.
Points to remember

• •

  The evolution of the wave function (=state of the system) for a quantum particle of mass $m$ moving in one dimension is determined by the Schrödinger equation

  $$i\mathrm{\hslash }\frac{\partial \mathrm{\Psi }(x;t)}{\partial t}=-\frac{{\mathrm{\hslash }}^2}{2m}\frac{{\partial }^2\mathrm{\Psi }(x;t)}{\partial x^2}+V(x)\mathrm{\Psi }(x;t),$$

  (24)

  or

  $$\widehat{E}\mathrm{\Psi }(x;t)=\widehat{H}\mathrm{\Psi }(x;t),$$

  (25)

  where $\widehat{H}={\widehat{p}}^2/2m+\widehat{V}(x)$ is the Hamiltonian.

If $\widehat{H}$ does not explicitly depend on time [a counter example would be a driven particle sitting in a modulated potential $V(x;t)$], energy is classically conserved. In quantum mechanics, this results in a simplification of the Schrödinger equation, since we can separate out the time variable $t$. Let us write

$${\mathrm{\Psi }}_E(x;t)=\exp (-iEt/\mathrm{\hslash })\psi (x),$$

(26)

where the time dependence $\propto \exp (-iEt/\mathrm{\hslash })=\exp (-i\omega t)$ is the same as for a propagating plane wave, but the spatial dependence has not been specified yet. With the energy operator defined in Eq. (14), we then have $\widehat{E}{\mathrm{\Psi }}_E(x;t)=E{\mathrm{\Psi }}_E(x;t)$ [see Eq. (12)]. It now remains to find an equation for the spatial dependence $\psi (x)$. We insert ${\mathrm{\Psi }}_E(x;t)$ into the time-dependent Schrödinger equation (22) and obtain

$$E\psi (x)\exp (-iEt/\mathrm{\hslash })=\widehat{H}\psi (x)\exp (-iEt/\mathrm{\hslash }).$$

(27)

Divide by the time-dependent factor $\exp (-iEt/\mathrm{\hslash })$, and we end up with the stationary Schrödinger equation

$$E\psi (x)=\widehat{H}\psi (x),$$

(28)

where $E$ is now just a real number. This equation is independent of time. For a particle of mass $m$ moving in a one-dimensional potential $V(x)$, its explicit form is

$$E\psi (x)=-\frac{{\mathrm{\hslash }}^2}{2m}\frac{d^2\psi }{d{x}^2}+V(x)\psi (x).$$

(29)

This is the most important equation in this module, and we will study it for a wide range of problems. Hence, remember it well!
Points to remember

• •

  For a particle of mass $m$ moving in one dimension, the states with fixed energy $E$ are determined by the stationary Schrödinger equation

  $$E\psi (x)=-\frac{{\mathrm{\hslash }}^2}{2m}\frac{d^2\psi }{d{x}^2}+V(x)\psi (x).$$

  (30)