This Quantum World/Serious illnesses/Schroedinger

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Schrödinger
If the electron is a standing wave, why should it be confined to a circle? After de Broglie's crucial insight that particles are waves of some sort, it took less than three years for the mature quantum theory to be found, not once, but twice. By Werner Heisenberg in 1925 and by Erwin Schrödinger in 1926. If we let the electron be a standing wave in three dimensions, we have all it takes to arrive at the Schrödinger equation, which is at the heart of the mature theory.

Let's keep to one spatial dimension. The simplest mathematical description of a wave of angular wavenumber $$k=2\pi/\lambda$$ and angular frequency $$\omega = 2\pi/T = 2\pi\nu$$ (at any rate, if you are familiar with complex numbers) is the function
 * $$\psi(x,t) = e^{i(kx-\omega t)}.$$

Let's express the phase $$\phi(x,t) = kx-\omega t$$ in terms of the electron's energy $$E=h\nu=\hbar\omega$$ and momentum $$p=h/\lambda=\hbar k:$$
 * $$\psi(x,t)=e^{i(px-Et)/\hbar}.$$

The partial derivatives with respect to $$x$$ and $$t$$ are

{\partial\psi\over\partial x}={i\over\hbar}p\psi\quad\hbox{and}\quad {\partial\psi\over\partial t}=-{i\over\hbar}E\psi. $$ We also need the second partial derivative of $$\psi$$ with respect to $$x$$:

{\partial^2\psi\over\partial x^2} = \left({ip\over\hbar}\right)^2\psi. $$ We thus have

E\psi=i\hbar{\partial\psi\over\partial t},\quad p\psi=-i\hbar{\partial\psi\over\partial x},\quad\hbox{and}\quad p^2\psi=-\hbar^2{\partial^2\psi\over\partial x^2}. $$ In non-relativistic classical physics the kinetic energy and the kinetic momentum $$p$$ of a free particle are related via the dispersion relation
 * $$E=p^2/2m.$$

This relation also holds in non-relativistic quantum physics. Later you will learn why.

In three spatial dimensions, $$p$$ is the magnitude of a vector $$\textbf{p}$$. If the particle also has a potential energy $$V (\textbf{r},t)$$ and a potential momentum $$\textbf{A}(\textbf{r},t)$$ (in which case it is not free), and if $$E$$ and $$\textbf{p}$$ stand for the particle's total energy and total momentum, respectively, then the dispersion relation is

E-V = (\textbf{p}-\textbf{A})^2/2m. $$ By the square of a vector $$\textbf{v}$$ we mean the dot (or scalar) product $$\textbf{v} \cdot \textbf{v}$$. Later you will learn why we represent possible influences on the motion of a particle by such fields as $$V(\textbf{r},t)$$ and $$\textbf{A}(\textbf{r},t).$$

Returning to our fictitious world with only one spatial dimension, allowing for a potential energy $$V(x,t)$$, substituting the differential operators $$i\hbar{\partial\over\partial t}$$ and $$-\hbar^2 {\partial^2\over\partial x^2}$$ for $$E$$ and $$p^2$$ in the resulting dispersion relation, and applying both sides of the resulting operator equation to $$\psi,$$ we arrive at the one-dimensional (time-dependent) Schrödinger equation: In three spatial dimensions and with both potential energy $$V(\textbf{r},t)$$ and potential momentum $$\textbf{A}(\textbf{r},t)$$ present, we proceed from the relation $$E-V = (\textbf{p}-\textbf{A})^2/2m,$$ substituting $$i\hbar{\partial\over\partial t}$$ for $$E$$ and $$-i\hbar{\partial\over\partial\textbf{r}}$$ for $$\textbf{p}.$$ The differential operator $${\partial\over\partial\textbf{r}}$$ is a vector whose components are the differential operators $$\left({\partial\psi\over\partial x},{\partial\psi\over\partial y},{\partial\psi\over\partial z}\right).$$ The result:

i\hbar{\partial\psi\over\partial t} = \frac{1}{2m} \left(-i\hbar{\partial\over\partial\textbf{r}} - \textbf{A}\right)^2\psi + V\psi, $$ where $$\psi$$ is now a function of $$\textbf{r}=(x,y,z)$$ and $$t.$$ This is the three-dimensional Schrödinger equation. In non-relativistic investigations (to which the Schrödinger equation is confined) the potential momentum can generally be ignored, which is why the Schrödinger equation is often given this form: The free Schrödinger equation (without even the potential energy term) is satisfied by $$\psi(x,t) = e^{i(kx-\omega t)}$$ (in one dimension) or $$\psi(\textbf{r},t) = e^{i(\mathbf{k}\cdot\mathbf{r}-\omega t)}$$ (in three dimensions) provided that $$E=\hbar{\omega}$$ equals $$p^2/2m=(\hbar k)^2/ 2m,$$ which is to say: $$\omega(k)=\hbar k^2/2m.$$ However, since we are dealing with a homogeneous linear differential equation — which tells us that solutions may be added and/or multiplied by an arbitrary constant to yield additional solutions — any function of the form

\psi(x,t) = {1\over\sqrt{2\pi}}\int \overline{\psi}(k)\,e^{i[kx-\omega(k)t]}dk= {1\over\sqrt{2\pi}}\int \overline{\psi}(k,t)\,e^{ikx}dk $$ with $$\overline{\psi}(k,t) = \overline{\psi}(k) e^{-i\omega(k)t}$$ solves the (one-dimensional) Schrödinger equation. If no integration boundaries are specified, then we integrate over the real line, i.e., the integral is defined as the limit $$\lim_{L\rightarrow\infty}\int_{-L}^{+L}.$$ The converse also holds: every solution is of this form. The factor in front of the integral is present for purely cosmetic reasons, as you will realize presently. $$\overline{\psi} (k,t)$$ is the Fourier transform of $$\psi(x,t),$$ which means that

\overline{\psi}(k,t)={1\over\sqrt{2\pi}}\int \psi(x,t)\,e^{-ikx}dx. $$ The Fourier transform of $$\psi(x,t)$$ exists because the integral $$\int|\psi(x,t)|dx$$ is finite. In the next section we will come to know the physical reason why this integral is finite.

So now we have a condition that every electron "wave function" must satisfy in order to satisfy the appropriate dispersion relation. If this (and hence the Schrödinger equation) contains either or both of the potentials $$\textbf{V}$$ and $$\textbf{A}$$, then finding solutions can be tough. As a budding quantum mechanician, you will spend a considerable amount of time learning to solve the Schrödinger equation with various potentials. NEXT >