# This Quantum World/Serious illnesses/Born

(Redirected from This quantum world/Serious illnesses/Born)

## Born

In the same year that Erwin Schrödinger published the equation that now bears his name, the nonrelativistic theory was completed by Max Born's insight that the Schrödinger wave function $\psi(\mathbf{r},t)$ is actually nothing but a tool for calculating probabilities, and that the probability of detecting a particle "described by" $\psi(\mathbf{r},t)$ in a region of space $R$ is given by the volume integral

$\int_R|\psi(t,\mathbf{r})|^2\,d^3r=\int_R\psi^*\psi\,d^3r$

— provided that the appropriate measurement is made, in this case a test for the particle's presence in $R$. Since the probability of finding the particle somewhere (no matter where) has to be 1, only a square integrable function can "describe" a particle. This rules out $\psi(\mathbf{r}) = e^{i\mathbf{k}\cdot\mathbf{r}},$ which is not square integrable. In other words, no particle can have a momentum so sharp as to be given by $\hbar$ times a wave vector $\mathbf{k}$, rather than by a genuine probability distribution over different momenta.

Given a probability density function $|\psi(x)|^2$, we can define the expected value

$\langle x\rangle=\int |\psi(x)|^2\,x\,dx=\int \psi^*\,x\,\psi\,dx$

and the standard deviation  $\Delta x = \sqrt{\int |\psi|^2(x-\langle x\rangle)^2}$

as well as higher moments of $|\psi(x)|^2$. By the same token,

$\langle k\rangle=\int \overline{\psi}\,^*\,k\,\overline{\psi}\,dk$  and  $\Delta k=\sqrt{\int |\overline{\psi}|^2(k-\langle k\rangle)^2}.$

Here is another expression for $\langle k\rangle:$

$\langle k\rangle=\int \psi^*(x)\left(-i\frac d{dx}\right)\psi(x)\,dx.$

To check that the two expressions are in fact equal, we plug  $\psi(x)=(2\pi)^{-1/2}\int \overline{\psi}(k)\,e^{ikx}dk$  into the latter expression:

$\langle k\rangle=\frac1{\sqrt{2\pi}}\int \psi^*(x)\left(-i\frac d{dx}\right)\int \overline{\psi}(k)\,e^{ikx}dk\,dx=\frac1{\sqrt{2\pi}}\int \psi^*(x)\int \overline{\psi}(k)\,k\,e^{ikx}dk\,dx.$

Next we replace $\psi^*(x)$ by $(2\pi)^{-1/2}\int \overline{\psi}\,^*(k')\,e^{-ik'x}dk'$  and shuffle the integrals with the mathematical nonchalance that is common in physics:

$\langle k\rangle= \int\!\int \overline{\psi}\,^*(k')\,k\,\overline{\psi}(k) \left[\frac1{2\pi}\int e^{i(k-k')x}dx \right]dk\,dk'.$

The expression in square brackets is a representation of Dirac's delta distribution $\delta(k-k'),$ the defining characteristic of which is  $\int_{-\infty}^{+\infty} f(x)\,\delta(x)\,dx = f(0)$  for any continuous function $f(x).$ (In case you didn't notice, this proves what was to be proved.)

## Heisenberg

In the same annus mirabilis of quantum mechanics, 1926, Werner Heisenberg proved the so-called "uncertainty" relation

$\Delta x\,\Delta p \geq \hbar/2.$

Heisenberg spoke of Unschärfe, the literal translation of which is "fuzziness" rather than "uncertainty". Since the relation $\Delta x\,\Delta k \geq 1/2$ is a consequence of the fact that $\psi(x)$ and $\overline{\psi}(k)$ are related to each other via a Fourier transformation, we leave the proof to the mathematicians. The fuzziness relation for position and momentum follows via $p=\hbar k$. It says that the fuzziness of a position (as measured by $\Delta x$ ) and the fuzziness of the corresponding momentum (as measured by $\Delta p=\hbar\Delta k$ ) must be such that their product equals at least $\hbar/2.$