# This Quantum World/Serious illnesses/Born

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## 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
    $\int _{R}|\psi (t,\mathbf {r} )|^{2}\,d^{3}r=\int _{R}\psi ^{*}\psi \,d^{3}r$
— 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:
    $\langle k\rangle ={\frac {1}{\sqrt {2\pi }}}\int \psi ^{*}(x)\left(-i{\frac {d}{dx}}\right)\int {\overline {\psi }}(k)\,e^{ikx}dk\,dx={\frac {1}{\sqrt {2\pi }}}\int \psi ^{*}(x)\int {\overline {\psi }}(k)\,k\,e^{ikx}dk\,dx.$
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:
    $\langle k\rangle =\int \!\int {\overline {\psi }}\,^{*}(k')\,k\,{\overline {\psi }}(k)\left[{\frac {1}{2\pi }}\int e^{i(k-k')x}dx\right]dk\,dk'.$
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 $\hbar /2.$
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