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^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.$$ NEXT >