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Quantum mechanics in the movies

Part I: A Serious Man

In the Coen brothers' 2009 film "A Serious Man" the protagonist lectures on the uncertainty principle. In one scene we see that $ \Delta p= \sqrt{ \langle p \rangle ^{2} - \langle p \rangle^{2}}$.



However in the next shot it has changed to $ \Delta p= \sqrt{ \langle p ^{2} \rangle - \langle p \rangle^{2}}$.



I'd like to think that this is demonstrating how the momentum of the '$2$' is perfectly known. Hence its position is completely undetermined and it can appear anywhere on the board.

Although a dream sequence in the film, it shows correct calculations (allowing for typos) of the uncertainties for the first excited-state of a particle in a box.

For a box of length $a$ we choose the infinite barriers to be at $-\frac{a}{2}$ and $\frac{a}{2}$. The wavefunction for the first excited-state is then $$\Psi(x)=\sqrt{\frac{2}{a}}\sin(\frac{2\pi}{a}x).$$ $|\Psi(x)|^{2}$ may be interpreted as the probability of finding the particle at $x$. This wavefunction has been constructed (normalized) so that $$ \int_{-\frac{a}{2}}^{\frac{a}{2}} |\Psi|^{2} dx =1.$$ This means the particle must be found somewhere.

The average, or expectation, value of the position $x$ with regards to this probability distribution is $$ \langle x \rangle =\int_{-\frac{a}{2}}^{\frac{a}{2}} x |\Psi|^{2} dx.$$ Now $x$ is an odd function while here $|\Psi(x)|^{2}$ is even therefore $$\langle x \rangle=0.$$ Calculating $$ \langle x^{2} \rangle =\int_{-\frac{a}{2}}^{\frac{a}{2}} x^{2} |\Psi|^{2} dx$$ is a little more involved, but trigonometric identities and integration by parts result in $$ \langle x^{2} \rangle =a^{2} \left( \frac{1}{12}-\frac{1}{8\pi^{2}} \right )\approx 0.071a^{2}.$$ The standard deviation, or uncertainty, of $x$ is $$\Delta x= \sqrt{ \langle x ^{2} \rangle - \langle x \rangle^{2}}$$ $$\Delta x\approx\sqrt{0.071a^{2}-0 }$$ $$\approx\sqrt{0.071a^{2}}$$

For the momentum we have $\hat{p}=-i\hbar\frac{d}{dx}$ in quantum mechanics. The average momentum must be real, but $\hat{p}$ is purely imaginary and our wavefunction is real so $$ \langle \hat{p} \rangle =0.$$ Next by taking derivatives of $\Psi$ we see that $$ \hat{p}^{2} \Psi= \frac{4\hbar^{2}\pi^{2}}{a^{2}}\Psi$$ so $$ \langle \hat{p}^{2} \rangle= \frac{4\hbar^{2}\pi^{2}}{a^{2}}\int_{-\frac{a}{2}}^{\frac{a}{2}} |\Psi|^{2} dx$$ $$ = \frac{4\hbar^{2}\pi^{2}}{a^{2}}$$ as $\Psi$ is normalized. Writing this in terms of $h=2\pi\hbar$ gives $$\langle \hat{p}^{2} \rangle=\left(\frac{h}{a}\right)^{2}.$$ We then have $$ \Delta p= \sqrt{ \langle p ^{2} \rangle - \langle p \rangle^{2}}$$ $$ = \sqrt{\left(\frac{h}{a}\right)^{2}}$$ and $$ \Delta x \Delta p \approx \sqrt{0.071a^{2}\left(\frac{h}{a}\right)^{2}}$$ $$ \approx 0.266 h $$ $$ \approx 1.67 \hbar .$$ Assuming that $0.071$ was accidentally written as $0.077$ then the film demonstrates uncertainties for the first-excited state of the particle in the box that conform (as they must) to the uncertainty principle $\Delta x \Delta p \geq \frac{\hbar}{2}$.

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