Subfield: Modern Physics

4. A complex dance 

In this problem, we will solve a number of differential equations corresponding to very different physical phenomena that are unified by the idea of oscillation. Oscillations are captured elegantly by extending our notion of numbers to include the imaginary unit number $i$, strangely defined to obey $i^{2}=-1$. In other words, rather than using real numbers, it is more convenient for us to work in terms of complex numbers.

Exponentials are usually associated with rapid growth or decay. However, with the inclusion of complex numbers, imaginary "growth" and "decay" can be translated into oscillations by the Euler identity:

$$
e^{i \theta}=\cos \theta+i \sin \theta
\tag{1}
$$
Context question:
(a) The usual form of Newton's second law $(\vec{F}=m \vec{a})$ breaks down when we go into a rotating frame, where both the centrifugal and Coriolis forces become important to account for. Newton's second law then takes the form

$$
\vec{F}=m(\vec{a}+2 \vec{v} \times \vec{\Omega}+\vec{\Omega} \times(\vec{\Omega} \times \vec{r}))
\tag{2}
$$

For a particle free of forces confined to the $x-y$ plane in a frame which rotates about the $z$ axis with angular frequency $\Omega$, this becomes the complicated-looking system of differential equations,

$$
\begin{aligned}
& 0=\ddot{x}+2 \Omega \dot{y}-\Omega^{2} x \\
& 0=\ddot{x}-2 \Omega \dot{x}-\Omega^{2} y
\end{aligned}
\tag{3}
$$

where dots represent time derivatives.

Defining $\eta=x+i y$, show that Equations 3 are equivalent to the following single (complex) equation:

$$
0=\ddot{\eta}-2 i \Omega \dot{\eta}-\Omega^{2} \eta
\tag{4}
$$
Context answer:
\boxed{证明题}


Context question:
(b) Equation 4 is a version of the damped harmonic oscillator, and can be solved by guessing a solution $\eta=\alpha e^{\lambda t}$.

Plugging in this guess, what must $\lambda$ be?
Context answer:
\boxed{$\lambda=i \Omega$}


Context question:
(c) Using your answer to part (b), and defining $\alpha=A e^{i \phi}$ where $A$ and $\phi$ are real, find $\mathbf{x}(\mathbf{t})$ and $\mathbf{y}(\mathbf{t})$.

This is the trajectory for a particle which is stationary with respect to the symmetry axis. While not required for this problem, an additional guess would reveal that $\eta=\beta t e^{\lambda t}$ is also a solution.
Context answer:
\boxed{$x(t)=A \cos (\Omega t+\phi)$ , $y(t)=A \sin (\Omega t+\phi)$}


Context question:
(d) The one-dimensional diffusion equation (also called the "heat equation") is given (for a free particle) by

$$
\frac{\partial \psi}{\partial t}=a \frac{\partial^{2} \psi}{\partial x^{2}}
\tag{5}
$$

A spatial wave can be written as $\sim e^{i k x}$ (larger $k$ 's correspond to waves oscillating on smaller length scales). Guessing a solution $\psi(x, t)=A e^{i k x-i \omega t}$, find $\omega$ in terms of k. A relationship of this time is called a "dispersion relation."
Context answer:
\boxed{$\omega=-i k^{2} a$}


Context question:
(e) The most important equation of non-relativistic quantum mechanics is the Schrödinger equation, which is given by

$$
i \hbar \frac{\partial \psi}{\partial t}=-\frac{\hbar^{2}}{2 m} \frac{\partial^{2} \psi}{\partial x^{2}}
\tag{6}
$$

Using your answer to part (d), what is the dispersion relation of the Schrödinger equation?
Context answer:
\boxed{$\omega=\frac{\hbar k^{2}}{2 m}$}


Context question:
(f) If the energy of a wave is $E=\hbar \omega$ and the momentum is $p=\hbar k$, show that the dispersion relation found in part (e) resembles the classical expectation for the kinetic energy of a particle, $\mathrm{E}=\mathrm{mv}^{2} / \mathbf{2}$.
Context answer:
\boxed{证明题}


Question: (g) The theory of relativity instead posits that the energy of a particle is given by $E=\sqrt{p^{2} c^{2}+m^{2} c^{4}}$. In accordance with this, we can try to guess a relativistic version of the Schrödinger equation:

$$
\frac{1}{c^{2}} \frac{\partial^{2} \phi}{\partial t^{2}}-\frac{\partial^{2} \phi}{\partial x^{2}}+\frac{m^{2} c^{2}}{\hbar^{2}} \phi=0
\tag{7}
$$

This is called the Klein-Gordon equation. Using the same guess as before, find $\omega$ in terms of $\mathrm{k}$.

Hint: If you are careful, you should find that there is an infinite continuum of energy states extending down to negative infinity. This apparently mathematical issue hints at the existence of antimatter, and ultimately demonstrates to us that we must formulate quantum field theory to properly describe relativistic quantum physics.

Solution:
We consider the Klein-Gordon equation:

$$
\frac{1}{c^{2}} \frac{\partial^{2} \phi}{\partial t^{2}}-\frac{\partial^{2} \phi}{\partial x^{2}}+\frac{m^{2} c^{2}}{\hbar^{2}} \phi=0
\tag{35}
$$

We can, as before, plug in a guess $\phi(x, t)=A e^{i k x-i \omega t}$. This yields

$$
-\frac{\omega^{2}}{c^{2}} A e^{i k x-i \omega t}+k^{2} A e^{i k x-i \omega t}+\frac{m^{2} c^{2}}{\hbar^{2}} A e^{i k x-i \omega t}=0
\tag{36}
$$

Cancelling out common terms, we see that

$$
\omega^{2}=k^{2} c^{2}+\frac{m^{2} c^{4}}{\hbar^{2}}
\tag{37}
$$

or

$$
\omega= \pm \sqrt{k^{2} c^{2}+\frac{m^{2} c^{4}}{\hbar^{2}}}
\tag{38}
$$

Note that both positive and negative $\omega$ solve the Klein-Gordon equation.

Final answer: $\sqrt{k^{2} c^{2}+\frac{m^{2} c^{4}}{\hbar^{2}}}$, $-\sqrt{k^{2} c^{2}+\frac{m^{2} c^{4}}{\hbar^{2}}}$