Exploring Quantum Physics - Week 2 Question 6

Problem:

Which of the following satisfy the diffusion equation in one dimension (for $t > 0$) (including, if applicable, unbounded solutions which however satisfy the equation at any finite $x$ and $t$)? (Check all that apply)

• $e^{\frac{−x^2}{4 D t}}$
• $\frac{1}{\sqrt{2\pi D t}}e^{\frac{−x^2}{4 D t}}$
• $A⋅(t+\frac{1}{2D}x^2)$
• $e^{i D ( p x − E t )}$ where $p$ and $E$ are non-zero real numbers.

Solution:

From the lecture, we see the diffusion equation is $\frac{\partial \rho}{\partial t} = D \nabla^2 \rho$, reducing to one dimension, we have $\frac{\partial \rho}{\partial t} = D \frac{\partial^2}{\partial x^2} \rho$, so we need to differentiate the equation once by $t$ and twice by $x$, then we can know the answer by simply checking if they are equals (of course, with the multiplicative constant).

Differentiating the options, we have



$\begin{eqnarray*} \frac{\partial}{\partial t} e^{\frac{−x^2}{4 D t}} & = & \frac{x^2}{4 D t^2} e^{\frac{−x^2}{4 D t}} \\ \frac{\partial^2}{\partial x^2} e^{\frac{−x^2}{4 D t}} & = & \frac{\partial}{\partial x} \frac{-x}{2 D t} e^{\frac{−x^2}{4 D t}} \\ & = & \frac{-1}{2 D t} e^{\frac{−x^2}{4 D t}} + \frac{-x}{2 D t} \frac{-x}{2 D t} e^{\frac{−x^2}{4 D t}} \\ & = & \frac{-1}{2 D t} e^{\frac{−x^2}{4 D t}} + \frac{x^2}{4 D^2 t^2} e^{\frac{−x^2}{4 D t}} \\ & = & e^{\frac{−x^2}{4 D t}} (\frac{x^2}{4 D^2 t^2} - \frac{1}{2 D t}) \end{eqnarray*}$

So the first options is incorrect, but it does help the computation of the below, see

$\begin{eqnarray*} \frac{\partial}{\partial t} \frac{1}{\sqrt{2\pi D t}} e^{\frac{−x^2}{4 D t}} & = & \frac{1}{\sqrt{2\pi D t}} \frac{x^2}{4 D t^2} e^{\frac{−x^2}{4 D t}} + \frac{-1}{2t\sqrt{2\pi D t}} e^{\frac{−x^2}{4 D t}} \\ & = & \frac{1}{\sqrt{2\pi D t}} e^{\frac{−x^2}{4 D t}} (\frac{x^2}{4 D t^2} - \frac{1}{2t}) \\ \frac{\partial^2}{\partial x^2} \frac{1}{\sqrt{2\pi D t}} e^{\frac{−x^2}{4 D t}} & = & \frac{1}{\sqrt{2\pi D t}} \frac{\partial^2}{\partial x^2} e^{\frac{−x^2}{4 D t}} \\ & = & \frac{1}{\sqrt{2\pi D t}} e^{\frac{−x^2}{4 D t}} (\frac{x^2}{4 D^2 t^2} - \frac{1}{2 D t}) \end{eqnarray*}$

So the second option is right - with the complex expressions behind us, the next two is less complex.

$\begin{eqnarray*} \frac{\partial}{\partial t} A (t+\frac{1}{2D}x^2) & = & A \\ \frac{\partial^2}{\partial x^2} A (t+\frac{1}{2D}x^2) & = & \frac{\partial}{\partial x} \frac{A}{D}x \\ & = & \frac{A}{D} \end{eqnarray*}$

So the third option is also right! Last but not least for the fourth option, we have

$\begin{eqnarray*} \frac{\partial}{\partial t} e^{i D ( p x − E t )} & = & -i D E e^{i D ( p x − E t )} \\ \frac{\partial^2}{\partial x^2} e^{i D ( p x − E t )} & = & \frac{\partial}{\partial x} i D p e^{i D ( p x − E t )} \\ & = & (i D p )(i D p )e^{i D ( p x − E t )} \\ & = & - D^2 p^2 e^{i D ( p x − E t )} \end{eqnarray*}$

So the fourth option is wrong, as if there are equal, as $D$, $E$ and $p$ cannot be real number at the same time.

Phew, that's a lot of calculation!