Today we extend our understanding of the modeling from last day. There we argued that measuring the density of an atomic gas gives us direct access to the square amplitude of the real-space wavefunction. We also argued that a time-of-flight experiment gives access to the square amplitude of the momentum-space wavefunction. Finally I tried to convince you that time-of-flight correlates the momentum and position of the particles. I thought that was pretty neat, but the first step in our analysis involved calculating a time dependent wavefunction. Finding the wavefunction requires solving a partial differential equation. That sounds like hard work to me, and I am intrinsically lazy. In this lecture I will show you a couple tricks which make things easier.

Correlation between position and momentum

The classical argument also gives an interesting observation.  Classically, all of the particles at position $x$ at time $t$ have momentum $p=mx/t$.  This is because only the particles moving with velocity $x/t$ would be at that location.  The position and momentum are correlated for the classical distribution of particles.  The same must be true of the quantum mechanical wavefunction.  What does this mean when you cannot simultaneously specify the momentum and position?  The solution is that if you look at a range of positions of width $\delta x$, there are a range of momenta of width $\delta p$, and the uncertainty $\delta p$ grows when you make $\delta x$ smaller.  Despite the breadth of the distribution, the average momentum in that region has some definite value, which is essentially independent of the region size:

Work through the Continuity handout.

A useful framework for understanding this local velocity, is to write the complex wavefunction as an amplitude and a phase
\begin{equation}
\psi(x,t)= f(x,t) e^{i\phi(x,t)},
\end{equation}
where both $f$ and $\phi$ are real.  The Schrodinger equation reads
\begin{eqnarray}
i\partial_t \psi(x,t) &=& -\frac{1}{2m} \partial_x^2 \psi(x,t)\\
i \partial_t f(x,t) e^{i \phi(x,t)} &=&  -\frac{1}{2m} \partial_x^2  f(x,t) e^{i\phi(x,t)}
\end{eqnarray}
We now apply the chain rule to simplify this.  The trickiest part is the second derivative with respect to $x$.  I like to do this in steps
\begin{equation}
\partial_x \left[ f e^{i \phi} \right] =( \partial_x f) e^{i \phi} + i (f\partial_x \phi) e^{i \phi}.
\end{equation}
We then apply a second derivative
\begin{eqnarray}
\partial_x^2  \left[ f e^{i \phi} \right]  &=& \partial_x \left[
( \partial_x f) e^{i \phi} + i (f\partial_x \phi) e^{i \phi}
\right]\\
&=&
\left(\partial_x^2 f - f (\partial_x \phi)^2 + 2 i \partial_x f \partial_x\phi+i f \partial_x^2 \phi \right) e^{i\phi}.
\end{eqnarray}
This brings us to
\begin{eqnarray}
i(\partial_t f) e^{i \phi} -(f \partial_t \phi) e^{i\phi} &=&
 -\frac{1}{2m}\left(\partial_x^2 f - f (\partial_x \phi)^2 \right)e^{i\phi}
 \\ \nonumber&&
  -\frac{i}{2m}\left(
 2 \partial_x f \partial_x\phi + f \partial_x^2\phi\right) e^{i\phi}
\end{eqnarray}
We then multiply by $e^{-i\phi}$ and take the real and imaginary parts.  Both are interesting, but for our purposes, we will get the most out of the imaginary part, which is
\begin{eqnarray}
\partial_t f +\frac{1}{2m} \left[2 \partial_x f \partial_x \phi + f \partial_x^2\phi\right] &=&0.
\end{eqnarray}
The canonical way to write this is to multiply by $2f$, and note
\begin{equation}
2 f\partial_t f = \partial_t f^2 = \partial_t n,
\end{equation}
where $n$ is the density.  Thus
\begin{eqnarray}
\partial_t n + \frac{1}{m} \left[\partial_x n \partial_x \phi + n \partial_x^2 \phi \right]&=&0\\
\partial_t n + \partial_x \left[ n \frac{\partial_x \phi}{m}\right] &=&0.
\end{eqnarray}
The form of equation is very common in physics.  You see it in fluid mechanics, electromagnetism...  It is known as the continuity equation, and is typically written as
\begin{equation}
\partial_t n + \partial_x j =0.
\end{equation}
If we went through the same argument in 3D we would have found
\begin{equation}
\partial_t n + \nabla\cdot {\bf \vec j} =0,
\end{equation}
where the particle current is
\begin{equation}
\nabla\cdot \vec {\bf j} = \frac{n(r,t)}{m} \nabla \phi(r,t). 
\end{equation}
The physics of the continuity equation is that the rate of change of the local density is given by the difference between the flow into some volume and the flow out.  Mathematically the differences between flows is given by the divergence.  The current can be thought of as a density times an average velocity.  Thus we see that the average velocity at location $\bf r$ is
\begin{equation}
{\bf v}({\bf r},t) = \frac{1}{m} \nabla \phi({\bf r},t),
\end{equation}
which gives us physical meaning to the phase of the wavefunction.  If you want to use physical units (where $\hbar\neq 1$), you can use dimensional analysis to write
\begin{equation}
{\bf v}({\bf r},t) = \frac{\hbar}{m}{ \nabla \phi({\bf r},t)}.
\end{equation}
We will talk more about dimensional analysis in the future, and you will become more comfortable with this glibness with $\hbar$'s.

It turns out that you have seen this local current before in PHYS 3316.  You may remember discussing "probability current," defined by
\begin{equation}
{\bf j} = \frac{\hbar}{2 m i} \left[\psi^*\nabla \psi - \psi \nabla \psi^*\right].
\end{equation}
I will leave it as an exercise to show that this is the same as the $\bf j$ that we defined in terms of the phase of the wavefunction.

Regardless, we can now take our solution to the time dependent Schrodinger equation
\begin{equation}
\psi(x,t)\approx A(t) \exp\left[ \frac{ i m x^2}{2\hbar t}\left(1+ \frac{i d^2 m}{i\hbar t}+\cdots \right)\right].
\end{equation}
What is $v(x,t)$?  How does it compare with the classical prediction?

Heisenberg's equations of motion

The first trick is to only calculate what you really want to know. Lets think about this. I am doing an experiment. I have a bunch of atoms trapped in a tight trap. I turn off the trap and it expands. I have the ability to take pictures of the cloud as it expands. I want to model the experiment, and prove to my colleagues that I understand what is going on.

Last day we argued that one way to model the experiment was to solve the time-dependent Schrodinger equation. The density I measure is exactly $|\psi|^2$. I could then make a bunch of plots which show the measured density, and $|\psi|^2$, and see if they agree. Suppose it is a two-dimensional experiment. I then have a three-dimensional data set (because I have time). How am I going to plot this? How am I going to organize this data?

One approach is to extract one number from each density profile which characterizes it. The natural thing to look at is the size of the cloud. There are many ways to characterize the size, but from experience I know that the easiest one to calculate will be the root-mean-square width:
\begin{equation}
\sigma(t)=\langle x^2\rangle = \int\, dx\, x^2\, |\psi(x,t)|^2.
\end{equation}
Here we will try to calculate equations of motion for this quantity without calculating the wavefunction. I believe this trick was invented by Dirac when he reconciled Heisenberg's matrix mechanics and Schrodinger's wave mechanics (though Schrodinger may have came up with it first). The idea is to simply take the derivative of $\sigma(t)$
\begin{equation}
\partial_t \sigma(t) = \partial_t \int dx\, x^2\, |\psi(x,t)|^2.
\end{equation}
We then use the linearity of the integral to take the derivative inside \begin{equation}
\partial_t \sigma(t) = \int dx\, x^2\, \left[\psi^*\partial_t \psi +(\partial_t \psi^*) \psi\right].
\end{equation}
We now use the Schrodinger equation,
\begin{equation}
i\partial_t \psi = H \psi
\end{equation}
and its conjugate
\begin{equation}
-i \partial_t \psi^* = (H \psi)^*
\end{equation}
to write
\begin{equation}
\partial_t \sigma(t) = \int dr\, r^2\, \frac{1}{i}\left[\psi^*H \psi -(H \psi)^* \psi\right].
\end{equation}
Note, that I am being a little obtuse by writing $(H\psi)^*$ instead of $(H \psi^*)$. The reason is that for an arbitrary operator, these things are in fact different.
In the case at hand
\begin{equation}
H=-\frac{1}{2m} \partial_x^2
\end{equation}
this distinction is meaningless, and we have
\begin{equation}
\partial_t \sigma(t) = \frac{-1}{2m} \int dx\, x^2\, \frac{1}{i}\left[\psi^* \partial_x^2 \psi -(\partial_x^2 \psi^*) \psi\right].
\end{equation}
We then integrate the second terms by parts twice,
\begin{eqnarray}
\partial_t \sigma(t) &=& \frac{-1}{2m} \int dx\, \frac{1}{i}\left[ \psi^* \partial^2_x \psi -\psi^* \partial_x^2 (x^2 \psi)\right]\\
&=& \int dx\, \psi^* \frac{1}{i} (x^2 H - H x^2) \psi
\end{eqnarray}
where by $H x^2 \psi$ I mean $H (x^2\psi)$. Following this logic, one writes
\begin{equation}
\partial_t \langle x^2\rangle = \langle \frac{1}{i}( x^2 H - H x^2) \rangle = \frac{1}{i} \langle [x^2,H]\rangle.
\end{equation}
It is clear that we can replace $x^2$ by any reasonable linear operator, and we would get the same structure. This logic will also go through for any $H$ which obeys the property
\begin{equation}
\int dx\, (H\psi)^*(x) \phi(x) = \int dx\, \psi^*(x) (H \phi)(x),
\end{equation}
for any functions $\psi$ and $\phi$.
We call such operators Hermitian. This Hermitian property is necessary for the Schrodinger equation to make sense.