# Wave packet expansion

1. Feb 6, 2016

### ognik

1. The problem statement, all variables and given/known data
The text states:
"Let us consider a wave packet whose Fourier inverse $\phi (\vec{k})$ is appreciably different from zero only in a limited range $\Delta \vec{k}$ near the mean wave vector $\hbar \vec{\bar{k}}$. In coordinate space, the wave packet $\psi(\vec{r}, t)$ must move approximately like a classical free particle with mean momentum $\hbar \vec{\bar{k}}$. To see this behavior we expand $\omega (\vec{k})$ about $\vec{\bar{k}}$:

$\omega (\vec{k}) = \omega(\vec{\bar{k}}) + (\vec{k} - \vec{\bar{k}}) \cdot (\vec{\nabla_k} \omega)_{k= \vec{\bar{k}}} + ...$

$=\bar{\omega} + (\vec{k} - \vec{\bar{k}})\cdot \vec{\nabla_{\bar{k}}} \bar{\omega} +...$ with obvious abbreviations"

I'm afraid not much of the above is obvious to me ... have searched all over but no progress ... please help me understand -

What expansion are they using and where do they get it from?
What are the abbreviations mentioned?

2. Feb 6, 2016

### George Jones

Staff Emeritus
A multivariable Taylor expansion has been used.

$\omega \left( \bar{k} \right)$ has been abbreviated to $\bar{\omega}$.

3. Feb 7, 2016

### ognik

Thanks George, a couple of things still puzzle me:
1. what is the 2nd variable? Is it $\vec{\bar{k}}$ .... how can they expand about a variable ($\vec{\bar{k}}$) if it is?
2. the text talks earlier about the width/spread of the wave packet as $k_x - \bar{k}$, shouldn't that be $|k_x - \bar{k}|$ to keep both sides positive? (its a length)

4. Feb 7, 2016

### George Jones

Staff Emeritus
No.

If you post the Taylor expansion of a function of three variables in the form that appears in a calculus text, then we can discuss how the form in the original post is exactly the same.

5. Feb 7, 2016

### ognik

Thanks, I'm assuming we are talking about 2 variables, then the expansion is:

$f(x,y,z) = f(a,b,c) + (x-a)f_x(a,b,c) + (y-b)f_y(a,b,c) + (z-c)f_z(a,b,c) + \frac{1}{2!}$ [quadratic and higher terms - which are ignored bu the text]

Last edited: Feb 7, 2016
6. Feb 7, 2016

### George Jones

Staff Emeritus
No, three variables are needed. Don't worry about retyping the expansion.

One more thing: Are you familiar with vector calculus and the gradient, divergence, and curl operations?

7. Feb 7, 2016

### ognik

Thanks George, Edited original, and am familiar with those operators

8. Feb 7, 2016

### George Jones

Staff Emeritus
The three variables $x$, $y$, and $z$ can be thought of as comprising one vector, i.e., $\vec{r} = \left( x,y,z \right)$ and $\vec{r_0} = \left( a,b,c \right)$. Then $f\left( x,y,z \right)$ and $f\left( a,b,c \right)$ become $f \left(\vec{r}\right)$ and $f \left(\vec{r_0}\right)$.

Also, $\vec{r} - \vec{r_0} = \left( x - a,y-b,z-c \right)$ and the gradient of $f$ evaluated (gradient first, then evaluation) at $\vec{r_0} = \left( a,b,c \right)$ is the vector

$$\left(\nabla f\right)_{\vec{r_0}} = \left( f_x \left( a,b,c \right), f_y \left( a,b,c \right), f_z \left( a,b,c \right) \right).$$

Consequently,

$$\left( \vec{r} - \vec{r_0} \right) \cdot \left(\nabla f\right)_{\vec{r_0}} = \left(x - a\right) f_x \left( a,b,c \right) + \left(y - b\right) f_y \left( a,b,c \right) +\left(z - c\right) f_z \left( a,b,c \right),$$

and, in new mod attire, the Taylor expansion for $f$ that you posted becomes

$$f \left(\vec{r}\right) = f \left(\vec{r_0}\right) + \left( \vec{r} - \vec{r_0} \right) \cdot \left(\nabla f\right)_{\vec{r_0}} + ...$$

If anything in the above is unclear, let me know.

9. Feb 8, 2016

### ognik

Very helpful, thanks George, the chapter is now clear to me - I had been thinking only in 1-D along the x-axis

10. Feb 8, 2016

### ognik

afterthought question please, is my notation correct when I wrote $\vec{\bar{k}}$ ? Or should it be $\bar{\vec{k}}$ or something else?

11. Feb 8, 2016

### ognik

Apologies, spoke too soon.
The text has a common fourier transform equation $\psi (\vec{r},t) = \frac{1}{(2 \pi)^{\frac{3}{2}}} \int \phi (\vec{k}) e^{i(\vec{k} \cdot \vec{r} - \omega t)} d^3k$
They substitute $\omega(\vec{k})=\bar{\omega} + (\vec{k} - \vec{\bar{k}})\cdot \vec{\nabla_{\bar{k}}} \bar{\omega} +...$ into the above, and get a result I can't get - $\psi (\vec{r},t) = exp(-i \bar{\omega}t + i \bar{\vec{k}} \cdot \nabla_{\bar{k}} \bar{\omega}t) \psi(\vec{r} - \nabla_{\bar{k}}t, 0)$

I instead get as far as $\frac{e^{-i \bar{\omega}t}} {(2 \pi)^\frac{3}{2}} \int \phi(\vec{k}) e^{i (\vec{k} \cdot \vec{r} - \vec{k} \nabla_{\bar{k}} \bar{\omega}t + \bar{k} {\nabla}_{\bar{k}} \bar{\omega}t) } d^3k$

Could you help me bridge that gap please?

12. Feb 9, 2016

### ognik

I have reviewed this and made more progress, still can't quite see all of it.
If I extract $exp(-i \bar{\omega}t + i \bar{\vec{k}} \cdot \nabla_{\bar{k}} \bar{\omega}t)$ I am left with something close to the original FT, except it looks like

$\frac{1}{(2 \pi)^{\frac{3}{2}}} \int \phi (\vec{k}) e^{i(\vec{k} \cdot \vec{r} - \vec{k}\nabla_{\bar{k}} \bar{\omega}t)} d^3k$

The text says this latter $= \psi(\vec{r} - \nabla_{\bar{k}} \bar{\omega}t, 0)$ - but the $\vec{k}$ wasn't dotted with the $\nabla$ term?

Also they set t=0, but then leave the t in the $\nabla$ term?

Please try and explain this to me? Thanks

13. Feb 12, 2016

### blue_leaf77

So, you have
$$\exp(-i \bar{\omega}t + i \bar{\vec{k}} \cdot \nabla_{\bar{k}} \bar{\omega}t) \frac{1}{(2 \pi)^{\frac{3}{2}}} \int \phi (\vec{k}) e^{i(\vec{k} \cdot \vec{r} - \vec{k}\cdot \nabla_{\bar{k}} \bar{\omega}t)} d^3k$$
Note that $\vec{k}$ is dotted with $\nabla_{\bar{k}} \bar{\omega}$ in the integrand. The integral can further be written as
$$\int \phi (\vec{k}) e^{i\vec{k}\cdot(\vec{r} - \nabla_{\bar{k}} \bar{\omega}t)} d^3k$$
At the same time you have the Fourier transform equation
$$\psi (\vec{r},t) = \frac{1}{(2 \pi)^{\frac{3}{2}}} \int \phi (\vec{k}) e^{i(\vec{k} \cdot \vec{r} - \omega t)} d^3k$$
If, in the above equation, you replace $\vec{r}$ with $\vec{r} - \nabla_{\bar{k}} \bar{\omega}t$ and $t$ with $0$, you should see that
$$\int \phi (\vec{k}) e^{i\vec{k}\cdot(\vec{r} - \nabla_{\bar{k}} \bar{\omega}t)} d^3k = \psi (\vec{r} - \nabla_{\bar{k}} \bar{\omega}t,0).$$

14. Feb 12, 2016

### ognik

Thanks blue_leaf.

1. Why can we set the t in (r, t) to 0, but leave the t in $\Delta_{\bar{k}} \bar{\omega}t$ ?

2. is my notation correct when I wrote $\vec{\bar{k}}$ ? Or should it be $\bar{\vec{k}}$, or something else?

15. Feb 12, 2016

### blue_leaf77

You are simply comparing
$$\int \phi (\vec{k}) e^{i\vec{k}\cdot(\vec{r} - \nabla_{\bar{k}} \bar{\omega}t)} e^{-i\omega (0)}d^3k$$
with
$$\int \phi (\vec{k}) e^{i\vec{k} \cdot \vec{r}} e^{-i\omega t} d^3k$$
I don't know if that really matters, but I personally would use the second one.