The Group Velocity in a One-Dimensional Material

[soekdi]

Homework Statement

A one-dimensional material has an applied time varying e-field as shown below:

$$\epsilon(t)=\left\{\begin{array}{cc}A_1,&0\le t \le 2<br /> \\0,& 2\le t \le 4<br /> \\-A_1,& 4\le t \le 6<br /> \end{array}\right$$

The band structure of the material is $E=\hbar\nu\|k\|$, where $\nu$ is a constant with units of velocity. What is the electron gorup velocity as a function of time from 0 to 6?

Homework Equations

$$v=\frac{d\omega}{dk}=\frac{dE}{dp}$$

The Attempt at a Solution

$$v_g=\frac{1}{\hbar}\frac{dE}{dk}$$

I'm not sure on how to include the e-field into this equation. Any suggestions?

 

[olgranpappy]

Homework Statement

A one-dimensional material has an applied time varying e-field as shown below:

$$\epsilon(t)=\left\{\begin{array}{cc}A_1,&0\le t \le 2<br /> \\0,& 2\le t \le 4<br /> \\-A_1,& 4\le t \le 6<br /> \end{array}\right$$

The band structure of the material is $E=\hbar\nu\|k\|$, where $\nu$ is a constant with units of velocity. What is the electron gorup velocity as a function of time from 0 to 6?

Homework Equations

$$v=\frac{d\omega}{dk}=\frac{dE}{dp}$$

The Attempt at a Solution

$$v_g=\frac{1}{\hbar}\frac{dE}{dk}$$

I'm not sure on how to include the e-field into this equation. Any suggestions?

Probably you just want to account for the fact that k depends on time via
$$\frac{d\bold{k}}{dt}=-e\bold{E}(t)\;,$$
where
$$\bold{E}(t)$$
is the electric field and $-e$ is the charge of the electron.

 

[soekdi]

Probably you just want to account for the fact that k depends on time via
$$\frac{d\bold{k}}{dt}=-e\bold{E}(t)\;,$$
where
$$\bold{E}(t)$$
is the electric field and $-e$ is the charge of the electron.

Now I'm having trouble with the math part:
$$dE=\hbar \nu dk$$

$$dk=-e\epsilon(t)dt$$

$$v_g=\frac{1}{\hbar}\frac{dE}{dk}=\frac{\nu dk}{-e\epsilon(t)dt}$$

where $\epsilon$ is the e-field. This doesn't really look right?

 

[olgranpappy]

Now I'm having trouble with the math part:
$$dE=\hbar \nu dk$$

$$dk=-e\epsilon(t)dt$$

$$v_g=\frac{1}{\hbar}\frac{dE}{dk}=\frac{\nu dk}{-e\epsilon(t)dt}$$

where $\epsilon$ is the e-field. This doesn't really look right?

mmm... if you just forget about the electric field for a second and just look at
$$E=\hbar v |k|\;,$$
what is the group velocity?