The Group Velocity in a One-Dimensional Material

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Homework Statement


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

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

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


<br /> v=\frac{d\omega}{dk}=\frac{dE}{dp}<br />

The Attempt at a Solution


<br /> v_g=\frac{1}{\hbar}\frac{dE}{dk}<br />

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

Homework Statement


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

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

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


<br /> v=\frac{d\omega}{dk}=\frac{dE}{dp}<br />

The Attempt at a Solution


<br /> v_g=\frac{1}{\hbar}\frac{dE}{dk}<br />

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

Now I'm having trouble with the math part:
<br /> dE=\hbar \nu dk<br />

<br /> dk=-e\epsilon(t)dt<br />

<br /> v_g=\frac{1}{\hbar}\frac{dE}{dk}=\frac{\nu dk}{-e\epsilon(t)dt}<br />

where \epsilon is the e-field. This doesn't really look right?
 
soekdi said:
Now I'm having trouble with the math part:
<br /> dE=\hbar \nu dk<br />

<br /> dk=-e\epsilon(t)dt<br />

<br /> v_g=\frac{1}{\hbar}\frac{dE}{dk}=\frac{\nu dk}{-e\epsilon(t)dt}<br />

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
<br /> E=\hbar v |k|\;,<br />
what is the group velocity?
 
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