Calculate Current Homework: Show Proportionality to df/dE

[aaaa202]

Homework Statement

In an exercise I am given the attached expression for the current. The functions f1 and f2 are fermi functions centered around the chemical potentials μ1 and μ2.
I am then asked to study this expression in the limit μ1-μ2 -> 0 and γ1->0,γ2->0.
Specifically I am supposed to show that the current is proportional to df/dE. I am unsure of how to do it though. I approximated the integral to be the integral of a delta function and then taylorexpanded f around the small parameter μ1-μ2 as attached. Do you think this is the way to do it, or what is meant by df/dE?

Homework Equations

The Attempt at a Solution

attached

 

[mighty2000]

expression: I = (e/h) * integral from -inf to inf of [f1(E) - f2(E)] * [γ1*f1(E) + γ2*f2(E)] * dE

Hello,

Yes, your approach of approximating the integral to a delta function and using a Taylor expansion is a good start. To show that the current is proportional to df/dE, you need to rewrite the integral in terms of the derivative of the fermi function, rather than the fermi function itself.

You can use the chain rule to rewrite the integral as:

I = (e/h) * integral from -inf to inf of [f1(E) - f2(E)] * [γ1*f1(E) + γ2*f2(E)] * (df1/dE - df2/dE) * dE

Now, when μ1-μ2 -> 0 and γ1->0,γ2->0, the fermi functions f1 and f2 will also approach the same value. This means that the difference (f1-f2) will approach 0, and the product (γ1*f1 + γ2*f2) will also approach 0. This allows you to simplify the integral to:

I = (e/h) * integral from -inf to inf of (df1/dE - df2/dE) * dE

Now, since the only term left in the integral is the derivative of the fermi function, you can see that the current is indeed proportional to df/dE.

I hope this helps. Let me know if you have any further questions or if you need clarification.