Effective mass, Taylor's theorem

1. May 26, 2015

chimay

Hi,

I'm starting to study conduction in semiconductor and I have a doubt about the concept of effective mass. Let's suppose to deal with an dispersion relation in two dimensions; basically, the effective masses along the two axes are defined by means of the Taylor's expansion of the function around its minimun. During lessons, we used the following formula:
$$E= E_{min} + \frac{{\hbar}^2}{2{m_{1}}^*} {k_{1}}^2 + \frac{{\hbar}^2}{2{m_{2}}^*} {k_{2}}^2$$

In particular my question is the following: Taylor's expansion for a two variables function includes also cross product like K1 * K2. Are we simply neglecting it?[/PLAIN] [Broken]

http://en.wikipedia.org/wiki/Taylor's_theorem

Thank you

Last edited by a moderator: May 7, 2017
2. May 26, 2015

The two dimensions are orthogonal, hence $$\overbar{k_1} \text{ and } \overbar{k_2} [\tex] are also orthogonal, so the scalar product of the two would be zero. Last edited by a moderator: May 7, 2017 3. May 26, 2015 chimay Thank you for your answer. Can you please explain me where does the following formula need to be changed as a consequence of the orthogonality of the two vectors? [tex] f(x_{0} + h, y_{0}+k) = f(x_{0},y_{0}) + f_{x}(x_{0},y_{0})h+ f_{y}(x_{0},y_{0})k + \frac{1}{2}[f_{xx}(x_{0},y_{0}){h}^2 + 2f_{xy}(x_{0},y_{0}) hk+ f_{yy}(x_{0},y_{0}){k}^2]+ R(h,k) = f(x_{0},y_{0}) + \frac{1}{2}[f_{xx}(x_{0},y_{0}){h}^2 + 2f_{xy}(x_{0},y_{0}) hk+ f_{yy}(x_{0},y_{0}){k}^2]+ R(h,k)$$

Thank you

4. May 27, 2015

DrDu

Of course there can be mixed terms, but you can eliminate these with a suitable choice of your axes.

5. May 27, 2015

chimay

I don't understand how, can you please explain it to me by means of an example?
In my case $$h = k_{1} , k = k_{2}$$

How can $$h k = k_{1}k_{2}$$ be zero?

Thank you a lot

Edit: I'm sorry for the layout, how can I write an equation in Latex alligned to the text?

6. May 27, 2015

Vagn

$k_1, k_2$ are vectors, you therefore have to take the scalar product between them. If, by sensible choice of axis, the wave vectors are orthogonal, then the scalar product is zero, in which case there are no cross terms.

7. May 27, 2015

chimay

But in the relationI posted in #3 they are just the increments along the two orthogonal axes x and y, and they are still included in the formula.
It is obvious there is a misunderstanding by my side, but I don't get what is the point.

8. May 27, 2015

DrDu

Ok, so maybe you can find an orthogonal transformation of the axes so that the mixed term vanishes?

9. May 27, 2015

chimay

But in the case of the formula in #1, defining the effective masses, K_1 and K_2 are orthogonal and, from a mathematical standpoint, they behave like the axes x and y...

10. May 27, 2015

DrDu

Then someone has already used the freedom to orient the coordinate system so that the mixed terms vanish.

11. May 27, 2015

chimay

So, basically, should I trust that someone, should I?

12. May 27, 2015

DrDu

13. May 27, 2015

chimay

Thank you very much!