Effective mass, Taylor's theorem

[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]

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

Thank you

 

[Vagn]

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]
http://en.wikipedia.org/wiki/Taylor's_theorem[/URL]

Thank you

The two dimensions are orthogonal, hence $$\overbar{k_1} \text{ and } \overbar{k_2} [\tex] are also orthogonal, so the scalar product of the two would be zero.$$

 

[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?
$$f(x_{0} + h, y_{0}+k) = f(x_{0},y_{0}) + f_{x}(x_{0},y_{0})h+ f_{y}(x_{0},y_{0})k + \frac{1}{2}[f_{xx}(x_{0},y_{0}){h}^2 + 2f_{xy}(x_{0},y_{0}) hk+ f_{yy}(x_{0},y_{0}){k}^2]+ R(h,k)<br /> = f(x_{0},y_{0}) + \frac{1}{2}[f_{xx}(x_{0},y_{0}){h}^2 + 2f_{xy}(x_{0},y_{0}) hk+ f_{yy}(x_{0},y_{0}){k}^2]+ R(h,k)$$

Thank you

 

[DrDu]

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

 

[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?

 

[Vagn]

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?

$k_1, k_2$ 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.

 

[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.

 

[DrDu]

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

 

[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...

 

[DrDu]

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

 

[chimay]

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

 

[DrDu]

I suppose this someone didn't wanted to show you the general mathematical beauties of Fourier expansions, but had a concrete objective in mind. So he used the freedom to start out with the principal axis form: http://en.wikipedia.org/wiki/Principal_axis_theorem
and http://en.wikipedia.org/wiki/Effective_mass_(solid-state_physics)

 

[chimay]

Thank you very much!