How to evaluate the effective mass tensor (band structure)

[LesterTU]

Homework Statement

The energy-band dispersion for a 3D crystal is given by

$$E(\mathbf k) = E_0 - Acos(k_xa) - Bcos(k_yb) - Ccos(k_zc)$$
What is the value of the effective mass tensor at $\mathbf k = 0$?

Homework Equations

The effective mass tensor is given by

$$ \left( \frac{1}{m^*} \right)_{ij} = \frac{1}{\hbar ^2} \frac {\partial^2 E(\mathbf k)} {\partial k_i \partial k_j}$$
where $i,j = x, y, z.$

The Attempt at a Solution

I guess I'm supposed to carry out the second order derivative of the expression for the energy in order to find the effective mass, but I don't know how to actually evaluate it. Can someone tell me how to do it please!

 

[nrqed]

Homework Statement

The energy-band dispersion for a 3D crystal is given by

$$E(\mathbf k) = E_0 - Acos(k_xa) - Bcos(k_yb) - Ccos(k_zc)$$
What is the value of the effective mass tensor at $\mathbf k = 0$?

Homework Equations

The effective mass tensor is given by

$$ \left( \frac{1}{m^*} \right)_{ij} = \frac{1}{\hbar ^2} \frac {\partial^2 E(\mathbf k)} {\partial k_i \partial k_j}$$
where $i,j = x, y, z.$

The Attempt at a Solution

I guess I'm supposed to carry out the second order derivative of the expression for the energy in order to find the effective mass, but I don't know how to actually evaluate it. Can someone tell me how to do it please!

You just need to actually apply the equation. First pick two values for $i,j$, let's say $i=j=x$ and apply the formula exactly as it is written. Let us know if it is still unclear.

 

[LesterTU]

You just need to actually apply the equation. First pick two values for $i,j$, let's say $i=j=x$ and apply the formula exactly as it is written. Let us know if it is still unclear.

So you take the second derivative of $E(\mathbf k)$ with respect to all nine combinations of $i, j$, and add them together? The cross terms would vanish, and we are left with

$$ \frac {\partial^2 E(\mathbf k)} {\partial k_x \partial k_x} = Aa^2cos(k_xa)$$

and similarly for yy and zz. Thus

$$ \left( \frac{1}{m^*} \right)_{ij} = \frac{1}{\hbar ^2} \left( Aa^2cos(k_xa) + Bb^2cos(k_yb) + Cc^2cos(k_zc) \right) $$

Is this correct?

 

[nrqed]

So you take the second derivative of $E(\mathbf k)$ with respect to all nine combinations of $i, j$, and add them together? The cross terms would vanish, and we are left with

$$ \frac {\partial^2 E(\mathbf k)} {\partial k_x \partial k_x} = Aa^2cos(k_xa)$$

and similarly for yy and zz.

This is correct.

Thus

$$ \left( \frac{1}{m^*} \right)_{ij} = \frac{1}{\hbar ^2} \left( Aa^2cos(k_xa) + Bb^2cos(k_yb) + Cc^2cos(k_zc) \right) $$

Is this correct?

No. You must pick values of i,j to be able to write an explicit expression on the right. From your calculation above, you get that
$$\left( \frac{1}{m^*} \right)_{xx} = \frac{1}{\hbar ^2} \left( Aa^2cos(k_xa)\right) $$
Now you have to write separately the other expressions for yy and zz.

 

[LesterTU]

So at $\mathbf k = 0$, we get

$$\left( m^* \right)_{xx} = \frac{\hbar ^2}{Aa^2} \ \ \ \ \ \ \left( m^* \right)_{yy} = \frac{\hbar ^2}{Bb^2} \ \ \ \ \ \ \left( m^* \right)_{zz} = \frac{\hbar ^2}{Cc^2}$$

Does this answer the question "What is the value of the effective mass tensor $\left( m^* \right)_{ij}$ where $(i, j = x, y, z)$ at $\mathbf k = 0$"?

 

[nrqed]

So at $\mathbf k = 0$, we get

$$\left( m^* \right)_{xx} = \frac{\hbar ^2}{Aa^2} \ \ \ \ \ \ \left( m^* \right)_{yy} = \frac{\hbar ^2}{Bb^2} \ \ \ \ \ \ \left( m^* \right)_{zz} = \frac{\hbar ^2}{Cc^2}$$

Does this answer the question "What is the value of the effective mass tensor $\left( m^* \right)_{ij}$ where $(i, j = x, y, z)$ at $\mathbf k = 0$"?

Yes. Good job.

 

[LesterTU]

Thank you very much!

They introduced the concept of effective mass in the lectures by stating the definition and giving us an intuitive feel for what it is based on the curvature of the bands, but they never told us how to actually apply it in practice or showed a concrete example, yet this was on an old exam for an introductory course in solid state. I have never encountered a tensor before...