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Homework Help: Math question in k dot p derivation

  1. Feb 5, 2013 #1
    (maybe this should go in the math section?)
    1. The problem statement, all variables and given/known data

    I'm supposed to derive the k.p form of the Schroedinger equation by plugging in the Bloch wavefunction expansion. But my actual question is just about the math.

    2. Relevant equations
    So when I plugged Bloch into the SE, one of the terms has this Laplacian:
    [itex]\nabla^2[u_{n\mathbf{k}}(\mathbf{r})\exp(i\mathbf{k}\cdot \mathbf{r})][/itex]

    The Laplacian is the divergence of the gradient, so first I need to show this:
    [itex]\nabla[u_{n\mathbf{k}}(\mathbf{r})\exp(i\mathbf{k}\cdot \mathbf{r})]={([\nabla u_{n\mathbf{k}}]+i\mathbf{k}u_{n\mathbf{k}}(\mathbf{r})})\exp{(i \mathbf{k} \cdot \mathbf{r})}[/itex]

    Then the next step would be this:
    [itex]\nabla^2[u_{n\mathbf{k}}(\mathbf{r})\exp(i\mathbf{k}\cdot \mathbf{r})]={{\nabla^2 u_{n\mathbf{k}}(\mathbf{r})+2i\mathbf{k}\cdot \nabla u_{n\mathbf{k}}(\mathbf{r})-k^2 u_{n\mathbf{k}}(\mathbf{r})}}\exp(i\mathbf{k}\cdot \mathbf{r})[/itex]

    I'm also given the following hints:
    • k vector is independent of position
    • u_nk(r) is a function of position only
    • so divergence of k vector is 0
    • (the above is used when calculating the laplacian)

    In addition, I'll probably need these:
    [itex]\text{Gradient } \nabla V=\mathbf{\hat{x}}\frac{\partial V}{\partial x}+\mathbf{\hat{y}}\frac{\partial V}{\partial y}+\mathbf{\hat{z}}\frac{\partial V}{\partial z}\\

    \text{Divergence } \nabla \cdot V=\frac{\partial A_x}{\partial x}+\frac{\partial A_y}{\partial y}+\frac{\partial A_z}{\partial z}\\

    \text{Laplacian } \nabla^2 V=\frac{\partial^2 V}{\partial x^2}+\frac{\partial^2 V}{\partial y^2}+\frac{\partial^2 V}{\partial z^2} [/itex]

    3. The attempt at a solution
    So I start by taking the gradient:
    [itex]\nabla[u_{n\mathbf{k}}(\mathbf{r})\exp(i\mathbf{k}\cdot \mathbf{r})]=u_{n\mathbf{k}}(\mathbf{r}) \nabla [\exp(i\mathbf{k}\cdot \mathbf{r})] +\exp(i\mathbf{k}\cdot \mathbf{r}) \nabla u_{n\mathbf{k}}(\mathbf{r})[/itex]

    Apparently, the first term on the right hand side works out to be this:
    [itex]i\mathbf{k}u_{n\mathbf{k}}(\mathbf{r})\exp{(i \mathbf{k}\cdot \mathbf{r})}[/itex]
    but how do you come up with that? It would really help me if someone could please work it out explicitly.
    Last edited: Feb 5, 2013
  2. jcsd
  3. Feb 5, 2013 #2
    To be blunt, going from the first term on the right hand side to the term below it is actually just a simple derivative. However, I think you can't see that in your current notation, so try the following: expand k dot r into the sum of the three components of each vector, then try calculating one component of the gradient (if you do that, it will be obvious what the other two are). Remember, you can factor out two of those three terms as a constant for each derivative you take.

    I would work it out for you explicitly, but I believe that's against the forum rules. On the other hand, once you've put the expression in the proper form, the work is just taking a derivative of the exponential, which is something I refuse to believe you don't know how to do.

    Is that all you had a question about?
    Last edited: Feb 5, 2013
  4. Feb 5, 2013 #3
    you will have to operate ∇ on eik.r,write it as ikrcosθ by choosing z axis along [itex]\vec{k}[/itex],then involve definition of ∇ in polar coordinates with[itex] \phi[/itex] part omitted.you will get the result.
  5. Feb 5, 2013 #4
    I think cartesian coordinates should be fine. That "r" is a vector, not the coordinate r (or at least, that's what I take the boldface to mean).
  6. Feb 5, 2013 #5
    That is a ridiculously roundabout way of doing a very straightforward derivative. There is no reason to switch from Cartesian to polar coordinates.
  7. Feb 5, 2013 #6
    That is how it is done.The cartesian way is just too simple.
  8. Feb 5, 2013 #7
    Thanks DimReg, andrien and LastOneStanding.

    I think that's the part I don't know how to do. Is that [itex]k_x\mathbf{x}+k_y\mathbf{y}+k_z\mathbf{z}[/itex]?

    If it's a sum, we couldn't just factor though, right? The constant terms would go to 0.
    Last edited: Feb 5, 2013
  9. Feb 5, 2013 #8
    Thanks DimReg, andrien and LastOneStanding.

    I think that's the part I don't know how to do. Is that [itex]\exp(i\mathbf{k}\cdot \mathbf{r})=\exp[(ik_x\mathbf{x}+ik_y\mathbf{y}+ik_z\mathbf{z})][/itex]?
    Then, the derivative of the first term is [itex]ik_x \exp[(ik_x\mathbf{x}+ik_y\mathbf{y}+ik_z\mathbf{z})] [/itex] ?

    If it's a sum, we couldn't just factor though, right? The constant terms would go to 0. Anyway thanks so much.
    Last edited: Feb 5, 2013
  10. Feb 5, 2013 #9
    The x,y and z should not be bolded; they're vector components, not vectors themselves. Otherwise, yes.

    The sum is in an exponential. What does ##e^{ik_xx + ik_yy + ik_zz}## equal? Remember your power rules.
  11. Feb 5, 2013 #10
    Remember, [itex] exp(a + b) = exp(a)exp(b) [/itex]

    So the x component will be a derivative of the form [itex] A(r)\frac{\partial}{\partial x} exp(a + bx) = A(r)exp(a)\frac{\partial}{\partial x}exp(bx) [/itex] because "a" is independent of x. You use the definition of the dot product [itex]k_x x+k_y y+k_z z[/itex] to put your original expression into that form.

    If you don't like the factoring, or you don't remember that formula, you can also just do a straight forward application of the chain rule for derivatives. (What is [itex] \frac{\partial}{\partial x} exp(f(x,y)) ? [/itex]
    Last edited: Feb 5, 2013
  12. Feb 5, 2013 #11
    OK so the x component is [itex]\exp(ik_y y+ik_z z) i k_x \exp(ik_x x)[/itex] and similar for the y and z term, and after a bunch of factoring I end up with [itex]\exp(i\mathbf{k}\cdot \mathbf{r}) i\mathbf{k}[/itex]
    And then there's the u_nk in front.
  13. Feb 5, 2013 #12
    sounds like you got it.
  14. Feb 5, 2013 #13
    Thanks. Now I have to take the divergence of [itex]\exp[(ik_x\mathbf{x}+ik_y\mathbf{y}+ik_z\mathbf{z})][/itex]
    any hints?
  15. Feb 5, 2013 #14
    Do exactly what you just did a second time.
  16. Feb 5, 2013 #15
    But last time it was the gradient. The divergence is the same thing? And whoops, didn't mean to bold the x,y,z.
  17. Feb 5, 2013 #16
    You have the formula for the divergence above. Just take derivatives the same way we just showed you.

    Also, that isn't a vector, so what exactly do you mean by take the divergence of it?
  18. Feb 5, 2013 #17
    So I managed to get this:
    [itex]\nabla[u_{n\mathbf{k}}(\mathbf{r})\exp(i\mathbf{k}\cdot \mathbf{r})]={([\nabla u_{n\mathbf{k}}]+i\mathbf{k}u_{n\mathbf{k}}(\mathbf{r})})\exp{(i \mathbf{k} \cdot \mathbf{r})}[/itex]
    Now I need to take the divergence of both sides, in order to get the laplacian.
    Just working with that first term again, like last time, I get this:
    [itex]i\mathbf{k}u_{nk}(\mathbf{r})\nabla \cdot \exp(i\mathbf{k} \cdot \mathbf{r})+\exp(i\mathbf{k} \cdot \mathbf{r})i \mathbf{k} \nabla \cdot u_{nk}(\mathbf{r})[/itex]
    So you're saying the divergence will give the exact same result as the gradient did? (namely, [itex] i\mathbf{k}\exp{(i\mathbf{k} \cdot \mathbf{r})}[/itex] )
  19. Feb 6, 2013 #18
    You have to remember that the divergence acts of vector functions, not scalars. But the result of a divergence is a scalar. On the other hand, the gradient acts on scalars and gives you a vector. So you wont get exactly the same thing for the two (since one will be a vector and the other a scalar). However, in both cases you are taking derivatives of the same kind of expression, so as long as you keep from getting confused you should be able to compute the divergence correctly.

    Look carefully at the divergence formula. You need to know what the components of the vector you are taking the divergence of. Once you figure that out, simply plug those components in to the correct place in the divergence formula, and it should be straight forward algebra/calculus.

    I get the feeling that you are confusing yourself somehow. If you had some vector (a(r), b(r), c(r)) you would know how to take it's divergence right? What about d(r)(a(r), b(r), c(r))?
  20. Feb 6, 2013 #19
    << Complete solution deleted by Moderators >>
    Last edited by a moderator: Feb 6, 2013
  21. Feb 6, 2013 #20
    andrien: DimReg and I have been working with snickersnee step-by-step so that he or she can ultimately work out the solution for him or herself. Having come this far, it is extremely frustrating to have someone jump in at the end and just give the answer. So far, your contributions to this thread have been to give an unhelpful suggestion that overcomplicated the problem, and then, against forum rules, just giving away the answer. In other words, you have had nothing but a negative impact.

    Please just go away.
    Last edited: Feb 6, 2013
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