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Time reversal symmetry in Topological insulators of HgTe quantum Wells

  1. Apr 2, 2012 #1
    Hi everyone,

    While reading about the BHZ model used to describe HgTe quantum well topological insulators, I read at many places that the effective Hamiltonian (which is a 4 x 4 matrix) can be written in block diagonal form and the lower 2x2 block can be derived from upper 2x2 block as follows:
    [H(k)][/lower]=[H(-k)][/*]

    This effective Hamiltonian is said to be Time reversal symmetric and then using Cramer's degeneracy, it is said that the dispersion relations for upspin and down spin should intersect at [k][/x]=0.

    I want to just show this through simple mathematical steps, but I am unable to get this result. In order to show time reversal invariance, I tried the following equation:
    [T][/-1]HT=H, where T is the Time reversal symmetry operator.
    but I am not sure what form of T should be used. I tried to use the following form:
    T=-i x [0 [σ][/y];[σ][/y] 0]K {K is complex conjugation which is a 4x4 matrix with [0][/2x2] in the diagonals and Pauli matrix in y as off diagonal elements.}

    But this is not giving me that BHZ Hamiltonian is time reversal symmetric.
    Can anybody help me where I am going wrong?

    Thanks

    Regards
    Minato
     
    Last edited: Apr 2, 2012
  2. jcsd
  3. Apr 2, 2012 #2
    Can you show us the explicit form of the Hamiltonian you start out with?

    Time reversal inverts the sign of momentum k and of spin/magnetic moment s.

    In the Schroedinger equation, complex conjugation of a wave function it will result in time reversal.

    With that you practically have your relation.

    T H(k) T psi = T H(k) psi* = H*(-k) psi

    btw, I have trouble reading your notation with []. Can you try to use [itex]?
     
  4. Apr 2, 2012 #3
    I am sorry for the formatting in the previous post.

    The original Hamiltonian for BHZ model used to describe HgTe quantum well Topological insulators is
    [tex]H=\left(\begin{array}{cc}h_{+}(k)&0\\0&{h_{-}(k)}\end{array}\right)[/tex]

    [tex]{h_{-}(k)}=h_{+}^{*}(-k)[/tex]

    here the meaning of * is to take the complex conjugate of the matrix.
    [tex]h_{+}(k)=\left(\begin{array}{cc}M-(B_{+})(k_{x}^{2}-\frac{\partial ^{2}}{\partial y^{2}}) & {A(k_x-\frac{\partial}{\partial y})}\\{A(k_x+\frac{\partial}{\partial y})} & {-M+B_{-}(k_{x}^{2}-\frac{\partial ^{2}}{\partial y^{2}}) }\end{array}\right)[/tex]

    where [tex] M, A, B_{+},B_{-} [/tex] are various system parameters.
    The form of Time reversal operator which I have used is:
    [tex]T=-i\left(\begin{array}{cc}0&0&0&-i\\0&0&i&0\\0&-i&0&0\\i&0&0&0\end{array}\right)K[/tex]
    where K is the conjugation operator
    I am trying to prove the following equation to show that the above Hamiltonian is Time reversal symmetric:
    [tex]H=T^{-1}HT[/tex]


    Regards
    Minato
     
  5. Apr 2, 2012 #4
    Should that not be [itex]A(k_x \pm i \frac{\partial}{\partial y})[/itex]???

    Also, with the time reversal operator you write, I get [itex]T^2 = -1[/itex] instead of [itex]T^2=1[/itex], so there are too many "i"s.
     
  6. Apr 2, 2012 #5

    DrDu

    User Avatar
    Science Advisor

    He wrote something about the system showing Cramers degeneracy. Then I would expect T^2=-1.
     
  7. Apr 2, 2012 #6
    Regarding the first point, it is [itex]A(k_x \pm i k_{y})[/itex] which will give the form I have earlier written.([itex]k_{y}=-i \frac{\partial}{\partial y}[/itex])

    Regarding the second point, the system is fermionic. That is why, [itex]T^2=-1[/itex] is required.

    Regards
    Minato
     
  8. Apr 3, 2012 #7
    Thanks for clarifying that.

    Going with the 2x2 block motif, let's write [itex]
    T = -i \left( \begin{array}{cc} 0 & t \\ t & 0 \end{array} \right)K
    [/itex] with [itex]
    t = \left( \begin{array}{cc} 0 & -i \\ i & 0 \end{array} \right)
    [/itex] such that [itex] t^\star t = -1 [/itex]

    We already know that [itex] T^2 = -1[/itex] and therefore [itex]T^{-1} = -T[/itex]

    Then

    [itex] T^{-1} H T = i \left( \begin{array}{cc} 0 & t \\ t & 0 \end{array} \right) K
    \left( \begin{array}{cc} h_+(k) & 0 \\ 0 & h_-(k) \end{array} \right)
    (-i) \left( \begin{array}{cc} 0 & t \\ t & 0 \end{array} \right) K
    =
    -\left( \begin{array}{cc} t h^*_-(k) t^* & 0 \\ 0 & t h_+^*(k) t^* \end{array} \right)
    [/itex]

    We still have to show [itex] h_{\pm}(k) = -t h_{\mp}^*(k) t^*[/itex], but at least we're down to 2x2 matrices.
     
  9. Apr 3, 2012 #8
    [itex]
    -t h_+^* t^*
    [/itex] gives [itex]
    \left(
    \begin{array}{cc}
    -M^* + B_-^* (k_x^2 - \frac{\partial^2}{\partial y^2})
    &
    -A^*(k_x + \frac{\partial}{\partial y})
    \\
    -A^*(k_x - \frac{\partial}{\partial y})
    &
    M^* - B_+^* (k_x^2 - \frac{\partial^2}{\partial y^2})
    \end{array}
    \right)
    [/itex]
     
  10. Apr 3, 2012 #9
    You are right regarding this. I forgot to tell that all the parameters are real so you can remove the conjugation. But by no means, I have [itex]B_{+}=\pm B_{-}[/itex].

    I have come to know 2 ways to solve this problem.
    (1) First is, I am probably choosing wrong matrix for Time reversal transformations. As my equation is for massless Dirac fermions, I should use proper relativistic quantum mechanics to calculate the transformation matrix for time reversal.
    (2)Second is to use CPT symmetry. The argument goes as : if I apply Parity operation, [itex]h(k)-> h(-k)[/itex] and applying Conjugation operation, it should go to [itex]h(-k)-> h^{*}(-k)[/itex] which is the lower 2 χ 2 matrix of the Hamiltonian. These 2 are equivalent to applying [itex]T^{-1}[/itex]. I know that there are some loop holes in this derivation also, but I just want to give a general idea on how it can be solved.

    I am trying these methods if they work.

    Regards
    Minato
     
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