# Time reversal symmetry in Topological insulators of HgTe quantum Wells

1. Apr 2, 2012

### Minato

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. Apr 2, 2012

### M Quack

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 $? 3. Apr 2, 2012 ### Minato 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 $$H=\left(\begin{array}{cc}h_{+}(k)&0\\0&{h_{-}(k)}\end{array}\right)$$ $${h_{-}(k)}=h_{+}^{*}(-k)$$ here the meaning of * is to take the complex conjugate of the matrix. $$h_{+}(k)=\left(\begin{array}{cc}M-(B_{+})(k_{x}^{2}-\frac{\partial ^{2}}{\partial y^{2}}) & {A(k_x-\frac{\partial}{\partial y})}\\{A(k_x+\frac{\partial}{\partial y})} & {-M+B_{-}(k_{x}^{2}-\frac{\partial ^{2}}{\partial y^{2}}) }\end{array}\right)$$ where $$M, A, B_{+},B_{-}$$ are various system parameters. The form of Time reversal operator which I have used is: $$T=-i\left(\begin{array}{cc}0&0&0&-i\\0&0&i&0\\0&-i&0&0\\i&0&0&0\end{array}\right)K$$ where K is the conjugation operator I am trying to prove the following equation to show that the above Hamiltonian is Time reversal symmetric: $$H=T^{-1}HT$$ Regards Minato 4. Apr 2, 2012 ### M Quack Should that not be [itex]A(k_x \pm i \frac{\partial}{\partial y})$???

Also, with the time reversal operator you write, I get $T^2 = -1$ instead of $T^2=1$, so there are too many "i"s.

5. Apr 2, 2012

### DrDu

He wrote something about the system showing Cramers degeneracy. Then I would expect T^2=-1.

6. Apr 2, 2012

### Minato

Regarding the first point, it is $A(k_x \pm i k_{y})$ which will give the form I have earlier written.($k_{y}=-i \frac{\partial}{\partial y}$)

Regarding the second point, the system is fermionic. That is why, $T^2=-1$ is required.

Regards
Minato

7. Apr 3, 2012

### M Quack

Thanks for clarifying that.

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

We already know that $T^2 = -1$ and therefore $T^{-1} = -T$

Then

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

We still have to show $h_{\pm}(k) = -t h_{\mp}^*(k) t^*$, but at least we're down to 2x2 matrices.

8. Apr 3, 2012

### M Quack

$-t h_+^* t^*$ gives $\left( \begin{array}{cc} -M^* + B_-^* (k_x^2 - \frac{\partial^2}{\partial y^2}) & -A^*(k_x + \frac{\partial}{\partial y}) \\ -A^*(k_x - \frac{\partial}{\partial y}) & M^* - B_+^* (k_x^2 - \frac{\partial^2}{\partial y^2}) \end{array} \right)$

9. Apr 3, 2012

### Minato

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 $B_{+}=\pm B_{-}$.

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, $h(k)-> h(-k)$ and applying Conjugation operation, it should go to $h(-k)-> h^{*}(-k)$ which is the lower 2 χ 2 matrix of the Hamiltonian. These 2 are equivalent to applying $T^{-1}$. 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