Time reversal symmetry in Topological insulators of HgTe quantum Wells

[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

 

[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 ?

 

[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

 

[M Quack]

Should that not be 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.

 

[DrDu]

Should that not be 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.

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

 

[Minato]

Should that not be 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.

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

 

[M Quack]

Thanks for clarifying that.

Going with the 2x2 block motif, let's write <br /> T = -i \left( \begin{array}{cc} 0 &amp; t \\ t &amp; 0 \end{array} \right)K<br /> with <br /> t = \left( \begin{array}{cc} 0 &amp; -i \\ i &amp; 0 \end{array} \right)<br /> 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 &amp; t \\ t &amp; 0 \end{array} \right) K<br /> \left( \begin{array}{cc} h_+(k) &amp; 0 \\ 0 &amp; h_-(k) \end{array} \right)<br /> (-i) \left( \begin{array}{cc} 0 &amp; t \\ t &amp; 0 \end{array} \right) K<br /> =<br /> -\left( \begin{array}{cc} t h^*_-(k) t^* &amp; 0 \\ 0 &amp; t h_+^*(k) t^* \end{array} \right)<br />

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

 

[M Quack]

<br /> -t h_+^* t^*<br /> gives <br /> \left(<br /> \begin{array}{cc}<br /> -M^* + B_-^* (k_x^2 - \frac{\partial^2}{\partial y^2})<br /> &amp;<br /> -A^*(k_x + \frac{\partial}{\partial y})<br /> \\<br /> -A^*(k_x - \frac{\partial}{\partial y})<br /> &amp;<br /> M^* - B_+^* (k_x^2 - \frac{\partial^2}{\partial y^2})<br /> \end{array}<br /> \right)<br />

 

[Minato]

<br /> -t h_+^* t^*<br /> gives <br /> \left(<br /> \begin{array}{cc}<br /> -M^* + B_-^* (k_x^2 - \frac{\partial^2}{\partial y^2})<br /> &amp;<br /> -A^*(k_x + \frac{\partial}{\partial y})<br /> \\<br /> -A^*(k_x - \frac{\partial}{\partial y})<br /> &amp;<br /> M^* - B_+^* (k_x^2 - \frac{\partial^2}{\partial y^2})<br /> \end{array}<br /> \right)<br />

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)-&gt; h(-k) and applying Conjugation operation, it should go to h(-k)-&gt; 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