Time reversal symmetry in Topological insulators of HgTe quantum Wellsby Minato Tags: hgte, insulators, quantum, reversal, symmetry, time, topological, wells 

#1
Apr212, 02:31 AM

P: 4

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 



#2
Apr212, 03:30 AM

P: 640

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]? 



#3
Apr212, 04:10 AM

P: 4

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 



#4
Apr212, 06:43 AM

P: 640

Time reversal symmetry in Topological insulators of HgTe quantum Wells
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. 



#5
Apr212, 07:30 AM

Sci Advisor
P: 3,379





#6
Apr212, 09:10 AM

P: 4

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



#7
Apr312, 07:32 AM

P: 640

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. 



#8
Apr312, 07:43 AM

P: 640

[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] 



#9
Apr312, 08:28 AM

P: 4

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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