# What are the eigenstates of the sum of the exchange and hyperfine hamiltonian in a quantum dot?

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I am trying to find the eigenstates for the Hamiltonian for a quantum dot (QD). Here, I am considering the exchange and hyperfine Hamiltonian. The Zeeman Hamiltonian is not included.

$\hat{H}$ = $\hat{H}_{exchange}$ + $\hat{H}_{hf}$

This post is a little long, so I will post my questions here, and also at the end of this post:

What I want to ask is that are my solutions correct? I am wondering if I missed anything in the Hamiltonians (given this simplified case of only considering heavy holes), or if I made a mistake somewhere with regards to the basis states.

I am also studying how to extend this to include light hole states. Can anyone give me some pointers as to how I can do this?

Here is what I did: I followed largely the formulation as described in this paper by Bayer.

http://journals.aps.org/prb/abstract/10.1103/PhysRevB.65.195315

The general form of the spin Hamiltonian for the electron-hole exchange interaction of an exciton formed by a

hole with spin $J_h$ and by an electron with spin $S_e$ is given by

$\hat{H}_{exchange}$ = $-\sum_{i=x,y,z}$ $(a_i$  $J_{h,i}$  $S_{e,i}$ + $b_i$  $J^3_{h,i}$  $S_{e,i})$.

For the following analysis the z direction is chosen to point along the heterostructure growth direction.

Since the heavy and light hole states are split in energy due to strain, only the heavy hole is consider here.

The single-particle basis from which the excitons are constructed therefore consists of a heavy hole with $J_h = 3/2$, $J_{h,z}= ±3/2$ and an electron $S_e= 1/2$, $S_{e,z} = ±1/2$.

From these states four excitons are formed and they are characterized by the angular momentum projections  $M=$ $S_{e,z}$ $+$  $J_{h,z}$. Only $|M| = 1$ states can couple with the light field (bright states) while $|M| = 2$ states are dark states.

Using the exciton states of $|+1>$, $|-1>$ as basis (here I am only considering the bright states for simplicity), the following matrix representation is obtained:

$\hat{H}_{exchange} =1/2\bigl(\begin{matrix} δ_0 & δ_1\\ δ_1 & δ_0 \end{matrix} \bigr)$

where $δ_0=1.5 (a_z+ 2.25 b_z )$ and $δ_1=0.75 (b_x-b_y )$.

The Hamiltonian for the hyperfine interaction if given by:

$\hat{H}_{hf} = \frac{\nu_0}{8} \sum_i A_i |\psi(R_i)|^2 [\hat{I}^i_z \hat{S}_z + \frac{1}{2} (\hat{I}^i_+ \hat{S}_- + \hat{I}^i_- \hat{S}_+)]$

where $\nu_0$ is the volume of a unit cell (of InAs in my case), $A_i$ is the hyperfine constant, $\psi(R_i)$ is the electron wavefunction at the position of the $ith$ nucleus at position $R_i$. $\hat{I}$ and $\hat{S}$ are the nuclear and electron spin operator respectively.

The first term on the RHS is the static part and the second term the flip-flop part.

The static part describes the Overhauser shift while the flip-flop part is responsible for the electron-nuclear spin flip.

I take $|\psi(R_i)| \propto \sqrt{\frac{8}{\nu_0}}$.

Since $\hat{H}_{hf}$ is written for the interaction of single spins, to generalise this for an ensemble of nuclear spins, I took average values for  $A_i \rightarrow \bar{A}$ and $I^i_z \rightarrow \bar{I_z}$

Using the same basis as above and again considering only the bright states hamiltonian then becomes

$\hat{H}_{hf} = \bar{A_i} \bar{I^i_z} S_{e,z}$

$\hat{H}_{hf} = \frac{1}{2}\bar{A_i} \bar{I^i_z}\bigl(\begin{matrix} -1 & 0\\ 0 & -1 \end{matrix} \bigr)$

Solving for the sum of these two Hamiltonians, I obtained two eigenenergies

$\lambda_\pm = \frac{1}{2} [\delta_0 \pm \sqrt{\bar{A}^2\bar{I_z}^2 +\delta_1^2}$ $]$

which are completely analogous to the solutions for the case of a non-zero magnetic field in the Faraday geometry as described in the paper.

What I want to ask is that are my solutions correct? I am wondering if I missed anything in the Hamiltonians (given this simplified case of only considering heavy holes), or if I made a mistake somewhere with regards to the basis states.

I am also studying how to extend this to include light hole states. Can anyone give me some pointers as to how I can do this?

Thank you very much!

asked Dec 21, 2014
edited Dec 21, 2014

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