One- and two-dimensional spectral diffusion of type-II excitons in InP/InAs/InP core-multishell nanowires

Physical Review B - Condensed Matter and Materials Physics

Volumn 82, Issue 7, 2010, Pages

  Masumoto, Yasuaki ^a   Goto, Ken ^a   Yoshida, Seitaro ^a   Sakuma, Yoshiki ^b   Mohan, Premila ^c   Motohisa, Junichi ^c   Fukui, Takashi ^c  

^a UNIVERSITY OF TSUKUBA   (Japan)

^b NATIONAL INSTITUTE FOR MATERIALS SCIENCE   (Japan)

^c HOKKAIDO UNIVERSITY   (Japan)

Author keywords

[No Author keywords available]

Indexed keywords

EID: 77957598670     PISSN: 10980121     EISSN: 1550235X     Source Type: Journal    
DOI: 10.1103/PhysRevB.82.075313     Document Type: Article

References (26)

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15. We can simply understand that the energy is lower at the corners than that at the sides, by considering that InAs well can be 2 / √ 3 times thicker at the corners than at the sides. Electronic states in a hexagonal tube with rounded edges were calculated by Ref.. The lowest energy state is the bonding sum of wave functions localized at the corners and six lowest-energy states are displayed. If wurtzite InP/InAs/InP CMNs have perfect hexagonal symmetry in the atomic level, linear combinations of wave functions shown in Fig. with sixfold rotation are the eigenfunctions for holes. Then hole bands are formed. The linear combinations of wave functions in CMNs are shown on a simplified model in Ref..
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26. Because the PL spectrum is well fitted by the Gaussian density of localized states, we use the Gaussian density of localized states and calculate Eq. numerically. To give the intuitive and instructive picture to the dimensionality-dependent energy loss, we can use the exponential tail for the localized states to calculate Eq. analytically. Tentatively assuming the density of the localized states as the exponential tail g d (E) = g 0 exp (- E / ε) for large E, we obtain g 0 ε exp (- Δ E / ε) for the integral in Eq.. Then Eq. leads to Δ E = ε ln (2 g 0 N 1 ε) + (ε / 6) ln (A t) for one dimension and Δ E = ε ln (π g 0 N 2 ε) + (ε / 3) ln (A t) for two dimensions. Then d Δ E / d ln (A t) = ε d / 6. It analytically shows the slope of two-dimensional energy loss versus ln (t) is twice of the slope of one-dimensional energy loss versus ln (t), as is seen in Fig..