2.8 Conclusions
3.1.1 QUIC in semiconductor
Traditionally, photocurrents are produced by a two stage process in which photon absorption promotes an electron to the semiconductor conduction band and then a bias voltage applied across the semiconductor accelerates the electron in the desired direction. The vector of the photocurrent is there- fore dependent on the voltage applied (and, to a lesser extent, on the proper- ties of the semiconductor). With no voltage, the free electrons are generated with random momenta and no net current can result. Even if anisotropy is present in the semiconductor, the probabilities of photocurrent in the forward or backward direction of a given crystal axis will still be equal and the net current zero. In investigating QUIC for photocurrents, Kurizkiet al [22] con- sidered a superposition of the two bound states 1s and 2p0 in a donor atom.
These states were then photoionised by two mutually phase-locked lasers op- erating at slightly different frequencies but with the same polarisation axes. This was therefore photoionisation of the superposition state
|ψi=c1|1i+c2|2i (3.5)
by a z-polarised, two-colour source with electric field
Ez(t) =E1cos(ω1t+φ1) +E2cos(ω2t+φ2) (3.6)
They found that the magnitude and the sign of the photocurrent in the
z direction was dependent upon E1 and E2, N (the donor concentration),
ω1 and ω2 and, most significantly, on the phase difference φ1 −φ2. Thus
for a given piece of semiconductor and two suitable phase-locked lasers a photocurrent could be created and controlled simply by adjusting the relative phase difference between the lasers without the need for any external bias whatsoever. Using some reasonable numbers for a piece of Si measuring
0.1 ×10× 10µm they predicted currents of 10 to 100 mA, enough to be of practical use and prompting suggestions of a number of possible applications, including relative phase measurement and information transmission by phase modulation.
Other suggestions for QUIC in semiconductors were put forward [23] but all did their best to consider small, atomically-analogous systems such as quantum wells or, as above, donor atoms. This seemed reasonable as all QUIC work had been carried out in the context of relatively simple quan- tum systems such as atoms or molecules. However, in their 1996 paper [24] Atanasov et al calculated one- and two-photon photocurrent injection for bulk intrinsic semiconductor. They found that when both processes were present simultaneously, the probability of an electron transition to the con- duction band was asymmetric in momentum spacek. This asymmetry meant a net photoelectron momentum in a given direction and hence a net pho- tocurrent without an external bias. The current density injection rate was calculated to be
˙
JI = ˆη...E(ω)E(ω)E(−2ω) +c.c. (3.7) where ˆη is a four-tensor calculated from the semiconductor band structure. This meant the possibility of QUIC-generated photocurrents in standard semiconductor devices.
Calculation of injected QUIC current in GaAs
Writing Equation 3.7 in terms of its general components gives ˙
Jµ=ηµαβγEαωEβωEγ−2ω (3.8)
using Eω to denote E(ω) for clarity. According to [24] the only non-zero
values of ηin LT-GaAs areηxxxx,ηxyxy =ηxxyy = 12ηxxxxandηxyyx≈0, these
relations holding true for all exchanges ofx,yandzdue to the cubic symmetry of GaAs. Sinceηxxxxis the largest component the authors recommended use
of both beams polarised in the (100) direction in GaAs for optimum QUIC current generation, claiming that the (110) direction would result in only half
the injected current since ηxyxy = 12ηxxxx. However, here it is demonstrated
that this is not in fact the case, and that the (110) direction yields the same
injected current as the (100) direction.
If the two beams are polarised in thexy plane so that Eω
z and Ez−2ω are
both zero, calculating the injected current density in component form obtains ˙
Jx =ηxxxx(Exω)2Ex−2ω+ηxxyyExωEyωEy−2ω+ηxyxyEyωExωEy−2ω+ηxyyx(Eyω)2Ex−2ω
(3.9) ˙
Jy =ηyyyy(Eyω)2Ey−2ω+ηyyxxEyωExωEx−2ω+ηxyxyExωEyωEx−2ω+ηyxxy(Exω)2Ey−2ω
(3.10) Taking the coordinate system where (100) is the x direction and (010) they
direction, having both beams polarised along (100) gives ˙
Jx =ηxxxx(Eω)2E−2ω (3.11)
˙
Jy = 0 (3.12)
(As an aside, having the beams cross-polarised, the ω beam being polarised along the (010) direction, gives
˙
Jx =ηxyyx(Eω)2E−2ω (3.13)
˙
Jy = 0 (3.14)
Since ηxyyx = 12ηxxxx, cross-polarising the beams clearly results in half the
injected current, but it is interesting to note that a current is still present.) If the beams are both polarised in the (110) direction
Eω x = 1 √ 2E ω, Eω y = 1 √ 2E ω (3.15) Ex−2ω = √1 2E −2ω, E−2ω y = 1 √ 2E −2ω (3.16)
Then, taking for GaAs
ηxxyy =ηxyxy = η
2 (3.18)
ηxyyx= 0 (3.19)
and inserting into Equation 3.9 gives ˙ Jx =η 1 2(E ω)2√1 2E −2ω+η 2 1 2(E ω)2√1 2E −2ω+η 2 1 2(E ω)2√1 2E −2ω (3.20)
which condenses down to ˙ Jx = √ 2 2 η(E ω)2E−2ω (3.21)
Equivalent expressions will be found for ˙Jy, and so the resultant injected
current will be in the xy (110) direction and have magnitude given by
¯ ¯ ¯J˙ ¯ ¯ ¯= r ˙ Jx 2 + ˙Jy 2