Question 10.5: The potential seen by an electron with effective mass me^* i......
The potential seen by an electron with effective mass m^{*}_{e} in a GaAs quantum well is approximated by a one-dimensional rectangular potential well of width 2L in such a way that
V(x)=0 for 0 <x<2L
and
V(x)=\inftyelsewhere.
(a) Find the eigenvalues E_{n}, eigenfunctions \psi _{n}, and parity of \psi _{n}.
(b) The system is subject to a perturbation in the potential energy so that \hat{W} (x)=e\left|E\right| \hat{x}, where E is a constant electric field in the x direction. Find the value of the new energy eigenvalues to first order (the linear Stark effect) for a quantum well of a width of 10 nm subject to an electric field of 10^{5} Vcm^{-1}. Compare the change in energy value with thermal energy at room temperature.
(c) Find the expression for the second-order correction to the energy eigenvalues for the perturbation in (b).
(a) It is given that the potential seen by an electron with effective mass m^{*}_{e} in a GaAs quantum well is approximated by a one-dimensional, rectangular potential well of width 2L in such a way that
V(x)=0for 0 <x<2L and
V(x)=\inftyelsewhere, we find the eigenfunctions and eigenvalues by solving the time independent Schrödinger equation
\hat{H}^{(0)} \psi ^{(0)}_{n} = E^{(0)}_{n}\psi ^{(0)}_{n}where the Hamiltonian for the electron in the potential is
\hat{H}^{(0)}=\frac{\hat{p}^{2}}{2m^{*}_{e} }+V(x)The solution for the eigenfunctions is
\psi ^{(0)}_{n} =\frac{1}{\sqrt{L} } \sin \left(\frac{n\pi x}{2L} \right) where n = 1, 2, · · ·
and the parity of the eigenfunctions is even for odd-integer and odd for even-integer values of n.
The solution for the eigenvalues is
E^{(0)}_{n} =\frac{\hbar ^{2} k^{2}_{n} }{2m^{*}_{e} } =\frac{\hbar ^{2}n^{2} \pi^{2} }{8m^{*}_{e}L^{2} } where k_{n} =\frac{n2\pi }{2\times 2L} =\frac{n\pi }{2L}
(b) First-order correction to energy eigenvalues is
E^{(1)}_{n} =\left\langle n\left|\hat{W} \right|n \right\rangle =\left\langle n\left|e\right|E\left|\hat{x} \right| n \right\rangle =W_{nn} =\frac{e\left|E\right| }{L} \int\limits_{x=0}^{x=2L}{x\sin ^{2}\left(\frac{n\pi x}{2L} \right) dx }Using the relation 2 \sin(x) \sin(y)=\cos(x−y)−\cos(x+y) with x=y =nπx/2L allows us to rewrite the integrand, giving
W_{nn} =\frac{e\left|E\right| }{2L} \int\limits_{x=0}^{x=2L}{x\left(1-\cos \left(\frac{n\pi x}{L} \right) \right) dx}W_{nn} =\frac{e\left|E\right| }{2L}\left[\frac{x^{2} }{2} \right]^{x=2L}_{x=0} =e\left|E\right| L
which is the linear Stark effect. The energy-level shift for an electric field \left|E\right| =10^{5} Vcm^{-1} in the x direction across a well of width 2L = 10 nm is
\Delta E=10^{7} \times 5\times 10^{-9} =50 me VAt temperature T = 300 K and \Delta E\gt k_{B} T=25, meV which indicates that the Stark effect produces a large-enough change in energy eigenvalue to be of potential use in a room-temperature device.
(c) The new energy levels to second order are found using
E_{n} =E^{(0)}_{n} +E^{(1)}_{n} +E^{(2)}_{n}where E^{(1)}_{k} = W_{kk} and E^{(2)}_{k}=\sum\limits_{j\neq k}{W_{kj}W_{jk} /(E^{(0)}_{k}-E^{(0)}_{j}) }.
The second-order matrix elements are the off-diagonal terms
W_{kj} =\left\langle k\left|\hat{W} \right|j \right\rangle =\left\langle k\left|e\right|E\left|\hat{x} \right| j \right\rangle =\frac{e\left|E\right| }{L}\int\limits_{x=0}^{x=2L}{x\sin \left(\frac{k\pi x}{2L} \right) \sin \left(\frac{j\pi x}{2L} \right)dx }Using the relation 2 sin(x) sin(y) = cos(x − y) − cos(x + y) with x = kπx/2L and y = jπx/2L allows us to rewrite the integrand, giving
W_{kj} =\frac{e\left|E\right| }{2L} \int\limits_{x=0}^{x=2L}{x\left(\cos \left(\frac{(k-j)\pi x}{2L} \right) -\cos \left(\frac{(k+j)\pi x}{2L} \right) \right) dx}For (k±j) odd we integrate by parts, using UV^{\prime } dx=UV-\int{U^{\prime }Vdx } with U=x and V^{\prime } =\cos ((k\neq j)\pi x/2L). This gives
W_{kj} =\frac{e\left|E\right| }{2L} \left(\frac{4L^{2}(\cos ((k-j)\pi )-1) }{\pi ^{2}(k-j)^{2} }-\frac{4L^{2}(\cos ((k+j)\pi )-1) }{\pi ^{2} (k+j)^{2} } \right)W_{kj} =\frac{-4e\left|E\right|L}{\pi ^{2} } \left(\frac{1}{(k+j)^{2} }-\frac{1}{(k-j)^{2} } \right)= \frac{-16eE_{x}L }{\pi ^{2} } \frac{kj}{(k^{2}-j^{2} )^{2} }
For (k±j) even, symmetry requires that
W_{kj} =0So, the perturbing potential only mixes states of different parity.