Beräkningsverktyg, lösning till övn. 3.4c i ‘Kvantvärldens fenomen’

In regions 1 and 3:

\frac{−h̄^{2}}{2m} ϕ′′ + 0ϕ = Eϕ ⟹

ϕ′′ = −\frac{2mE}{h̄^{2}} ϕ = −k^{2}ϕ (k^{2} = \frac{2mE}{h̄^{2}} )

Solutions of the form (with no returned wave in region 3):

ϕ_{1} = Ae^{ikx} + Be^{−ikx} ⟹ ϕ′_{1} = Aike^{ikx} + B(−ik)e^{−ikx}
ϕ_{3} = F e^{ikx} ⟹ ϕ′_{1} = F (ik)e^{ikx}

In region 2:

\frac{−h̄^{2}}{2m} ϕ′′ − V_{0}ϕ = Eϕ ⟹ ϕ′′ = −\frac{2m(E + V_{0}}{h̄^{2}} ϕ = −K^{2}ϕ (k^{2} = \frac{2m(E + V_{0})}{h̄^{2}})

with the same type of solution as in regions 1 and 3:

ϕ_{2} = Ce^{iKx} + De^{−iKx} ⟹ ϕ′_{2} = C(iK)e^{iKx} + D(−iK)e^{−iKx}

Continuity of the wave-function and its first derivative at x = 0 and x = a:

A + B = C + D
kA − kB = kC − kD
Ce^{iKa} + De^{−iKa} = F e^{ika}
C(iK)e^{iKa} + D(−iK)e^{−iKa} = F (ik)e^{ika}

With the amplitudes in units of A

\{ \begin{matrix}+ (B/A) & − (C/A) & − (D/A) & & = −1 \\ − k(B/A) & − K(C/A) & + K(D/A) & & = −k \\ & + e^{iKa}(C/A) & + e^{−iKa}(D/A) & − e^{ika}(F/A) & = 0 \\ & + iKe^{iKa}(C/A) & − iKe^{−iKa}(C/A) & − ike^{ika}(E/A) & = 0 \end{matrix}

corresponding to the system Ax = y

with

A = [ \begin{matrix}1 & −1 & −1 & 0 \\ − k & − K & K & 0 \\ 0 & e^{iKa} & e^{−iKa} & −e^{ika} \\ 0 & + Ke^{iKa}) & − Ke^{−iKa} & − ke^{ika} \end{matrix} ] , x = [ \begin{matrix}B/A \\ C/A \\ D/A \\ F/A \end{matrix} ] , y = [ \begin{matrix}−1 \\ k \\ 0 \\ 0 \end{matrix} ]

The probability current of the incoming wave in region 1 is given as (h̄k/m)|A|^{2} and for the outcoming wave in region 3 (h̄k/m)|F |^{2}. This gives the transmission coefficient:

T = T (E) = \frac{(h̄k/m)|F |^{2}}{(h̄k/m)|A|^{2}} = |\frac{F}{A}|^{2}

Constants:

k = \sqrt{\frac{2mE}{h̄^{2}}} = \sqrt{\frac{2mc^{2}E}{(h̄c)^{2}}}

With h̄c = 197.3271 eV nm and m_{e}c^{2} = 510.99910^{3} eV and with E in eV, k will be in units of [1/nm]. The only difference for K compared with k is that E is replaced by (E + V_{0}), where V_{0} = 5 eV.

It is possible to find an analytic solution (see p. 201, G. Ohlén, Kvantvärldens fenomen:

T = (1 + \frac{V_{0}}{4E (E + V_{0})} sin^{2} (ka))^{−1}

Matlab-fil för lösning av uppgiften:

% transmission.m  inlamningsuppgift 2, deluppgift 1, ht 2005
clear all
close all
L=50; % antal punkter i energivektorn.
E_vector = linspace(eps,10,L);
T_vector=zeros(L,1);
a = 1; % [nm]
mC2=510999; % [eV]
hbarC=197.3271; % [eV nm]
V0=5; % [eV]
for j = 1:L
    E=E_vector(j);
    K = sqrt( (2*mC2*(E+V0))/(hbarC^2) );
    k = sqrt( (2*mC2*(E))/(hbarC^2) );
    y = [-1 -k 0 0]'; % ekvationssystemets högerled
    A = [            % matris for koefficienterna till de fyra obekanta "B, C, D, E".
        1 -1 -1 0;
        -k -K K 0;
        0 exp(i*K*a) exp(-i*K*a) -exp(i*k*a);
        0 K*exp(i*K*a) -K*exp(-i*K*a) -k*exp(i*k*a)
    ];
   % x = A\y; % losningen till ekvation systemet.
   % x = inv(A)*y;
   x = A\y
    T_vector(j) = abs(x(4))^2;
   % 4.e positionen i x vektorn är amplituden för den transmitterade vagen.
end
plot (E_vector, T_vector,'-')
hold on
% Jämförelse med analytisk beräkning
Ein_vector = linspace(eps,10,200);
k_vector=sqrt( 2*mC2*(Ein_vector+V0)/hbarC^2 );
analytic=1./(1 + V0^2./(4*Ein_vector.*(Ein_vector+V0)).* sin(k_vector*a).^2 );
% sidan 201 i 'Kvantvärldens fenomen'.
plot(Ein_vector,analytic,'r--')
axis([0 10 0 1.4])
legend('Numeriskt beraknad','Analytisk formel, sid. 201','Location','NW')
xlabel('Energi [e.V.]','fontsize',15)
ylabel('Transmitans','fontsize',15)