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UNIVERSITY OF CALIFORNIA, SANTA CRUZ
BOARD OF STUDIES IN COMPUTER ENGINEERING
CMPE 240: INTRODUCTION TO LINEAR DYNAMICAL SYSTEMS
Gabriel Hugh Elkaim Fall 2005
Quantum Mechanics Example
• wave function and Schrodinger equation
• discretization
• preservation of probability
• eigenvalues & eigenstates
• example
1 Quantum mechanics
• single particle in interval [0, 1], mass m
• potential V : [0, 1] → R
Ψ : [0, 1] × R+ → C is (complex-valued) wave function
interpretation: |Ψ(x, t)|2 is probability density of particle at position x, time t
1
(so |Ψ(x, t)|2 dx = 1 for all t)
0
evolution of Ψ governed by Schrodinger equation:
h2
¯
h˙
i¯ Ψ = V − 2
x Ψ = HΨ
2m
√
where H is Hamiltonian operator, i = −1
1
2 Discretization
let’s discretize position x into N discrete points, k/N , k = 1, . . . , N
wave function is approximated as vector Ψ(t) ∈ CN
2
x operator is approximated as matrix
−2 1
1 −2 1
2 2
=N 1 −2 1
.. .. ..
. . .
1 −2
2
so w = v means
(vk+1 − vk )/(1/N ) − (vk − vk−1 )(1/N )
wk =
1/N
(which approximates w = ∂ 2 v/∂x2 )
discretized Schrodinger equation is (complex) linear dynamical system
˙
Ψ = (−i/¯ )(V − (¯ /2m)
h h 2
h
)Ψ = (−i/¯ )HΨ
where V is a diagonal matrix with Vkk = V (k/N )
hence we analyze using linear dynamical system theory (with complex vectors & matrices):
˙ h
Ψ = (−i/¯ )HΨ
h
solution of Shrodinger equation: Ψ(t) = e(−i/¯ )tH Ψ(0)
h
(−i/¯ )tH
matrix e propogates wave function forward in time t seconds (backward if t < 0)
3 Preservation of probability
d 2 d ∗
Ψ = ΨΨ
dt dt
= ˙ ˙
Ψ∗ Ψ + Ψ ∗ Ψ
= h h
((−i/¯ )HΨ)∗ Ψ + Ψ∗ ((−i/¯ )HΨ)
= h
(i/¯ )Ψ HΨ + (−i/¯ )Ψ HΨ
∗
h ∗
= 0
(using H = H T ∈ RN ×N )
hence, Ψ(t) 2 is constant; our discretization preserves probability exactly
h
U = e−(i/¯ )tH is unitary, meaning U ∗ U = I
unitary is extension of orthogonal for complex matrix: if U ∈ CN ×N is unitary and
z ∈ CN , then
U z 2 = (U z)∗ (U z) = z ∗ U ∗ U z = z ∗ z = z 2
2
4 Eigenvalues & eigenstates
H is symmetric, so
• its eigenvalues λ1 , . . . , λN are real (λ1 ≤ · · · ≤ λN )
• its eigenvectors v1 , . . . , vN can be chosen to be orthogonal (and real)
h h
from Hv = λv ⇔ (−i/¯ )Hv = (−i/¯ )λv we see:
h
• eigenvectors of (−i/¯ )H are same as eigenvectors of H, i.e., v1 , . . . , vN
h h h
• eigenvalues of (−i/¯ )H are (−i/¯ )λ1 , . . . , (−i/¯ )λN (which are pure imaginary)
• eigenvectors vk are called eigenstates of system
• eigenvalue λk is energy of eigenstate vk
h
• for mode Ψ(t) = e(−i/¯ )λk t vk , probability density
2
h
|Ψm (t)|2 = e(−i/¯ )λk t vk = |vmk |2
doesn’t change with time (vmk is mth entry of vk )
5 Example
Potential Function V (x)
1000
900
800
700
600
500
V
400
300
200
PSfrag replacements
100
0
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
x
• potential bump in middle of infinite potential well
¯
• (for this example, we set h = 1, m = 1 . . . )
3
lowest energy eigenfunctions
0.2
0
−0.2
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
0.2
0
−0.2
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
0.2
0
−0.2
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
0.2
0
PSfrag replacements
−0.2
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
x
• potential V shown as dotted line (scaled to fit plot)
• four eigenstates with lowest energy shown (i.e., v1 , v2 , v3 , v4 )
now let’s look at a trajectory of Ψ, with initial wave function Ψ(0)
• particle near x = 0.2
• with momentum to right (can’t see in plot of |Ψ|2 )
• (expected) kinetic energy half potential bump height
4
0.08
0.06
0.04
0.02
0
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
x
0.15
PSfrag replacements
0.1
0.05
0
0 10 20 30 40 50 60 70 80 90 100
eigenstate
• top plot shows initial probability density |Ψ(0)|2
• bottom plot shows |vk Ψ(0)|2 , i.e., resolution of Ψ(0) into eigenstates
∗
time evolution, for t = 0, 40, 80, . . . , 320:
|Ψ(t)|2
PSfrag replacements
x
cf. classical solution:
5
• particle rolls half way up potential bump, stops, then rolls back down
• reverses velocity when it hits the wall at left
(perfectly elastic collision)
• then repeats
1
0.9
0.8
0.7
0.6
0.5
0.4
0.3
0.2
PSfrag replacements 0.1
0
0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2
4
x 10
t
N/2
plot shows probability that particle is in left half of well, i.e., |Ψk (t)|2 , versus time t
k=1
6 MATLAB code
% quantum mechanics example for cmpe240
% -----------------------------------
%
% Symmetric Potential
% define matrix that is discretized
% version of Laplacian, ie, del squared operator
% ----------------------------------------------
n=100; deltax=1/n; x=0:deltax:1;
delsq = toeplitz([-2 1 zeros(1,n-1)]);
%not including 1/deltax^2 term
6
% define potential function
% -------------------------
Vmax=.1; v1=0*ones(1,40); v2=Vmax*sin((0:20)/20*pi).^2;
v3=0*ones(1,40); v=[v1,v2,v3];
do=1; if do
% hamiltonian operator
% --------------------
hbar=1; m=1; H = -hbar^2/(2*m)*delsq + diag(v); A = 1/(i*hbar)*H;
% finding energies of eigenstates
[V,D]=eig(H);
[E,index]=sort(diag(D)); %sort energies
V=V(:,index); %sort eigenfunctions
for m=1:(n+1); %normalizing eigenfunctions
V(:,m)=V(:,m)/norm(V(:,m));
end;
% initial wave fct
% ----------------
% stationary wave packet;
% Psi0 = exp(-(x-0.2).^2/0.1^2);
% moving wave packet;
k=2*pi/0.2;
Psi0=exp(-(x-0.2).^2/0.1^2).*exp(i*k*x); Psi0=conj(Psi0’);
Psi0 = Psi0/norm(Psi0); meanE=Psi0’*H*Psi0; Alpha=V’*Psi0;
absAlpha=abs(Alpha).^2;
fprintf(’Mean energy is %f\n’, meanE);
fprintf(’Vmax is %f \n\n’, Vmax);
% time evolution
% --------------
deltat=40; M=expm(deltat*A); PsiMat=Psi0; Psi=Psi0;
nsteps=500; for z=1:nsteps; Psi=M*Psi; PsiMat=[PsiMat Psi]; end;
PDense=abs(PsiMat).^2; % Probability Density.
PDensel=PDense(1:50,:); Pl=sum(PDensel);
%deltat=200; M=expm(deltat*A);
7
%PsiMat=Psi0; Psi=Psi0;
%nsteps=10;
%for z=1:nsteps; Psi=M*Psi; PsiMat=[PsiMat Psi]; end;
%PDense2=abs(PsiMat).^2; % Probability Density.
end;
% some plots
% ----------
close all; figure(1); plot(x,v);
xlabel(’x’); ylabel(’V’); title(’potential function’);
grid on;
if do
figure(2);
m=4;
for l=1:m
subplot(m,1,l);
plot(x, real(V(:,l)),x,v/Vmax*.3,’:’);
axis([0 1 -.3 .3]);
end;
subplot(m,1,1); title(’ Low energy eigenfunctions’);
subplot(m,1,m); xlabel(’x’);
figure(3);
subplot(2,1,1);
plot(x,PDense(:,1)); grid on;
xlabel(’x’); ylabel(’|Psi0|^2’);
title(’Wavefunction at t=0’);
subplot(2,1,2);
plot(0:100,absAlpha,’o’,0:100,absAlpha);
grid on;
xlabel(’n’);
ylabel(’|Alphasubn|^2’);
title(’|Alphasubn|^2 vs. n’);
figure(4);
m=8;
for l=1:m;
subplot(m,1,l);
plot(x,PDense(:,l),x,v,’:’);
8
axis off;
end;
subplot(m,1,1); title(’Psi at intervals of 40 t.u.’);
subplot(m,1,m); xlabel(’x’);
figure(5);
m=8;
for l=1:m;
subplot(m,1,l);
plot(x,PDense(:,l+8),x,v,’:’);
axis off;
end;
subplot(m,1,1); title(’Psi at intervals of 40 t.u.’);
subplot(m,1,m); xlabel(’x’);
figure(6); plot(0:nsteps,Pl)
grid on; zoom on;
axis([0 nsteps 0 1]);
xlabel(’time’); ylabel(’Pleft’);
title(’Pleft vs. t’);
% momentum operator
% -----------------
%P= hbar/(2*deltax)*toeplitz([0, i,zeros(1,n-1)]);
end;
courtesy of Stephen Boyd @ Stanford University
9
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