# Quantum Mechanics Example 1 Quantum mechanics

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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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