Computational Physics - Department of Physics

7.8 Schrödinger’s Equation Through Diagonalization 227

Using the fact thatQˆQˆT=Iˆ, we can rewrite

Tˆ=QˆTAˆQˆ,

as
QˆTˆ=AˆQˆ,

and if we equate columns (recall from the previous slide)

Tˆ=











α 1 β 1 0 ... ... 0

β 1 α 2 β 2 0 ... 0

0 β 2 α 3 β 3 ... 0

... ... ... ... ... 0

... βn− 2 αn− 1 βn− 1

0 ... ... 0 βn− 1 αn











we obtain
Aˆqˆk=βk− 1 qˆk− 1 +αkqˆk+βkqˆk+ 1.

We have thus
Aˆqˆk=βk− 1 qˆk− 1 +αkqˆk+βkqˆk+ 1 ,

withβ 0 qˆ 0 = 0 fork=1 :n− 1. Remember that the vectorsqˆkare orthornormal and this implies

αk=qˆTkAˆqˆk,

and these vectors are called Lanczos vectors. We have thus

Aˆqˆk=βk− 1 qˆk− 1 +αkqˆk+βkqˆk+ 1 ,

withβ 0 qˆ 0 = 0 fork=1 :n− 1 and
αk=qˆTkAˆqˆk.

If
rˆk= (Aˆ−αkIˆ)qˆk−βk− 1 qˆk− 1 ,

is non-zero, then
qˆk+ 1 =rˆk/βk,

withβk=±||rˆk|| 2. These steps can then be written in terms of the following simple algorithm:

r_0 = q_1; beta_0=1; q_0=0;intk = 0;

while(beta_k != 0)

q_{k+1}= r_k/beta_k

k = k+1

alpha_k = q_k^T A q_k

r_k = (A-alpha_k I) q_k -beta_{k-1}q_{k-1}

beta_k = || r_k||_ 2

endwhile

7.8 Schrödinger’s Equation Through Diagonalization

Instead of solving the Schrödinger equation as a differential equation, we will solve it through
diagonalization of a large matrix. However, in both cases weneed to deal with a problem with
boundary conditions, viz., the wave function goes to zero atthe endpoints.