The role of quantum mechanics in signal processing - Signal Processing in MRS with Biomedical Applications - page 50

the basis{Q r }is numerically generated (or chosen from the family of classical

polynomials), the assembly of the coe cients{η n (E)}will remain the same
for different systems if they are taken at the same energy E [125]-[131].
Alternatively, the integral representation (2.59) of the polynomial P n (z) can
also be employed to obtain the expansion coe cient η n (z). This is achieved
by replacing z by z ∗ in (2.59) yielding P n (z ∗ ) which is, by reference to(2.60),
equal to P ∗ n (z)
P n (z ∗ ) = P ∗ n (z) = β 1
0

b b

dσ(x)
z ∗ −x

dσ(x)
z ∗ −x Q n (x)

Q n (z ∗ ) −

a a

∗

b b

= β 1
0

dσ(x)
z−x

dσ(x)
z−x Q ∗ n (x)

Q n (z) −

a a

∗

b

= β 1
0

dσ(x)
z−x

Q n (z) −η n (z)

a

η n (z) = S(z)Q n (z)− 0
β 1

∴ P n (z) (2.94)

where S(z) is the Stieltjes integral (2.62). Employing (2.32) in a direct com
parison of (2.66) with (2.92) or inserting (2.92) into (2.94), we obtain

η n (z) = Q n (z)E n (z). (2.95)

Thus, the expansion coe cient η n (z) in the series for the Green operator
(2.92) is given by the product of the polynomial Q n (z) and the error term
E

n (z) from the representation of the Stieltjes integral (2.62) by the Pade
approximantR

n (z) or from its equivalent Heaviside partial fractions (2.76).
In the method of propagation of wave packets in the Schrodinger picture of
quantum mechanics, the evolution operator U(t) is employed. Following the
outlined procedure for H, the evolution operator U(t) can also be expanded
in the polynomial basis{Q n (x)}via

∞

U(t) =

η n (t)Q n ( H).

(2.96)

n=0

Here, the expansion coe cients η n (t) are related to η n (E) from (2.91) by the
Fourier integral

∞

dtθ(t)η n (t)e iEt .

η n (E) =−i (2.97)

−∞

When the set{Q n (x)}is taken to be within the classical polynomials [77,
132], the Stieltjes integral (2.62) with a known weight function W(x) can be
calculated explicitly in the analytical form.
The expressions (2.49) and (2.54) can be used in the corresponding operator
forms

β n+1 Q n+1 ( H) = ( H−α n 1)Q n ( H)−β n Q n−1 ( H)
Q −1 ( H) = 0

. (2.98)

Q 0 ( H) = 1