(E – E2 – 1)z(k, l) = 0 (5.51) or (E1 + V1+ E2)(E1 – V1+ E2)z(k, l) = 0. (5.52) We will obtain a solution using the first form. Our main reason for not attempt- ing to proceed with the second operator form is that, in general, the operator V1+ E2 does not have a well-defined meaning in terms of an infinite-series expansion in E2. From equation (5.51), we obtain z(k, l) = (-)*(1 – E²)°¢(k), (5.53)

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Chapter4: Eigenvalues And Eigenvectors
Section4.6: Applications And The Perron-frobenius Theorem
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5.2.4 Example D
The equation
2(k + 2, e) = z(k, l+ 1) + z(k, l)
is second order in k and first order in l. In operator form, it can be written
(5.50)
as either
(E – E2 – 1)z(k, l) = 0
(5.51)
or
(E1 + V1+ E2)(E1 – /1+ E2)z(k, l) = 0.
(5.52)
We will obtain a solution using the first form. Our main reason for not attempt-
ing to proceed with the second operator form is that, in general, the operator
V1+ E2 does not have a well-defined meaning in terms of an infinite-series
expansion in E2.
From equation (5.51), we obtain
2(k, l) = (-)*(1 – E)'¢(k),
(5.53)
where ø(k) is an arbitrary function of k. Now,
() 대 + () 대
(6)
(1 – E)' = 1 –
E?
()E-1) +(-)'E?", (5.54)
)(-)-'E?“-1) + (-)'E“, (5.54)
l –1
and, therefore,
2(,
Ф(k + 2) +
(k + 4) + · · ·
(5.55)
+(,,)(-)-olk + 2(e – 1)] + (-)'ø(k + 20)
Again, we point out that equation (5.44) is first order in l and thus the single
arbitrary function that appears in equation (5.55).
Transcribed Image Text:5.2.4 Example D The equation 2(k + 2, e) = z(k, l+ 1) + z(k, l) is second order in k and first order in l. In operator form, it can be written (5.50) as either (E – E2 – 1)z(k, l) = 0 (5.51) or (E1 + V1+ E2)(E1 – /1+ E2)z(k, l) = 0. (5.52) We will obtain a solution using the first form. Our main reason for not attempt- ing to proceed with the second operator form is that, in general, the operator V1+ E2 does not have a well-defined meaning in terms of an infinite-series expansion in E2. From equation (5.51), we obtain 2(k, l) = (-)*(1 – E)'¢(k), (5.53) where ø(k) is an arbitrary function of k. Now, () 대 + () 대 (6) (1 – E)' = 1 – E? ()E-1) +(-)'E?", (5.54) )(-)-'E?“-1) + (-)'E“, (5.54) l –1 and, therefore, 2(, Ф(k + 2) + (k + 4) + · · · (5.55) +(,,)(-)-olk + 2(e – 1)] + (-)'ø(k + 20) Again, we point out that equation (5.44) is first order in l and thus the single arbitrary function that appears in equation (5.55).
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