Fitts D.D. - Principles of Quantum Mechanics[c] As Applied to Chemistry and Chemical Physics (1999)(en)
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Appendix C |
H(x, å) |
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2å/2 |
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H(x) |
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Figure C.1 The Heaviside unit step function H(x), de®ned as the limit as å ! 0 of H(x, å).
and condition (C.2) is satis®ed.
We next assume that the derivative ä9(x) of ä(x) with respect to x exists. If we integrate the product f (x)ä9(x) by parts and note that the integrated part vanishes, we
obtain |
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f (x)ä9(x) dx ÿ f 9(x)ä(x) dx ÿf 9(0)
ÿ1 ÿ1
where f 9(x) is the derivative of f (x). From equations (C.5a) and (C.5d), it follows that ä9(ÿx) ä9(x)
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xä9(x) ÿä(x) |
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We may also evaluate the Fourier transform ä(k) of the Dirac delta function |
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ä(x ÿ x0) |
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ä(k) p2ð …ÿ1ä(x ÿ x0)eÿikx dx p2ð eÿikx0 |
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The inverse Fourier transform•••••• |
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then gives an integral representation of the delta |
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function |
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(C:6) |
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ä(x ÿ x0) p2ð |
…ÿ1ä(k) eikx dk |
2ð |
…ÿ1eik(xÿx0) dk |
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•••••• |
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The Dirac delta function may be readily generalized to three-dimensional space. If r represents the position vector with components x, y, and z, then the three-dimensional delta function is
ä(r ÿ r0) ä(x ÿ x0)ä(y ÿ y0)ä(z ÿ z0) and possesses the property that
Dirac delta function |
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… f (r)ä(r ÿ r0) dr f (r0) |
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or, equivalently |
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… … … f (x, y, z)ä(x ÿ x0)ä( y ÿ y0)ä(z ÿ z0) dx d y dz f (x0, y0, z0) |
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where the range of integration includes the points x0, y0, and z0. The integral |
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representation is |
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ä(r ÿ r0) |
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…ÿ1eik(rÿr0) dk |
(C:7) |
(2ð)3 |
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where k is a vector with components kx, ky, and kz and where dk dkx dky dkz
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Hermite polynomials |
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1 k |
( 1)á2kÿáîkÿás k á |
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g(î, s) |
k |
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ÿ |
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0 á 0 |
á!(k ÿ á)! |
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XXn |
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We next collect all the coef®cients of s |
for an arbitrary n, so that k á n, and |
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replace the summation over k by a summation over n. When k n, the index á equals zero; when k n ÿ 1, the index á equals one; when k n ÿ 2, the index á equals two; and so on until we have k n ÿ M and á M. Since the index á runs from 0 to k so that á < k, this ®nal term gives M < n ÿ M or M < n=2. Thus, for k á n, the summation over á terminates at á M with M n=2 for n even and M
(n ÿ 1)=2 for n odd. The result of this resummation is
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1 M |
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ÿ |
1)á2nÿ2áînÿ2á |
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g(î, s) |
n 0 á |
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á!(n ÿ |
2á)! |
sn |
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XX |
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n |
in the |
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Since the Hermite polynomial Hn(î) divided by n! is the coef®cient of s |
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expansion (D.1) of g(î, s), we have |
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M |
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(ÿ1)á |
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înÿ2á |
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(î) |
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2n n! |
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22á |
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(D:4) |
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á!(n ÿ |
2á)! |
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á |
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We note that Hn(î) is an odd or even polynomial in î according to whether n is odd or even and that the coef®cient of the highest power of î in Hn(î) is 2n.
Expression (D.4) is useful for obtaining the series of Hermite polynomials, the ®rst
few of which are |
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H0(î) 1 |
H3(î) 8î3 ÿ 12î |
H1(î) 2î |
H4(î) 16î4 ÿ 48î2 12 |
H2(î) 4î2 ÿ 2 H5(î) 32î5 ÿ 160î3 120î |
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Recurrence relations
We next derive some recurrence relations for the Hermite polynomials. If we differentiate equation (D.1) with respect to s, we obtain
2(î ÿ s)e2îsÿs |
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Hn(î) |
s nÿ1 |
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1)! |
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The ®rst term (n 0) in the summation on the right-hand side vanishes because it is the derivative of a constant. The exponential on the left-hand side is the generating function g(î, s), for which equation (D.1) may be used to give
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sn |
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s nÿ1 |
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2(î ÿ s) n |
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0 Hn(î) |
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n 1 Hn(î) |
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n! |
(n ÿ 1)! |
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Since this equation is valid for all values of s with jsjn, 1, we may collect terms |
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corresponding to the same power of s, for example s , and obtain |
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2îHn(î) |
2 H nÿ1(î) |
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H n 1(î) |
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n! |
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or |
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Hn 1(î) ÿ 2îHn(î) 2nH nÿ1(î) 0 |
(D:5) |
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This recurrence relation may be used to obtain a Hermite polynomial when the two preceding polynomials are known.
298 Appendix D
Another recurrence relation may be obtained by differentiating equation (D.1) with
respect to î to obtain |
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2se2îsÿs |
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d Hn sn |
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n 0 |
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dî n! |
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Replacing the exponential on the left-hand side using equation (D.1) gives |
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sn |
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dHn sn |
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2s n 0 Hn(î) |
n! |
n 0 |
dî |
n! |
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If we then equate the coef®cients of s |
, we obtain the desired result |
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dHn |
2nH nÿ1(î) |
(D:6) |
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dî |
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The relations (D.5) and (D.6) may be combined to give a third recurrence relation. |
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Addition of the two equations gives |
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Hn 1(î) 2î ÿ |
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Hn(î) |
(D:7) |
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dî |
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With this recurrence relation, a Hermite polynomial may be obtained from the |
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preceding polynomial. By applying the relation (D.7) to Hn(î) k times, we have |
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H n k (î) 2î ÿ |
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(D:8) |
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dî |
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Differential equation
To ®nd the differential equation that is satis®ed by the Hermite polynomials, we ®rst differentiate the second recurrence relation (D.6) and then substitute (D.6) with n replaced by n ÿ 1 to eliminate the ®rst derivative of H nÿ1(î)
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2n |
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4n(n ÿ 1)H nÿ2(î) |
(D:9) |
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dî |
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Replacing n by n ÿ 1 in the ®rst recurrence relation (D.5), we have |
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Hn(î) ÿ 2îHnÿ1(î) 2(n ÿ 1)H nÿ2(î) 0 |
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which may be used to eliminate Hnÿ2(î) in equation (D.9), giving |
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d2 Hn |
2nHn(î) |
ÿ 4nîH nÿ1(î) 0 |
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dî2 |
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Application of equation (D.6) again to eliminate H nÿ1(î) yields |
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d2 Hn |
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ÿ 2î |
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2nHn(î) 0 |
(D:10) |
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dî2 |
dî |
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which is the Hermite differential equation.
Integral relations
To obtain the orthogonality and normalization relations for the Hermite polynomials, we multiply together the generating functions g(î, s) and g(î, t), both obtained from equation (D.1), and the factor eÿî2 and then integrate over î
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1 1 sntm |
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Hn(î)Hm(î) dî (D:11) |
I …ÿ1eÿî |
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g(î, s) g(î, t) dî n 0 m 0 n!m! |
…ÿ1eÿî |
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X X |
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300 |
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Appendix D |
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The function ön(î) as de®ned by equation (D.15) then becomes |
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ö |
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(ÿi)n |
eî2=2 1 |
eÿ(s2=4) iîs sn ds |
(D:17) |
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2(2nðn!)1=2ð1=4 |
…ÿ1 |
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We now evaluate the summation
X1
ön(î)ön(î9)
n 0
by substituting equation (D.17) twice, once with the dummy variable of integration s and once with s replaced by t. Since the functions ön(î) are real, they equal their complex conjugates. These substitutions give
1 ö |
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(î)ö |
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(î9) |
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e(î2 î92)=2 |
1 1 eÿ[(s2 t2)=4] i(îs î9t) |
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(st)n ds dt |
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4ð3=2 |
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n 0 |
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…ÿ1…ÿ1 |
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n 0 |
2n n! |
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2n |
(ÿ1) |
n |
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since (ÿi) |
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. The summation on the right-hand side is easily evaluated using |
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equation (A.1) |
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1)n |
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(ÿ |
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eÿst=2 |
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Noting that |
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s2 t2 |
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we have |
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ön(î)ön(î9) |
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e(î |
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…ÿ1…ÿ1eÿ[(s t) |
=4] i(îs î9t) ds dt |
(D:18) |
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4ð3=2 |
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The double integral may be evaluated by introducing the new variables u and v |
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ÿ |
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ds dt 2 du dv |
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The double integral is thereby factored into |
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…1 |
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…1 |
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2eÿu2 i(î î9)u du ei(îÿî9)v dv 2 3 ð1=2eÿ(î î9)2=4 3 2ðä(î ÿ î9)
ÿ1 ÿ1
where the ®rst integral is evaluated by equation (A.8) and the second by (C.6). Equation (D.18) now becomes
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ön(î)ön(î9) e[(î2 î92)=2]ÿ[(î î9)2=4]ä(î ÿ î9) e(îÿî9)2=4ä(î ÿ î9)
n 0
Applying equation (C.5e), we obtain the completeness relation for the functions ön(î)
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ön(î)ön(î9) ä(î ÿ î9) |
(D:19) |
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n 0
demonstrating, according to equation (3.31), that the set ön(î) is a complete set.
