Unformatted text preview: 74 CHAPTER 5. MULTIPLEPARTICLE SYSTEMS
all r1 all r 2 ψl (r1 )ψr (r2 ) ψr (r1 )ψl (r2 ) d3 r1 d3 r2 can according to the rules of calculus be factored into threedimensional integrals as
SS
ψ1 ψ2 = all r 1 ψl (r1 ) ψr (r1 ) d3 r1 all r 2 ψr (r2 ) ψl (r2 ) d3 r2 = ψl ψr ψr ψl
which is zero if ψl and ψr are orthonormal.
Also, do not try to ﬁnd actual values for H11 , H12 , H21 , and H22 . As section 5.2 noted, that
can only be done numerically. Instead just refer to H11 as J and to H12 as −L:
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H11 ≡ ψ1 Hψ1 ≡ ψl ψr Hψl ψr ≡ J
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H12 ≡ ψ1 Hψ2 ≡ ψl ψr Hψr ψl ≡ −L. Next note that you also have
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H22 ≡ ψ2 Hψ2 ≡ ψr ψl Hψr ψl = J
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H21 ≡ ψ2 Hψ1 ≡ ψr ψl Hψl ψr = −L because they are the exact same inner product integrals; the diﬀerence is just which electron
you number 1 and which one you number 2 that determines whether the wave functions are
listed as ψl ψr or ψr ψl .
Answer: Using the abbreviations J and L, the matrix eigenvalue problem becomes
Ja1 − La2 = Ea1
−La1 + Ja2 = Ea2
or taking everything to the left hand side,
(J − E ) a1 − La2 = 0
.
−La1 + (J − E ) a2 = 0
For this homogeneous system of equations to have a solution other than the trivial one a1 =
a2 = 0, the determinant of the matrix must be zero:
(J − E )
−L
=0
−L
(J − E ) =⇒ (J − E )2 − L2 = 0 which allows for the two possibilities
J −E =L or J − E = −L. ...
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 Fall '11
 Dr.DanielArenas
 mechanics

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