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Fund Quantum Mechanics Lect &amp; HW Solutions 92

# Fund Quantum Mechanics Lect &amp; HW Solutions 92 - 74...

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Unformatted text preview: 74 CHAPTER 5. MULTIPLE-PARTICLE 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 three-dimensional 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: S S H11 ≡ ψ1 |Hψ1 ≡ ψl ψr |Hψl ψr ≡ J S S H12 ≡ ψ1 |Hψ2 ≡ ψl ψr |Hψr ψl ≡ −L. Next note that you also have S S H22 ≡ ψ2 |Hψ2 ≡ ψr ψl |Hψr ψl = J S S 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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