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A Born-Oppenheimer Expansion in a Neighborhood of a Renner ...

A Born-Oppenheimer Expansion in a Neighborhood of a Renner ...

where ( h P⊥) (0) ψ

where ( h P⊥) (0) ψ (k) ⊥ + + + + for j ≥ 4. k� j=4 k� j=3 k� j=3 �k−3 j=2 Hj = j� l=4 j� l=3 j� l=3 1 � 2 � � = �k−3 j=2 1 2 � (j−2) P ⊥ ∆X, Y ψ (k−j) � ⊥ P (j−l) � � (l−4) ⊥ (∆x,y Ψ1) P (j−l) ⊥ P (j−l) ⊥ ��∂Ψ1 � � (l−3) ∂x ��∂Ψ2 � � (l−3) ∂x ∂ ∂X ∂ ∂X − �k−3 j=1 ( hP⊥) (j) ψ (k−j) ⊥ f (k−j) + P (j−l) � � (l−4) ⊥ (∆x, y Ψ2) + P (j−l) ⊥ + P (j−l) ⊥ ��∂Ψ1 � � (l−3) ∂y ��∂Ψ2 � � (l−3) ∂y ∂ ∂Y ∂ ∂Y � � � (k−j) g f (k−j) g (k−j) E (j) ψ (k−j) ⊥ , (3.25) � − 1 � 2 ⎛ ⎝ ⎛ ⎜ ⎝ T (j−4) 11 T (j−4) 21 − A (j−3) 21 T (j−4) 12 T (j−4) 22 ∂ ∂X ⎞ ⎠ + ⎛ ⎝ h(j) 11 h (j) 12 h (j) 21 h (j) 22 ⎞ 0 − A (j−3) 12 − B(j−3) 21 ∂ ∂Y ⎠ + ∂ ∂X 0 − B(j−3) 12 Following what we have seen through order 4, assume from previous orders that ⎛ ⎞ ⎝ f(j) g (j) ⎠ for j = 0,1,... ,k − 3, E (j) and ψ (j) ⊥ ∂ ∂Y for j = 0,1,... ,k − 1, are already determined. Then, we can solve (3.24) and (3.25) for f (k−2) , g (k−2) , ψ (k) ⊥ , and E(k) . From (3.24) we obtain: � ⎛ E (k) = �k−1 j=3 ⎝ f(0) − 1 2 g (0) �k−3 j=2 ⎞ ⎠ , � ⎛ ⎛ � Hj − E (j)� ⎝ f(k−j) g (k−j) ⎝ f(0) g (0) ⎞ ⎠ , ⎛ ⎝ ⎞ � ⎠ H + � ⎛ ⎝ f(0) 〈 Ψ(j−2) 1 , ∆X, Y ψ (k−j) ⊥ 〉el 〈 Ψ (j−2) 2 , ∆X, Y ψ (k−j) ⊥ 〉el 16 g (0) ⎞ ⎞ � ⎠ ⎠ , Hk H ⎛ ⎞ ⎟ ⎠, ⎝ f(0) g (0) ⎞ � ⎠ H (3.26)

and ⎛ ⎝ f(k−2) g (k−2) ⎞ ⎠ = − From (3.25) we get ψ (k) ⊥ = + Hk � ( h P⊥) (0)� −1 + + + + k� j=4 k� j=3 k� j=3 �k−3 j=2 j� l=4 j� l=3 j� l=3 r 1 � 2 � � � H2 − E (2)� −1 ⎛ ⎝ f(0) ⎡ ⎣ g (0) �k−3 j=2 ⎞ r Q⊥ ⎠ − 1 2 1 2 ⎡ ⎣ �k−3 j=2 �k−1 j=3 ⎛ ⎝ � (j−2) P ⊥ ∆X, Y ψ (k−j) � ⊥ P (j−l) � � (l−4) ⊥ (∆x, y Ψ1) P (j−l) ⊥ P (j−l) ⊥ E (j) ψ (k−j) ⊥ ��∂Ψ1 � � (l−3) ∂x ��∂Ψ2 � � (l−3) ⎤ ∂x ⎛ � Hj − E (j)� ⎝ f(k−j) g (k−j) 〈 Ψ(j−2) 1 , ∆X, Y ψ (k−j) ⊥ 〉el 〈 Ψ (j−2) 2 , ∆X, Y ψ (k−j) ⊥ 〉el ∂ ∂X ∂ ∂X − �k−3 j=1 ⎞ ⎠ ⎞ ⎤ (hP⊥) (j) ψ (k−j) ⊥ ⎠ ⎦ . (3.27) f (k−j) + P (j−l) � � (l−4) ⊥ (∆x, y Ψ2) + P (j−l) ⊥ + P (j−l) ⊥ � �∂Ψ1 � � (l−3) ∂y � �∂Ψ2 � � (l−3) ∂y ∂ ∂Y ∂ ∂Y � � � (k−j) g f (k−j) g (k−j) ⎦ . (3.28) So we can proceed in this manner to obtain Ψ(ǫ) and E(ǫ) up to any order in ǫ. 4 Properties of the Leading Order Hamiltonian We adopt the following notation throughout: � � 1 0 1. We let I2 = 0 1 � � A ⊗ I2, is the operator on H given by A 0 0 A . . If A is an operator on the Hilbert space L 2 (IR 2 , dX dY ), then 2. If D(A) is the domain of the operator A on the Hilbert space L 2 (IR 2 , dX dY ), then D(A ⊗ I2) = D(A) ⊕ D(A) ⊂ H. In what follows, we prove various needed properties for the expansion to all orders. Let H2 = − 1 2 ∆X, ⎛ a + b ⎜ Y ⊗ I2 + ⎜ 2 ⎝ X2 a − b + Y 2 2 bX Y a − b bX Y 2 X2 a + b + Y 2 2 ⎞ ⎟ ⎠ . 17

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