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

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

## Note that if we let ( ˜

Note that if we let ( ˜ X, ˜ Y ) = (a 1/4 X, a 1/4 Y ) and ˜ b = b H2 = √ a �� − 1 2 ∆ ˜ X, ˜Y + ˜ b + 1 2 ( ˜ X 2 + ˜ Y 2 ) ⎛ ⎜ ⎝ 1 2 , then a � ⊗ I2 � ˜X 2 − ˜ Y 2 � ˜X ˜ Y − 1 2 ˜X ˜ Y � ˜X 2 − ˜ Y 2 � We now use the Kato-Rellich Theorem [18] to prove self-adjointness of H2. ⎞ ⎤ ⎟ ⎥ ⎠ ⎦ . (4.1) Theorem 4.1. If a > b > 0, then H2 is self-adjoint on DHO ⊕ DHO , where DHO is the usual Harmonic oscillator domain in L 2 (IR 2 ,dXdY ), and essentially self-adjoint on ˜ DHO ⊕ ˜ DHO , where ˜ DHO is any core for the usual Harmonic oscillator. Proof: and Define HHO = V ( ˜ b) = ˜ b � ⎛ ⎜ ⎝ − 1 2 ∆X, Y + 1 2 (X2 + Y 2 � ) 1 2 � X 2 − Y 2 � X Y − 1 2 X Y � X 2 − Y 2 � We prove that for 0 < ˜ b < 1, V ( ˜ b) is relatively bounded with respect to HH0, with relative bound ˜ b. The conclusion then follows from the Kato-Rellich theorem [18] and (4.1). ⊗ I2 For each fixed X and Y , the eigenvalues of V (b) are ± ˜b 2 (X2 + Y 2 ). It follows that � � � V ( ˜ � � b)v � ≤ e ˜ �� � b � 1 � 2 (X2 + Y 2 � � � ) ⊗ I2 v � , where v ∈ C 2 is any two component vector, and we use the usual Euclidean norm. This inequality implies the L2 (R2 , dX dY ; C2 ) = H norm estimate � � �V ( ˜ � � b)ψ � ≤ H ˜ � � � b � 1 � 2 (X2 + Y 2 � � � ) ⊗ I2 ψ � � , H where ψ(X, Y ) ∈ H is a two-component vector-valued function. We now show that �V ( ˜b)ψ �H ≤ ˜ �� � b � 1 � 2 (X2 + Y 2 � � � ) ⊗ I2 ψ � � H � e ⎞ ⎟ ⎠. ≤ ˜ b �HHO ψ �H + ˜ b �ψ �H. (4.2) for all ψ ∈ DHO ⊕DHO . We have already shown the first inequality. The hard part is the second estimate, which follows from � � �� X 2 + Y 2 � � � ⊗ I2 ψ � ≤ � � �� − ∆X,Y + X 2 + Y 2 � � � ⊗ I2 ψ � + 2 �ψ �. 18

This easily follows from � � �� X 2 + Y 2 � � � ⊗ I2 ψ �2 � �� ≤ � − ∆X,Y + X 2 + Y 2 � � � ⊗ I2 ψ �2 2 + 4 �ψ � . (4.3) Rather than proving this directly, let us first prove a simpler relative bound estimate for the operators on L2 � (IR,dx). We show that for φ ∈ D − ∂2 ∂x2 + x2 � , � �x 2 φ � � 2 ≤ � � � � � − ∂2 � � � + x2 φ� ∂x2 � To prove this, let p = −i ∂ ∂x , and calculate the commutators We have [x, p ] = i and [x, p 2 ] = 2ip. � � x 2 φ � � 2 = 〈φ, x 4 φ 〉 = 〈φ, � (p 2 + x 2 ) 2 − x 2 p 2 − p 2 x 2 − p 4� φ 〉 2 + 2 �φ� 2 . (4.4) ≤ � � � p 2 + x 2� φ � � 2 − 〈φ, � x 2 p 2 + p 2 x 2� φ 〉. (4.5) In this last expression, we use the commutators above to write 〈φ, � x 2 p 2 + p 2 x 2� φ 〉 = 〈φ, � xp 2 x + x[x, p 2 ] + xp 2 x + [p 2 , x]x � φ 〉 = 2 〈φ, xp 2 xφ〉 + 2i〈φ, (xp − px)φ〉 = 2 〈φ, xp 2 xφ〉 − 2 〈φ, φ 〉. In this last expression, the first inner product is the expectation of a positive operator (since xp 2 x has the form A ∗ A with A = px). Using this and (4.5), we see that and (4.4) is proved. � � x 2 φ � � 2 ≤ � � � p 2 + x 2� φ � � 2 + 2 �φ� 2 , Now we simply mimic the proof of (4.4) to prove (4.3). We write � � X 2 + Y 2� φ � 2 = � �� φ, − ∆X,Y + X 2 + Y 2�2 2 − ∆X,Y + ∆X,Y � X 2 + Y 2 � + � X 2 + Y 2� ∆X,Y � � φ The operator ∆ 2 X,Y is positive. The operator − ∆X Y 2 = − Y 2 ∆X is also positive since it equals A ∗ A with A = pXY . Similarly, − ∆Y X 2 = − X 2 ∆Y is positive. By the commutator tricks we 19

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