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Attractive contact energy

Example

Assume the Axiom of Choice and Countable Choice (The Axiom of Choice, The Axiom of Countable Choice (ACω)). Let A−1 be the delta interface operator of The one-dimensional delta interface operator with α=−1: D(A−1)={u∈H1(R): u∣(0,∞),u∣(−∞,0)∈H2, u′∗(0+)−u′∗(0−)=−u∗(0)}, and put ψ(x):=12e−∣x∣/2,∣ψ(x)∣2=12e−∣x∣. Then:

(a) ψ∈D(A−1), ∥ψ∥L2(R)=1, and A−1ψ=−14ψ; the negative eigenvalue −14 is the only negative eigenvalue of A−1 and its eigenspace is Cψ;

(b) more generally, for every α<0 the operator Aα has the single negative eigenvalue −α2/4 with normalized eigenvector −α/2 eα∣x∣/2, and for α≥0 there is none (The unique negative eigenvalue of the attractive delta interface operator);

(c) if α<0, with the scaling of part (c) of The unique negative eigenvalue of the attractive delta interface operator, if one assigns positive constants ℏ,m and sets H:=(ℏ2/2m)Aα and g:=ℏ2α/(2m), then the bound energy is −mg2/(2ℏ2); for the case α=−1 this is EH=−ℏ2/(8m).

Verification

Given: the delta interface operator A−1 with α=−1 and the function ψ(x)=2−1/2e−∣x∣/2.

Proof technique: direct verification of the interface conditions and the eigenvalue equation for the explicit function.

[F1] u∈D(Aα) means u∈H1(R) with H2 restrictions to the half-lines, the interface condition u′∗(0+)−u′∗(0−)=αu∗(0), and Aαu=−D2u on each half-line (The one-dimensional delta interface operator).

[F2] The representative of ψ is C∞ on each open half-line with ψ′′(x)=14ψ(x) there, and ψ′∗(0+)=−12ψ(0), ψ′∗(0−)=+12ψ(0); these are direct computations from the exponential (The complex exponential by its power series, The exponential function is smooth and (exp⁡)′=exp⁡), and the traces exist by Absolutely continuous representative of an L2 function with L2 weak derivative.

1.1F2algebra

[Normalisation] ∫R∣ψ∣2=12∫Re−∣x∣dx=∫0∞e−xdx=1.

2.1step 1.1F1F2algebra

[Membership and eigenvalue] By [F2], ψ is H1 with H2 half-line restrictions, and its derivative jump is ψ′∗(0+)−ψ′∗(0−)=−ψ∗(0), so ψ∈D(A−1) by [F1] with α=−1. On each half-line, A−1ψ=−D2ψ=−14ψ, so A−1ψ=−14ψ. This is the verification of (a) for the explicit function; uniqueness of the negative eigenvalue and of its eigenspace is part (b) of The unique negative eigenvalue of the attractive delta interface operator, specialised to α=−1: −α2/4=−1/4 and −α/2=1/2.

3.1step 2.1∎

[The general and scaled statements] Part (b) is exactly The unique negative eigenvalue of the attractive delta interface operator (b), and part (c) is its (c): for α<0, EH=−ℏ2α2/(8m)=−mg2/(2ℏ2) with g=ℏ2α/(2m), and at α=−1 this equals −ℏ2/(8m).

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