Calculus for delta function based on wave function

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SUMMARY

The discussion focuses on calculating the integrals <ψ|δ(r12)|ψ> and <ψ|δ(r1)|ψ> for a two-electron wavefunction defined as ψ(r1,r2,r12) = exp(-Ar1-Br2-Cr12), where r12 = |r1-r2|. The user, sandf, correctly sets up the integrals but encounters confusion in the final steps. A key insight provided is that when r1 is set to zero, r12 simplifies to |r2|, which is crucial for evaluating the integrals accurately. The integrals are deemed straightforward once this adjustment is made.

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sandf
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Homework Statement



Two-electron Wavefunction: ψ(r1,r2,r12) = exp(-Ar1-Br2-Cr12), r12 = |r1-r2|
A, B, and C are coefficients

Calculate <ψ|δ(r12)|ψ> and <ψ|δ(r1)|ψ>

Homework Equations



NO

The Attempt at a Solution



<ψ|δ(r12)|ψ>

= ∫∫dv1dv2ψ2(r1,r2,r12)δ(r12)

=4π∫dr2ψ2(r2,r2,0)

<ψ|δ(r1)|ψ>

= ∫∫dv1dv2ψ2(r1,r2,r12)δ(r1)

=4π∫dr2ψ2(0,r2,-r2)

Could you tell me where I am wrong?

Best regards.
sandf
 
Last edited:
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sandf said:

Homework Statement



Two-electron Wavefunction: ψ(r1,r2,r12) = exp(-Ar1-Br2-Cr12), r12 = |r1-r2|
A, B, and C are coefficients

Calculate <ψ|δ(r12)|ψ> and <ψ|δ(r1)|ψ>

Homework Equations



NO

The Attempt at a Solution



<ψ|δ(r12)|ψ>

= ∫∫dv1dv2ψ2(r1,r2,r12)δ(r12)

=4π∫dr2ψ2(r2,r2,0)

<ψ|δ(r1)|ψ>

= ∫∫dv1dv2ψ2(r1,r2,r12)δ(r1)

=4π∫dr2ψ2(0,r2,-r2)

Could you tell me where I am wrong?

Best regards.
sandf
To be precise, in the last step, one should use that ##r_{12} = |r_1-r_2|= r_2 ## upon setting ##r_1=0##. But other than that, everything is good, one simply has to calculate the integrals which are easy.
 

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