By P. Dirac
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Extra resources for Development of Quantum Theory: J. Robert Oppenheimer Memorial Prize Acceptance Speech
Ni , . . 11) |. . , ni − 1, . . ni . Here, we have introduced the factor ni , since, for ni = 0, the Kronecker delta δni +1,ni = 0 always gives zero. The factor ni also ensures that the right-hand side cannot become equal to the state |. . , ni − 1, . . = |. . , −1, . . To summarize, the eﬀects of the creation and annihilation operators are P a†i |. . , ni , . . = (1 − ni )(−1) P ai |. . , ni , . . = ni (−1) j
The potential energy in ﬁrst-order perturbation theory4 reads: E (1) = e2 2V k,k ,q,σ,σ 4π φ0 | a†k+q,σ a†k −q,σ ak σ akσ |φ0 . 6) The prime on the summation sign indicates that the term q = 0 is excluded. The only contribution for which every annihilation operator is compensated by a creation operator is proportional to δσσ δk ,k+q a†k+qσ a†kσ ak+qσ akσ , thus: 4π e2 nk+q,σ nk,σ E (1) = − 2V q2 k,q,σ =− =− 2 e 2V σ 4π Θ(kF − |q + k|)Θ(kF − k) q2 k,q 2 4πe V (2π)6 d3 k Θ(kF − k) d3 k 1 2 Θ(kF |k − k | − k ) .
Take the continuum limit k ,σ = R 3 3 0 2V d k/(2π) and calculate S (q) explicitly. Hint: Consider the cases q = 0 and q = 0 separately. 2 Prove the validity of the following relations, which have been used in the evaluation of the energy shift ∆ (k) of the electron gas, Eq. 21): Z 2e2 d3 k 1 kF F (k/kF ) , Θ(k − k ) = − a) − 4πe2 F (2π)3 | k − k |2 π with F (x) = b) ˛ ˛ 1 1 − x2 ˛˛ 1 + x ˛˛ . + ln ˛ 2 4x 1 − x˛ ˛– ˛ » d3 k kF2 − k2 ˛˛ kF + k ˛˛ ln 1 + ˛ kF − k ˛ Θ(kF − k) (2π)3 2kkF „ «1/3 e2 3N 3 e 2 kF 9π N =− , =− 4 π 2a0 rs 4 2π E (1) = − e 2 kF V π Z where rs is a dimensionless number which characterizes the mean particle separation in units of the Bohr radius a0 = 2 /me2 .