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1 Introduction 2 Resolvents and Green's Functions

1 Introduction 2 Resolvents and Green's Functions

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z<br />

.<br />

. . . . . . .<br />

q 4<br />

C 4<br />

Figure 1: The complex z plane, with eigenvalues of Q indicated on the real<br />

axis. A contour is shown encircling one of the eigenvalues.<br />

That is,<br />

∫<br />

1<br />

G(z) dz = 1 ∫<br />

2πi C 4 2πi C 4<br />

∫<br />

= 1<br />

2πi<br />

∞ ∑<br />

k=1<br />

C 4<br />

|φ 4 〉〈φ 4 |<br />

q 4 − z<br />

|k〉〈k|<br />

dz (29)<br />

q k − z<br />

dz (30)<br />

=<br />

( z − q4<br />

|φ 4 〉〈φ 4 | lim z→q4 q 4 − z<br />

(31)<br />

= −|φ 4 〉〈φ 4 |. (32)<br />

|φ 4 〉〈φ 4 | = − 1 ∫<br />

G(z) dz. (33)<br />

2πi C 4<br />

The contour integral of G around an eigenvalue of Q gives the projection<br />

onto the one-dimensional subspace of the corresponding eigenvector of Q.<br />

Now suppose that the spectrum of Q is bounded below, i.e., there exists<br />

an α > −∞ such that q k > α, ∀k. Of particular interest is the Hamiltonian,<br />

which has this property. In this case, we may consider a contour which<br />

encircles all eigenvalues, as in Fig. 2: Then we have<br />

I = − 1 ∫<br />

G(z) dz, (34)<br />

2πi C ∞<br />

as may be proven by a limiting process, <strong>and</strong> noting that convergence is all<br />

right.<br />

According to Cauchy’s integral formula, we can express analytic functions<br />

of Q in terms of contour integrals: Let f(z) be a function which is analytic<br />

)<br />

5

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