05.03.2015 Views

1 Introduction 2 Resolvents and Green's Functions

1 Introduction 2 Resolvents and Green's Functions

1 Introduction 2 Resolvents and Green's Functions

SHOW MORE
SHOW LESS

You also want an ePaper? Increase the reach of your titles

YUMPU automatically turns print PDFs into web optimized ePapers that Google loves.

defined according to:<br />

‖f(Q)‖op ≡ sup ‖f(Q)φ‖ (6)<br />

‖φ‖=1<br />

= sup |f(q)| (7)<br />

q∈Σ(Q)<br />

< ∞, if f(q) is bounded. (8)<br />

Now define an operator-valued function G(z), called the resolvent of Q,<br />

of complex variable z, for all z not in Σ(Q) by: 1<br />

G(z) = 1 , z /∈ Σ(Q). (9)<br />

Q − z<br />

For any such z the operator G(z) is bounded, <strong>and</strong> we have<br />

‖G(z)‖op = sup<br />

k<br />

1<br />

|q k − z| . (10)<br />

The resolvent satisfies the identities<br />

G(z) − G(z 0 ) = ∑ ( 1<br />

k<br />

q k − z − 1 )<br />

|k〉〈k|<br />

q k − z 0<br />

= ∑ z − z 0<br />

k<br />

(q k − z)(q k − z 0 ) |k〉〈k|<br />

z − z 0<br />

=<br />

(Q − z)(Q − z 0 )<br />

= (z − z 0 )G(z)G(z 0 ), (11)<br />

<strong>and</strong><br />

G(z) =<br />

G(z 0 )<br />

1 + (z 0 − z)G(z 0 ) , (12)<br />

Q = z + 1/G(z). (13)<br />

If the eigenvectors are written as functions of x ∈ R 3 (assuming that<br />

the Hilbert space is L 2 (R 3 )), we can represent the resolvent as an integral<br />

transform on the wave functions. That is, with<br />

G(z) = 1<br />

Q − z = ∑ k<br />

|k〉〈k|<br />

q k − z , (14)<br />

1 The resolvent is sometimes defined with the opposite sign. We shall see eventually<br />

that G(z) also finds motivation in terms of Cauchy’s integral formula.<br />

2

Hooray! Your file is uploaded and ready to be published.

Saved successfully!

Ooh no, something went wrong!