Operators in denominators

1. Jun 18, 2010

rgoerke

I am led to believe (because it is in a paper I am reading) that
$$\frac{1}{H - z} \left|\phi\rangle = \frac{1}{E - z}\left|\phi\rangle$$
where $$H$$ is the hamiltonian, $$\left|\phi\rangle$$ is an energy eigenstate with energy $$E$$, and $$z$$ is a complex variable.
In attempting to understand this expression, I have realized I do not know what is meant by
$$\frac{1}{A}\left|\phi\rangle$$
for some operator $$A$$. Is this the same thing as the inverse of A?

Thanks.

2. Jun 18, 2010

strangerep

Welcome to the wonderful world of the spectral theorem(s)! :-)

The basic idea is this: let A is a self-adjoint operator on a Hilbert space,
and suppose it has a (continuous) spectrum from 0 to infinity.
Then it is possible to find a basis in the Hilbert space where each basis
state corresponds to a particular value in the spectrum, and in fact
a complex-analytic function of A, written "f(A)", can be expressed as

$$f(A) ~\leftrightarrow~ \int_0^\infty\! dk\; f(k) \; |k\rangle\langle k|$$

where each |k> is one of the eigenstates of A, with eigenvalue k.
(The set of all such k is called the "spectrum" of A.)

In your case, we're dealing with the Hamiltonian operator H, and your
particular complex-analytic function is f(H) := 1/(H-z), so we can
express it as

$$\frac{1}{H - z} ~\leftrightarrow~ \int_0^\infty\! dk\; \frac{ |k\rangle\langle k|}{k - z}$$

where now each |k>is an eigenstate of H with eigenvalue k, and these states are
"normalized" according to $\langle k|k'\rangle = \delta(k - k')$.

Now apply this to your $|\phi\rangle$, but first let's rewrite $|\phi\rangle$
as $|E\rangle$, since this is a more helpful notation. This is ok because
$|\phi\rangle$ is (by definition) the eigenstate of H with eigenvalue E. We get:

$$\int_0^\infty\! dk\; \frac{ |k\rangle\langle k|}{k - z} ~ |E\rangle ~=~ \int_0^\infty\! dk\; \frac{ |k\rangle}{k - z} ~ \delta(k - E) ~=~ \frac{1}{E - z} \; |E\rangle$$

To understand the spectral theorem(s) in more detail, it's probably best
to start with the finite-dimensional matrix versions in linear algebra
textbooks, and then progress to the functional-analytic versions,
then to the versions for rigged Hilbert space which is what I've been
using above. Or, for a more QM-flavored overview, try Ballentine ch1,
in particular sections 1.3 and 1.4.

HTH.

Last edited: Jun 18, 2010
3. Jun 19, 2010

Hurkyl

Staff Emeritus
While it's good to know there's a generality, this particular case can be handled by lower-powered means -- so long as z ranges over the non-eigenvalues* of H, the operator (H-z) has a multiplicative inverse. 1/(H-z) is the unique operator that you can substitute for T in
$$T \left( (H-z) | \psi \rangle \right) = | \psi \rangle$$​
which makes the equation hold for all kets.

If you restrict to the space generated by $|\phi\rangle$, then z merely has to avoid the value E. (Because, on this subspace, E is the only eigenvalue of H)

*: Really, I mean the points that aren't in the spectrum. But I think your H has a discrete spectrum... and so they are all eigenvalues too.

4. Jun 20, 2010

rgoerke

Thanks to both of you! That's a big help.