What is the meaning of \frac{1}{A}\left|\phi\rangle in quantum mechanics?

[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.

 

[strangerep]

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.

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

<br /> f(A) ~\leftrightarrow~ \int_0^\infty\! dk\; f(k) \; |k\rangle\langle k|<br />

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

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

where now each |k>is an eigenstate of H with eigenvalue k, and these states are
"normalized" according to \langle k|k&#039;\rangle = \delta(k - k&#039;).

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:

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

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.

 

[Hurkyl]

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.

 

[rgoerke]

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