Significance of zero eigenvalues??

[Mike Fontenot]

If the state |a> of a system (at some instant) is an eigenvector, with
the non-zero eigenvalue "a", of some particular observable property
(whose operator corresponding to the measurement of that property is
denoted by A), then if A operates on |a>, the result of that operation
is the vector a|a>, which corresponds to the same state existing before
the measurement.

But instead, suppose that the system is in the state |b> corresponding
to an eigenvector of A whose eigenvalue is zero. The result of
operating with A on |b> is then the "zero vector", which (according to
Dirac) corresponds to "no state at all". So after the measurement, the
system apparently "has no state at all". What does that really mean?
Has the system ceased to exist (perhaps forever)? Or does the system
still exist, but simply has NO properties of any kind (and perhaps can
never again have any properties of any kind)?

--
Mike Fontenot

[Jim Heckman]

On 17-Jan-2012, Mike Fontenot <mlf...@comcast.net>
wrote in message <4F15B7D1...@comcast.net>:

You're confusing (1) the mathematical result of acting with a
linear operator A on a vector |a>, and (2) the physical effect of
measuring the observable corresponding to A on the state
represented by |a>.

If |a> is an eigenvector of A (which by the definition of an
eigenvector means |a> /= 0 -- see below) then any measurement of
the observable corresponding to A does *not* change the state of
the system, which continues to be represented by |a> (or an
unimportant *nonzero* scalar multiple thereof -- again, see below).
This is true regardless of the eigenvalue "a", and in particular it
doesn't matter if a = 0.

What Dirac means by the zero vector |x> = 0 not being able to
represent a physical state follows from the fact that it can't be
normalized: <x|x> = 0 whatever constant you multiply |x> by. Also,
for *every* operator B, B|x> = 0 = b*0 = b|x> for every constant b,
so |x> would be an eigenvector of every operator, with an
indeterminate eigenvalue (this is why the zero vector is excluded
by definition from being an eigenvector). Again in particular, an
eigenvector |0> with an eigenvalue of 0 *cannot* be the zero
vector: |0> /= 0.

BTW, I intend to reply to your "two-particle QM" posts, but may not
be able to get around to it right away.

--
Jim Heckman

[Mike Fontenot]

On 01/19/2012 07:13 AM, Jim Heckman wrote:
>
> You're confusing (1) the mathematical result of acting with a
> linear operator A on a vector |a>, and (2) the physical effect of
> measuring the observable corresponding to A on the state
> represented by |a>.
>

Thanks for your explanation, Jim.

--
Mike Fontenot

[Mike Fontenot]

On 01/19/2012 07:13 AM, Jim Heckman wrote:
>

> On 17-Jan-2012, Mike Fontenot<mlf...@comcast.net>
> wrote in message<4F15B7D1...@comcast.net>:
>

>> [...]

>>
>> But instead, suppose that the system is in the state |b> corresponding
>> to an eigenvector of A whose eigenvalue is zero. The result of
>> operating with A on |b> is then the "zero vector", which (according to
>> Dirac) corresponds to "no state at all". So after the measurement, the
>> system apparently "has no state at all".
>

> You're confusing (1) the mathematical result of acting with a
> linear operator A on a vector |a>, and (2) the physical effect of
> measuring the observable corresponding to A on the state
> represented by |a>.

> [...]

Yeah, that's a hole I've fallen into before ... in fact, you rescued me
from that hole in a previous thread I started, a few months back. The
fact that an operator is defined as being a thing that operates on a
vector, and produces a vector, seems to make me want to give more
significance than there actually is, to the association between the
state left by a measurement, and the vector which results from operating
with the operator on the state vector existing before the measurement
The role of the operator seems to be limited to just defining the
eigenvalues and eigenvalues corresponding to the measurement, with the
vector resulting from the operator operating on the initial state
playing a role only in the calculation of the mean value of the result
of many identical measurements. The vector resulting from the operator
operating on the initial vector seems to play no role at all in any
single measurement.

Thanks for pulling me out of that hole once again. I'll try to avoid
falling into it again.

--
Mike Fontenot