vanesch
Staff Emeritus
Science Advisor
Gold Member
- 5,102
- 20
I would like to come back to a very fundamental reason why
the detector correlation in the case of single-photon states
must be zero, and this even INDEPENDENTLY of any consideration
of photon number operators, finite-size detectors, hamiltonians
etc...
It is in fact a refinement of the original argument I put here,
before I was drawn in detailled considerations of the photon
detection process.
It is the following: we have a 1-photon state inpinging on a
beam splitter, which, if we replace it by a full mirror, gives
rise to the incoming state |R> and if we remove it, gives
rise to the state |T> ; with the beam splitter in place, the
ingoing state is 1/sqrt(2) {|R> + |T>}
Up to now, there is no approximation, no coarse graining.
You can replace |T> and |R> with very complicated expressions
describing explicitly the beams. It is just their symbolic expression.
Next, we consider the entire setup, with the two detectors and the
coincidence counter, as ONE SINGLE MEASUREMENT SYSTEM. Out of it can
come 2 results: 0 or 1. 0 means that there was no coincident clicks. 1
means that there was a coincident click detected. It doesn't really
matter exactly how everything is wired up.
As this is an observable measurement, general quantum theory (of which,
as I repeated often, QED is only a specific application) dictates that
there is a hermitean operator that corresponds to this measurement, and
that it is an operator with 2 eigenvalues: 0 and 1.
THIS is the actual content which is modeled by the normal ordering of
the 2 number operators, but for the argument here, there is no need
to make that link.
You can, if you want, analyse in detail, the operator that will give us
the correlation is the above operator, which we will call C.
In Mandelian QO this is :nv1 nv2: but it doesn't matter: if you want to
construct it yourself, based upon interaction hamiltonians, finite size
detectors, finite size beams etc...be my guest. Write down an operator
expression 250 pages long if you want.
At the end of the day, you will have to come up with an operator, that
corresponds to the correlation measurement, and that measurement, for each
individual measurement in a single time frame,
has 2 possible answers: 0 and 1. (no coincidence,or coincidence).
What we want to calculate, is 1/2 (<R|+<T|) C (|R> + |T> )
What do we know about C ?
If we remove the beamsplitter, we have a pure |T> state.
And for |T> we know FOR SURE that we will not see any coincidence.
Indeed, nothing is incident on the R detector.
This means, by general quantum theory (if we know an outcome for sure),
that |T> is an eigenstate of C with eigenvalue 0.
If we put in place a full mirror, we have a pure |R> state.
Again, we know for sure that we will not see any coincidence.
This time, nothing is incident on the T detector.
So |R> is an eigenstate of C, also with eigenvalue 0.
From this it follows that 1/sqrt(2) (|R>+|T>) is also an eigenstate with
eigenvalue 0 of C (a linear combination of eigenvectors with same
eigenvalue).
1/2 (<R|+<T|) C (|R> + |T> ) = 0
This follows purely from general quantum theory, its principal
point being the superposition principle and the definition, in
quantum theory, of what is an operator related to a measurement.
Now, exercise:
If you misunderstood the above explanation, you could think that
this is an absurd result that means that if you have two intensive
beams on 2 detectors, you could never have a coincidence, which is
clearly not true. The answer is that the above reasoning
(put in a mirror, remove the splitter) only works for 1-photon
states. So here is the exercise:
Why does this only work if the incoming states on the beam splitter
are 1-photon states ?
If you understood this, and found the answer, you will have gained
a great insight in quantum theory in general, and in quantum optics
in particular :-)
cheers,
Patrick.
the detector correlation in the case of single-photon states
must be zero, and this even INDEPENDENTLY of any consideration
of photon number operators, finite-size detectors, hamiltonians
etc...
It is in fact a refinement of the original argument I put here,
before I was drawn in detailled considerations of the photon
detection process.
It is the following: we have a 1-photon state inpinging on a
beam splitter, which, if we replace it by a full mirror, gives
rise to the incoming state |R> and if we remove it, gives
rise to the state |T> ; with the beam splitter in place, the
ingoing state is 1/sqrt(2) {|R> + |T>}
Up to now, there is no approximation, no coarse graining.
You can replace |T> and |R> with very complicated expressions
describing explicitly the beams. It is just their symbolic expression.
Next, we consider the entire setup, with the two detectors and the
coincidence counter, as ONE SINGLE MEASUREMENT SYSTEM. Out of it can
come 2 results: 0 or 1. 0 means that there was no coincident clicks. 1
means that there was a coincident click detected. It doesn't really
matter exactly how everything is wired up.
As this is an observable measurement, general quantum theory (of which,
as I repeated often, QED is only a specific application) dictates that
there is a hermitean operator that corresponds to this measurement, and
that it is an operator with 2 eigenvalues: 0 and 1.
THIS is the actual content which is modeled by the normal ordering of
the 2 number operators, but for the argument here, there is no need
to make that link.
You can, if you want, analyse in detail, the operator that will give us
the correlation is the above operator, which we will call C.
In Mandelian QO this is :nv1 nv2: but it doesn't matter: if you want to
construct it yourself, based upon interaction hamiltonians, finite size
detectors, finite size beams etc...be my guest. Write down an operator
expression 250 pages long if you want.
At the end of the day, you will have to come up with an operator, that
corresponds to the correlation measurement, and that measurement, for each
individual measurement in a single time frame,
has 2 possible answers: 0 and 1. (no coincidence,or coincidence).
What we want to calculate, is 1/2 (<R|+<T|) C (|R> + |T> )
What do we know about C ?
If we remove the beamsplitter, we have a pure |T> state.
And for |T> we know FOR SURE that we will not see any coincidence.
Indeed, nothing is incident on the R detector.
This means, by general quantum theory (if we know an outcome for sure),
that |T> is an eigenstate of C with eigenvalue 0.
If we put in place a full mirror, we have a pure |R> state.
Again, we know for sure that we will not see any coincidence.
This time, nothing is incident on the T detector.
So |R> is an eigenstate of C, also with eigenvalue 0.
From this it follows that 1/sqrt(2) (|R>+|T>) is also an eigenstate with
eigenvalue 0 of C (a linear combination of eigenvectors with same
eigenvalue).
1/2 (<R|+<T|) C (|R> + |T> ) = 0
This follows purely from general quantum theory, its principal
point being the superposition principle and the definition, in
quantum theory, of what is an operator related to a measurement.
Now, exercise:
If you misunderstood the above explanation, you could think that
this is an absurd result that means that if you have two intensive
beams on 2 detectors, you could never have a coincidence, which is
clearly not true. The answer is that the above reasoning
(put in a mirror, remove the splitter) only works for 1-photon
states. So here is the exercise:
Why does this only work if the incoming states on the beam splitter
are 1-photon states ?
If you understood this, and found the answer, you will have gained
a great insight in quantum theory in general, and in quantum optics
in particular :-)
cheers,
Patrick.