Natural Science Forum / Physics / Research / March 2008

Thus spake Oz <Oz@farmeroz.port995.com>

I found the calculation I did, showing that qed predicts that different
photons may interfere, but different electrons may not. Dirac was
thinking, quite naturally, of the structure of one particle Hilbert
space in which the quantum superposition, |f>+|g>, appears in the
addition of states of one particle. It makes no sense to add the state
of one particle to the state of another, because they are strictly
described by states in different Hilbert spaces. However in quantum
electrodynamics we have to think of the wave function not in terms of
the probability for where a photon is, but rather the probability for
where a photon is annihilated (detected) and we get a different result.

As a concession to ASCII I shall ignore spin and simplify the formula as
much as possible, and just show the bones of the argument. The photon
field operator is,

       A(x) = |x> + <x|

where I am using ket notation for operators, so that |x> creates the
state |x>, and <x| annihilates the same state.

The photon wave function in the state |f> is

       <|A(x)|f> = <x|f>

so that |<x|f>|^2 is the probability for detection of a photon at x.

Now consider a two photon state |f;g> = |f>|g> + |f>|g>. For simplicity
I am assuming no entanglement, and that the photons are distinguishable
at some time in the past (they come from different sources) so that
<f|g>=0.

When one of the photons is detected the other remains as it was, but we
don't know which one is detected and which one remains. Then the final
state is either |f> or |g>, i.e. it is |f> + |g>, so the probability for
detection at x is given by

       (<f|+<g|) A(x) |f;g> = (<f|+<g|)(|x> + <x|)|f;g>)

                            = (<f|+<g|)(<x|f>|g> + <x|g>|f>)

where I have lost the inner product between states of different numbers
of particles (it is zero). Using <f|g> =0 and <f|f>=<g|g>=1 this comes
down to

       <x|f> + <x|g>

Which shows the superposition between states of different photons.

+++++

This argument does not work for electrons. I shall ignore spin. For
electrons the field operator is

       Psi(x) = |x> + <x^|

where  |x> creates an electron and <x^| annihilates a positron. The
thing we detect is the current

       j(x) =  :Psi+(x)Psi(x): = :(|x^> + <x|)(|x> + <x^|):

Colons mean normally ordering, so that <x||x> is replaced by  |x><x|,
otherwise the theory is divergent before you start.

Ignoring positrons and pair creation j boils down to

       j(x) = |x><x|

which is the position operator (more strictly j^0 is charge, but here I
have lost the spin indices).

Now when an electron is detected both particles remain in the final
state, so the probability of detection at x is given by

       <f;g| j(x) |f;g> = <f;g| |x><x| |g;f>

                        = (<f|x><g| - <g|x><f|)(<x|g>|f> - <x|f>|g>)

Again  the <f|g>=0 and <f|f> = <y|y> =1 so this reduces to

                        = |<f|x>|^2  + |<g|x>|^2

So in this case we do not have interference.

Regards