| MadSci Network: Physics |
Dominic,
VERY interesting question! The simple answer is that no, the Pauli exclusion principle requires no special intermediate bosons in order to work. To explain why this is so takes a bit of background information, and perhaps restating some of what we already know in a new way, so please bear with me...
Let's start by looking at an electromagnetic interaction between two electrons, e1 and e2. The first order Feynman diagrams for this interaction are:
e1 -------------------- e1' e1 ------------ / e1'
\ \ \ /
/ / \ /
gamma\ and gamma\ X
/ / / \
\ \ / \
e2 -------------------- e2' e2 ------------ \ e2'
What does all this mean? Before the interaction, the electrons fill two
quantum states described by the wavefunction phi(e1,e2). After the
interaction, the two electrons are in different quantum states, having had
their momenta, spins, etc. changed by the interaction between them. The
final two quantum states are described by the wavefuntion phi(e1',e2').
To find the probability of this interaction taking place requires
us to calculate the integral:
/ *
M = | phi (e1,e2) H phi(e1',e2')
/
where phi(e1,e2) and phi(e1',e2') represent the probability functions of the
initial and final states, and H is the Hamiltonian, or the energy function
of the interaction between the two particles, in this case the exchange of a
photon (gamma) described by the two Feynman diagrams.
There are three important points to understand here. The first is that for this interaction to occur, both the initial and final wavefunctions must be non-zero,or in other words, it must actually be possible for the two particles to exist in the initial and final states. Otherwise, the integral (and thus, the probablity of the interaction) is zero. The second point is that the Hamiltonian (represented by the photon exchange diagrams) serves to produce changesin the final states compared to the initial ones. One final point is that there are two diagrams describing the interaction, in one of which the two electrons "change places". It is important not to forget that since all electrons are exactly the same, it is not possible to tell whether they have changed places or not.
With this background in mind, let's go back to the neutron star, and take a look at two of the neutrons being held apart by the Pauli exclusion principle. The initial state is phi(n1,n2) and the final state is phi(n1',n2'). We can draw a pair of "pseudo"-Feynman diagrams:
n1 ------- n1' n1 n1'
\ /
\ /
and X
/ \
/ \
n2 ------- n2' n2 n2'
which lead to the following integral:
/ *
M = | phi (n1,n2) phi(n1',n2')
/
Note that there is no boson-mediated interaction going on here, so the final
probability function is identical to the initial one. This leads to a rather
boring integral, with answer "1". Now for fun's sake, let's imagine another
interaction where one of the two neutrons joins the other, so that they end
up in the same quantum state:
n1 n1
\
\
\
\
n2 ---- n2
The resulting integral looks something like:
/ *
M = | phi (n1,n2) phi(n2',n2')
/
But as you already know, the probability of two fermions in the same quantum
state, phi(n2,n2), is equal to zero. Therefore, the probability for this
transition to take place is zero, plain and simple. The neutrons are always
held apart in their own separate quantum states by the simple fact that it
is forbidden for them ever to be in one together.
I hope this makes things a little clearer. For more information on Feynman diagrams and particle interactions, I recommend the book Introduction to Elementary Particles by David Griffiths. If you have further questions, please feel free to email me at silver@physto.se
Regards,
Sam Silverstein
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