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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 *changes*in 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 ---- n2The 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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