Understanding the Impossibility of a Bound State of Two Identical Nucleons

[CarlB]

Yes it does because the PEP is defined like that and the "no two fermions in the same quantum state" follows from that. It is that simple and this also explains why your examples were not correct. The wavefunction you used is NOT a fermion wavefunction by above definition.

We may or may not be in agreement, depending on what the definition of "that" is. Why don't you spell out what you think the PEP is, and what follows from it, and try to avoid language that can have multiple interpretations.

 

[Hurkyl]

May I jump in? I think y'all are talking past each other: you are each giving a different meaning to "in the same state".

I think you will all agree that if there is only one quantum state, then the antisymmetrization of that state is zero, and thus a pair of fermions cannot exist in the antisymmetrization of that state.

If we have two quantum states (say, |0> and |1>), then we can have a pair of fermions that exists in the antisymmetrization of those states. For example,

|01> - |10>.

To speak of the state of a single particle in a two-particle system, you must take the partial trace to eliminate the other particle. For both particles, the result is the same statistical mixture of |0> and |1>. Thus, the individual fermions are in identical (mixed) states.


Maybe this phrasing will please everyone?

If you have N indistinguishable fermions, then they are all in an the same mixture of M distinct one-particle states, with M no smaller than N.

 

[CarlB]

Hurkyl,

I'm in agreement with what you're saying. But my motivation for grinding this into the ground gets back, of course, to my attempting to understand density matrices as the foundation for QM. Here's the problem from the DM point of view.

If one characterizes fermion states by \psi(r_1,r_2) = -\psi(r_2,r_1), then the statement that no two fermions exist in the same state follows in that one has that \psi(r,r) = 0. The problem for the density operator formalism is that this way of characterizing fermion states is written in spinor formalism, not density matrix formalism.

If I could instead characterize fermion states by "no two fermions can occupy the same state", then the definition is written in terms of quantum states, rather than spinors, and one can immediately apply it to the density matrix / density operator formalism.

The problem is that I don't see how to get from "no two fermions can occupy the same state" to \psi(r_1,r_2) = -\psi(r_2,r_1) without making a bunch of other assumptions. One of those assumptions would be linear superposition, which is already iffy in the density operator formalism since linear superposition is normally applied to spinors instead of density matrices. Another assumption is that the wave function must treat all identical particles equivalently. This assumption, CAN be written into the density matrix formalism.

I do think that there is a way to elegantly obtain the PEP / fermion wave function restriction completely inside the density operator formalism, and that is by assuming that the PEP follows as a force of constraint. (And when spin is generalized to bigger Clifford algebras, the constraint force implied could be used to make preons clump together.) To do it, we enforce the restriction by assigning a very high potential energy to violations of the PEP. This turns out to be very natural for fermions described as density operators.

The difference between bosons and fermions then becomes whether or not they have the charge for the force associated with that potential energy. Fermions have the charge and are forced away from each other. Bosons don't. By the way, I've seen a paper claiming that the quantum zeno effect, when applied to bosons, causes them to act like fermions, without the quantum zeno effect:
http://arxiv.org/abs/quant-ph/0408097

Here are some interesting papers. They demonstrate that the pros get into fights about this in the peer reviewed literature that have to do with how the PEP should be written:

Phys. Rev. A 67, 042102 (2003)
http://www.arxiv.org/abs/quant-ph/0207017

http://www.arxiv.org/abs/quant-ph/0304088

Phys. Rev. A 68, 046102 (2003)
http://www.arxiv.org/abs/quant-ph/0402118

 

[marlon]

May I jump in? I think y'all are talking past each other: you are each giving a different meaning to "in the same state".

I think you will all agree that if there is only one quantum state, then the antisymmetrization of that state is zero, and thus a pair of fermions cannot exist in the antisymmetrization of that state.

If we have two quantum states (say, |0> and |1>), then we can have a pair of fermions that exists in the antisymmetrization of those states. For example,

|01> - |10>.

CLEAR.

To speak of the state of a single particle in a two-particle system, you must take the partial trace to eliminate the other particle. For both particles, the result is the same statistical mixture of |0> and |1>. Thus, the individual fermions are in identical (mixed) states.

What you mean is that if you calculate the trace "for particle one", you will get a mixed state of |0> and |1> and when you calculate the trace "for particle two", you will get that SAME mixed state of |0> and |1>. Got that, but are the coefficients of |0> and |1> also the same after taking the two traces ?

Ok, but "being in the same state" means for me "having the exact same quantumnumbers". No two fermions can have the exact same quantumnumbers.

marlon

 

[dextercioby]

The problem is that I don't see how to get from "no two fermions can occupy the same state" to \psi(r_1,r_2) = -\psi(r_2,r_1) without making a bunch of other assumptions. One of those assumptions would be linear superposition, which is already iffy in the density operator formalism since linear superposition is normally applied to spinors instead of density matrices. Another assumption is that the wave function must treat all identical particles equivalently. This assumption, CAN be written into the density matrix formalism.

Why would anyone try to "get from from "no two fermions can occupy the same state" to \psi(r_1,r_2) = -\psi(r_2,r_1)" ?

The way i see it and the way it's taught in school is that the symmetrization postulate implies \psi(r_1,r_2) = -\psi(r_2,r_1) for fermions in the same spin state and that automatically means "no two fermions can occupy the same (quantum) state".

Daniel.

P.S. Since i haven't been taught that, can one give a mathematical formulation of the symmetrization postulate in the density operator formalism ??

 

[Hurkyl]

What you mean is that if you calculate the trace "for particle one", you will get a mixed state of |0> and |1> and when you calculate the trace "for particle two", you will get that SAME mixed state of |0> and |1>. Got that, but are the coefficients of |0> and |1> also the same after taking the two traces ?

The result is "50% |0> and 50% |1>": it's a statistical mixture, not a pure state. It cannot be written as "a|0> + b|1>".