# Two Neutron Bound State

1. Jan 3, 2007

### Worzo

Hello,

Can someone explain to me exactly why a bound state of two identical nucleons is not possible? I have a feeling its something to do with antisymmetric wavefunction, but haven't found a satisfactory explanation in any book.

Cheers.

2. Jan 3, 2007

3. Jan 3, 2007

### CarlB

Not quite true, or at least not so simple. Such a bound state could exist if the position wave function has odd parity, or if the two nucleons have opposite spin (in the presence of a sufficient attractive mechanism). For example, see:

In analogy with the ground state of the hydrogen atom it is reasonable to assume that the ground state of the deuteron also has zero orbital angular momentum. Since the total angular momentum is one unit of (h-bar) it follows that this comes from the spin angular momentum which means that the proton and neutron spins are parallel. The implication is that two nucleons are not bound together if their spins are antiparallel and this is consistent with there being no proton-proton or neutron-neutron bound states. In the case of identical particles the bound parallel spin state is forbidden by the Pauli exclusion principle.
http://www.shef.ac.uk/physics/teaching/phy303/phy303-2.html [Broken]

The reason that PP and NN aren't stable is because of energy considerations. As an example of two fermions in a bound state, consider the two electrons that make a Cooper pair in superconductors. Here's an article that discusses even parity and odd parity Cooper pairs:
http://www.sciencedaily.com/releases/2004/11/041117003519.htm

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4. Jan 3, 2007

### Worzo

How did you work this out?

5. Jan 3, 2007

### CarlB

I didn't. It's an observed fact about deuterons. As far as the question of the original poser, it's off topic, but I included it because the same paragraph commented on how Pauli exclusion only applies to parallel spin states.

The Pauli exclusion principle is easier to understand if you think of it this way:

Two identical fermions cannot exist at the same point in space (i.e. they can't be in the same position eigenstate). However, they can exist at two different points in space. On the other hand, in general, quantum states are spread around, they are not concentrated at a single state.

To understand particles that are spread around in space, you have to look at linear superpositions. Then the Pauli exclusion principle says that two identical fermions can't be in the same linear superposition of position eigenstates.

The phrase "identical fermion" is interpreted to allow us to distinguish between two opposite spin particles. For example, two electrons, one with spin +1/2 in the x-direction, the other with spin -1/2 in the x-direction, are not identical, and the Pauli exclusion principle does not apply to them. Two electrons with opposite spin can share the same 1S wave state of an atom.

Two identical fermions that are in two incompatible spin states are a more interesting problem. For example, one electron with spin +1/2 in the x direction, and another electron with spin +1/2 in the y direction. Two such electrons are not exactly identical, but neither are they exactly distinguishable. Such a case requires that they not be in the same linear superposition of position eigenstates, but it is messier than the usual situation.

Last edited: Jan 3, 2007
6. Jan 3, 2007

### Severian

Then they wouldn't be in an identical state.

7. Jan 3, 2007

### Worzo

8. Jan 3, 2007

There are "dineutron halos" [N-N] that exist in some isotopes:http://en.wikipedia.org/wiki/Dineutron
For example Li-11--see this recent paper for literature review:
http://arxiv.org/PS_cache/physics/pdf/0509/0509265.pdf [Broken]

And from this paper it is concluded that 2 neutron halos [N-N] + [N-N] bound to an alpha [NNPP] provide a model to explain ground state of He-8. I give the paper conclusion below:
http://arxiv.org/PS_cache/nucl-th/pdf/0006/0006001.pdf [Broken]

"In this paper, we have investigated the ground state properties of 8He within the threecluster microscopic model. The three-cluster configuration 4He+2n+2n was used to simulate the dynamics of the eight-nucleon system. The model suggested describes reasonably well parameters of the ground state: binding energy, mass, proton and neutron root-mean-square radii. The analysis of the system shows, that valent neutrons move at a large distance from α-particle, forming a neutron halo in 8He."

A weak subthreshold resonance (pairing) for the neutron+neutron interaction[N-N] was estimated at 0.067 MeV--see Sofianos et al. 1997. J. Phys G:N Part. Phy 23:1619-1629. Perhaps this pairing energy of 0.067 MeV is what allows neutron halos [N-N] to exist for a short time in isotopes like Li-11 and He-8 ?--but this is not clear to me--perhaps someone can comment.

There are also studies on diproton halos [P-P]. The first published documentation of diproton halos was in 1995 by Schwab, et al., Z. Phys A350(1995): 283 for the isotope Boron-8. See this link for recent research on two-proton emisson studies:
http://gdr.ifj.edu.pl/Zakopane_articles/pfuetzner.pdf [Broken]

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9. Jan 3, 2007

### CarlB

Wrong. Two identical fermions in a bound state are most assuredly in the same state.

see Landau & Lifshiitz, Quantum Mechanics, Chapter IX, page 227:

Let us start by considering a system of only two particles. Because of the identity of the particles, the states of the system obtained from each other by merely interchanging the two particles must be completely equivalent physically. This means that, as a result of this interchange, the wave function of the system can change only by an unimportant phase factor.

Last edited: Jan 3, 2007
10. Jan 4, 2007

### Severian

Each field is represented by an operator. For fermions, the operators anticommute, eg. $$\psi_1 \psi_2 = -\psi_2 \psi_1$$, and if they are identical, ie. $$\psi_1 = \psi_2 = \psi$$, then $$\psi \psi = - \psi \psi =0$$ which is not a valid state.

Of course, as I said, they have to be identical, which means also having the same position space wavefunction. I don't see how you can get around this.

Unfortunately, I don't have Landau and Lifshiitz, but the quotation you give sounds like it is for bosons.

11. Jan 4, 2007

### Worzo

It's only true that these two neutrons in the bound state have to have an antisymmetric TOTAL wavefunction. I thought it would be analagous to two electrons with the same spatial wavefunction, but opposite spins.

12. Jan 4, 2007

### CarlB

The quote from L&L applies to both fermions and bosons.

In quantum mechanics, wave functions are not operators, they're just functions. For a two-particle state, one has only a single wave function, $$\psi(r_1,r_2)$$ where $$r_1$$ is the position variable associated with the first particle, etc. So your comment can't be quantum mechanics because the wave functions are not correctly described.

On the other hand, in quantum field theory, the creation and annihilation operators do anticommute (with rules too complicated to go into here). But if you're writing in the QFT language, you've written the annihilators. The creation operators are written with daggers, i.e. $$\psi^{\dag}$$. And you need to deal with the vacuum, which you haven't included.

In general, when you have a situation where there are no particles being created nor destroyed, and for bound states in general, you will find that QM is easier to apply than QFT.

My comment was with respect to an "odd wave function". An odd wave function is one which satisfies $$\psi(r_1,r_2) = -\psi(r_2,r_1)$$. Identical fermions use these kinds of functions. Now it is possible to get wave functions in the QFT language, but to do this, you are going to have to use creation operators, and apply them to the vacuum appropriately.

In applying the creation operators to the vacuum, to produce an "odd wave function" you will have to apply them following the anticommutation rules that apply to fermions so there is no contradiction. Both formalisms give the same result, though they can be confusing.

This website discusses the facts:

http://galileo.phys.virginia.edu/classes/752.mf1i.spring03/IdenticalParticlesRevisited.htm

Getting back to the subject at hand, the odd wave function is fully antisymmetric in its entries and treats the two particles identically. Swapping them brings in a minus sign, but that minus sign is, as L&L notes, a phase change which cannot be physically detected.

Furthermore, the density matrix formalism contains all the physical information that is contained in the wave function formalism, and in the density matrix formalism, unphysical phases are cancelled, so the minus sign from swapping coordinates doesn't even make an appearance.

Last edited: Jan 4, 2007
13. Jan 4, 2007

### marlon

But if two particles have opposite spin, how can they be an identical quantum state ?

marlon

14. Jan 4, 2007

### CarlB

I agree that if two particles have opposite spin they are not in an identical quantum state. Let me say it again more clearly. There are two cases for the relative spins of two identical fermions. Two identical fermions in an odd wave function have to have identical spin. Two identical fermions in an even wave function have to have opposite spin.

Getting back to "two identical fermions with identical spin must have an odd wave function", what matters is that when the two particles are swapped, the overall wave state has to get multiplied by -1. $$\psi(r_1,r_2) = - \psi(r_2,r_1)$$ This is the same thing as saying that they are in an odd wave function.

The subtle point here is that in the usual state vector formalism, phase does not matter at all. One can multiply a state by any phase at all, and one gets the same state. The state $$e^{i\theta}\psi$$ is a representation of the same state as the state that $$\psi$$ represents. In particular, as L&L very clearly states, $$-\psi$$ is the same state as $$\psi$$.

Don't feel bad. The difficulty in understanding this is fairly universal in physics students. "Two fermions can't have the same quantum state" is a simplification of the actual situation and is confusing. The correct statement of the principle is that "the wave state of two identical fermions must take a minus sign when the fermions are swapped." (But even this is a little inaccurate. While it is a correct description of the mathematics of state vector formalism, it does not accurately describe the density operator formalism. But that would take us too far off track.)

My experience in teaching is that difficult points have to be repeated about three times before they will take hold in most people's minds. Let me say this again a slightly different way.

Half the problem with this is that spin-0 wave functions are too easy to understand. Two identical fermions in the same spin-0 wave functions have to have opposite spin. Easy to understand. Two identical fermions in a spin-1 wave function have to have the same spin. Harder to understand.

Here's a reference from the peer reviewed physics literature that applies these things to a practical problem, the collision of two identical fermions. Two identical fermions can't have an even wave function, instead, they must have an odd wave function. The paper notes this by stating:

For identical fermions the Pauli exclusion principle forbids s-wave collisions. This means the dominant interaction is via p-wave collisions (l = 1).
http://www.arxiv.org/PS_cache/cond-mat/pdf/0209/0209071.pdf [Broken]

s-waves are even, p-waves are odd. Now in a collision, there are only two particles involved, and since they are identical, we cannot distinguish between them. Therefore the wave function must treat the two particles identically, and therefore the two particles have identical quantum numbers. For the case of two identical fermions, the wave function must take a -1 on swapping the two particles. For the case of two identical bosons, the wave function must take a +1 on swapping the two particles.

I agree it's quite confusing and I apologize if I've added to the confusion. There is a practice GRE physics test out there that the student can take to see if they understand this. Out of around 100 questions, 1 or 2 cover these details.

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15. Jan 4, 2007

### CarlB

In general, bound states of identical neutrons are possible. They're called "neutron stars", and it is the force of gravity that holds them together.

I guess two neutrons could be bound by a gravitational interaction, but gravity is so weak that it would be a huge state. And of course neutrons decay in only 20 minutes.

16. Jan 5, 2007

### Severian

OK, I went to our library and dug out a copy. I think you are being very disingenuous, and quoting them out of context. The statement in L&L is correct, but is not what you are implying. It is made at the beginning of chapter IX as a prelude to demonstrating that "in a system consisting of identical fermions, no two (or more) particles can be in the same state at the same time" (a couple of pages after your quote, just after Eq 61.3 - unfortunately we must have different editions, since the statement you quote is on page 209 of my book, with my quote on 211).

They are using the statement you quote to point out that interchanging 'identical' particles can only lead to an overall + or - sign (ie. symmetric or antisymmetric wavefunctions) and thus when the two 'identical' particles are in the same state, the antisymmetric one must vanish.

In other words their use of the word 'identity' in the passage you quote is not the usual sense implied when one says 'fermions cannot be in identical states'. The latter definition would insist that the fermions have the same state vector (eg. be in the same place, have the same momentum etc) while L&L are meaning that they are just indistinguishable particles (eg two neutrons with the same spin). Obviously, if the neutrons are spatially separated, for example, then there is no issue from fermi statistics.

Wavefunctions are simply projections of the state vector onto whatever basis you choose (eg position or momentum). The state vectors are the more fundamental objects, and they are perfectly well defined in QFT, just the operator acting on the vacuum. QM is inferior to QFT, and is only useful (as you say) when there is no creation or annihilation in the process under study; all processes you might describe with QM can equally well be described using QFT, but not vice versa.

What label you use for the operator is convention. However, $$\psi$$ contains both creation and annihilation operators. See for example, Eq 3.99 of Peskin and Schroeder. Admitedly, the $$\psi$$ traditionally contains the annihilation operator for the particle, and the creation operator for the antiparticle, but this is inconsequential.

This would not be very tricky (just sandwich them between the vacuum and the position basis) but is pointless. The operator anticommutation identity is more fundamental and illustrates the same principle.

Last edited: Jan 5, 2007
17. Jan 5, 2007

### CarlB

My version is that of 2002 and may be newer than your library's copy. It was reprinted "with corrections" in 1991. My version does not contain your quote just after Eq. (61.3), but I don't doubt that it was included in a previous version. Something similar is stated in my version a paragraph after Eq. (61.5).

Look about 3/4 of a page earlier than where you found that quote, at the start of the third paragraph earlier (top of page 229, my copy, before equation (61.1)). The authors write: "Let us consider a system composed of N identical particles, whose mutual interaction can be neglected." Equation (61.3) is part of this analysis, which appears correct. However, a bound state is not a state where mutual interaction can be neglected.

Let me try and describe what L&L is doing in their analysis. They are considering a system of non interacting particles. Since the particles do not interact, they write each one as having its own, separate, wave function. Under this circumstance, to define the wave state for the system one anticommutes the wave states for the individual fermions. Because fermions can't occupy the same state, one builds up the wave function from individual wave functions.

But after you've done this antisymmetrization, the individuality of the wave states is gone. It has been antisymmetrized away and the final wave state shows that all the fermions (or bosons) are identical and all occupy all the individual wave states. The inability to distinguish between quantum particles is quite central to quantum mechanics.

When you're considering non interacting particles, the analysis that implies that you can distinguish between the fermions according to "which state it is in" makes mathematical sense. A great example of this is the case when you have one fermion on the earth and the other on the moon. Since they are well separated, they do not interact and you can label them. And then, sure, you can say that they occupy different states. But this does not apply to a bound state.

The basic problem here is that you are insisting on the particles having individual existences that are distinguishable from each other by quantum numbers that only exist in the non interacting case.

The error here is as above. Bound states can't be split into non interacting individual particle states that you can compare between to see whether they are in the same state or not. Both identical particles are in the same bound wave state. L&L is quite explicit in stating that antisymmetric wave states are not "zero", but instead are suitable for identical fermions. They only state that you get zero when you use non interacting wave states to model the particles, as would be suitable for a non interacting gas, or several widely separated particles.

Sigh. Back at the beginning of all this you wrote to the original poser: "Yes, it is because they are fermions, so they obey the Pauli Exclusion Principle" as an explanation for why neutrons don't bind together. Do you still believe this? How do you explain how neutron stars bind together?

http://hitoshi.berkeley.edu/221B/scattering4.pdf

The hypothetical bound state between two neutrons turns out to be called a "dineutron" in the literature. There are extensive references in the literature. The reader is invited to look through them and find references to the fact that dineutrons are not bound because of the Pauli exclusion principle. It would be nice if it were that simple, we'd have one true statement reinforcing another. Too bad the situation is more complicated:

By the way, the hypothetical bound state of three neutrons is called the "trineutron" and is discussed in the peer reviewed literature. A paper that discusses four neutron ("tetraneutron") bound states is here:
http://arxiv.org/abs/nucl-th/0511006

Last edited: Jan 5, 2007
18. Jan 5, 2007

### Severian

This makes no odds. L&L make this statement only to simplify the analysis for the reader - one can also apply the Pauli Exclusion principle to interacting particles. Indeed, in the same chapter L&L go on to discuss interacting particles. If one could not apply the Pauli Exclusion Priniple to interacting fermions then one could never apply it at all since all particles are interacting!

The reason why their simplified analysis requires no interaction is that one cannot write the two particle wavefunction $$\psi(\xi_1,\xi_2)$$ as a product of one particle wavefunctions when they are interacting. However, even in pure QM (ie. no QFT), the Pauli Exclusion Principle follows from the antisymmetry they derive on the first page of this chapter. Setting $$\xi_1=\xi_2$$ in the antisymmetric wavefunction causes the wavefunction to vanish.

Firstly, these two fermions are still interacting - it is just that their interaction is vanishingly small. Secondly, this does not effect the Pauli Exclusion Principle - it only affects the ability to write the two particle wavefunction as a product of one particle wavefunctions, which is not the same thing.

Well, now you are just misrepresenting. The original question was why a bound state of two identical neutrons does not exist. This is simply due to taking L&L's equation $$\psi(\xi_1,\xi_2)=-\psi(\xi_2,\xi_1)$$ and setting $$\xi_1=\xi_2$$ (since they are identical), i.e. the Pauli Exclusion Principle. There is no requirement for non-interaction in this equation. (Indeed, as I pointed out earlier, it is easy to see in QFT, where one generates the two particle wavefunction by applying the field operators. Since one applies two field operators to create the two particle state, one can explicitly use the anticommutation relation of the operators to derive the antisymmetry property of the two particle wavefunction, with or without interactions.)

Neutron stars can still bind together because the neutrons are not in identical states.

In response I could ask you, do you think the Pauli Exclusion Principle does not apply to quarks?

And again, these are bound systems of neutrons in different states, so the Pauli Exclusion Principle does not apply. It has nothing to do with whether or not they are interacting.

Last edited: Jan 5, 2007
19. Jan 5, 2007

### CarlB

First you quote L&L's analysis of non interacting fermions, now you claim that non interacting fermions do not exist. Try to present a consistent target.

Yes, that's the original poser's question. But the question you now are claiming you answered was why a bound state of two neutrons in identical states does not exist. If that's really what you meant, and you're not being disingenuous, then you should explain why you didn't say it at first. But in any case your answer doesn't make sense because the only way items in a bound state can be thought of as having "states" is by considering something other than their wave state, which is a joint object.

Another reasonable interpretation of "identical neutrons" would be that they are in the same spin state. And two neutrons can most definitely exist in the same spin state. In fact, if the neutrons are in an antisymmetric (odd) wave function state, their spin states must be identical.

This is the crux of our difference. We probably agree that the wave function for a bunch of neutrons needs to be antisymmetrized. After doing this, which is necessary to produce a wave function that represents the physical object, I claim that all the neutrons are in the same, antisymmetrized state. You claim that all the neutrons are in different states. Why do you think we went to the trouble of antisymmetrizing them? If they were all in different states we could apply classical principles to the problem. One cannot distinguish between interacting identical particles, and distinguishing them according to their (interacting) states is a case of this.

Let's put this to a test. We take two neutrons, not bound, perhaps in a collision. Their combined wave state is:
$$\psi(r_1,r_2) = - \psi(r_1,r_2)$$
where $$\psi$$ is some odd function of the coordinates. You claim that these two particles are in different states. It is clear that this wave function satisfies the Pauli exclusion principle.

Now tell me what the individual neutron wave states are.

You could ask me that, but you'd be making a complete change of topic, and putting words in my mouth. You wouldn't use that sort of empty debate tactic, would you. Of course the PEP applies to quarks.

20. Jan 9, 2007

### Severian

I think I have already explained this to you. L&L's analysis assumes non-interacting fermions to simplify the situation for the reader - they are not saying that the fermions must be non-interacting for the PEP to work.

I will condeed that the OP's question is ambiguous, and it is feasible that you misunderstood. I certainly interpreted the phrase 'two identical neutrons' to mean that the two neutrons were identical, having the same representation in position space, momentum space etc, rather than just having the same spin (which seems to be your intepretation).

I do admit that this can be quite misleading for those not familiar with the terminology. Indeed, I have been guilty myself of using the word 'identical' when taking into account two fermions of the same flavour and spin in the final state of Feynman diagrams (where one must include the diagram with the fermions switched and a relative minus sign).

As to why I "didn't say it at first", was "Then they wouldn't be in an identical state" not enough for you?

It is a comparison of $$\psi_1 \psi_2 |0 \rangle$$ with $$\psi_2 \psi_1 |0 \rangle$$ which is perfectly well defined.

I would regard this as a trivial statement. Obviously two neutrons can exist in the same spin state. Since they can only be spin up or spin down, you would otherwise be implying that there are only 2 neutrons in the universe.

No, actually i do not think this is the crux of our difference at all. I don't particularly care for your semantic issues. What I do care about is that you made the claim that the PEP was invalid if the particles are interacting ("this does not apply to a bound state"). This is nonsense, and I suspect that you realize this now and a trying to muddy the argument with semantics. Indeed, the reason I pointed out that no particles are non-interacting is because then, according to your claim, no particles anywhere would be constrained by the PEP!

The Pauli Exclusion principle is independent of whether or not the particles can interact.