# Mass eigenstates and occupied physical volume?

I recall that some isotopes of some elements can behave as bosons, such as helium-4, under certain conditions. I was consistently confused by the concept of an ensemble of bosons being able to occupy only one quantum state. So here are my questions.

1) If bosons can occupy the same quantum state, does that say anything about the physical volume which they occupy? I ask because of things like white dwarf and neutron stars, which are physically maintained by degeneracy pressure. So could an ensemble of bosons in one quantum state occupy the same physical volume as one boson? [I am inclined to say no from my knowledge of liquid helium experiments, but I thought I would ask]

2) If the answer to (1) is yes, a large ensemble of bosons could occupy much less volume than the sum of their individual constituents. Then would it be possible to use something like an atom laser to assemble something akin to a mass eigenstate (perhaps this is the wrong term, but I am envisioning a large number of bosons in the same state) composed of a massive number of bosons in one physical location?

3) I ask with the idea in mind of trying to create a manmade device that could locally manipulate gravity [kind of a sci-fi question]. Is this ridiculous or have I missed a critical point from my quantum classes?

Sometimes when people try to simplify physics they go too far and oversimplify. An example would be if people tell you that two atoms of $He^4$ can be in the same state because they "are bosons."

First let's look at one fermion, say an electron. It has a spin and a spatial extent, the general state of the electron is a product of the spin and the spatial states for that one electron.

Now let's look at two particles without spin, they each have a spatial part so you can (and must) use a 6d space to describe where both particles are, so the wave is $\Psi=\Psi(x_1,y_1,z_1,x_2,y_2,z_2,t)=\Psi(\vec r_1,\vec r_2,t)$ where $\vec r_i=(x_i,y_i,z_i)$ says where the the ith particle is. If the particles are identical species (and bosons) then you insist that for any two locations $\vec r_1$ and $\vec r_2$ that $\Psi(\vec r_1,\vec r_2,t)$= $\Psi(\vec r_2,\vec r_1,t)$ in other words you use that to describe the frequency of finding a particle at $\vec r_1$ and another at $\vec r_2.$ Not the frequency of finding "the first" at $\vec r_1$ and "the second" at $\vec r_2$ (whatever that could mean since they are of the same species).

So we can do one fermion and two spin zero bosons.

OK, what about two spin 1 bosons? Now the general single particle state has a spatial part and a spin part and looks like $\Psi_i(\vec r_1)S_j$ where the $\Psi$ is a spatial wave and the $S$ is a spin. So you could consider an orthonormal basis $B_{ij}=\Psi_i(\vec r_1)S_j$ where the $\{B_{ij}=\Psi_iS_j\}$ are orthonormal. Note that the spatial part is a function of coordinates so for a vector $\vec r$ you get a number $\Psi_i(\vec r).$

Now for a single particle $\{B_{ij}=\Psi_iS_j\}$ spans the space. For two particles it might look like $\{\Psi_i(\vec r_1)S_{j1}\otimes\Psi_k(\vec r_2)S_{l2}\}$ spans the space, but it turns out it spans too large a space when the two bosons are of an identical species. You could start with a state like $\Sigma\alpha^{ijkl}\Psi_i(\vec r_1)S_{j1}\otimes\Psi_k(\vec r_2)S_{l2}$ and replace it it with $\Sigma\alpha^{ijkl}(\Psi_i(\vec r_1)S_{j1}\otimes\Psi_k(\vec r_2)S_{l2}+\Psi_i(\vec r_2)S_{j2}\otimes\Psi_k(\vec r_1)S_{l1}).$ So just like spin 0 there isn't a frequency of finding the first particle someplace with a certain spin, just of finding one at a location with one of them having a certain spin. The set of such symmetric states is actually the space from which we pick our models of systems with multiple bosons.

OK we can do two bosons of the same species and we can do one fermion. How about two fermions of the same species?

Instead of just using states like $\{F_{ij}=\Psi_iS_j\}$ that spans the single particle space for two fermions of the same species we can't just use $\{\Psi_i(\vec r_1)S_{j1}\otimes\Psi_k(\vec r_2)S_{l2}\}$ to span the space, because it turns out it also spans too large a space when the two fermions are of an identical species. You can instead start with a state like $\Sigma\alpha^{ijkl}\Psi_i(\vec r_1)S_{j1}\otimes\Psi_k(\vec r_2)S_{l2}$ and replace it it with $\Sigma\alpha^{ijkl}(\Psi_i(\vec r_1)S_{j1}\otimes\Psi_k(\vec r_2)S_{l2}-\Psi_i(\vec r_2)S_{j2}\otimes\Psi_k(\vec r_1)S_{l1}).$ So just like bosons there isn't a frequency of finding the first particle someplace with a certain spin, just of finding one at a location with one of them having a certain spin. The set of such antisymmetric states is actually the space from which we pick our models of systems with multiple fermions.

Finally if you have more than two of the same species you can start with a state from the "too large" space and then consider an arbitrary state from that "too large" space and then consider every possible way of repeatedly swapping fermions or bosons of the same species. When you have every possible repeated swapping you give it an overall (-1) if there were an overall odd number of swaps of fermions involved in the related swapping, otherwise you give it an overall (+1) and then you take all those possible repeated swappings with the $\pm 1$ factor and then add them all up.

So whenever you see a particular multiparticle state just replace each term with a sum of versions that contain every version that looks just like it except you replaced some of the particles having a state (spatial and spin) with a different particle of the same species so that each one now just sits in a different spot for one from the same species, with a $\pm 1$ factor.

So what's all this talk about the same state? If you want to have fermions of the same species have the same spatial state then you are out of luck because you you take the original and the version where just those two are swapped you get two things that are negatives of each other and when you add together you get zero. Which means if you start with such a state from the "too large" space then you get zero, which isn't a valid state. So it just doesn't happen. So it is just forbidden as even being allowed for a basic building block. Your basic building blocks can start with states where every fermion of the same species is in a different state (at least one of spatial or spin is different) and then you antisymmetrize it so each has an equal frequency to be in those states, then that is the basic thing you take superpositions of.

As for that something must be different if the spatial parts are the same then the spin needs to be different, and there is only a size 2 basis for spin states for spin 1/2 (size 4 for spin 3/2, size 6 for spin 5/2, etcetera). Its not saying something else beyond the fact that the minus restricts which kinds of states are possible.

OK, so now we know everything. But wait, what about all that talk about sometimes treating groups of fermions as if they were fermions?

Fine. First you have to treat a pair of fermions as if they were a single composite bject, which they are not. Then you ask what happens when you swap the two composite objects.

So imagine you have two muons and two positrons, each muon can bind with a positron to make a hydrogen like atom. The atom technically is a muon and a positron, so you should specify the location of both. But that is true of the hydrogen as well, but sometimes people like to describe it as a state for the center of mass and as a relative position state of the electron as displaced from the center of mass. So now you get used to thinking of the muon-positron system as a single composite object and might think of your four particle system as like a 2 atom system.

At which point you might ask what happens when you consider the configuration where each of your atoms is swapped. When you swap one entire atom with another entire atom it means you swapped two muons (so got a minus sign) and you also swapped two positrons (so got another minus sign). So when you swapped one whole atom for another whole atom you got a $+1.$ Just like when you swapped two bosons. But that only happened when you treated the two particle system like it was a single particle.

There still really are four particles and you do pick up a minus sign when you swap two muons and leave everything else the same and you really do pick up a minus sign when you swap two positrons and leave everything else the same. It is still a system of fermions.

So can the two muons be in the same state?

No.

So can the two positrons be in the same state?

No.

Can the two atoms have their interactions behave differently if there is a region of coherent overlap if their waves?

Most definitely. That over simplification did come from somewhere and that is the effect the were trying to describe. The effect even already happens with three fermions. You can get a constructive interference between three fermions if they are mutually swapped a to b, b to c and c to a (because that is done with two swaps, e.g. first swap a and b as one swap and then swap c with what is now b).

The antisymmetry of swapping single pairs of same species particles can lead to symmetry in even swappings. And from the muon positron example you could swap a piar of muons and a pair of positrons and get the symmetry. But the basic state you are allowed to start with already had same species fermions in different states (when you take spin and space into a account) and that doesn't change over time.

OK and fine. So what is this effect of symmetry and antisymmetry? Specifically what does it do if these fermions are still in different states just like always?

It can affect how easily they interact with other things. For instance if it is hard to affect the center of mass of the muon-positron system then interactions try to change the relative displacement of the, say, positron. If that is then higher symmetrized with other things in a near spatial region is can be hard to interact with all of them in an antisymmetric way.

P.S. A muon-positron system is a poor choice since the muon decays on its own and the positron annihilates with electrons. I picked them since they bind, are elementary, have easy to say names and are clearly different species.

• Thanks for the detailed answer. It will take me a bit to digest this... Aug 9 '15 at 13:33
• P.S. Why do you suppose Plato chose the cube for Earth? Aug 9 '15 at 13:36