Particles of matter as Bosons?

[bbbl67]

Now from my basic understanding of particle physics, matter is supposed to be fermions, while particles involved in force interactions are bosons (photons, gluons, W/Z, and Higgs). Now, apparently there are also some composite particles of matter that are considered to be bosons too. For example, the Helium nucleus, or a Bose-Einstein Condensate.

Now, also according my basic understanding of Quantum Physics, fermions cannot occupy the same state as each other, which in my interpretation means that no two fermions can be in the exact same location at the same time as each other. Thus this gives us the familiar spacing between particles that we notice all of the time.

Bosons on the other hand, are able to occupy the same states as each other, meaning that they can occupy the same space at the same time as each other. This to me sounds like a fancy way of saying that bosons can go right through each other.

So do matter particles which are supposed to be bosons, go through each other?

 

[PeterDonis]

You are not correctly stating the difference between bosons and fermions. The difference, heuristically, is that if you exchange two bosons (i.e., imagine taking two bosons and swapping their states), the total state of the system is unchanged; whereas if you exchange two fermions, the total state of the system gets a minus sign. The Pauli exclusion principle--that no two fermions can be in the same state--is a consequence of this (if two fermions were in the same state, then swapping their states could not change the total state of the system--but giving the total state a minus sign is a change). But there is no complementary principle for bosons--just knowing that swapping them doesn't change the total state of the system doesn't tell you anything about whether bosons can be in the same place.

do matter particles which are supposed to be bosons, go through each other?

No, because those matter particles--things like He atoms--are composite objects made of fermions, and that restricts what they can do.

 

[dextercioby]

Plus that for massless bosons, their "position" in space-time is an ill-defined concept, therefore to say "tho bosons can occupy the same place at the same time" makes no sense.

 

[vanhees71]

You are not correctly stating the difference between bosons and fermions. The difference, heuristically, is that if you exchange two bosons (i.e., imagine taking two bosons and swapping their states), the total state of the system is unchanged; whereas if you exchange two fermions, the total state of the system gets a minus sign.

To make it really right, the statement is that the many-body Hilbert spaces of identical bosons and bosons are different.

Bosons:

Let $\mathcal{H}$ denote the single-particle state and let $|u_n \rangle$ be a complete orthonormal basis of this Hilbert space. Then for bosons the Hilbert space of $N$ is given by the span of all totally symmetrized product states, i.e.,
$$\langle |u_{n_1},u_{n_2},\ldots, u_{n_N} \rangle^{(+)}=\frac{1}{\sqrt{N!}} \sum_{P \in S_N} |u_{n_{P(1)}},u_{n_{P(2)}},\ldots,u_{n_{P(N)}} \rangle.$$

Fermions:

Let $\mathcal{H}$ denote the single-particle state and let $|u_n \rangle$ be a complete orthonormal basis of this Hilbert space. Then for fermions the Hilbert space of $N$ is given by the span of all totally antisymmetrized product states, i.e.,

$$\langle |u_{n_1},u_{n_2},\ldots, u_{n_N} \rangle^{(-)}=\frac{1}{\sqrt{N!}} \sum_{P \in S_N} \sigma(P) |u_{n_{P(1)}},u_{n_{P(2)}},\ldots,u_{n_{P(N)}} \rangle.$$
Here, $S_N$ denotes the permutation group of $N$ elements, and $\sigma(P)$ the sign of the permutation (i.e., if you need an even or odd number of interchanges of pairs to get the indices from the lexical order $(1,2,\ldots,N)$ to the order given by the permatuation $P(1),P(2),\ldots,P(n)$.

For fermions this implies indeed that whenever two indices $n_i=n_k$ in the set are equal, the antisymmetrized product vanishes identically.

The states, represented by statistical operators (or for pure states equivalently by unit rays in Hilbert space) are of course invariant under reordering of particles, as it must be. Boson and fermion $N$-particle spaces are the most common realizations of this principlce of indistinguishability of identical particles. As far as we know on a fundamental level these are the only realizations relevant in nature. Of course, in effective theories of 2D problems you can have less trivial realizations, socalled "anyons".

 

[bbbl67]

You are not correctly stating the difference between bosons and fermions. The difference, heuristically, is that if you exchange two bosons (i.e., imagine taking two bosons and swapping their states), the total state of the system is unchanged; whereas if you exchange two fermions, the total state of the system gets a minus sign. The Pauli exclusion principle--that no two fermions can be in the same state--is a consequence of this (if two fermions were in the same state, then swapping their states could not change the total state of the system--but giving the total state a minus sign is a change). But there is no complementary principle for bosons--just knowing that swapping them doesn't change the total state of the system doesn't tell you anything about whether bosons can be in the same place.

Okay, so that's the math of it, fulfills David Mermin's "just shut up and calculate" quotation. But what does it mean physically? What does the state function measure physically, if not time and position? I suppose, you could also put quantum spin in as one of the states, as two electrons can occupy the same orbital, as long as they are opposite spins from each other. What else is in the state function?

No, because those matter particles--things like He atoms--are composite objects made of fermions, and that restricts what they can do.

So at some large scale, they act like bosons, but at the smaller scales they are still fermions?

 

[bbbl67]

Plus that for massless bosons, their "position" in space-time is an ill-defined concept, therefore to say "tho bosons can occupy the same place at the same time" makes no sense.

Well, if a massless particle are said to occupy all positions of time at the same time, then that would prove that massless bosons can occupy the same positions in time, at the very least.

 

[weirdoguy]

if a massless particle are said to occupy all positions of time at the same time, then (...)

But they don't, so there is no "then".

 

[Demystifier]

So do matter particles which are supposed to be bosons, go through each other?

They don't. The composite "bosons" made of an even number of fermions are not really true bosons. For instance, if you start from fermion creation operators $a_1^{\dagger}$, $a_2^{\dagger}$ and from them construct a composite "boson" creation operator
$$b^{\dagger}=a_1^{\dagger}a_2^{\dagger}$$
you can see that it is not really a boson creation operator, in the sense that
$$[b,b^{\dagger}]\neq 1$$

For the statistics of Cooper pairs see also Ashcroft and Mermin, Solid State Physics, page 741.

 

[PeterDonis]

What does the state function measure physically, if not time and position?

The state is a vector in a Hilbert space. It is not anything like the "state" in classical physics, which is what "time and position" implies.

Physically, the state is a collection of probability amplitudes for different possible measurements that could be made on the system.

 

[PeterDonis]

if a massless particle are said to occupy all positions of time at the same time

They don't.

 

[ZapperZ]

Now from my basic understanding of particle physics, matter is supposed to be fermions, while particles involved in force interactions are bosons (photons, gluons, W/Z, and Higgs). Now, apparently there are also some composite particles of matter that are considered to be bosons too. For example, the Helium nucleus, or a Bose-Einstein Condensate.

Now, also according my basic understanding of Quantum Physics, fermions cannot occupy the same state as each other, which in my interpretation means that no two fermions can be in the exact same location at the same time as each other. Thus this gives us the familiar spacing between particles that we notice all of the time.

Bosons on the other hand, are able to occupy the same states as each other, meaning that they can occupy the same space at the same time as each other. This to me sounds like a fancy way of saying that bosons can go right through each other.

So do matter particles which are supposed to be bosons, go through each other?

There is an issue with the whole starting premise here which I don't think anyone has touched upon.

To be able to determine if "bosons go through each other", it has to have an implicit assumption that one can actually tag and follow a boson as it passes through another one. This runs contradictory to the notion of INDISTINGUISHABLE particles, and this indistinguishibility is the whole principle on why the Fermi-Dirac and Bose-Einstein statistics come into play. They are no longer identical particles obeying the Maxwell-Boltzmann statistics. These are not just identical particles. These are indistinguishable particles. It makes a lot of difference.

Because of this, one actually cannot say that a boson goes through another boson. This is because when they are in such a state, you can no longer able to pick out one boson against the other. All you can say is the symmetry of the total wavefunction.

Zz.

 

[vanhees71]

Yes, indeed, and that I've stressed in #4. It is also of utmost importance, as also stressed in #4, not to claim that (pure) states are represented by vectors in Hilbert space but by unit rays in Hilbert space. This makes both the Bose AND Fermi many-body Hilbert spaces realizations of the idea of indistinguishability (let alone the important aspect of the possibility to have half-integer spin and the close relation between spin and statistics), which is the key point in quantum many-body theory.

The most simple point is indeed the resolution of the Gibbs paradox of classical statistical physics, which was cured by Boltzmann in an ad-hoc manner by multiplying a $1/N!$ into the $N$-body microcanonical or canonical partition sum (or equivalently in each term of the grand-canonical partition sum), and is through QT's indistinguishability natural in the classical approximation.

 

[Gigaz]

In principle, two non-interacting indistinguishable bosons should be able to occupy the same state, which includes position eigenstates. So they can can go through each other, even if they are composite.
The problem here is that we are not really dealing with non-interacting bosons. Neutral atoms do interact quite strongly at close range. This ultimately prevents them from having the same position, but they can still share other quantum states.

 

[Demystifier]

In principle, two non-interacting indistinguishable bosons should be able to occupy the same state, which includes position eigenstates. So they can can go through each other, even if they are composite.

That requires clarification. Suppose that we have 4 fermionic particles (of the same mass) with position coordinates ${\bf x}_1$, ${\bf x}_2$, ${\bf x}_3$ and ${\bf x}_4$. Suppose that the particles 1 and 2 comprise the first composite with the CM coordinate
$${\bf y}_1=\frac{{\bf x}_1+{\bf x}_2}{2}$$
and that the particles 3 and 4 comprise the second composite with the CM coordinate
$${\bf y}_2=\frac{{\bf x}_3+{\bf x}_4}{2}$$
Then the two composites can have the same position in the sense that
$${\bf y}_1={\bf y}_2$$
Nevertheless, the positions of constituent particles must be different, so we must have
$${\bf x}_1\neq{\bf x}_2, \;\;\; {\bf x}_1\neq{\bf x}_3, \;\;\; {\bf x}_1\neq{\bf x}_4, \;\;\; ... $$

Of course, I was talking about particle positions as if they were classical, but the above can easily be put into a proper quantum language by translating it into the language of Dirac $\delta$-functions.

 

[vanhees71]

There's no state of precise single-particle position. Thus all speculations about having particles exactly at the same position are pointless. Quantum theory is a consistent theory, at least on this very elementary level. Otherwise one would have to find a better theory already from purely theoretical considerations, which is however not the case, and it's even empirically successful, i.e., there's no hint at any failure of QT yet!

 

[Demystifier]

There's no state of precise single-particle position.

How about $\delta$ functions? And if they are too pathological, how about narrow Gaussians? What I said above can also be translated into a language on narrow Gaussians.

If you prefer second-quantization language, take a look at Fetter and Walecka (Quantum Theory of Many-Particle Systems). By acting with the second operator in Eq. (2.1) on the vacuum, you create a particle at a well-defined position.

 

[vanhees71]

It's $\delta$ distributions not functions! They are not representing states. Narrow Gaussians are, of course, fine.

The same holds in 2nd quantization. The field operators are distributionvalued operators and don't create states but generalized states.

 

[PeterDonis]

In principle, two non-interacting indistinguishable bosons should be able to occupy the same state, which includes position eigenstates.

With some caveats about whether position eigenstates are actually possible states (see the subthread that @vanhees71 and @Demystifier are having), yes.

So they can can go through each other

No, because if the bosons are indistinguishable, you can't distinguish them, which means you can't assign them definite trajectories, which you would have to in order to say they "can go through each other". All you can say is that we have, for example, two bosons at the same position A, and then, an instant later, two bosons at positions A and B. You can't say either of them "moved" from A to B.