# Electrons distinguishable ?

1. Nov 25, 2009

### hardphysics

hi, i have a question.

are electrons distinguishable ?

my personal view is they are not. because distinguishable means different quantum numbers. okay one electron can have spin up and one can have spin down, but this property is not lorentz invariant, and so this property is not qualified to distinguish.

that means all electrons are not distinguishable, although they might have different quantum numbers (like spin, or impulse) ?

is this right ? thanks in advance.

2. Nov 25, 2009

### Neo_Anderson

A bunch of non-entangled electrons are about as distinguishable as a bunch of basketballs lying on a basketball court.
Or a bunch of tennis balls on a tennisball court. Or a bunch of golf balls on a golf course. Or a bunch of baseballs on a baseball field. Or...

3. Nov 26, 2009

### hardphysics

ok, but regarding quantum numbers. is my view correct ?

4. Nov 26, 2009

### mikeph

I think so, all electrons are the same and are only distinguishable by the state they occupy. They have no fundamental fixed "index quantum number" that you can use to track individual electrons or label them, like a bank note or record in a database. I always thought this had some sort of philosophical importance but have never really thought about it deep enough to work out what that importance is.

This is to the best of my knowledge and I am quite interested to read more educated replies!

edit- I would also wonder how this extends to bosons...

5. Nov 26, 2009

### hardphysics

hmm. regarding the lorentz invariant i think i was wrong. yes the eigenvalue m of S_z is not lorentz invariant. but when you have two electrons, and they would have different m's in one frame, they would have different m's in another frame.
therefore the quantum number m is sufficient to distinguish between two photons! otherwise the pauli principle would have no sense? is that right ?
therefore electrons are distinguishable, but only if they have different spins (different m's)?
and what if they have different impulses but the same spins (m's), are they considered the same or different, i would say they are still the same, but impulse and enegry could be considered as quantum numbers which allow distinguishment ?

how is distinguishable difined ?

6. Nov 26, 2009

### ZapperZ

Staff Emeritus
Why can't we define this via the applicability of the Maxwell-Boltzmann statistics (distinguishable) and Fermi-Dirac statistics (indistinguishable)?

Electrons under different conditions can be described via each of those. The Drude model of conduction electrons and free electrons in particle accelerators are adequately described via the MB statistics. Here, the electrons are far apart enough from each other that such classical distribution are valid. You can, in theory, track each of them, and you are aware when two electrons are exchanged.

But if you invoke more stringent circumstances where the wavefunction of the individual electrons start to overlap, then FD statistics kicks in and the indistinguishibility criteria becomes valid. This is where MB statistics breaks down and you know that you can no longer tell when two electrons swap states.

Zz.

7. Nov 27, 2009

### BobGom

Does the use of the MB distribution actually imply that? Surely the reason why the MB distribution sometimes works is because the issue of distinguishability does not arise since the occupancy is low enough that the chance of finding two electrons in the same state is small.

8. Nov 27, 2009

### gulsen

They are, as a matter of fact, indistinguishible. Of course their dynamical properties differ, but indistinguishible means you cannot a priori know which one has that particular property. That is to say, if you somehow know that one electron has spin up, and other has spoin down, the wavefunction is (let the first term denote the state for electron A, and second term electron B)

$$|\psi \rangle = |+ \rangle |- \rangle - |+ \rangle |- \rangle$$

which is antisymetric (dont worry about the minus sign, it will flushed away when you compute the probabilities) under swapping the electrons, which means, altough you know an electron has spin, say down, you cannot tell which. Hence they're indistinguishible.

Obviously, if you could say electron A has spin down, the composite wavefunction would be

$$|\psi \rangle = |- \rangle |+ \rangle$$

9. Nov 27, 2009

### hardphysics

ok this is very close to my last post.

my problem arised, when i was using the wick theorem, say on Yukawa theorie (fermion with scalar field). if you evalute the S matrix elements you come across such terms like
$$\langle p',k' \mid \bar\psi(x)\psi(x)\bar\psi(y)\psi(y) \mid p,k\rangle$$ with both $$\phi$$'s already contracted. now to fully contract this term you have mainly two possibilities. they only differ in the exchange of p' with k' (or p and k, but this is results in the same expressions). that means from such a term follow two diagramms.

now if you can distinguish between p and k particles, only one diagramm contributes. my question was, in which way the formalism eliminates one of the diagramms. my guess is that when both particles differ, the field operators can act only on one particle and not on the other (for example on the p particle), and as a result only one contraction (diagramm) is possible. i dont know if this is true, but i have the feeling that this is correct.

what do you think ?

10. Nov 28, 2009

### ZapperZ

Staff Emeritus
If you look at MB statistics, you can, in principle distinguish one ball versus the other, even though they look identical (i.e. you have 10 identical red balls, for example). So that is why MB statistics is not identical to FD statistics.

The fact that the electrons are far enough apart that "the chance of finding two electrons in the same state is small" IS the reason why, under that situation, electrons behave like classical particle and become distinguishable.

Zz.