# Fermions

Heraclitis said that you can not step in the same river twice.
Wolfgang Pauli said that you can not be an identical fermion twice
that is that there can’t be two identical fermions.
Some Heraclitian said that you can’t step in the same river once.
This is clearly silly.

Similarly, I think that an honest application of the basic assumptions of quantum mechanics (as listed by Von Neuman) would imply that there can’t be one fermion.

A Fermion is a particle with spin + or minus hbar/2. For example, electrons are fermions. The word spin suggests that it refers to well spin as in the Earth spins around its axis and elementary particles spin around. Spin has the logical implication that charged particles with spin have magnetic fields. The magnetic properties of atoms are logically implied by the spin of their electrons, the orbital angular momentum of electrons and confusing stuff going on in the nucleus making extremely weak magnetic fields.

The spin of photons makes perfect sense if it is interpreted as an intrinsic angular momentum. I mean you know spinning. The strongest reason to interpret spin as angular momentum is that, if it is so interpreted, total momentum is conserved as particles are created and destroyed.

The Pauli exclusion principal (no two identical fermions) could have been accepted as a mere fact. However, physicists aim to explain results in terms of more fundamental principles. In this case, they note that if there are two identical fermions a and b and one switches a and b, then the wave form describing the system of particles is multiplied by -1 (I think the proof is hard in any case I never made head nor tail of it). Now clearly switching two identical particles changes nothing. That means the wave is equal to minus itself, which means it is zero. That means the probability of their being two identical fermions is zero.

OK back to spin and angular momentum. I argued at length that spin is, as the name suggests, a kind of angular momentum, because quantum mechanics has a clear interpretation of momentum and therefore angular momentum. The momentum operator is the gradient with respect to space of the wave form. Thus angular momentum is the derivative with respect to the angle.

Now what sort of wave form does a particle with angular momentum hbar/2 have ? well the derivative of the wave form with respect to the angle measured from a point is angular momentum around that point divided by h. That means that if one rotates the waveform of a fermion 360 degrees it is mulitplied by e to the power i times Pi which is equal to -1.

As far as I can tell, the only natural quantum mechanical interpretation of spin implies that the wave form of a fermion is multiplied by -1 if it is rotated 360 degrees. Clearly nothing changes if it is rotated 360 degrees. That seems to me to mean that the wave form of a fermion must be equal to zero, that is, the probability that there is a fermion is zero.

So why is the same argument a proof of the true Pauli exclusion principle and not a proof of the false claim that electrons don’t exist ?

One way to avoid the conclusion that quantum mechanics can prove that electons don’t exist is to say that the spin operator can not be interpreted as any known operator operating on a wave form. That is one can conclude that, if the core hypotheses of quantum mechanics as listed by Von Neuman are indeed the core hypotheses of quantum mechanics, then quantum mechanics is false. This is the approach of David Saxon in “Elementary Quantum Mechanics” except he doesn’t note that he is concluding that Von Neuman or quantum mechanics is wrong.

The other way is to just ignore it. The same reasoning leads to a result which is interpreted as proof that something is impossible in one case and which is interpreted as proof of nothing in the other.

I think that physicists pretend that they have reduced their theories to a few core principles when they haven’t. The reasoning from the core principles to correct predictions does not seem to me to follow fixed rules (that would be among the core principles). Instead arguments are accepted because their conclusions are empirically correct.

If I am right, this is nothing horrible. Even if the Pauli exclusion principle is an empirical fact not explained by deaper theories it is very deep, that is a very simple rule which explains a huge amount of data. Still I think one ought to admit which things one has figured out from other discoveries and which one has learned empirically but not explained with deaper theories.

Of course I never got past elementary quantum mechanics, so I’m probably just confused.

## 12 thoughts on “Fermions”

1. Clearly nothing changes if it is rotated 360 degrees.

Well, clearly there are some somethings that do, isn’t it? In particular the somethings known as “electrons”.

If you must write stuff like this, could you do us all a favour and change the font to green, the universally-accepted trademark of the crank physiciste?

I was afraid this post would get no comments at all. I am not familiar enough with movable type to make the font green. I’m sure you have good reason to consider me a crank, but it would be nice if you pointed out something wrong with my argument.

I admit I should have added that the EPR experiment only makes sence if spin is angular momentum, and that the astounding implication of quantum mechanics noted by E, P and R has been confirmed experimentally.

3. Mark Amerman on said:

Robert Waldmann,

I really don’t understand this post and it’s obvious
that you know much more about physics than I do.
Nevertheless I took some enjoyment from reading it and speculating
to myself what it might mean.

I particularly enjoyed this line:

“The momentum operator is the gradient with respect
to space of the wave form.”

I’m not sure quite why. Particularly since I don’t
know what “momentum operator,” “gradient,” or even
“wave form” really mean.

Here’s what I think I understand about what you are doing.

“Spin” is one of the handful of properties used to
describe and discriminate between elementary particles.
But it is not spin in the physical sense that say a
ball would spin. Instead of coining a new word — which
is what I think should have been done — the physicists in
question gave this new property, rigoursly described
mathematically, an old name, that in some small sense
may overlap in meaning, but not really.

Here you attempt to treat that quantum mechanical “spin”
as if it literally were “spin” in the macroscopic world
and then combine it with some of the mathematics of
quantum mechanics to come to the conclusion that ergo there
are no such particles.

The real point of the exercise seems the last line:

“So why is the same argument a proof of the true Pauli
exclusion principle and not a proof of the false claim
that electrons don’t exist ?”

Where you argue that Pauli used a similar logic, ie.
treating “spin” as if it were literally spin, to deduce
that no two fermions can be in the same place — which
idea I know from what physics I did learn goes some distance
to explain why the universe is the way it is.

Although actually you state this as: “…there can’t
be two identical fermions.” Which seems a different
statement, and I don’t recall reading that Wolfgang Pauli
wrote such. Is it in some sense an equivalent statement?

Is it true that Wolfgang Pauli used a similar logic
to deduce his exclusion principle? I don’t know
enough to evaluate that. Or is it instead that you
are laying a new metaphysical interpretation of quantuum
mechanics on top the old mathematics and coming up

4. Walt Pohl on said:

I’m not sure what you’re trying to do here. The _definition_ of a fermion is that if you switch two identical particles, you introduce a minus sign in the wave function. From this you can derive the Pauli exclusion principle along the lines you discuss.

The fact that any half-integer spin particle is a fermion (the Spin-Statistics Theorem) is much harder, and it requires the use of 4 dimensional space-time. In the quantum-mechanical version of Abbott’s Flatland, the result no longer holds.

5. Daniel on said:

Here’s your mistake. “Clearly nothing changes if it is rotated 360 degrees.” Indeed, and what doesn’t change is the absolute value squared of the wave function – which is physically the probability which you can actually detect. Note that the change of sign in the wavefunction has no effect on the observed probabilities. More generally a change of phase (exp(i alpha)) is allowed.

Apart from that fact, the strange behaviour
of fermions is just that – strange. What you are really after is the “spin-statistics theorem”. I know of no simple proof.

“The reasoning from the core principles to correct predictions does not seem to me to follow fixed rules (that would be among the core principles).” Look – as soon as a theory is expressed mathematically the rules are fixed. There is no ambiguity left. However, physicists do argue about the interpretation of these empirically tested rules. Certainly quantum mechanics has its difficulties in that regard – though not the ones you mention.

6. Daniel

Either the claim “as soon as a theory is expressed mathematically the rules are fixed. There is no ambiguity left” is simply false, or physicists do not express quantum mechanics mathematically. I don’t care which.

You say that the phase of a wave doesn’t matter sicine all that matters is the magnitude psi*’psi. However, in the spin statistics theorem the result
psi=-psi is interpreted as proving psi=0 so psi*’psi=0 so the probability is zero.

The position of quantum mechanics is that they start with core hypotheses and the rest is math so that “as soon as a theory is expressed mathematically the rules are fixed. There is no ambiguity left.”

This claim is correct only if you accept both that the phase of psi doesn’t matter (as you assert) in the context of rotating a the psi of a fermion 360 degrees AND that psi=-psi implies psi=0 so psi*’psi=0 in the ocntext of the spin statistics theorem.

That is either basic quantum mechanics has not been stated mathematically or math allows logically inconsistent claims.

I conclude that physicists claim that basic quantum mechanics is a set of mathematically rigorous assertions which form a testable hypothesis is fraudulent.

Rather, I think, the unreduced (unexplained) empirical results are more numerous than quantum mechanics generally admit and the “mathematical demonstrations” of resuts from core principles are nothing of the kind.

To me the 4 axioms (core assertions) of quantum mechanics described by Von Neuman, the spin statistics theorem and its interpretation as deriving the Pauli exclusion principle from more fundamental hypotheses, the angular momentum/ rotation result and the claim that electrons exist are logically inconsistent.

The fact that there appear to be two physicists contributing to this thread (you and Walt Pohl) and that you completely disagree with the implications of the claim psi = -psi tends to support my claim.

Sorry, I am no longer trying to be polite, but I don’t want to write overlong comments.

7. Thinkmarble on said:

The Pauli Exclusion principle does not follow from psi=-psi.
Instead it follows from the anticommuating relationship of creation operators.
Which follows from the fact that the wavefunction of an fermion has to be the eigenfunction of any transposition operator to the eigenvalue -1.
You have your totally antisymmetrized wavefunction
|psi^->
To this wavefunctuion you apply the creation operator C^+_alpha and then the creation operator C^+_alpha’.
So you get something like:
C^+_alpha’C^+_alpha|psi^->=|psi^-_(alpha’,alpha)>
Now you apply the Transposition operator which exchanges alpha and alpha’.

C^+_alphaC^+_alpha’|psi^->=(-1)*|psi^-_(alpha’,alpha)>

Which holds due to:
P*psi=-psi
With P being an Transposition operator and psi an fermion wavefunction.

Now add the two lines and get the follwing anticommutator:
C^+_alphaC^+_alpha’+C^+_alpha’C^+_alpha = 0

We you know set alpha’ equal to alpha you get:
C^+_alphaC^+_alpha= 0
And that is the Pauli-Exclusion principle.

In words:
Creating two identical fermions results in an wavefunction which is equal to zero.

To reiterate:
If psi is the wavefunction of an fermion
psi=-psi does not hold.
What holds is
P*psi=-psi
With P being an Transposition operator.

Which is a slightly different relation.

There are a couple of other mistakes in your post (photon-spin,defition angular momentum operator, fermion spin(that’s (2n+1)/2)), but there are not as important.

I would suggest that the next time you want to talk about an topic where you have no clue you be a little less arrogant.

8. Daniel on said:

Look, I was only trying to help. Contrary to what you have deduced, I think I agree with the substance of what Walt Pohl said – I merely thought your problem was rooted at a different place.

“This claim is correct only if you accept both that the phase of psi doesn’t matter (as you assert) in the context of rotating a the psi of a fermion 360 degrees AND that psi=-psi implies psi=0 so psi*.psi=0 in the context of the spin statistics theorem.”

Yes. You accept both, and it is entirely consistent and rigorous when formulated precisely. But you have put the latter half very sloppily. I will try to explain in gory detail – bear with me. It is of course entirely possible you have heard this all before, in which case I apologise for not understanding what your problem is.

1) If rotating the spin of a spin 1/2 particle 360 degrees leads to no observable changes, |psi| should not change (|psi_before|=|psi_after|).

This condition is satisfied by psi going to -psi under such a rotation. This is not the same as psi=-psi, which would of course imply psi=0.

(Just in case, I mention in passing that rotating spin is not at all the same as rotating in space – because you cannot write J = x /cross p for some fiduciary “internal” coordinate x and “internal” momentum p, since that would mathematically imply integer spin. Not sure if that was your intuition).

2) If you have two identical particles in some system, what is meant by “identical”? That they cannot be distinguished from one another (mathematically, the Hamiltonian of the
combined system is invariant under particle exchange).

So if you have two such particles A and B in the same system, any observables (probabilities!) you calculate must be invariant under exchanging A and B, because you won’t be able to tell the difference.

What does that mean for the wavefunction describing the two particles? In the simplest case we build the combined wavefunction for the two particles in the system out of the wavefunctions for a single particle in the system by forming the product (more precisely the state of the combined system is in the tensor product of the spaces of states of the individual particles). Say as a toy model the wavefunction for A on its own is given by A(x) and that of B by B(x). Now the naive guess for the wavefunction of two particles, one in state A(x) and one in state B(x), is psi(x1,x2)=A(x1)B(x2). But this will not in general give probabilities invariant under exchanging A and B, because |psi(x1,x2)| is not equal to |psi(x2,x1)|!

So we must symmetrise to get the correct wavefunction for the 2-particle system. Obviously psi(x1,x2) = A(x1)B(x2)+B(x1)A(x2) is a possibility, but is it the only one? No! Precisely because the overall phase doesn’t matter, there is another possibility, namely A(x1)B(x2)-B(x1)A(x2), for which psi -> -psi under particle exchange. (Maybe it is clearer now why I replied to your post as I did.)

Which one is realised in nature? That is an experimental question, despite the fact that the overall phase has no observable consequences, because the *relative* phase does, as follows.

*If* the single particle wavefunctions combine as psi(x1,x2) = A(x1)B(x2)-B(x1)A(x2), then for x1=x2=x it follows from the equation that psi(x,x)=0. Hence there is zero probability that the particles are both in the same place at the same time. (That’s not the same as psi(x1,x2)=0.)

Furthermore, the particles cannot ever be in the same state, i.e. A(x)=B(x), because then psi(x1,x2)=0 and that of course has zero probability. This, the Pauli exclusion principle, is therefore a *consequence* of the statistics (i.e. the choice of sign in the symmetrisation). We call particles obeying the ‘+’ choice of sign bosons and those obeying the ‘-‘ choice of sign fermions.

So, are two fermions identical? Well, yes according to my definition above (they are not distinguishable). No in the sense that they cannot both be in the same state.

The resolution to this puzzle lies in the observation that it makes no sense to talk about the state of each particle individually, because the combined wavefunction does not factorise as a consequence of the symmetrisation (itself a consequence of indistinguishability). The best one can do is say that the two states A and B are occupied, and that no state can be occupied twice.

3) It turns out, experimentally, that half-integer spin particles are always fermions, and integer-spin particles are always bosons. This connection can be derived mathematically in the context of quantum field theory (the spin-statistics theorem). But I now believe this has nothing to do with your questions anyway(?).

I hope this has been more helpful (at least I may win the prize for longest blog comment ever ;-). May I add you are rather quick to suspect deep inconsistencies.

9. One first thing
Daniel “May I add you are rather quick to suspect deep inconsistencies.” I do not believe that there is a deep inconsistency here. I only suspect that physicists get from stated core hypotheses to true predictions by using assumptions which are not stated as hypotheses and thus overstate the simplicity of their theory making its success more surprising.

There seem to be three physicists here. None seems impressed with my thought. I find their criticisms strikingly contradictory.

Pohl “The _definition_ of a fermion is that if you switch two identical particles, you introduce a minus sign in the wave function. From this you can derive the Pauli exclusion principle along the lines you discuss.”

Pohl agrees that I have correctly stated the connection between the antisymmetry of the wave function and the Pauli exclusion principle.

Thinkmarble “The Pauli Exclusion principle does not follow from psi=-psi.”

Daniel 2nd comment) “Contrary to what you have deduced, I think I agree with the substance of what Walt

Daniel 1) “Here’s your mistake. “Clearly nothing changes if it is rotated 360 degrees.” Indeed, and what doesn’t change is the absolute value squared of the wave function – which is physically the probability which you can actually detect. Note that the change of sign in the wavefunction has no effect on the observed probabilities.”

Daniel argues here that a change in the sign of a wave function is a matter of no interest. In making this argument he notes (correctly but irrelevantly) that “the change of sign in the wavefunction has no effect on the observed probabilities.” Now I think he did not mean that nothing can be deduced from the calculation that an operation has changed the sign of the wave function. However, the word “indeed” only makes sense if the sign of a wave function changing is interpreted as nothing changing and hence having the same implications as no change at all in the wave function. Pohl argues in plain English and Daniel (2nd comment) and Thinkmarble argue with technical language that the antisymmetry of the wave function of a system of indistinguishable fermions implies the Pauli exclusion principle. I deduce that Daniel’s first comment contradicts Pohl’s comment because it does. I conclude that Daniel did not mean what he wrote in his first comment.

Waldmann “In this case, they note that if there are two identical fermions a and b and one switches a and b, then the wave form describing the system of particles is multiplied by -1 (I think the proof is hard in any case I never made head nor tail of it).

In my post I used “identical” to mean “in the same state” and not to mean “indistinguishable”.
Note that “if” does not imply “if and only if”. I did not assert (and never thought) that the two fermions have to be in the same state (identical). My claim is a special case which follows from the claim “if there are two fermions a and b and one switches a and b, then the wave form describing the system of particles is multiplied by -1”

Thinkmarble The Pauli Exclusion principle does not follow from psi=-psi.
Instead it follows from the anticommuating relationship of creation operators.
Which follows from the fact that the wavefunction of an fermion has to be the eigenfunction of any transposition operator to the eigenvalue -1.

I think that Thinkmarble meant “a system of fermions” not “an fermion”. Given his definition of a transposition operator as an operator which exchanges alpha (a creation operator) and alpha’ (a creation operator), it doesn’t make sence to talk about the effect of a transposition operator on a single fermion.

So I believe I am correcting when I rewrite
” The Pauli Exclusion principle does not follow from psi=-psi.
Instead it follows from the anticommuating relationship of creation operators.
Which follows from the fact that the wavefunction of a system of indistinguishable fermions has to be the eigenfunction of any transposition operator to the eigenvalue -1.

Now that my claim is made more general (that the specific case from which one obtains the Pauli exclusion principle) and Thinkmarble’s claim is modified, I think they are identical. The differences are two — Thinkmarble uses technical terms, which I think are un-necessary and inappropriate for a general interest blog, and Thinkmarble claims that I said something different and thus made a mistake and show I am talking about a ” topic where you have no clue”.

10. Thinkmarble on said:

>I think that Thinkmarble meant “a system of
>fermions” not “an fermion”.

My mistake.

>There seem to be three physicists here.

Aint?t a physicists.

>Given his definition of a transposition operator
>as an operator which exchanges alpha (a creation
>operator) and alpha’ (a creation operator), it
>doesn’t make sence to talk about the effect of a
>transposition operator on a single fermion.

alpha and alpha? are sets of quantum number wich are exchanged by the transposition operator.
One of the effects it, that creations operators are exchanged.
Straight by the book the transposition operator is exchanging particle labels. The effect is the same for this case.

>Thinkmarble claims that I said something
>different and thus made a mistake and show I am
>talking about a ” topic where you have no clue”.

You did not say that P|psi^->=-|psi^->.
You said that |psi^->=-|psi^->
And that is a huge honking difference.

>You say that the phase of a wave doesn’t matter
>sicine all that matters is the magnitude
>psi*’psi. However, in the spin statistics
>theorem the result
>psi=-psi is interpreted as proving psi=0 so
>psi*’psi=0 so the probability is zero.

Applying an transposition operator to an antisymmetrized wavefunction is equal to multiplying the wavefunction with minus 1.
The resulting wavefunction (-|psi^->) is not the same as the original wavefunction (|psi^->), but the physical content is the same.
So this statment

>However, in the spin statistics theorem the
>result psi=-psi is interpreted as proving psi=0
>so psi*’psi=0 so the probability is zero.

is wrong.
You can?t just ignore operators because you want
to.

I won?t be able to answer for the next week.

11. Thinkmarble on said:

Been too fast, so I have to apologize.

You are correct that in the case of two fermions in the same state the permutations of the labels of these two fermions the relationship
P|psi^->=|psi^->=-|psi^-> holds.
And that is a very special case.

>As far as I can tell, the only natural quantum
>mechanical interpretation of spin implies that
>the wave form of a fermion is multiplied by -1
>if it is rotated 360 degrees.

Correct.

>Clearly nothing changes if it is rotated 360
>degrees.

Nothing we can measure.
But we can only measure the absolute square of an wavefunction, not the wavefunction itself.
The sign of the wavefunction obviously changes.
Spinors have to be rotated by 720? to get back to the same function.

Posting in a hurry

12. Daniel on said:

“There seem to be three physicists here. None seems impressed with my thought. I find their criticisms strikingly contradictory.”

They are not contradictory, everybody is trying to explain the same thing, but each on a slightly different technical level and from a different angle, in the hope that one of them unravels the knot in your thinking.

“Daniel (2nd comment) and Thinkmarble argue with technical language that the antisymmetry of the wave function of a system of indistinguishable fermions implies the Pauli exclusion principle. I deduce that Daniel.s first comment contradicts Pohl.s comment because it does. I conclude that Daniel did not mean what he wrote in his first comment.”

No, I meant what I wrote, and my second post was meant to elucidate what I had said. I hope you tried to follow the “technical language” I used, since that is the most basic formulation possible (it only uses multiplication of wavefunctions, and two particles, the minimum necessary to exchange them ;-). I was trying to show you how when your “psi=-psi” shorthand is expressed precisely, the contradictions you see disappear.

Again, and again from a different angle:

As Thinkmarble is also trying to explain to you, “You did not say that P|psi^->=-|psi^->. You said that |psi^->=-|psi^->. And that is a huge honking difference.”.

The act of exchanging particles (or rotating spin by 360 degrees), is mathematically expressed as an operator P, a gadget that takes a wavefunction as input and gives you as output the wavefunction after the operation has been performed: P(psi_before)=psi_after.

Now for fermionic particle exchange, and for rotating spin 360 degrees, the following equation holds: P(psi) = -psi.

ONLY if it is ALSO true that P(psi) = psi (which you have to prove independently, and which I did in my second comment for the case of two fermionic particles in identical states, and which does NOT hold for rotating half-integer spin by 360 degrees) does it follow that psi = -psi. From *this* equation you can then indeed, and indeed always, deduce that psi=0.

You write “I think that Thinkmarble meant ‘a system of fermions’ not ‘a fermion’.” Well particle exchange cannot really be investigated well with a single particle, can it? ‘A fermion’ is a kind of particle that has particular properties when you have more than one of them.

The rest of what I said was trying to explain to you that P(psi)=-psi makes sense even if you know that the act P has no observable consequences, since for the probabilities |P(psi)| = |psi| as desired.

“In making this argument he notes (correctly but irrelevantly) that ‘the change of sign in the wavefunction has no effect on the observed probabilities’. Now I think he did not mean that nothing can be deduced from the calculation that an operation has changed the sign of the wave function.”

It is not irrelevant as that fact is the essential reason why the condition P(psi)=-psi is a meaningful possibility for wavefunctions. IF psi were directly observable, and the act of P changes nothing observable, then it would HAVE to be true that P(psi)=psi, and then if you also imposed P(psi)=-psi then you would indeed be describing nothing (since that can only hold for psi=0). In quantum mechanics that is NOT the case, which, as I tried to explain in great detail, allows the possibility of such things as fermions and spin 1/2 particles. I deduced some observational consequences of P(psi)=-psi for you in my second post. Note the key difference between the statement that “performing P on a system has no observable consequences” and the statement that “particles obeying P(psi)=-psi can be distinguished by observation from particles obeying P(psi)=psi”: The first is a statement about one system, the second a comparative statement between two different systems.