| Topic: |
Science > Physics |
| User: |
"Henning Makholm" |
| Date: |
08 Mar 2006 04:23:13 PM |
| Object: |
Fermions |
The archetypical massive boson, if many popular texts are to be
believed, is the He-4 atom. Being bosons means that two He-4 atoms are
allowed to occupy the same state simultaneously. But these atoms have
internal parts, so if two atoms are in the same states, then e.g. the
spin-up electron in one atom must be in the same state as the spin-up
electron in the other atom. And _electrons_ are not allowed to do
that. An apparent paradox.
This has bothered me for a while, and I have concluded that I need to
understand what's up with bosons and fermions at a more qualified
level than the popular books can reach. Unfortunately I have not been
lucky in my (trial-and-error-based) attempts to acquire textbooks that
could clear up the mystery. I have the Feynman Lectures and
P.C.W.Davies & D.S.Betts _Quantum Mechanics_. The latter dedicates a
single page to identical particles, but immediately loses me when it
apparently tries to conclude from |a|²=|b|² that either a=b or a=-b
must hold, when a and b may be complex(!). Feynman provides more
words, but unfortunately only in one of the early informal chapters,
from which I have trouble extracting general formulas.
Combining with allusions from popular texts, I gather that that the
ferminess of electrons means that "the amplitude changes sign if two
electrons are swapped". This implies a formula of the general shape
(*) f(P(a)) = -f(a)
where P is some "position exchange" operator, but I am not quite sure
what P operates on, or what f and a are in the equation (except that f
must be something that produces an amplitude), or even if (1) should
be understood as a constraint on f, a constraint on a, or both.
My best (but unsuccesful) attempt to make sense of (1) is that f must
be a state vector in a multi-particle system - i.e. f maps base states
to amplitudes, and a must be a base state. Assuming for simplicity
that there is only one kind of particles in the system, a base state a
would be identified with a multiset of positions, e.g.
a = | 1 electron at position x and 1 electron at position x' >
However, already here things break down, because how would P operate
on such an a? No matter how much I swap the electrons, the multiset
of their positions is still unchanged, i.e.
P(a) = | 1 electron at position x and 1 electron at position x' > = a
and then (*) reduces to f(a) = 0 for arbitrary a, obviously nonsense.
Alternatively, this could be avoided if a were somehow an _ordered_
sequence of particle positions, a = |x;x'>, P(a) = |x';x>. Then (*)
could be read as a restriction on the set of f's that are considered
physical.
However, this also breaks down if I am to follow the rule of adding
the amplitudes of all physically indistinguisable states before
squaring to get a probability. Even if I maintain a distinction
between |x;x'> and |x';x> internally in the theory, they are surely
indistinguishable by observation, and as (*) says that adding them
gives zero, this option leads to every observation having zero
probability - not a meaningful outcome either.
I am obviously missing something, but what?
(In lieu of explanations here, I would also welcome references to a
book I could buy to see this explained slowly and patiently. I am
looking for something suitable for self-study towards an "advanced
laymans' understanding", *with* mathematics but preferably without
endless sequences of particular calculations that do not serve to
illustrate general principles).
--
Henning Makholm "Den nyttige hjemmedatamat er og forbliver en myte.
Generelt kan der ikke peges på databehandlingsopgaver af
en sådan størrelsesorden og af en karaktér, som berettiger
forestillingerne om den nye hjemme- og husholdningsteknologi."
.
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| User: "Edward Green" |
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| Title: Re: Fermions |
09 Mar 2006 06:20:02 PM |
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Henning Makholm wrote:
The archetypical massive boson, if many popular texts are to be
believed, is the He-4 atom. Being bosons means that two He-4 atoms are
allowed to occupy the same state simultaneously. But these atoms have
internal parts, so if two atoms are in the same states, then e.g. the
spin-up electron in one atom must be in the same state as the spin-up
electron in the other atom. And _electrons_ are not allowed to do
that. An apparent paradox.
I wonder if they are only "bosons" at some degree of approximation:
that degree being, to the extent we ignore internal structure. But let
me not tarnish the things I am more confident of with undue
speculation.
This has bothered me for a while, and I have concluded that I need to
understand what's up with bosons and fermions at a more qualified
level than the popular books can reach. Unfortunately I have not been
lucky in my (trial-and-error-based) attempts to acquire textbooks that
could clear up the mystery. I have the Feynman Lectures and
P.C.W.Davies & D.S.Betts _Quantum Mechanics_. The latter dedicates a
single page to identical particles, but immediately loses me when it
apparently tries to conclude from |a|=B2=3D|b|=B2 that either a=3Db or a=
=3D-b
must hold, when a and b may be complex(!).
I know what you mean! The argument seems to go something like this:
If we interchange the coordinates of identical particles the final
situation is physically indistinguishable from the initial. Hence,
they argue, the squared magnitude of the wave function -- corresponding
to probability (density) of physical measurement -- must be identical
under this swap. Hence, they argue (in your terminology) either a =3D b,
or a =3D -b. But, you say, why couldn't a =3D ib !? We must add one more
condition to avoid this possibility: that the multiplicand of the wave
function after the swap is a constant depending only on the species of
particle, and not on the particle coordinates. In case "a =3D ib" then,
taking the final coordinates as our starting coordinates and swapping
again, we would have instead "a =3D -ib": so this condition is not met.
Feynman provides more
words, but unfortunately only in one of the early informal chapters,
from which I have trouble extracting general formulas.
Combining with allusions from popular texts, I gather that that the
ferminess of electrons means that "the amplitude changes sign if two
electrons are swapped". This implies a formula of the general shape
(*) f(P(a)) =3D -f(a)
where P is some "position exchange" operator, but I am not quite sure
what P operates on, or what f and a are in the equation (except that f
must be something that produces an amplitude), or even if (1) should
be understood as a constraint on f, a constraint on a, or both.
Your f is the wave function, usually written "psi" (the Greek letter).
Psi is a function of some complete set of coordinates, such as (for a
single spinless particle) position coordinates. For two such particles
psi is a function of two such sets of position coordinates, and the
"position exchange" operator swaps the values of these coordinates
appearing in the first-particle-slot and the second-particle-slot:
e=2Eg.,
f(1,1,1,0,0,0) --> f(0,0,0,1,1,1)
My best (but unsuccessful) attempt to make sense of (1) is that f must
be a state vector in a multi-particle system - i.e. f maps base states
to amplitudes, and a must be a base state. Assuming for simplicity
that there is only one kind of particles in the system, a base state a
would be identified with a multiset of positions, e.g.
a =3D | 1 electron at position x and 1 electron at position x' >
However, already here things break down, because how would P operate
on such an a? No matter how much I swap the electrons, the multiset
of their positions is still unchanged, i.e.
P(a) =3D | 1 electron at position x and 1 electron at position x' > =3D=
a
and then (*) reduces to f(a) =3D 0 for arbitrary a, obviously nonsense.
Well, you've already put the "physical indistinguishability" of the
interchange of the electrons in here, so this formulation can't even
inquire what happens if we transpose the positions of electron 1 and 2.
The actual theory, on the other hand, hedges the bet: we label the
electrons individually, and then require that the probability density
for finding the particles must be identical for
electron 1 at position x and electron 2 at position x',
and
electron 1 at position x' and electron 2 at position x.
However, we still allow an "unobservable" change in the internal
working of the theory: the wave function goes to minus itself under
this symmetry operation. "Unobservable" is in quotes because while we
say, on the one hand, that we can't tell the difference between these
two states, we apparently _can_ tell the difference between a theory
which allows for such internally distinct states and one which does
not. We can "observe" the internal presence of the sign change in the
other, observable, consequences of the theory.
Alternatively, this could be avoided if a were somehow an _ordered_
sequence of particle positions, a =3D |x;x'>, P(a) =3D |x';x>. Then (*)
could be read as a restriction on the set of f's that are considered
physical.
Yes.
However, this also breaks down if I am to follow the rule of adding
the amplitudes of all physically indistinguishable states before
squaring to get a probability.
Hmm... where did you get this rule? It sounds like something Feynman
might have said, but it is not correctly applied here. What you say
applies when we have different histories which end in identical final
states -- for example: a particle arriving at a particular location via
multiple possible paths.
Even if I maintain a distinction
between |x;x'> and |x';x> internally in the theory, they are surely
indistinguishable by observation, and as (*) says that adding them
gives zero, this option leads to every observation having zero
probability - not a meaningful outcome either.
I am obviously missing something, but what?
Well, we don't add these amplitudes. Otherwise, this is a correct
interpretation.
Going back to the original argument above, I'd say on reflection that
the whole thing is rather weak. If, on the one hand, we argue that
physically indistinguishable outcomes should be indistinguishable in
theory, we are required to take the states as you first suggested: 1
electron at position x, 1 electron at position x'. But this state
space turns out not to be too poor to capture physical behavior.
If, on the other hand, we accept that any manner of seemingly invisible
internal machinery may operate in our theory so long as, in the end,
we make physical predictions, then there seems to be no reason to limit
ourselves to +/- 1 as the multiplicands when we swap coordinates. Why
not a transformation of the form:
psi -> exp(i phi(x,x')) psi ?
The only consistency requirement on the phase factor would seem to be
that
phi(x,x') =3D - phi(x',x)
that is, flipping back gets us back to where we started.
For that matter, why is there any physical requirement that the modulus
be the same after the swap? There isn't. If all we can measure is "1
electron at x and 1 electron at x" it doesn't follow that the
probability density at the two internal conditions consistent with this
observation must be equal. It only follows that the _sum_ of these
densities (adding densities now, not amplitudes) must agree with our
observed probability density for the space of physically
distinguishable outcomes. The two possibilities could be internally
asymmetric.
The attempt at logical requirement seems weak, and maybe all we can say
is that there seem to be classes of particles acting like the fermions
and bosons suggested by this development, and that more "unobservable"
structure seems redundant whereas less seems inadequate, so that the
choice of multiplying the wave function by +/- 1 on swapping the
coordinates of identical particles seems to be just right.
(In lieu of explanations here, I would also welcome references to a
book I could buy to see this explained slowly and patiently. I am
looking for something suitable for self-study towards an "advanced
laymans' understanding", *with* mathematics but preferably without
endless sequences of particular calculations that do not serve to
illustrate general principles).
Your motivation is precisely my motivation, sir, but I have no book
recommendations. Maybe we should become coauthors.
.
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| User: "Henning Makholm" |
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| Title: Re: Fermions |
10 Mar 2006 06:52:25 PM |
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Scripsit "Edward Green" <spamspamspam3@netzero.com>
Henning Makholm wrote:
(*) f(P(a)) = -f(a)
where P is some "position exchange" operator, but I am not quite sure
what P operates on, or what f and a are in the equation
Your f is the wave function, usually written "psi" (the Greek letter).
I suspected as much, but did not want to presuppose a particular
answer when I stated my problem, hence the more anonymous "f".
However, we still allow an "unobservable" change in the internal
working of the theory: the wave function goes to minus itself under
this symmetry operation. "Unobservable" is in quotes because while we
say, on the one hand, that we can't tell the difference between these
two states, we apparently _can_ tell the difference between a theory
which allows for such internally distinct states and one which does
not.
I fail to see quite how. I reason thus: Okay, let us take base states
of the form |x;x'>, that is, with ordered pairs of positions, and then
for all x and x' impose the additional condition
<psi|x;x'> = -<psi|x',x>
on the psi's we're willing to contemplate. These additional conditions
are all _linear_ relations in the vector space of possible states, so
we might as well quotient them out of the space once and for all.
This gives us a quotient space that is quite isomorphic to the
multiset representation I first had in mind (except sets rather than
multiset, because those multisets with more than one particle
somewhere get quotiented out of existence too). And if the physics
is symmetric under which order we choose to label the electrons in,
we can pull any Hamiltonian back through the quotienting operation
too, such that every theory in the "redundant" ordered-pair
representation gives rise to an _equivalent_ theory with multiset
base states.
Hmm.. there would be a peculiarity with the equivalent theory if we
have more than one dimension, namely that one can find a path through
the "ordered" state space that _continuously_ exchanges two particles
without them passing through each other. The image of this path in the
quotiented space would go from {x,x'} to itself but the amplitude
would phase shift by 180° along it - so a nontrivial state in the
quotient space cannot be continuous.
However, I am reluctant to believe that this is the full reasoning.
Namely, a popular example of Bose statistics in action concerns the
distribution of photons among modes within a resonant cavity, and
those modes are _discrete_. If continuity of the wave function were a
primary ingredient in distinguising bosons from fermions, the photon
cavity example wouldn't work very well.
However, this also breaks down if I am to follow the rule of adding
the amplitudes of all physically indistinguishable states before
squaring to get a probability.
Hmm... where did you get this rule? It sounds like something Feynman
might have said,
Yes.
but it is not correctly applied here. What you say applies when we
have different histories which end in identical final states -- for
example: a particle arriving at a particular location via multiple
possible paths.
That makes sense.
If, on the other hand, we accept that any manner of seemingly invisible
internal machinery may operate in our theory so long as, in the end,
we make physical predictions, then there seems to be no reason to limit
ourselves to +/- 1 as the multiplicands when we swap coordinates.
I have found brief references to "anyons" that might exist in
two-dimensional worlds, and where the the phase may shift less than pi
(but more than 0) in a coordinate exchange. Apparently the idea is
that in 2D one can distinguish between a continuous clockwise exchange
and a continuous anticlockwise exchange, and two clockwise exchanges
in sequence is not homotopic to zero (in contrast to a double exchange
in 3D), so their combined phase shift need not be a multiple of 2pi.
I am looking for something suitable for self-study towards an
"advanced laymans' understanding", *with* mathematics but
preferably without endless sequences of particular calculations
that do not serve to illustrate general principles).
Your motivation is precisely my motivation, sir, but I have no book
recommendations. Maybe we should become coauthors.
Hm, I am not sure I would trust a text written by the likes of myself :-)
--
Henning Makholm "I ... I have to return some videos."
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| User: "Edward Green" |
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| Title: Re: Fermions |
10 Mar 2006 11:25:53 PM |
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Henning Makholm wrote:
I am torn between the desire to give a thoughtful answer, and the
eagerness to return an intelligible signal from cyberspace...
Eagerness wins. ;-)
Scripsit "Edward Green" <spamspamspam3@netzero.com>
<...>
However, we still allow an "unobservable" change in the internal
working of the theory: the wave function goes to minus itself under
this symmetry operation. "Unobservable" is in quotes because while we
say, on the one hand, that we can't tell the difference between these
two states, we apparently _can_ tell the difference between a theory
which allows for such internally distinct states and one which does
not.
I fail to see quite how. I reason thus: Okay, let us take base states
of the form |x;x'>, that is, with ordered pairs of positions, and then
for all x and x' impose the additional condition
<psi|x;x'> = -<psi|x',x>
on the psi's we're willing to contemplate.
I am a little bothered by your notation. This looks like the famous
bra-ket notation, but I'm not quite clever enough to see if you are
using it in a formally correct but possibly unexpected way, or not.
Ok... "|x,x'>" is an element of the "position basis" -- that much I
see. But are you saying the abstract inner product of such a state
with psi is precisely what the simple-minded would write "psi(x,x')"?
Yes, I guess I see this.
These additional conditions
are all _linear_ relations in the vector space of possible states, so
we might as well quotient them out of the space once and for all.
I don't quite see or understand this ... which of course is a result of
my colossal ignorance. Ok... once again I think I see on reflection:
our symmetry condition restricts us to a linear subspace of our
original space, so we may as well restrict our attention to elements of
this subspace at the beginning. Sounds reasonable.
This gives us a quotient space that is quite isomorphic to the
multiset representation I first had in mind (except sets rather than
multiset, because those multisets with more than one particle
somewhere get quotiented out of existence too).
Huh? Why?
You seem to be suggesting that we can take as a basis the set of all
states like
"|an electron at x and an electron at x'>"
Ok. I was in fact unfamiliar with "multiset", but on looking it up I
see it is a fancy way of describing a box which contains a collection
of unordered objects, some of which may be identical. This seems to be
an accurate description of bosons, particularly when occupying a
discrete set of states. Now, are you suggesting that for fermions,
since we can only have one of each "kind" of particle/electron, we
simply have a set?
And if the physics
is symmetric under which order we choose to label the electrons in,
we can pull any Hamiltonian back through the quotienting operation
too, such that every theory in the "redundant" ordered-pair
representation gives rise to an _equivalent_ theory with multiset
base states.
Lets see... I can't quite follow you, so I'm going to do the usual
thing in such cases, and go back and say something I'd like to say
instead; ahem, "approach the problem from a different angle".
Your "multiset" description is, I believe, precisely the accepted way
to write a (multi-particle) state space for bosons: so many particles
in (single particle basis) state 0, so many in state 1, ... , etc. So,
your idea that any theory in which particle ordering is unobservable
must be isomorphic to one with this set of basis states seems to be
correct: that's how it's done.
Now, given fermions, our "multiset" state vector is further restricted
to objects which look like "...,0,0,0,1,0,0,1,0,..." etc. I.e., each
(single particle) basis state is either occupied, or not. This in turn
seems to be a correct -- well, not totally incorrect -- way of
describing an accepted treatment of many-particle fermion systems.
So, your ansatz is correct, but you've just used it to prove such a
description leads to nonsense: am I correct? Well, we seem to have a
problem here. :-)
I'd have to think about this methodically -- and probably harder than I
have the time or ability to muster right now -- to sort this out, but,
just off the top of my head, I think the state descriptions we've just
given already have the symmetry in 'em, so to speak. We've already
"quotiented out" the superfluous information, as you put it, so we
can't go back in and "swap the coordinates of particle 1 with particle
2" anymore and make sense of the result -- at least in the case of
fermions.
Alright, that's not really an argument, but I think I'm sticking to my
guns for now: the symmetry of the full "un-quotiented" wave function
really does seem to have observable physical consequences -- even
though we later may use this symmetry to reduce the size of the state
space and make its original manifestation invisible.
You know, it may be widely hand-wavingly accepted that if two physical
states are experimentally indistinguishable, it is "meaningless" to
distinguish them in theory: but this bit of sloganeering conflates two
cases. We may indeed have different equivalent descriptions of a
single physical state, or we may have physically indistinguishable
physical states. This latter case has a special name: "symmetry".
Because, for example, we can't tell if somebody has rotated a hydrogen
molecule aligned with the x axis about the y axis while our back was
turned, doesn't mean that the rotation of hydrogen atoms about this
axis may not have observable physical consequences.
<...>
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| User: "Henning Makholm" |
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| Title: Re: Fermions |
16 Mar 2006 07:46:05 AM |
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Scripsit "Edward Green" <spamspamspam3@netzero.com>
Henning Makholm wrote:
Now, given fermions, our "multiset" state vector is further restricted
to objects which look like "...,0,0,0,1,0,0,1,0,..." etc. I.e., each
(single particle) basis state is either occupied, or not. This in turn
seems to be a correct -- well, not totally incorrect -- way of
describing an accepted treatment of many-particle fermion systems.
So, your ansatz is correct, but you've just used it to prove such a
description leads to nonsense: am I correct?
No, I don't think it least to nonsense, but I think I have
proved^H^H^H^H^H^Hasserted that it can lead to _the same_ results as
staring from a basis where "hedge the bet' as you put it, and take the
n electrons in some particular order.
My problem is, roughly:
1) *I* think that the "...,0,0,0,1,0,0,1,0,..." basis is a more
intuitive one to write down, given that the particles in question
are presumed identical.
2) In contrasts, physicists seem to prefer starting with ordered sets
of positions.
What I need to figure out is to choose correctly between various
explanations of this:
a) The two descriptions are mathematically equivalent, but the
physicists' one is somehow more convenient for further
calculations.
b) The two descriptions are *not* equivalent, and I have erred
somewhere when I thought I proved that they are.
c) Mainstream physics is all wrong, and THE FOOLS! THEY SHALL BOW TO ME!
While I don't think that option (c) should be entertained without
extraordinary evidence, I'm less certain of (a) versus (b).
Alright, that's not really an argument, but I think I'm sticking to my
guns for now: the symmetry of the full "un-quotiented" wave function
really does seem to have observable physical consequences --
That would be the same as asserting (b), because the equivalence in
(a) by definition excludes the possibility of observable differences.
--
Henning Makholm "En tapper tinsoldat. En dame i
spagat. Du er en lykkelig mand ..."
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| User: "Edward Green" |
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| Title: Re: Fermions |
18 Mar 2006 06:43:57 PM |
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Henning Makholm wrote:
Greetings.
<....>
Let's take it as an experimental fact that there seem to be two classes
of particles in the world. For one, labled bosons, we may label
multiple particle basis states such as:
"...0,0,0,2,0,0,3,1,0,0,..."
whereas for the other, labeled fermions, our basis states look like:
"...0,0,0,1,0,0,1,0,0,..."
i.e., given an otherwise unspecified set of "single particle basis
states" we have one kind of particle which can draw occupancies the
non-negative integers, whereas the other kind of particle may only use
{0,1,}.
Ok so far?
(I suddenly have a doubt these descriptions are physically complete,
but leave that aside a second).
Now, we have a second kind of description of such systems, which in
general looks like:
"psi(X_1;X_2;...;X_n)"
which is supposed to represent a multiple particle wave function : each
of the "X_i" represents a complete set of variables for a single
particle -- say, the position and the spin state.
Now, we make an argument that upon an operation like:
"psi(X_1;X_2;...;X_n)" -> "psi(X_2;X_1;...;X_n)"
only one of two things are reasonable: either the amplitude (value of
psi) is unchanged, or else it goes to minus itself. Leaving aside
whether these are the only two "reasonable" symmetries of the multiple
particle wave function, we use this assertion to heuristically deduce
the posibility of the two representations above: our _new_ "multiple
particle basis states" are either isomorphic to a string of 0's and 1's
or else to strings of non-negative integers.
It's clear from this brief and vague development that the "+/- 1" thing
applies to the _second_ representation, and not the first. The +/- 1
thing is part of a heuristic argument motivating the first
representation from the second: we can't ask again in the simplified
representation "what happens to the representation under particle
interchange" -- it's no longer a meaningful question in our theory.
My problem is, roughly:
1) *I* think that the "...,0,0,0,1,0,0,1,0,..." basis is a more
intuitive one to write down, given that the particles in question
are presumed identical.
2) In contrasts, physicists seem to prefer starting with ordered sets
of positions.
I don't know that that's true. You may be working from a very limited
selection of literature in determining what "physicists" do ... no
offense intended, my knowledge of the literature is even more scanty!
What I need to figure out is to choose correctly between various
explanations of this:
a) The two descriptions are mathematically equivalent, but the
physicists' one is somehow more convenient for further
calculations.
b) The two descriptions are *not* equivalent, and I have erred
somewhere when I thought I proved that they are.
Well, clearly mathematically the one description has more structure.
The question is, is this redundant structure? On the one hand, even if
the simpler structure is physically complete, we may have deduced what
appears as assumption in this structure from simpler -- or at least
different -- assumptions. That may have weight even if the models are
physically indistinguishable. On the other hand, I'm not sure if the
simpler description is really physically complete: what about the
relative phase of the single particle basis states? I can't quite
decide if that question makes sense right now.
c) Mainstream physics is all wrong, and THE FOOLS! THEY SHALL BOW TO ME!
Yeah, yeah, yeah. Go post this in that nutso ng "sci.physics". Oh...
this _is_ sci.physics. Oops. ;-)
While I don't think that option (c) should be entertained without
extraordinary evidence, I'm less certain of (a) versus (b).
Alright, that's not really an argument, but I think I'm sticking to my
guns for now: the symmetry of the full "un-quotiented" wave function
really does seem to have observable physical consequences --
That would be the same as asserting (b), because the equivalence in
(a) by definition excludes the possibility of observable differences.
This argument would be on the level of axiom depth, as above. Whether
that would be an observable difference is a matter of opinion, or
semantics.
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| User: "Gregory L. Hansen" |
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| Title: Re: Fermions |
12 Mar 2006 09:56:54 AM |
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In article <878xrko9ta.fsf@kreon.lan.henning.makholm.net>,
Henning Makholm <henning@makholm.net> wrote:
The archetypical massive boson, if many popular texts are to be
believed, is the He-4 atom. Being bosons means that two He-4 atoms are
allowed to occupy the same state simultaneously. But these atoms have
internal parts, so if two atoms are in the same states, then e.g. the
spin-up electron in one atom must be in the same state as the spin-up
electron in the other atom. And _electrons_ are not allowed to do
that. An apparent paradox.
This has bothered me for a while, and I have concluded that I need to
understand what's up with bosons and fermions at a more qualified
level than the popular books can reach. Unfortunately I have not been
....
In the first paragraph you seem to forget that there's another quantum
number involved-- which atom is the electron associated with? Both
electrons can be in the n=0, S=0 state if one is associated with atom one
and the second is associated with atom two.
You've suggested below that the wavefunction must be a vector in a
multiparticle system, and that's true. We really have
psi(x,y,z,x',y',z',Q)
where the first set of cordinates correspond to particle 1, the second set
to particle 2, and Q is the remaining parameters relevant to the system.
If we have two atoms separated enough that they can be treated separately,
then we can write
psi1(x,y,z)phi1(Q1)psi2(x',y',z')phi2(Q2) = A*psi1(x,y,z)
where on the right hand side everything else was summed and integrated
over and can be presumed unaffected by psi1(x,y,z).
If the atoms are closer, as in a hydrogen molecule, we can't pretend they
separate cleanly like that. We'd have to write out a wavefunction that
explicitly includes both electrons. But we can choose a basis of
single-atom states, and write something like
|psi> = \sum_i,j A_ij |one_i> (x) |two_j>
with (x) the Cartesian product that combines two vector spaces.
That's sort of a half-assed hand-wavy explanation. Ramamurti Shankar
devotes Chapter 10 to multi-particle systems, and gives enough attention
to things like Cartesian products and vector spaces that you should be
able to figure out what's really going on, if you don't mind ponying up
for another textbook or can borrow it from a library.
--
"Very well, he replied, I allow you cow's dung in place of human
excrement; bake your bread on that." -- Ezekiel 4:15
.
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| User: "Henning Makholm" |
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| Title: Re: Fermions |
16 Mar 2006 07:31:58 AM |
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Scripsit (Gregory L. Hansen)
Henning Makholm <henning@makholm.net> wrote:
The archetypical massive boson, if many popular texts are to be
believed, is the He-4 atom. Being bosons means that two He-4 atoms are
allowed to occupy the same state simultaneously. But these atoms have
internal parts, so if two atoms are in the same states, then e.g. the
spin-up electron in one atom must be in the same state as the spin-up
electron in the other atom. And _electrons_ are not allowed to do
that. An apparent paradox.
In the first paragraph you seem to forget that there's another quantum
number involved-- which atom is the electron associated with? Both
electrons can be in the n=0, S=0 state if one is associated with atom one
and the second is associated with atom two.
I think that this solution amounts to begging the question. - At least
it begs the question I was trying to express: The electron, by itself,
knows nothing of atoms - how can it know that it is supposed to behave
as if it had a "which atom" quantum number? Or in other words, how can
it be that the approximation where we asign electrons to particular
atoms is even approximately valid compared to what we'd get if we just
wrote down the Schrödinger equation for two alpha particles and four
electrons and investigated from first principles what would happen
then? The Schrödinger equation certainly don't presuppose any atomic
quantum numbers; rather it is supposed to _explain_ them.
Of course I'm not sure that the 18-dimensional case of 2 alpha + 4 e
is tractable even numerically, but if I manage to learn how to write
it down in the first place I might try to see which understanding
numeric solutions of lower-dimensional model systems could lead me to.
That's sort of a half-assed hand-wavy explanation. Ramamurti Shankar
devotes Chapter 10 to multi-particle systems,
I am unfamiliar with "Ramamurti Shankar". Do you have a fuller reference?
--
Henning Makholm "Jeg forstår mig på at anvende sådanne midler på
folks legemer, at jeg kan varme eller afkøle dem,
som jeg vil, og få dem til at kaste op, hvis det er det,
jeg vil, eller give afføring og meget andet af den slags."
.
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| User: "Gregory L. Hansen" |
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| Title: Re: Fermions |
17 Mar 2006 06:12:11 PM |
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In article <87slpisegh.fsf@kreon.lan.henning.makholm.net>,
Henning Makholm <henning@makholm.net> wrote:
Scripsit (Gregory L. Hansen)
Henning Makholm <henning@makholm.net> wrote:
The archetypical massive boson, if many popular texts are to be
believed, is the He-4 atom. Being bosons means that two He-4 atoms are
allowed to occupy the same state simultaneously. But these atoms have
internal parts, so if two atoms are in the same states, then e.g. the
spin-up electron in one atom must be in the same state as the spin-up
electron in the other atom. And _electrons_ are not allowed to do
that. An apparent paradox.
In the first paragraph you seem to forget that there's another quantum
number involved-- which atom is the electron associated with? Both
electrons can be in the n=0, S=0 state if one is associated with atom one
and the second is associated with atom two.
I think that this solution amounts to begging the question. - At least
it begs the question I was trying to express: The electron, by itself,
knows nothing of atoms - how can it know that it is supposed to behave
as if it had a "which atom" quantum number? Or in other words, how can
it be that the approximation where we asign electrons to particular
atoms is even approximately valid compared to what we'd get if we just
wrote down the Schrödinger equation for two alpha particles and four
electrons and investigated from first principles what would happen
then? The Schrödinger equation certainly don't presuppose any atomic
quantum numbers; rather it is supposed to _explain_ them.
Both electrons might be in 1S spin up states, but one has a 1S spin up
state that centered over here, and the other as a 1S spin up state that's
centered over there. Not the same state.
Of course I'm not sure that the 18-dimensional case of 2 alpha + 4 e
is tractable even numerically, but if I manage to learn how to write
it down in the first place I might try to see which understanding
numeric solutions of lower-dimensional model systems could lead me to.
That's sort of a half-assed hand-wavy explanation. Ramamurti Shankar
devotes Chapter 10 to multi-particle systems,
I am unfamiliar with "Ramamurti Shankar". Do you have a fuller reference?
Shankar, "Principles of Quantum Mechanics" 2nd ed., Plenum Press, 1994.
--
"Suppose you were an idiot... And suppose you were a member of
Congress... But I repeat myself." - Mark Twain
.
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| User: "Henning Makholm" |
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| Title: Re: Fermions |
18 Mar 2006 07:39:48 AM |
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Scripsit (Gregory L. Hansen)
Henning Makholm <henning@makholm.net> wrote:
Scripsit (Gregory L. Hansen)
Henning Makholm <henning@makholm.net> wrote:
The archetypical massive boson, if many popular texts are to be
believed, is the He-4 atom. Being bosons means that two He-4 atoms are
allowed to occupy the same state simultaneously. But these atoms have
internal parts, so if two atoms are in the same states, then e.g. the
spin-up electron in one atom must be in the same state as the spin-up
electron in the other atom. And _electrons_ are not allowed to do
that. An apparent paradox.
In the first paragraph you seem to forget that there's another quantum
number involved-- which atom is the electron associated with?
I think that this solution amounts to begging the question. - At least
it begs the question I was trying to express: The electron, by itself,
knows nothing of atoms - how can it know that it is supposed to behave
as if it had a "which atom" quantum number?
Both electrons might be in 1S spin up states, but one has a 1S spin up
state that centered over here, and the other as a 1S spin up state that's
centered over there. Not the same state.
But "over here" is the same place as "over there" when the two atoms
sit atop each other.
--
Henning Makholm "Hør, hvad er det egentlig
der ikke kan blive ved med at gå?"
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| User: "boson boss" |
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| Title: Re: Fermions |
08 Mar 2006 08:54:33 PM |
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Trial and error? Like, 250$ for a book in nuclear physics? Every
measurement is unrepeatable, so how would they verify same spin here
and there separately? But if electrons can't have same spin, it means
that electrons won't collapse through each other's orbitals and each
other. Instead whole of Big Monster Boson in single measurement is the
2 of He4. So how did they know it was 2 of He4 in the first place?
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