[math-fun] Bozonic boson question
"An important characteristic of bosons is that their statistics do not restrict the number that can occupy the same quantum state" https://en.wikipedia.org/wiki/Boson Ok, suppose that I want to _weigh_ one boson. I guess that most bosons are pretty light, so I'll have to weigh a bunch of identical bosons (in "the same" quantum state) & divide by their number. Q: Does this work? Is there any additional energy required to hold them together that would show up as additional mass? Or does an individual boson weigh more singly than when together in "the same" quantum state? How big is the biggest boson, anyway?
On Mon, Jun 23, 2014 at 7:53 PM, Henry Baker <hbaker1@pipeline.com> wrote:
How big is the biggest boson, anyway?
If you mean single particle, the largest one we know of is the Higgs boson, with a mass of ~125 GeV. Fermions can pair up to form bosonic quasiparticles, giving rise to superconductivity and superfluidity. Cores of neutron stars are in a superfluid state. -- Mike Stay - metaweta@gmail.com http://www.cs.auckland.ac.nz/~mike http://reperiendi.wordpress.com
Some bosons are quite massive. Like the W and Z gauge bosons. You measure their mass by reconstructing the kinematics of their relativistic decay. Some bosons are not so massive, but it's probably very difficult measure their mass directly, even though they are all in the same (or similar) quantum state. See Cooper Pairs. http://en.wikipedia.org/wiki/Gauge_boson#Massive_gauge_bosons http://en.wikipedia.org/wiki/Cooper_pair On Mon, Jun 23, 2014 at 7:53 PM, Henry Baker <hbaker1@pipeline.com> wrote:
"An important characteristic of bosons is that their statistics do not restrict the number that can occupy the same quantum state"
https://en.wikipedia.org/wiki/Boson
Ok, suppose that I want to _weigh_ one boson. I guess that most bosons are pretty light, so I'll have to weigh a bunch of identical bosons (in "the same" quantum state) & divide by their number.
Q: Does this work? Is there any additional energy required to hold them together that would show up as additional mass? Or does an individual boson weigh more singly than when together in "the same" quantum state?
How big is the biggest boson, anyway?
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I always cringe a little bit when people refer to particles being "in the same quantum state”. Were I a better citizen, I would edit wikipedia something along these lines: In a multi-particle quantum system the individual particles do not have a “state" assigned to them, in the same sense as some property like mass or charge. There is always just a single quantum “state”, and this state describes the entire system. A quantum state of two particles, for example, is a single wave function psi(x1,x2), where x1 and x2 are the positions of the particles. There is a small class of wave functions for which one can say there is an assignment of particles to states. This is when the multi-particle wave function factorizes; wave functions in this class, for two particles, would be written as psi1(x1) psi2(x2). However, wave functions with this structure are never the states of definite energy (solutions of the time-independent Schroedinger equation) when the particles interact. The property of being bosons or fermions does not restrict how particles are assigned to states, but restricts the permutation symmetry of the wave function for the system. The wave function for two bosons has the property psi(x1,x2) = psi(x2,x1), and for fermions, psi(x1,x2) = -psi(x2,x1). An arbitrary wave function can be symmetrized or antisymmetrized, as needed for bosons or fermions. Antisymmetrizing the product wave function psi(x1)psi(x2) yields no wave function at all (the null vector of linear algebra) — it is for this reason that it is sometimes said fermions cannot be in the same state. -Veit On Jun 23, 2014, at 10:53 PM, Henry Baker <hbaker1@pipeline.com> wrote:
"An important characteristic of bosons is that their statistics do not restrict the number that can occupy the same quantum state"
https://en.wikipedia.org/wiki/Boson
Ok, suppose that I want to _weigh_ one boson. I guess that most bosons are pretty light, so I'll have to weigh a bunch of identical bosons (in "the same" quantum state) & divide by their number.
Q: Does this work? Is there any additional energy required to hold them together that would show up as additional mass? Or does an individual boson weigh more singly than when together in "the same" quantum state?
How big is the biggest boson, anyway?
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I found Veit's description admirably clear, except for the sentence An arbitrary wave function can be symmetrized or antisymmetrized, as needed for bosons or fermions. I feel like I'm lacking context for the operation of symmetrization/antisymmetrization, but I'm not sure if I can explain what I mean by context. I'm guessing Veit means something like "If you symmetrize it, you'll get a wave-function that's just as descriptive of reality; the universe won't notice the difference." Or "If you symmetrize it, you'll still get a solution to the relevant equation, but if it isn't unitary then it won't correspond to anything we see in the universe." Veit, can you clarify? Jim Propp On Tuesday, June 24, 2014, Veit Elser <ve10@cornell.edu> wrote:
I always cringe a little bit when people refer to particles being "in the same quantum state”. Were I a better citizen, I would edit wikipedia something along these lines:
In a multi-particle quantum system the individual particles do not have a “state" assigned to them, in the same sense as some property like mass or charge. There is always just a single quantum “state”, and this state describes the entire system. A quantum state of two particles, for example, is a single wave function psi(x1,x2), where x1 and x2 are the positions of the particles. There is a small class of wave functions for which one can say there is an assignment of particles to states. This is when the multi-particle wave function factorizes; wave functions in this class, for two particles, would be written as psi1(x1) psi2(x2). However, wave functions with this structure are never the states of definite energy (solutions of the time-independent Schroedinger equation) when the particles interact.
The property of being bosons or fermions does not restrict how particles are assigned to states, but restricts the permutation symmetry of the wave function for the system. The wave function for two bosons has the property psi(x1,x2) = psi(x2,x1), and for fermions, psi(x1,x2) = -psi(x2,x1). An arbitrary wave function can be symmetrized or antisymmetrized, as needed for bosons or fermions. Antisymmetrizing the product wave function psi(x1)psi(x2) yields no wave function at all (the null vector of linear algebra) — it is for this reason that it is sometimes said fermions cannot be in the same state.
-Veit
On Jun 23, 2014, at 10:53 PM, Henry Baker <hbaker1@pipeline.com <javascript:;>> wrote:
"An important characteristic of bosons is that their statistics do not restrict the number that can occupy the same quantum state"
https://en.wikipedia.org/wiki/Boson
Ok, suppose that I want to _weigh_ one boson. I guess that most bosons are pretty light, so I'll have to weigh a bunch of identical bosons (in "the same" quantum state) & divide by their number.
Q: Does this work? Is there any additional energy required to hold them together that would show up as additional mass? Or does an individual boson weigh more singly than when together in "the same" quantum state?
How big is the biggest boson, anyway?
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On Jun 24, 2014, at 9:45 AM, James Propp <jamespropp@gmail.com> wrote:
I feel like I'm lacking context for the operation of symmetrization/antisymmetrization, but I'm not sure if I can explain what I mean by context. I'm guessing Veit means something like "If you symmetrize it, you'll get a wave-function that's just as descriptive of reality; the universe won't notice the difference." Or "If you symmetrize it, you'll still get a solution to the relevant equation, but if it isn't unitary then it won't correspond to anything we see in the universe."
Veit, can you clarify? One of the “rules”, when constructing a quantum state for a system of identical particles, is that it transform as one of the 1-dimensional representations of the symmetric group. If you are dealing with fermions, then the only valid multi-particle wave function has the property it changes sign when the particle labels are given an odd permutation. This is Nature telling you that the particles do not actually have labels!
If you are given a multi-particle wave function without this group representation property, say psi(x1,x2), then you can “symmetrize” or “antisymmetrize” it so it does: psi_B(x1,x2) = psi(x1,x2) + psi(x2,x1) psi_F(x1,x2) = psi(x1,x2) - psi(x2,x1) (These are not normalized, and you may end up with zero — no wave function.) In math I believe this is called “Fourier analysis with respect to the characters of a finite group”. -Veit
In two dimensions, when we can keep track of which way one particle winds around another one, we can get phase factors other than +1 and -1 --- in this case, we have a representation of the braid group rather than the symmetric group. These are the so-called "anyons". They can actually happen in the physics of two-dimensional systems. - Cris On Jun 24, 2014, at 8:13 AM, Veit Elser <ve10@cornell.edu> wrote:
On Jun 24, 2014, at 9:45 AM, James Propp <jamespropp@gmail.com> wrote:
I feel like I'm lacking context for the operation of symmetrization/antisymmetrization, but I'm not sure if I can explain what I mean by context. I'm guessing Veit means something like "If you symmetrize it, you'll get a wave-function that's just as descriptive of reality; the universe won't notice the difference." Or "If you symmetrize it, you'll still get a solution to the relevant equation, but if it isn't unitary then it won't correspond to anything we see in the universe."
Veit, can you clarify? One of the “rules”, when constructing a quantum state for a system of identical particles, is that it transform as one of the 1-dimensional representations of the symmetric group. If you are dealing with fermions, then the only valid multi-particle wave function has the property it changes sign when the particle labels are given an odd permutation. This is Nature telling you that the particles do not actually have labels!
If you are given a multi-particle wave function without this group representation property, say psi(x1,x2), then you can “symmetrize” or “antisymmetrize” it so it does:
psi_B(x1,x2) = psi(x1,x2) + psi(x2,x1)
psi_F(x1,x2) = psi(x1,x2) - psi(x2,x1)
(These are not normalized, and you may end up with zero — no wave function.) In math I believe this is called “Fourier analysis with respect to the characters of a finite group”.
-Veit
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It might be an advantage to have bosons (vs. fermions), where they can all be in the ground state at the same time. So, no extra energy in the mass calculation from being in a higher energy state. This ignores possible interaction energy, though, which is what you are asking. -----Original Message----- From: math-fun [mailto:math-fun-bounces@mailman.xmission.com] On Behalf Of Henry Baker Sent: Monday, June 23, 2014 8:54 PM To: math-fun@mailman.xmission.com Subject: [EXTERNAL] [math-fun] Bozonic boson question "An important characteristic of bosons is that their statistics do not restrict the number that can occupy the same quantum state" https://en.wikipedia.org/wiki/Boson Ok, suppose that I want to _weigh_ one boson. I guess that most bosons are pretty light, so I'll have to weigh a bunch of identical bosons (in "the same" quantum state) & divide by their number. Q: Does this work? Is there any additional energy required to hold them together that would show up as additional mass? Or does an individual boson weigh more singly than when together in "the same" quantum state? How big is the biggest boson, anyway? _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
participants (7)
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Cordwell, William R -
Cris Moore -
Henry Baker -
James Propp -
Mike Stay -
Rowan Hamilton -
Veit Elser