What's the Difference Between Fermions and Bosons?
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- Опубліковано 19 кві 2020
- Often, when physicists talk about particles, they classify them in two fundamentally different classes: fermions and bosons. But, outside of name, what's the difference between these? How different can particles really be?
As it turns out, the answer to this question comes from how the wavefunctions of systems change when exchanging two identical particles. This small difference has huge consequences in the physics of the two different types of particles.
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it's crazy to think how much further subatomic physics goes beyond the proton/neutron/electron most people were taught in school
Right? Particle physics is quite the rabbit hole to dive down...
@@zapphysics Rabbit hole is a good term for it.
I love that "atom" literally means indivisible. "This is it guys! We've found the most elementary particles!"
Particle physics: *I'm about to (ext)end this man's whole career.*
Particle physics is love
This video is a great mix between simple images and thorough math, I love it!
Thanks! Glad you liked it!
I am actually so lucky to have just taken this exact lesson in the course of quantum mechanics in my university and open youtube to see this video in my recommendations, just as if youtube had access to my zoom session :D. Good job!
You ever think about how lucky we are that fermions were named after Fermi, giving us the "fermions are firm" reminder.
Could you imagine if they were named after someone named something like Jon or Bob?
FANTASTIC video. Well done!
This is SO clear!!!
The most underatted video on the topic.... dude you are very accurate.... and your presentation are so so so good......
Loved it ✨
Great video, thanks :)
great video, thanks!
Great video! I'd love to see one on what this humble plus/minus sign has to do with integer/half-integer spin, if you want to make one! ;)
Great video, thanks!
Great video! You've got an interesting perspective on things other videos/articles usually talk about in a same way, again and again. As an enthusiast who didn't go for a degree in this area, it's nice to get some fresh air on these explanations about QM, really makes me understand it better.
Also, I saw you were answering some questions in the comments and I was hoping you could see this one about the video: Why is the "a" value different for bosons and fermions? I mean, on 4:42 you show that "a" has only 2 solutions (1, -1), but I didn't quite get why the bosons get a=1 and the fermions get -1
@Gabriel Cecatto I'm glad you are finding the videos useful! I would be more than happy to answer your question. It comes down to a small subtlety: it isn't so much that fermions get the value of a=-1 and bosons a=+1 as much as the value of "a" DEFINES fermions and bosons. In other words, if the wavefunction describing two identical particles gets a minus sign when exchanging, the particles are what we define as fermions. If the wavefunction gets a plus sign, the particles are defined to be bosons. So the value of "a" determines whether we have fermions or bosons, not the other way around. Really, it isn't anything much deeper than the names that we assign to the two cases. Hopefully that answers what you were asking!
@@zapphysics lemme see if I got this then... This swapping refers to some kind of experimental swapping? So, the particles that behave in a way described by the solution "-1" for "a" are fermions, and the ones that behave as described by an "a=1" are bosons?
@Gabriel Cecatto pretty much. The only thing is that this isn't an experimental swapping, it's purely mathematical. Remember, these are two identical particles so physically swapping them should make no difference in the "experimental" or "observable" system. This requirement is how we found that a could only take 2 values in the first place. The different values of a only show up in the wavefunction which isn't really an observable thing. Though, it does have observable consequences like the Pauli exclusion principle. So if you prefer an observable explanation, you could say that any particle which obeys the Pauli exclusion principle is a fermion and any particle which does not is a boson. However, whether or not a particle obeys the exclusion principle is determined by whether a=1 or a=-1.
@@zapphysics Thank you so much for the reply! I understand it better now
ZAP, I appreciate this video because it is one of the very few that uses real math rather than hand waving, and so far the only one that derives the Pauli Exclusion principle rather than just stating it (wow!, or in other words, WOW!!!), but I still have some questions, what does switching two particles have to do with anything, IIUC that isn't something you can actually do in the real physical world (the Uncertainty principle, and they don't make tweezers that small :-) ! ), and IIRC its the "spin" that differentiates fermions from bosons, and what I'd really like to understand is how spin does that. So far I've found videos that say the difference between fermions and bosons is their spin, and videos that say that quantum spin is different from classical spin, but no video so far that shows the difference between integer and half-integer spin and puts the two concepts together and derives the Pauli Exclusion principle from spin. IIUC it was Dirac who first did this, and that he felt it was his most profound contribution to physics, but where I read that didn't include the math, so I'm still searching...
Why is a the same for every coefficient? Couldn't some coefficients have a = 1 and some coefficient have a = -1 and still satisfy the requirement that the probabilities are the same? And I've read in a few places that bosons have integer spin and fermions have fractional spin. Can you shed any light on that?
Ahh I was waiting for someone to ask about that! What an excellent question! The answers to both questions are actually closely related and are surprisingly deep.
The answer actually is found by considering the topology of the Lorentz group. All the Lorentz group is is the collections of ways of transforming things in spacetime. This just consists of the rotations in 3D space as well as "boosts" which are just ways of changing a state's velocity. As it turns out, you can think of this as a sphere, but not just any sphere. This sphere is special because if we draw a point on one side, an identical point appears on the exact opposite side of the sphere. Now, the question that is of concern is: how many ways can we draw a closed loop on this sphere? Some rules: we can't double over a line that exists and if we can bend/stretch one path so that it looks like another one, they don't count as two different paths.
It's pretty easy to convince yourself that there are two possibilities. 1.) We just draw a loop and my pen ends up where it started. This will actually create two loops because of the doubling effect, the one that I drew and one on the opposite side of the sphere. 2.) I draw a path which ends on the opposite side of the sphere that it started on. This will only create a single loop since the doubling effect automatically completes the loop for me. As it turns out, that's it: those are the only unique ways of drawing a closed path on this special sphere.
Okay, what does this have to do with particles? So, these "loops" actually directly correspond to how particles (or more accurately, their wavefunctions) are allowed to change when one of these transformations act on them. So we only have two options, and it turns out the symmetry properties are nicely mirrored in this example. The case of a single loop where we ended on the opposite side of the sphere is fermionic: the negative sign is related to the fact that we end on the "negative" side of the sphere. The other case is bosonic, where we ended up exactly where we started (related to the +1). You can sort of think of the fact that bosons have integer spin coming from the fact that I drew a complete loop, whereas fermions are half-integer because I only drew half of a loop. I don't believe this is exactly correct, but it's a good way to think about it.
So now, it's easy to see why we can't write a wavefunction that has fermionic and bosonic pieces to it. The paths are completely incompatible with each other and will always necessarily be distinct (I can never deform the single loop into two loops without breaking it).
The proof of this is actually very beautiful, though it's pretty mathematically intense. If you are somewhat comfortable with topology and group theory, I would recommend Weinberg's QFT Vol I for the proof.
Thanks for the great question, and sorry for the really long explanation!
@@zapphysics Nicely done!
Video says only 2 electron can be on an orbital, but there are orbitals with up to 8 electrons. I dont get that.
Wait what????? WHAT????!!!!! THAT'S THE PAULI EXCLUSION PRINCIPLES' CAUSE???? YOOOOOOOOOOOOOO
wtf an intuitive video about QM :D
Lol thanks, man. Glad you liked it!
But the biggest question is why do these properties of wave functions depend on the spin of the particle (integer for bosons/half-integer for fermions)?! Dirac's equation should help or is it a fundamental law of nature?
Ok, so I get this in terms of abstractly talking about electrons in orbitals. There aren't really many states for them to take.
But how does this apply to free floating electrons? Does it mean that if I try to add the wave functions of two identical electrons in the same spot they cancel out? And if so, what about the wave function is inverted to cause this? Are they just out of phase? That doesn't seem like it makes sense because then you could add two electrons together if they had the same phase.
Thanks for the videos btw. This is amazing content.
This is an excellent question. Here's one way to think about it: we know that there is some minimum uncertainty to know both the position and momentum of a particle. In other words, the location of the particle is always smeared out a bit and the momentum is also smeared out over a range. So, say we minimize both, i.e. we know the position of the particle as well as we can while also simultaneously keeping the uncertainty in momentum from getting too large. What Pauli exclusion tells us is that, if we take two fermions (say, neutrons, so we don't have to consider electric charge) which are in the same state otherwise (say, both spin up) and try to cram them into the same "smeared" position, the ranges of momenta that they can take can't be the same and vice versa: if two spin up neutrons have the same momenta ranges, they can't be in the same position ranges. This is actually what keeps neutron stars and white dwarfs from collapsing. Gravity is trying to force all of our neutrons together into the same location and give them as small of momentum as possible (call it zero with error bars). But, as soon as the state where the position and momentum ranges of two neutrons (one spin up and one spin down) are the same, no other neutrons can fit in this state. This means that they either have a higher momentum or a different location. Either way, the result is the same: the neutrons feel a force pushing them away from the filled state. This force is known as "neutron/electron degeneracy pressure" and is a star's last hope before collapsing into a black hole.
@@zapphysics It's really interesting that the very unbreakable sounding mathematical rule that fermions can't occupy the same state is a force that can be overcome with enough energy. Does that mean that if I had a quantum mechanics simulator that it would actually be possible to encode two fermions in the same state, and that they'd just be super unstable? Does this force have a related boson?
@@docopoper What you would find if you tried to code two fermions in the same state is that the probability of this occurring would just always be exactly zero. Not even some small chance, just always zero, so it is actually impossible for this configuration to form to begin with.
To have a boson related to a force like in QED with the photon, there must be an underlying symmetry associated with the force. In this case, the "force" doesn't come from some coupling to fields or anything and there is no symmetry attached to it, it just simply comes from the nature of fermions themselves. In other words, one could have a completely free, non-interacting theory of fermions and it would still be impossible to put two in the same state.
why was i trying to learn quantum mechanics in 4th grade
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Lasers are a boson thing too, right?
Bosons (symmetry, waves) are dual to Fermions (anti-symmetry, particles) -- quantum duality.
"Always two there are" -- Yoda.
Bosons like to be in the same state (laser) and Fermions like to be in different states (Pauli exclusion principle).
Same is dual to different.
Duality creates reality!
Oh!
I think you deserve a Nobel Prize! 😀
i think i have always assumed there is only in and out in the entire universe...
I wish I knew what you know
I prefer old R n' B, like maybe Ray Charles, or Big Joe Turner.
is quantum a theory that has been deduced through probability?..
Love your videos but you edited this one terribly. When you flashed on the screen [ v bar is a superposition (scaled sum) of x hat and y hat]
This is a point that deserves its own 1 minute discussion of what you personally mean by superposition. Sabine hosenfelder seems to think that they're is nothing much more than mathematical convenience in the term superposition - what do you mean?