This reminds me of a science fiction book I just read called quantum xy, about a group of Silicon Valley engineers who sought to create an ai learning model with moralistic principles. The system learned from the input commands of known moralists and grew its network based on the networks of these individuals. I can’t remember the author but I believe it was released in 93’ early 93. Way ahead of its time.
Excellent! I think this time you took your time explaining something, which is not complicated in its nature, but counter-intuitive. Why not map 1 to 1 is a good question and the resulting transformation does look not that useful. But when I saw the spin flip equation I knew this will be very, very helpful with the annoying neighbors! Glad to see you back again, I needed my weekly-ish stat mech
Next up we will put it to use and solve a model! It's a nice playground to look at quantum phase transitions, thermal properties ect. :). Glad you enjoyed it!
Thanks a ton Jonathan could please provide the resources or textbooks or lecture notes where I can dive deeply into the concepts of these transformation and Quantum Many Body Hamiltonian.
so if I understand correctly, I can rewrite the Hubbard model to a Heisenberg model with a step spin operator, so maybe I can calculate the ground state energy and excitations of a 3*3 grid exactly on the computer?
Very well explained. I'm looking for someone to hold my hand through onsagers and Shultz/mattis/live solutions to 2D using. Shankar goes through it somewhat well in his book but some subtleties are missing. I'd be super pleased if there was a pristine video lecture on it
Hello !! Thanks for the video. The XY hamiltonian you presented at 0:45 is somehow disturbing. Written that way, it's an anisotropic system, how come the interaction strength(J) is not anisotropic for this system? Looking forwards to read you.
You could totally just write it as having J_x and J_y if you wanted to. However the way I wrote makes it convenient for the limits \gamma = 0 or \gamma = 1 to recover the XX or Ising model.
@@JonathonRiddell Thanks, sir. I understand that it is a convenient form to recover the isotropic model and the Ising one. But how do we understand the interaction strength (J) in this Hamiltonian with respect to the XY model we are describing?
Please also add how to interpret the results that comes from Jordan Wigner Transformation. Like, once I transformed to spinless fermions and found the eigen values but the Filling is not fixed as we can have any amount of up or down spin within Total spins, so how to find the ground state for our original Hamiltonian from this.( I am not able to find it anywhere)
I think the key here is identifying what the vacuum state is, and then constructing the ground state in the fermionic picture. This will of course depend on the Hamiltonian in the end. But the general procedure is to construct the ground state in the fermionic picture and then transform back to spin space. In our case here |all spins down> = |0> for the spins to the fermions in the transformation. If you are considering the XX model, you would then "turn on" any eigenmode that is negative by applying d_k^\dagger to the vacuum state. Hope that helps!
Are we talking about the eigenmode basis from the XX model video? Then that is just a number operator for the d space fermions. So it'd work just like f_j^\dagger f_j If your basis was instead in real space, then you'd expand each operator in f space.
@@JonathonRiddell yes sir it is in context to xx spin chain model. I want to write the final Hamiltonian that we got in terms of operator d_k^\dagger d_k in Matrix Form. It will be a 4×4 matrix. With states 00, 01, 10, 11. While acting d_k^\dagger d_k on these states what should be the value of subscript k?
@@alfiashaikh99 That would be a conventional choice. You can order k whichever way you like. If those basis states are in d space then the matrix H would just be a diagonal matrix with diagonal of \epsilon_k, order at whatever your convention ends up being. The d_k^\dagger d_k is just a number operator right? So if there is a 1, it pops out a 1, and if it is a zero, it gives back a zero. Hope that helps!
So here's how I think about it. What we just did is talk about how one type of microscopic system (Spin 1/2) can be transformed into another (spinless fermions). The story usually goes microscopic physics -> stat mech -> thermodynamics. So it turns out that when we do the second step, stat mech, in this case, a bunch of the microscopic physics is washed away in the sense that doing a bunch of calculations on either system (spins or fermions) would be equivalent. So this is somewhat expected. By the time we get to thermodynamics we have almost entirely removed the information about the microscopic physics. You can think of it as particle number and magnetization playing a similar role when they are introduced to thermodynamics.
@@JonathonRiddell Huh, that seems counterintuitive to me, since Bosons and Fermions have such different properties. By changing our perspective to the Fermion perspective in this case, is it correct to think about these 'new' Fermions that we're thinking about as quasiparticles or something similar? Another question, how would one think do come up with this transformation? Does it only work for the spin operators, or can this be done in general, when you're working on a system of Bosons and wish to think of it as an equivalent system of Fermions?
@@benburdick9834 They are definitely quite different. However at high temperature / when we expect classical thermodynamics to work, we would expect a lot of these differences to wash away. You can see this by taking the Fermi-Dirac statistic, and Bose-Einstein statistic at high temperatures, and see that the distribution becomes identical. Otherwise we need to be careful when proceeding with thermodynamics. Yeah it can absolutely be thought of as a quasi-particle picture. You could similarly take a real fermionic system and transform it to a spin model with this. There are other transformations like the Holstein-Primakoff transformation to take Bosons to spins (but you truncate the Fock space). This one works well because both are two level systems, all we needed to do is find a transformation that mapped the spin algebra onto the fermions. So it can be an exact transformation.
@@idessjiometio7707 you can think of it as a mathematical trick. We are just mapping one two level system to another, and we call the fermion one fermionic because of its algebra.
Hey bro, do you think Computational physics is a good career path?. I love math, physics and computer science but I'm confused between comp.physics and AI.
Is this for an undergrad program? I think both are great options. For comp physics you'll learn a lot of different techniques and ways to approach problems computationally. Your skill set will be pretty generic and it'll be easy to adapt to new problems as you go along. Once you get past undergraduate you'll end up specializing in a certain type of problem (specifically a physics sub field, like i do quantum lattice models). So transferring to industry and using those specific skills directly will be hard depending on what sub field you get into. Hope that helps!
Typo at 3:33, the bottom right identity should read f_j |1> = |0>.
This reminds me of a science fiction book I just read called quantum xy, about a group of Silicon Valley engineers who sought to create an ai learning model with moralistic principles. The system learned from the input commands of known moralists and grew its network based on the networks of these individuals. I can’t remember the author but I believe it was released in 93’ early 93. Way ahead of its time.
Excellent! I think this time you took your time explaining something, which is not complicated in its nature, but counter-intuitive. Why not map 1 to 1 is a good question and the resulting transformation does look not that useful. But when I saw the spin flip equation I knew this will be very, very helpful with the annoying neighbors! Glad to see you back again, I needed my weekly-ish stat mech
Next up we will put it to use and solve a model! It's a nice playground to look at quantum phase transitions, thermal properties ect. :).
Glad you enjoyed it!
Thanks a ton Jonathan could please provide the resources or textbooks or lecture notes where I can dive deeply into the concepts of these transformation and Quantum Many Body Hamiltonian.
so if I understand correctly, I can rewrite the Hubbard model to a Heisenberg model with a step spin operator, so maybe I can calculate the ground state energy and excitations of a 3*3 grid exactly on the computer?
Very well explained. I'm looking for someone to hold my hand through onsagers and Shultz/mattis/live solutions to 2D using. Shankar goes through it somewhat well in his book but some subtleties are missing. I'd be super pleased if there was a pristine video lecture on it
This was really helpful, thank you!
Hello !!
Thanks for the video. The XY hamiltonian you presented at 0:45 is somehow disturbing.
Written that way, it's an anisotropic system, how come the interaction strength(J) is not anisotropic for this system?
Looking forwards to read you.
You could totally just write it as having J_x and J_y if you wanted to. However the way I wrote makes it convenient for the limits \gamma = 0 or \gamma = 1 to recover the XX or Ising model.
@@JonathonRiddell Thanks, sir.
I understand that it is a convenient form to recover the isotropic model and the Ising one. But how do we understand the interaction strength (J) in this Hamiltonian with respect to the XY model we are describing?
hello, please can u extend this transformation to the 2D.
Very well explained! Thank you
The fermionic operator used here is actually spinless?
Please also add how to interpret the results that comes from Jordan Wigner Transformation.
Like, once I transformed to spinless fermions and found the eigen values but the Filling is not fixed as we can have any amount of up or down spin within Total spins, so how to find the ground state for our original Hamiltonian from this.( I am not able to find it anywhere)
I think the key here is identifying what the vacuum state is, and then constructing the ground state in the fermionic picture. This will of course depend on the Hamiltonian in the end. But the general procedure is to construct the ground state in the fermionic picture and then transform back to spin space. In our case here |all spins down> = |0> for the spins to the fermions in the transformation.
If you are considering the XX model, you would then "turn on" any eigenmode that is negative by applying d_k^\dagger to the vacuum state. Hope that helps!
Thanks, great video!😀
tks you so much, it helps me a lot
Amazing videos and channel! Do you follow any books or reviews in particular for your condensate matter videos?
Not in particular. Of course there is always Coleman :). If I use a specific resource it's usually in the video description.
Hey I have a question regarding this.. How to prove Qj Qj+1 = 1 - 2 fjDagger fj?
Use the identity at 8:00. Since all terms in the products commute you can rearrange everything and get a bunch of identities.
For N=2 spin chain system,
States are |00> , |01>, |10>, |11>
So how to find the action of operators (dk+)dk on these states? Please help me out.
Are we talking about the eigenmode basis from the XX model video? Then that is just a number operator for the d space fermions. So it'd work just like f_j^\dagger f_j If your basis was instead in real space, then you'd expand each operator in f space.
@@JonathonRiddell yes sir it is in context to xx spin chain model.
I want to write the final Hamiltonian that we got in terms of operator d_k^\dagger d_k in Matrix Form.
It will be a 4×4 matrix. With states 00, 01, 10, 11.
While acting d_k^\dagger d_k on these states what should be the value of subscript k?
@@alfiashaikh99 That would be a conventional choice. You can order k whichever way you like. If those basis states are in d space then the matrix H would just be a diagonal matrix with diagonal of \epsilon_k, order at whatever your convention ends up being. The d_k^\dagger d_k is just a number operator right? So if there is a 1, it pops out a 1, and if it is a zero, it gives back a zero.
Hope that helps!
thank you! great help
Is there an analogue to this in thermodynamics?
So here's how I think about it.
What we just did is talk about how one type of microscopic system (Spin 1/2) can be transformed into another (spinless fermions). The story usually goes microscopic physics -> stat mech -> thermodynamics. So it turns out that when we do the second step, stat mech, in this case, a bunch of the microscopic physics is washed away in the sense that doing a bunch of calculations on either system (spins or fermions) would be equivalent. So this is somewhat expected. By the time we get to thermodynamics we have almost entirely removed the information about the microscopic physics.
You can think of it as particle number and magnetization playing a similar role when they are introduced to thermodynamics.
@@JonathonRiddell Huh, that seems counterintuitive to me, since Bosons and Fermions have such different properties.
By changing our perspective to the Fermion perspective in this case, is it correct to think about these 'new' Fermions that we're thinking about as quasiparticles or something similar?
Another question, how would one think do come up with this transformation? Does it only work for the spin operators, or can this be done in general, when you're working on a system of Bosons and wish to think of it as an equivalent system of Fermions?
@@benburdick9834 They are definitely quite different. However at high temperature / when we expect classical thermodynamics to work, we would expect a lot of these differences to wash away. You can see this by taking the Fermi-Dirac statistic, and Bose-Einstein statistic at high temperatures, and see that the distribution becomes identical. Otherwise we need to be careful when proceeding with thermodynamics.
Yeah it can absolutely be thought of as a quasi-particle picture. You could similarly take a real fermionic system and transform it to a spin model with this.
There are other transformations like the Holstein-Primakoff transformation
to take Bosons to spins (but you truncate the Fock space). This one works well because both are two level systems, all we needed to do is find a transformation that mapped the spin algebra onto the fermions. So it can be an exact transformation.
@@JonathonRiddell Hello sir, How can a fermion be spinless knowing that a fermion is a half-integer spin particle?
@@idessjiometio7707 you can think of it as a mathematical trick. We are just mapping one two level system to another, and we call the fermion one fermionic because of its algebra.
Hey bro, do you think Computational physics is a good career path?. I love math, physics and computer science but I'm confused between comp.physics and AI.
Is this for an undergrad program? I think both are great options. For comp physics you'll learn a lot of different techniques and ways to approach problems computationally. Your skill set will be pretty generic and it'll be easy to adapt to new problems as you go along.
Once you get past undergraduate you'll end up specializing in a certain type of problem (specifically a physics sub field, like i do quantum lattice models). So transferring to industry and using those specific skills directly will be hard depending on what sub field you get into.
Hope that helps!