Quantum spin liquids and valence bond solids

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Quantum Condensed Matter Physics: Lecture 10
Theoretical physicist Dr Andrew Mitchell presents an advanced undergraduate / introductory Master's level lecture course on Quantum Condensed Matter Physics at University College Dublin. This is a complete and self-contained set of lectures, in which the theory is built up from scratch, and requires only a knowledge of basic quantum mechanics.
In this lecture I discuss quantum spin systems with antiferromagnetic interactions. The classical Neel state is not an eigenstate of the Heisenberg exchange Hamiltonian, but the ground state can be understood in terms of so-called resonating valence bonds. We discuss how magnetic frustration leads to macroscopic ground state degeneracy and the suppression of magnetic ordering, to produce a quantum spin liquid state. We also explore a class of exactly solvable spin models with valence bond solid ground states, such as the AKLT chains. This gives insight into topological defects and fractionalization in quantum condensed matter systems.
Navigate through the lectures of this course in order using the playlist:
• Quantum condensed matt...
Recommended course textbook: "Many-body quantum theory in condensed matter physics" by Bruss and Flensberg

Пікірлер: 16

@wathqankolthoum3929 3 жыл бұрын

Channels like yours are what people need on KZfaq, specially physics, material engineering and chemistry students .... Thank you , Sir

@element4element4 10 ай бұрын

I think something is missing in the AKLT example. The representation of spin-1 degrees of freedom in terms of two spin 1/2, will end up giving us a too large Hilbert space. We have 4 states per spin instead of 3, so in the end we need to project out the spin-0 singlet on each original site. It would have been interesting to do that, to see how the fractionalised spin-1/2 edges appear in the original spin-1 language. I will try to do that myself later to see what comes out of it.

@sanjeevkumarthakur3744 Жыл бұрын

Thanks for this lecture, Sir. It's really informative and easy to understand.

@lbyles3776 3 жыл бұрын

Really helpful video, thank you!

@mahmudshamim1437 Жыл бұрын

Amazing!

@element4element4 10 ай бұрын

I have a simple question regarding the triangle example. The system also has a S3 symmetry, corresponding to any permutation of the three spins. The two chosen ground-states transform under the even and odd representation under permuting 2-3. However, permuting 1-2 or 1-3 seems to mix the states in some strange way. Is it because the two state transform under the 2-dimensional irreducible representation of S3? I wonder whether one can trace the geometric frustration of the triangle, down to some group-theoretic property of the symmetry group of the triangle? Somehow S3=D3 requires a higher-dim irrep in the ground-state, while D4 (for square) does not?

@vikramravindranath3222 2 жыл бұрын

Thank you for the lecture! Is there a way of generically showing (without explicity constructing the product state and going through the motions) that a given electron can simultaneously be part of a triplet with the site to its left, and a singlet with the site to its right (as in AKLT)? It seems unituitive to me that the RVB can be constructed independently with each fictitious electron that forms the spin 1, since the constituent spin 1/2s aren't exactly 'free'. Thank you!

@crawlgandum806 2 жыл бұрын

What’s the meaning of quantum fluctuations? This terminology has been showing up in the last few lectures however I think I don’t fully understand it. Is it the same meaning in different situations?

@liukelyu6009 2 жыл бұрын

Dear Prof. Mitchell, since the MG Hamiltonian only has sum of three nearest neighbor spins, wouldn't Neel state also constitute a ground state? Why did we choose pairs of neighboring singlets as ground state?

@drmitchellsphysicschannel2955 2 жыл бұрын

Thanks for the interesting question. Actually The Neel state is not an eigenstate of this Hamiltonian. You can see this because the spin flip operators in the exchange interactions introduce domain walls in the pristine Neel order. It has net zero Sz on neighbouring sites and it has net Sz=+1/2 or -1/2 on groups of three spins. But the spin S is not well defined in the Neel state (it has broken SU(2) symmetry). But we can find an eigenstate with lowest energy that is the valence bond solid state.

@user-wu7yf2ht7d 10 ай бұрын

Thank you for this lecture professor. I have a question about the GS of the triangular lattice. I think the GS of the 2D triangular lattice is "120-degree canted spin" state. So the GS of the triangular lattice has an ordered magnetic state and thus the triangular lattice cannot be a candidate for a spin liquid. As regard of my question, i want to know your opinion. Thank you professor.

@drmitchellsphysicschannel2955 10 ай бұрын

The quantum system with spin-1/2 and heisenberg interactions on a triangle lattice is famously not a magnetically ordered state, but is characterised by quantum-disordered states with unconfined bosonic spinons. Large spin systems can be ordered by quantum fluctuations however. Some of the degeneracy of the classical frustrated system is lifted in the quantum spin-1/2 case, but my understanding is that degeneracy remains down to T=0

@user-wu7yf2ht7d 10 ай бұрын

@@drmitchellsphysicschannel2955 I appreciate for your comment. Now i understand. In the spin-1/2 and heisenberg interactions on a triangle lattice, the states are diordered. But in the case of the large spin system, for example spin-5/2 and heisenberg interactions on a triangle lattice, the spins can be aligned (120-degree canted). If my understanding is correct, what is the crucial role of the magnitude of the spin regarding to the ordered and disordered states? Thank you professor.

@sidharthachatterjee3373 Жыл бұрын

Can you kindly clarify 12:53 to 13:25? How two spin 1/2 states and one spin 3/2 state is coming?

@noism5429 9 ай бұрын

same question here

@sunberserker8276 4 ай бұрын

If you have three spin 1/2's , then two of them form a singlet (S=0) or a triplet state (S=1) whereas the third must remain unpaired und has S=1/2. Thus summing all three spins together gives either S = 1/2 where two form a singlet and the third is unpaired, or S = 3/2 when you have a triplet state with an unpaired third spin