Magnets, 1D quantum system, and quantum Phase transitions

Transcription

1 134 Phys620.nb 10 Magnets, 1D quantum system, and quantum Phase transitions In 1D, fermions can be mapped into bosons, and vice versa magnetization and frustrated magnets (in any dimensions) Consider a lattice of spins. The Heisenberg model has the following Hamiltonian Ø Ø H = J S i ÿ S j = J <ij> S x i S x j + S y i S y j + S z i S z j <ij> (10.1) where i,j stands for nearest neighbors. If J < 0, spins want to point in the same direction (ferromagnetic). If J > 0, neighboring spins want to point in the opposite direction to minimize the energy (anti-ferromagnetic). In real materials (which may be anisotropic), the coupling constants in different directions may be different, which can be described by the XYZ model: H = <ij> J x S i x S j x + J y S i y S j y + J z S i z S j z (10.2) If J x = J y, we get the XXZ model H = <ij> J x S i x S j x + J x S i y S j y + J z S i z S j z (10.3) If J x = J y and J z = 0, we get the XY model H = J <ij> S i x S j x + S i y S j y (10.4) If J x = J y = 0 and J z 0, the model is known as the Ising model H = J <ij> S i z S j z (10.5) Frustration and spin liquids The key question for a spin system is to determine the ground state at T = 0 K. For ferromagnetic couplings, the ground state is simple: all spin pointing in the same directions (a ferromagnetic state). For anti-ferromagnetic couplings, it depends on geometry. For example, on a square lattice, the spins can find a way to make all spins happy (all neighboring spins stay in opposite directions). But on a triangle, there is no way to make all the spin happy in an Ising model with J Z > 0. So the spins are frustrated. This type of systems are known as frustrated magnets. The reason that frustrated magnets are interesting is because their ground states are not obvious. Therefore, some of these system may have very strange ground state, with exotic properties, such as spin liquids. As T goes down, in most systems, some ordering will develop (e.g. liquid turns in to solids, paramagnetism turns into ferromagnetism, etc.). There, we will get a phase transition from an disordered state at high temperature to an ordered state at low temperature. The ordered states are often considered as solids, while the disordered states are sometimes called liquids, using the terminology from the solid-liquid transition. In most systems, the low-temperature phase is a solid state, in which some type of ordering is developed. However, occasionally, the system may not develop any ordering even at T = 0 K. There, we say that the system remains a liquid at 0 K (a quantum liquid that cannot be frozen by cooling down to zero temperature). For spin systems, if the spin doesn t form any magnetic ordering, the T = 0 K state is call a spin liquid

2 Phys620.nb 135 or a quantum spin liquid (quantum here stands for T = 0 K, at which the system is governed by quantum physics instead of classical statistics). Typically, in theoretical studies, one maps spins into a strongly-correlated Fermi gas (or boson gas), which is known as the slave fermion (or slave boson) approach. For the slave fermion approach, the fermion may form a insulator or superconductor (which is a gaped spin liquid), semimetal like graphene (which is a gapless spin liquid, known as the algebra spin liquid), or a metal (which is known as a Bose metal and the Fermi surfaces for this Bose metal is known as a Bose surface or a spinon Fermi surface or a spinon surface). The mapping from spin to fermions/bosons can be rigorously constructed in any dimensions. However, one needs to pay a price; these fermions/- bosons have strong interactions, and thus the problem cannot be solved in general. Very typically, one needs to make some approximations, which cannot be fully justified. Because these approximation are not well-controlled, it is not obvious that whether we can trust their predictions or not. Large scale numerical simulations are used to compare with these theories. Recently, some good news are obtained using density matrix renormalization group (DMRG). In 1D, there is a very straightforward mapping which can map the XY model into a free fermion problem. This is because in 1D, bosons and fermions are not fundamentally different from each other (one cannot exchange two particles in 1D) Spin-1/2 and hard-core bosons (in any dimensions) This section describes a baby version of the slave boson approach. Hard-core bosons are bosons hopping on a lattice, but we assume that there can be at most one boson per site. Hard-core bosons can be realized using ordinary bosons, if we introduce strong (short-range) repulsive interactions between these bosons. When the repulsion is infinitely strong, bosons will not want to stay together, so we will have at most one boson per site (i.e. hard-core bosons). Hard-core bosons has an obvious connection to spin-1/2. Let s consider one single lattice site. For a single site, the size (dimension) of the Hilbert space of hard core bosons is 2 (occupied 1 or empty 0 ). For spin-1/2, the Hilbert space has the same size (spin up up and spin down down ). Because the two Hilbert space has the same size, we can constructe a mapping between them. We map the spin up state for spin-1/2 to the occupied state for a hard core boson, and map the spin down state to the empty state. 0 = down 1 = up (10.6) For hard-core bosons, we can define creation and annihilation operators b i and b i, which satisfy b 0 = 1 b 0 = 0 b 1 = 0 b 1 = 0 (10.7) (10.8) For spins, we have the spin rising and lowering operators S i + = S x +ÂS y and S i - = S x -ÂS y, which satisfy S + down = up S i - down = 0 S + up = 0 S i - up = down (10.9) (10.10) Under this mapping, we find that b = S + and b = S -. The particle number operator for hard-core bosons is n = b b. For spin-1/2, this operator is mapped to n = S + S - = S x +ÂS y S x -ÂS y = S z n = S + S - = S x +ÂS y S x -ÂS y = (10.11) S 2 x + S 2 y -Â S x S y - S y S x = S Ø 2 - S 2 z -Â S x, S y = s s S 2 z + S z = S z = S z It is easy to check that S z up = 1μ up and S z down = 0μ down, which is the same as the particle number operator n = b b. One can further verify this mapping by checking the commutation relations b, b = b, b = 0 (10.12) and b, b = 1-2 n (10.13) The last fomular can be proved using b, b 0 = bb 0 - b b 0 = 0 = 1-2 n 0 (10.14)

3 136 Phys620.nb b, b 1 = bb 1 - b b 1 =- 1 = 1-2 n 1 (10.15) Note: b, b = 1-2 n is the commutation for hard-core bosons, which is a little bit different from ordinary bosons (without the hard-core constrain). For ordinary bosons, we have b, b = 1. For a lattice with many different sites, b i, b j = 1-2 n i d ij (10.16) For spins, we have the same commutation relations. S i -, S j - = S i +, S j + = 0 S i +, S j - = S ix +ÂS iy, S jx -ÂS jy =-Â S ix, S jy +Â S iy, S jx = 2 Â S iy, S ix d i, j = 2 S iz d ij = 2 n i - 1 d ij = b j, b i (10.17) (10.18) XXZ model Consider the XXZ model H = J x <ij> S i x S j x + S i y S j y + J z <ij> S i z S j z (10.19) It is easy to check that S + i S - j + S + j S - i = S x i +ÂS y i S x j -Â S y j + S x j +ÂS y j S x i -Â S y i = 2 S x i S x j + 2 S y i S y j +ÂS y i S x j -ÂS x i S y j +ÂS y j S x i -ÂS x j S y i = 2 S x i S x j + 2 S y y i S j (10.20) Therefore, we have H = J x 2 S i + S - j + S + j S - i + J z n i n j = <ij> <ij> J x 2 b i b j + b j b i + J z n i n j - J z <ij> <ij> n i = J x ij 2 b i + b - j + b + j b - i + J z n i n j - constant <ij> <ij> If we ignore the unimportant constant term, we get H = t b i b j + b j b i + V <ij> n i n j <ij> (10.21) (10.22) where t = J x 2 and V = J z. So, we mapped the XXZ model into hard-core bosons with nearest-neighbor hoppings (t) and nearest-neighbor repulsions (V). It is worthwhile to emphasize that these bosons are hard-core bosons, which are bosons with a huge on-site repulsion. Using ordinary bosons, this model becomes H = t <ij> a i a j + a j a i + V <ij> n i n j + U i n i n i - 1 (10.23) with U Ø. The U Ø limit prevents two bosons to stay on the same site. This boson model has infinite interactions, which cannot be solved Spin-1/2 and fermions (1D) This section is a baby version of the slave fermion approach. The problem of bosons is that we need infinite interactions to avoid many bosons occupying the same state. If we use fermions, the problem of multi-occupation can be avoided easily, thanks for the Pauli exclusive principle a single site Consider one single lattice site, the size (dimension) of the Hilbert space of spinless fermion is 2 occupied 1 or empty 0. For spin-1/2, the Hilbert space has the same size: spin up and spin down. So it seems possible to make a connection between spins and spin-less fermions. Again, We map the spin up state for spin-1/2 to the occupied state and map the spin down state to the empty state. Define fermion creation and annihilation operators c i and c i. c 0 = 1 c 0 = 0 (10.24)

4 Phys620.nb 137 c 1 = 0 c 1 = 0 (10.25) Define spin rising and lowering operator S i + = S x +ÂS y and S i - = S x -ÂS y. S + down = up S i - down = 0 S + up = 0 S i - up = down (10.26) (10.27) Under this mapping, we find that c is S + and c is S -. The particle number operator is n = c c. For spin-1/2, the particle number operator is mapped to n = S + S - = S x +ÂS y S x -ÂS y = S z One can further verify the anti-commutation relations c, c = c, c = 0 c, c = 1 S -, S - = S +, S + = 0 S +, S - = S x +ÂS y, S x -ÂS y = 2 S 2 x + S 2 y -Â S x S y +ÂS y S x -Â S y S x +ÂS x S y = 2 S 2 x + S 2 y = 2 s s S 2 z = So far so good. But we will run into trouble with more than one site = 1 (10.28) (10.29) (10.30) (10.31) two sites If we have two sites, the fermion satisfies anti-commutation relations c 1, c 2 = c 1, c 2 = c 1, c 2 = 0 (10.32) Here, the subindex 1 or 2 stands for site 1 or 2. However, for spin operators, we don t have this type of anti-commutation relations S 1 + -, S There, what we have are commutation relations S 1 + -, S = 0. Therefore, we CANNOT naively identify S +, S - with c and c Why is there a problem? For spin rising operators, if we start from the states that both spins are pointing down down, down, we can rise the spin on site 1 first, and then the spin on site 2, which is S + 2 S + 1 down, down = up, up (10.33) or, we can rise the two spins in opposite ordering, i.e. rise the spin on site 2 first, and then the spin on site 1, which is S 1 + S 2 + down, down = up, up (10.34) Regardless the ordering, we have the same final state. S 1 + S 2 + down, down = S 2 + S 1 + down, down (10.35) However, for fermionic operators, if we start from the empty-empty state, let s first create particle on site 1 and then on site two. c 2 + c 1 + 0, 0 (10.36) Now, let s do it in the opposite order, c 1 + c 2 + 0, 0 (10.37) Because c 1 +, c 2 + = 0, we know that c 2 + c 1 + =-c 1 + c 2 +. Therefore, c 2 + c 1 + 0, 0 =-c 1 + c 2 + 0, 0 (10.38) If we define c 1 + c 2 + 0, 0 = 1, 1 (10.39) then we get,

5 138 Phys620.nb c 2 + c 1 + 0, 0 =- 1, 1 (10.40) This - sign comes from the anti-commutation relation, but it is often ignored in textbooks on second quantization. This minus sign distinguishs bosons from fermions and this is the reason why we cannot simply identify S +, S - with c and c The correct way: The correct mapping for a system with two sites is S 1 + = c 1 S 2 + = c 2-1 n 1 (10.41) (10.42) where n 1 = c 1 c 1 is the particle number operator on site 1. This extra factor -1 n 1 will generate the correct anti-commutation/commutation relation, and we can check it easily. For example S 2 + S 1 + down, down = c 2-1 n 1 c 1 0, 0 = -1 n 1 c 2 c 1 0, 0 = -1 n 1-1, 1 (10.43) Here, because site 1 has one particle, -1 n 1 =-1. As a result, S 2 + S 1 + down, down = -1 n1-1, 1 = 1, 1 = up, up (10.44) Now, for S 1 + S 2 + down, down = c 1 c 2-1 n1 0, 0 (10.45) Because site 1 has no particle, n 1 = 0 here. As a result, S 1 + S 2 + down, down = c 1 c 2-1 n 1 0, 0 = c 1 c 2 0, 0 = 1, 1 = up, up (10.46) Using this correct mapping, we find that indeed, S 2 + S 1 + = S 1 + S Many-sites If we have many sites i=1,2,3 N, the correct mapping is S i + = c i -1 j=1 j=i-1 c j c j (10.47) This is known as the Jordan Wigner transformation. The inverse transformation is c i = S i + -1 j=1 j=i-1 S j + S j - (10.48) XXZ model For the XXZ model, we can map it to the following fermionic Hamiltonian using the Jordan Wigner transformation H = t <ij> c i c j + c j c i + V <ij> n i n j (10.49) with t = J x 2 and V = J z. Here, we have fermions with nearest-neighbor hoppings and nearest-neighbor repulsions. For the XY model (with J z = 0), this is just a free fermion problem. 1D is very special. In higher dimensions, the JW transformation will generate long range interactions (and the Chern-Simon s gauge field), making the problem unsolvable (at least analytically unsolvable). But in 1D there is no such problem Quantum phase transitions Classical phase transition: phase transition at finite temperature Quantum phase transition: phase transition at T = Quantum phase transition and level crossing Consider a quantum Hamiltonian: H = 1 - x H 1 + xh 2 (10.50)

6 Phys620.nb 139 where x varies between 0 and 1. At x = 0, H = H 1. So the ground state at x = 0 is the ground state of H 1 G 1. At x = 1 H = H 2, so the ground state of the system is the ground state of H 2 G 2. Q: At x increases from 0 to 1, how does the ground state wavefunction of H evolves from G 1 to G 2? A: there are two possibilities. (1) G 1 evolves adiabatically to G 2 or (2) there is a level crossing point at some critical x = x C. The second case is known as a quantum phase transition. Q: Why the level crossing? (Why not avoided level crossing?) A: If the symmetry properties of G 1 and G 2 are fundamentally different, there would be a level crossing at some 0 < x C < Example: transverse field Ising model Let s consider a quantum system (at T = 0 K) H =-Jg i s i x - J i, j s i z s j z (10.51) Here, we have an Ising model (the last term) along the z-direction. In addition, we have a transvers magnetic field (B = jg) pointing in the direction (x). For simplicity, we choose g > 0 and J > 0. x The first term -J g i s i is our H 1 and the second term -J i, j s z z i s j is our H 2. Two special limits are easy to study: At g =+, the first term dominates so H = H 1. Now, the ground state is: all spins point to the +x direction, which is our G 1. At very large g (but not infinite g), most of the spin will be pointing in the +x direction. At g = 0, the first term vanishes so H = H 2. Now, there two degenerate ground states: all spins point to the +z or -z direction, which we call G + 2 and G - 2. At 0 < g << 1, not all the spins will be pointing into the same direction, but we will still have two ground states, one has most of the spin pointing to +z the other have most spin pointing to -z. Symmetry properties: The Hamiltonian preserves the symmetry: z Ø-z. For the ground state G 1, the symmetry is preserved. But for G 2, this symmetry is broken spontaneously. If we change z Ø-z, G + 2 Ø G - 2. Because of the symmtry difference between G 1 and G 2, it is impossible for G 1 to evolve adiabatically into G 2 as g decreases from + to 0. So there must be (at least) one phase transition between g = 0 and g =+. For this model, the transition is at g = Quantum critical point and quantum criticality For some quantum phase transitions, the transition point shows scaling behavior. This type of quantum phase transitions are known as second order quantum phase transitions and the transition points for these transitions are known as quantum critical points. For example, the correlation functions: s z i s z j exp i - j a x (10.52) At long distance i - j >> 1, this correlation function is typically an exponential function. Here x is known as the correlation length and a is the lattice spacing. Near a quantum critical point, - x g - g c (10.53) these type of power-law behavior is known as scaling law. The energy gap near a quantum critical point also obey the scaling law D g - g c z =x -z (10.54) Notice that Ñ/D defines a time scale t t = Ñ D g - g c z =x z (10.55) So t x z (10.56) Time counts as z space dimension. This z is known as dynamic critical exponent. In fact, one can usually prove that for a system in d space dimensions, a quantum critical point with dynamic critical exponent z can be mapped to a classical phase transition in d + z space dimensions with the same symmetry breaking pattern. For example, the transverse Ising model has z = 1. Therefore, a d-dimensional transverse Ising model is Equivalent to a d+1 dimensional classical Ising model.

7 140 Phys620.nb Jordan-Wigner transformation First let's rotate the system around y-axis by 90 degrees (changing coordinates, which doesn't effect any physics). x Ø-z and z Ø x H = Jg i s i z - J i s i x s i+1 x (10.57) Now, let s perform the Jordan-Wigner transformation: s + i = -1 j=1 j=i-1 n j c i =P j=i-1 j=1-1 n j c i =P j=i-1 j=1 1-2 n j c i s - i = -1 j=1 j=i-1 n j c i P j=i-1 j=1-1 n j c i =P j=i-1 j=1 1-2 n j c i s z i = 2 S z = 2 n i = 2 n i - 1 s x i =s + i +s - i =P j=i-1 j=1 1-2 n j c i + c i (10.58) (10.59) (10.60) (10.61) We get H = Jg i s i z - J i s i x s i+1 x = Jg i 2 n i J i P j=1 j=i n j c i + c i P j=1 j=i 1-2 n j c i+1 + c i+1 = Jg i 2 c i c i J i P j=1 j=i n j c i + c i P j'=1 j'=i n j' 1-2 n i c i+1 + c i+1 = Jg i 2 c i c i J i P j=1 j=i-1 P j'=1 j'=i n j 1-2 n j' c i + c i 1-2 n i c i+1 + c i+1 = (10.62) Jg i 2 c i c i J i P j=1 j=i n j 2 c i - c i c i+1 + c i+1 The last step we used the fact that c i 1-2 n i = c i and c i 1-2 n i =-c i. This is because c i 1-2 n i 0 = c i 0 = 1 (10.63) c i 1-2 n i 1 =-c i 1 = 0 (10.64) c i 1-2 n i 0 = c i 0 = 0 c i 1-2 n i 1 =-c i 1 = 0 (10.65) (10.66) Compare with c i 0 = 1 (10.67) c i 1 = 0 (10.68) c i 0 = 0 -c i 1 = 0 (10.69) (10.70) We find that c i 1-2 n i = c i and c i 1-2 n i =-c i H = Jg i 2 c i c i J i P j=1 j=i n j 2 c i - c i c i+1 + c i+1 (10.71) Because 1-2 n j 2 = 1 2 = 1, H = Jg i 2 c i c i J i c i - c i c i+1 + c i+1 = -J i c i c i+1 + c i+1 c i - J i c i c i+1 + c i+1 c i + 2 Jg i c i c i - JgN (10.72) The last term is a constant number, which can be ignored. The first term describes the nearest-neighbor hoppings. The second term is a pairing term (Cooper pairs in a superconductor). The third term is chemical potential m=-2jg. The 1D transverse field Ising model can be mapped into a 1D superconductor. (We find Cooper pairs!) If we go to the momentum space, c k = 1 N j c j -Â kr j (10.73)

8 Phys620.nb 141 we get H =-J k c k c k  ka + c k c k - ka - J i c -k c k - ka + c -k c k - ka + 2 Jg k c k c k = J k 2 g - cos ka c k c k +Âsin ka c -k c k + c -k c k = J k c k c -k Eigenvalues of the 2μ2 matrix gives the energy spectrum of a quasi - particle. E = J g - cos ka 2 + sin ka 2 = J 1 + g 2-2 g cos ka g - cos ka  sin ka - sin ka -g + cos ka c k c -k (10.74) (10.75) (10.76) We get two bands with E + 0 and E - 0. At T = 0 K, the E - band is fully filled while the E + band is empty. The gap between the two bands is D k = E + - E - = 2 J 1 + g 2-2 g cos ka (10.77) The minimum of D(k) is at k = 0 D k D=2 J 1 + g 2-2 g = 2 J 1 - g (10.78) D is the gap between the ground state (the E - band is fully filled while the E + band is empty) and the first exacted states (the E - band is fully filled while the E + band has on particle). As one can see that D=0 at g = 1, i.e. g = 1 is the quantum phase transition point Degenerate ground states at g < 1. For g < 1, we know that there should be two degenerate ground states, one with all spins up, the other with all spins down. How can we see this degeneracy in the fermion model? This is because the 1D superconductor is a topological superconductor for -1 < g < 1, while for g > 1 the system is a topologically trivial superconductor. The topological superconductor has two Majorana fermions at the two ends of this 1D chain. We can prove that if there are two Majorana fermions, the ground state is 2-fold degenerate. If you want to know why, you can go through the following optional homework problem an optional homework problem: Majorana fermions In general a complex number z is usually different from its complex conjugate z *, z z *. For the special case where z = z *, we call them real numbers. Very similarly, for operators, in general, the Hermitian conjugate of an operator O is different from O itself, O O. For the special case O = O, these operators are known as Hermitian operators. For fermion creation and annihilation operators c and c, typically, they are NOT Hermitian, so c c. The fermions created and annihilated by these non-hermitian operators are known as complex fermions. Most of the fermions we studied in condensed matter physics are complex : e.g. Dirac fermions. In principle, their fermion creation and annihilation operators could be Hermitian with c = c. These fermions are known as Majorana fermions. For ordinary fermions, we know that c i, c j =d ij, where i and j are site indices indicating different lattice sites. For Majorana fermions c i = c i =g i (let us define g here for simplicity), the anti-commutation relation takes a very similar form g i, g j = 2 d ij. Here the normalization factor 2 is adopted for historical reasons. This normalization factor implies that g i, g i =g i g i +g i g i = 2 g i g i = 2. So we can write g i g i = 1. (The square of a Majorana operator is the identity) g i g j : Hermitian or anti-hermitian? Prove g i g j =-g i g j, assuming i j. This relation implies that g i g j is anti-hermitian. In other words, it is NOT a physical observable, which must be Hermitian. Prove Âg i g j =Âg i g j, assuming i j. This relation implies that  g i g j is Hermitian. In other words, for any physical observables that are the products of two Majorana fermions, the Coefficient must be pure imaginary.

9 142 Phys620.nb Majorana fermion as real and imaginary part of a complex fermion. Consider complex fermions, which we are very familiar with. We assume the creation and annihilation operators for the complex fermions are c and c. Define the real part of c as g 1 = c + c (10.79) Define the imaginary part of c as g 2 = c - c  Check that (10.80) c = g 1 +Âg 2 and c = g 1 -Âg 2 (10.81) 2 2 These g 1 and g 2 operators are two Majorana fermion operators. Check that they satisfies all the properties of Majorana fermions by showing the following relations 1 g 1 =g 1 and 2 g 2 =g 2 (10.82) 3 g 1, g 1 = g 2, g 2 = 2 and 4 g 1, g 2 = g 2, g 1 = 0 (10.83) The equations (1) and (2) here tell us that g 1 and g 2 are both Hermitian (they are real fermions ). The last two equations imply that they satisfy the anti-commutation relation for Majorana fermions: g i, g j = 2 d ij D superconducting chain Consider a 1D superconducting chain. H =-t j=1 N c j c j+1 + c j+1 c j -m j=1 N c j c j +D j=1 N c j c j+1 + c j+1 c j (10.84) The first term here is the kinetic energy (hoppings between nearest neighbors). The second term is the chemical potential and the last term is pairing (which breaks the particle number conservation). Here, for simplicity, we assume D is a positive real number. Define Majorana fermion operators: g 2 j-1 = c j + c j (10.85) and c j - c j g 2 j = (10.86)  Rewrite the Hamiltonian using the Majorana operators. This definition split each fermion sites into two Majorana fermion sites. If we start from a model with N sites, the mapping gives us a Majorana fermion Chain with 2N sites special case I: D=t=0 and m<0. Consider a special case with D=t = 0 and m<0. Prove that the Hamiltonian is H =- m 2 j=1n Âg2 j-1 g 2 j (10.87) Please notice that here each term of the Hamiltonian contains a complex  in the coefficients. Please compare it with the problem This complex  is crucial to keep the Hamiltonian a Hermitian operator special case II: D=t>0 and m=0. Consider a special case with D=t > 0 and m=0. Prove that the Hamiltonian is H = t j=1 N-1 Âg2 j g 2 j+1 (10.88)

10 Phys620.nb 143 Again we noticed the complex  here, which is important for the Hermitian condition. In addition, it is important to notice that g 1 and g 2 N do NOT appear in the Hamiltonian. In other words, g 1, H = g 2 N, H = Zero-energy Majorana modes at the end of the chain. Let s focus on the special case studied in Suppose the ground state wavefunction for our system is G with energy E 0. H G = E 0 G (10.89) Using the commutation relation we find in ( g 1, H = g 2 N, H = 0) to prove that H g 1 G = E 0 g 1 G H g 2 N G = E 0 g 2 N G (10.90) (10.91) These two formulas tell us that we can create one Majorana fermion at each of the two ends of this 1D superconducting chain. After we create this Majorana fermion, the new state is still the ground state with the same eigen energy as G. In other words, it costs no energy to create a Majorana fermion at any of the two ends of this chain More general statement (no question for this part) In fact, for this 1D superconducting chain, there are two different phases (assuming D>0). For 2 t <m, it is a topologically trivial phase with no extra modes at the two ends. For 2 t <m, it is a topological phase, which contains one zero-energy Majorana modes at each of the two ends. The two cases mentioned above in and are examples for the topologically trivial and topological phases, respectively. Here, the meaning of the word topological is similar to what we learned in higher dimensions. In 2D and 3D, topological insulator/superconductors have low-energy edge/surface modes. In 1D, they have zero energy modes near the two ends Topological degeneracy part I: How many ground states do we have for a 1D topological superconducting chain? For the 1D superconducting chain studied in 2.3.5, we can define a complex fermion operator using g 1 and g 2 N c = g 1 +Âg 2 N 2 and c = g 1 +Âg 2 N 2 Because H g 1 G = E 0 g 1 G and H g 2 N G = E 0 g 2 N G, it is easy to check that Hc G = E 0 c G Hc G = E 0 c G (10.92) (10.93) (10.94) (10.95) Therefore, the space formed by all the ground states of this Hamiltonian can be written using the operators c and c Define a state 0, which is a ground state of the Hamiltonian H 0 = E 0 0 and in addition it satisfies the condition c 0 =0. Another ground state 1 can be defined as 1 = c 0. Prove that c 1 = 0 and c 1 = 0, so that we cannot get any extra ground states from 0 using c and c operators. Prove that 1 0 =0, so that 1 and 0 are two different ground state. Therefore, we conclude that there are two degenerate ground state for the case studied in Topological degeneracy part II: How many ground states do we have, if we have many Majorana chains? Assume that we have n Majorana chains. For each chain, we can define c i and c i operator as we did in 2.3.8, with i=1,2, n. Now, we can start from a state , which is a ground state of the system and satisfies the condition c i = 0 (10.96) for any c i.

11 144 Phys620.nb Now, it is easy to check that any states like this is a new ground states. c i1 c i2 c i3 c im 00 0 (10.97) Show that there are 2 n ground states. This conclusion is very generic. For a system with 2n Majorana fermions., the number of ground states is 2 n. (Majorana fermions always appears in pairs, so the total number is always even. This is why we consider a system with 2n Majorana fermions here, instead of 2n+1).