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The presence of a metallic lead allows us to measure all the energies $E_n$ of the electronic states in the dot with respect to the Fermi level $E_F$ of the electrons in the metallic lead. In the following we will set $E_F=0$. Hence, all negative energies $E_n<0$ correspond to filled states in the dot, and all positive energies $E_n>0$ to empty states. In the sketch, the lead and the dot are separated by a potential barrier, such that they are only coupled very weakly. Thus, we can still consider the dot as an isolated system, to a good approximation.
Topology studies whether objects can be transformed continuously into each other. In condensed matter physics we can ask whether the Hamiltonians of two quantum systems can be continuously transformed into each other. If that is the case, then we can say that two systems are 'topologically equivalent'.
If we considered all Hamiltonians without any constraint, every Hamiltonian could be continuously deformed into every other Hamiltonian, and all quantum systems would be topologically equivalent. This changes drastically if we restrict ourselves to systems with an energy gap. This means that there is a finite energy cost to excite the system above its ground state.
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For simplicity, we have taken $H$ to be real. Let's now deform this Hamiltonian into another Hamiltonian $H'$, also real. We can imagine that this deformation describes the changes that occur to the dot as an external parameter, such as a gate voltage, is varied. We can parameterize the deformation by
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You may notice from the plot that as $\alpha$ varies, it can happen that an energy level crosses zero energy. When this happens, we break the condition that there should be an energy gap in the system. Notice, however, that this does not necessarily mean that there is no continuous transformation between $H$ and $H'$ such that the gap does not close. It simply means that this particular path has gap closings. Perhaps it is possible to find another path which does not.
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In order to know whether there is any path which connects $H$ and $H'$ without closing the gap, we can count the number of levels below zero energy, i.e. the number of filled energy levels. Indeed, for gapped Hamiltonians no energy level can move through zero, but otherwise they can move freely. Therefore continuous transformations exist exactly between Hamiltonians with the same number of energy levels below zero.
In order to know whether there is any path which connects $H$ and $H'$ without closing the gap, we can count the number of levels below zero energy, i.e. the number of filled energy levels. This is possible because the eigenvalues of gapped Hamiltonians can move freely as long as they don't cross zero energy. Therefore continuous transformations exist exactly between Hamiltonians with the same number of energy levels below zero.
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The plot makes it clear that we do not actually have to count the number of filled energy levels for both $H$ and $H'$, so it is enough to keep track of *zero energy crossings*. Whenever an energy level crosses zero energy, the number of levels below zero energy changes. Such a crossing therefore changes the topological invariant. We call that a *topological phase transition*.
On the other hand, if there are equally many levels crossing from below to above zero energy as the other way around, the number of levels below zero energy does not change. The topological invariant is therefore the same for the initial and final Hamiltonian. In this case, there must be a continuous transformation between the initial and final Hamiltonian which does not close the gap.
Once we have identified a topological invariant, we can *classify* all quantum Hamiltonians according to its value. In this way we create classes of Hamiltonians which are all topologically equivalent, and we can keep track of all the different *topological phases* that these Hamiltonians can support.
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In our previous examples, you might have wondered, whether there was anything special with choosing real matrices? Indeed there was something special. A real Hamiltonian is a manifestation of time-reversal symmetry. Time-reversal symmetry is represented by an anti-unitary operator, and as such it can always be written as the product $\mathcal{T}=U\mathcal{K}$ of a unitary matrix times complex conjugation. In the case above, we had simply $\mathcal{T}\equiv\mathcal{K}$. Our real Hamiltonians clearly obeyed time-reversal symmetry since $H=H^*$.
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While the spectrum looks quite similar to the previous ones, whenever a line crosses zero energy, our topological invariant makes a jump of two, and not one! In this case, time-reversal symmetry constrains the topological invariant to only take even values. This is an example of how topological properties can be influenced by discrete symmetries.
Let's now take a system where we can split all the degrees of freedom into two groups (say group $A$ and group $B$) such that the Hamiltonian only has nonzero matrix elements between two groups, and not inside each group. This situation arises naturally when the lattice has two sublattices, as in the hexagonal carbon lattice of graphene. So let's imagine our quantum dot is now a graphene dot:
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The $H$-term is the dynamics of the electrons in the dot, while $\Delta$ describes the pair creation and annihilation. The matrix $\Delta$ is antisymmetric because the fermion operators anticommute. Now $\mathcal{H}$ does not conserve the number of electrons, but still conserves the parity of the number of electrons, that is whether the number of electrons is even or odd. We can now group all the creation and annihilation operators in a vector, $C = (c_1, \dots, c_n, c^\dagger_1, \dots, c^\dagger_n)^T$. Then we write $\mathcal{H}$ in the form 'row multiplies matrix multiplies column':
Particle-hole symmetry is represented by an anti-unitary operator which anti-commutes with the Hamiltonian (compare this situation with that of time-reversal and sublattice symmetries). Because of the minus sign in the particle-hole symmetry, the spectrum of $H_\textrm{BdG}$ must be symmetric around zero energy (that is, the Fermi level). Indeed, for every eigenvector $\psi = (u, v)^T$ of $H_\textrm{BdG}$ with energy $E$, there will be a particle-hole symmetric eigenvector $\mathcal{P}\psi=(v^*, u^*)^T$ with energy $-E$.
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Let's think a bit more about these crossings. At first, they might look a bit mysterious. In general a crossing between energy levels happens in the presence of a conserved quantity, and our random Bogoliubov-de Gennes Hamiltonian does not seem to have an obvious one. Let's however recall what we said earlier: While the mean-field Hamiltonian of a superconductor does not conserve the number of particles, it conserves the parity of this number. In other words, forming and breaking Cooper pairs does not affect whether the superconducting quantum dots contains an even or odd number of electrons. In short, fermion parity is a conserved quantity (provided that isolated electrons do not enter or leave the dot, a possibility which we will disregard).
When a pair of levels crosses zero energy, the excitation energy $E$ of the Bogoliubov quasiparticle changes sign and it becomes favorable to add a Bogoliubov quasiparticle to, or remove it from the superconducting quantum dot. In other words, at each crossing the fermion parity in the ground state of the dot changes from even to odd, or vice versa. Hence these crossings are *fermion parity switches*.
Since the ground state fermion parity is preserved by the superconducting Hamiltonian if there are no Bogoliubov quasiparticles crossing zero energy, the ground state fermion parity is the topological invariant of this system. It is clear however that this invariant is of a different nature than the one of the non-superconducting systems, which is given by the number of negative eigenvalues of the Hamiltonian. The latter cannot change for a Bogoliubov-de Gennes Hamiltonian, which has a symmetric energy spectrum, and hence it is not suitable to describe changes in fermion parity. Is there a way to compute this new invariant directly from the Bogoliubov-de Gennes Hamiltonian?
There is a special number that we can compute for antisymmetric matrices, the [Pfaffian](http://en.wikipedia.org/wiki/Pfaffian). Its rigorous definition is not important for our course. The basic idea is simple: The eigenvalues of antisymmetric matrices always come in pairs. In the case of our $\tilde{H}_{BdG}$, these are the energy eigenvalues $\pm E_n$. By taking their product we obtain the determinant of the matrix, equal to $\prod_n (-E_n^2)$. The key property of the Pfaffian is that it allows to take a square root of the determinant, equal to $\pm i\prod_n E_n$, in such a way that the sign of the product is uniquely defined. At a fermion parity switch a single $E_n$ changes sign, so the Pfaffian changes sign as well (while the determinant stays the same).
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