Notes on Physics

Welcome! This is the place for me to write some notes on physics. Contents will be added indefinitely.

Classical Mechanics:

Phase space Lagrangian

Regular derivation

The climax part of classical mechanics lies in the Lagrangian and Hamiltonian form. It starts with the extremal principle, the real motion of a mechanical system is the one makes the variation of the action \(S\) vanish, i.e.,

\[\delta S=\delta \int L(q,\dot{q},t) dt=0\]

When the Lagrangian is not depend on time explicitly, we get

\[\delta L=\frac{\partial L}{\partial q}\delta q+\frac{\partial L}{\partial \dot{q}}\delta \dot{q}=\frac{d}{dt}\left(\frac{\partial L}{\partial \dot{q}}\delta {q}\right)+\left[\frac{\partial L}{\partial q}-\frac{d}{dt}\frac{\partial L}{\partial \dot{q}}\right]\delta q\]

which gives us the Lagrangian equation:

\[\frac{d}{dt}\frac{\partial L}{\partial \dot{q}}=\frac{\partial L}{\partial q}\]

Define the canonical momentum \(p=\frac{\partial L}{\partial \dot{q}}\), we have

\[\dot{p}=\frac{\partial L}{\partial q}\]

We can easily prove that energy is an integral of motion (which is a conserved quantity in motion) based on the homogeneity of time. When the Lagrangian does not depend on time explicitly, we found that its total derivative is

\[\begin{split}\frac{dL}{dt}=&\frac{\partial L}{\partial q}\dot{q}+\frac{\partial L}{\partial \dot{q}} \ddot{q}\\ =&\frac{d}{dt}\left(p\dot{q}\right)\end{split}\]

We have used Lagrangian equation in the above derivation,and we have now

\[\frac{d}{dt}\left(p\dot{q}-L\right)=0\]

indicating that \(H=p\dot{q}-L\) is conserved in the motion, which is called Hamiltonian with physical meaning of energy.

On the other hand, from:

\[\begin{split}dL(q,\dot{q})=&\frac{\partial L}{\partial q}dq+\frac{\partial L}{\partial \dot{q}}d\dot{q} \\ =&\dot{p}dq+pd\dot{q}\\ =&\dot{p}dq+d(p\dot{q})-\dot{q}dp\\ \Rightarrow d(p\dot{q}-L)=&-\dot{p}dq+\dot{q}dp\\ dH(q,p)=&-\dot{p}dq+\dot{q}dp\end{split}\]

which means \(dH\) is the total differential with respect to \(q\) and \(p\). And from:

\[dH=\frac{\partial H}{\partial q}dq+\frac{\partial H}{\partial p}dp\]

we get

\[\dot{p}=-\frac{\partial H}{\partial q} \qquad \dot{q}=\frac{\partial H}{\partial p}\]

This is the Hamilton canonical equations.

The total derivative with time is:

\[\begin{split}\frac{dH}{dt}=&\frac{\partial H}{\partial t}+\frac{\partial H}{\partial q}\dot{q}+\frac{\partial H}{\partial p}\dot{p} \\ =&\frac{\partial H}{\partial t}\end{split}\]

where Hamilton’s equations are used. It also indicates that if H is not depend on time explicitly, we have the conservation of energy.

Note

From Lagragian to Hamiltonian, we did a Legendre transformation, makes the dependence \(L(q,\dot{q},t)\) to \(H(q,p,t)\), i.e., \(L\) is a function in configuration space \((q,\dot{q})\), but \(H\) is in phase space \((q,p)\).

Dynamic system

Mathematically, a continuous-time dynamical system is defined to be a system of first order differential equations

\[\dot{\mathbf{z}} = \mathbf{f}(\mathbf{z}, t) , \quad t \in \mathbb{R}\]

where \(\mathbf{f}\) is known as the vector field and \(\mathbb{R}\) is the set of real numbers. The space in which \(\mathbf{z}\) is defined is called phase space.

Lagrange’s equations do not form a dynamical system, because they implicitly contain second-order derivatives, \(\ddot{q}\). However, there is a standard way to obtain a system of first-order equations from a second-order system, which is to double the size of the space of time-dependent variables by treating the generalized velocities \(u\) as independent of the generalized coordinates, so that the dynamical system is \(\dot{q} = u, \dot{u} = \ddot{q}(q,u,t)\). Then the phase space is of dimension \(2n\). This trick is used very frequently in numerical problems, because the standard numerical integrators require the problem to be posed in terms of systems of first-order differential equations.

In the particular case of Lagrangian mechanics, expanding out \(\frac{d}{dt}\frac{\partial L}{\partial \dot{q}}\) using the chain rule and moving all but the highest-order time derivatives to the right-hand side,

\[\sum\limits_{j=1}^{n}\frac{\partial^2 L}{\partial \dot{q}_i \partial \dot{q}_j}\ddot{q}_j=\frac{\partial L}{\partial q_i}-\frac{\partial^2 L}{\partial \dot{q}_i \partial t}-\sum\limits_{j=1}^{n}\frac{\partial^2 L}{\partial \dot{q}_i \partial {q}_j}\dot{q}_j\]

The matrix \(H\) acting on \(\ddot{q}\), whose elements are given by \(H_{i,j}\equiv\frac{\partial^2 L}{\partial \dot{q}_i \partial \dot{q}_j}\), is called the Hessian matrix . It is a kind of generalized mass tensor, and for our method to work we require it to be nonsingular, so that its inverse, \(H^{-1}\), exists and we can find \(\ddot{q}\). Then our dynamical system becomes:

\[\begin{split}\dot{q}=&u, \\ \dot{u}=&H^{-1}\cdot\left[\frac{\partial L}{\partial q}-\frac{\partial L}{\partial \dot{q}\partial t}-\frac{\partial^2 L}{\partial \dot{q}\partial q}\cdot \dot{q}\right]\end{split}\]

Momentum instead of velocity

We can achieve our aim of finding 2n first-order differential equations by using many choices of auxiliary variables other than \(u\). These will be more complicated functions of the generalized velocities, but the extra freedom of choice may also bring advantages.

In particular, Hamilton realized that it is very natural to use as the new auxiliary variables the set \(p=\left\{p_i|i=1,\cdots,n\right\}\) defined by

\[p_i\equiv\frac{\partial}{\partial \dot{q}_i}L(q,\dot{q},t)\]

where \(p_i\) is called the canonical momentum conjugate to \(q_i\).

At this moment, we shall assume that the above equation can be solved to give \(\dot{q}\) as a function of \(q\) and \(p\)

\[\dot{q}=u(q,p,t).\]

The Lagrange’s equations immediately give us

\[\dot{p}=\frac{\partial L(q,\dot{q},t)}{\partial q}\left.\right|_{\dot{q}=u(q,p,t)}\]

The above two equations do indeed form a dynamical system, but so far it looks rather unsatisfactory: now \(u\) is defined only implicitly as a function of the phase-space variables \(q\) and \(p\), yet the right-hand side of above equation involves a partial derivative in which the \(q\)-dependence of \(u\) is ignored!

We can fix the latter problem by holding \(p\) fixed in partial derivatives with respect to \(q\) (because it is an independent phase-space variable) but then subtracting a correction term to cancel the contribution coming from the \(q\)-dependence of \(u\). Applying the chain rule, we get

\[\begin{split}\dot{p}=&\frac{\partial L(q,u,t)}{\partial q}-\frac{\partial L}{\partial u}\frac{\partial u}{\partial q} \\ =&\frac{\partial L(q,u,t)}{\partial q}-p\frac{\partial u}{\partial q} \\ =&\frac{\partial}{\partial q}\left[L(q,u,t)-p\cdot u\right] \\ =&-\frac{\partial H}{\partial q}\end{split}\]

where we have already defined the Hamiltonian \(H(q, p, t)= p\cdot u − L(q,u, t)\) is a function of \((q, p)\).

Given the importance of \(\partial H/ \partial q\) it is natural to investigate whether \(\partial H/\partial p\) plays a significant role as well. Differentiating \(H(q, p, t)= p\cdot u − L(q,u, t)\) we get

\[\begin{split}\frac{\partial H}{\partial p}=&u(q,p,t)-\left[p-\frac{\partial}{\partial u}L(q,u,t)\right]\frac{\partial u}{\partial p} \\ =&\dot{q}\end{split}\]

the above two equations are just the Hamilton’s equations we derived before.

Phase space Lagrangian

The equation \(H(q,p,t)\equiv p\cdot\dot{q}-L(q,\dot{q},t)\) suggest we define \(L_{ph}(q,\dot{q},p,t)\equiv p\cdot\dot{q}-H(q,p,t)\). If \(\dot{q}=u(q,p,t)\) were identically satisfied, even on arbitrarily varied phase-space paths, then \(L_{ph}\) would simply be \(L\) expressed in phase-space coordinates.

However, one can easily construct a counter example to show that this is not the case: consider a variation of the path in which we can vary the direction of its tangent vector, at some point \(z\equiv (q, p)\), while keeping this point fixed. Then \(\dot{q}\) changes, but \(u\) remains the same. Thus \(L_{ph}\) and \(L\) are the same value only on the subset of paths (which includes the physical paths) for which \(p\cdot\dot{q} = p\cdot u(q, p, t)\) .

_images/1.jpg

Fig. 1 Phase-space variations of different paths.

Replacing \(L\) by \(L_{ph}\) in \(S = \int L dt\) we define the phase-space action integral

\[S_{ph}\left[q, p\right] = \int_{t_1}^{t_2}dt L_{ph}(q,p,\dot{q},t)=\int_{t_1}^{t_2}dt \left(p\cdot\dot{q}-H(q,p,t)\right)\]

We know from variational calculus that \(S_{ph}\) is stationary under arbitrary variations of the phase-space path (with endpoints fixed), explicitly, we get:

\[\begin{split}\delta S_{ph} =&\delta\int_{t_1}^{t_2}dt \left(p\cdot\dot{q}-H(q,p,t)\right) \\ =&\int_{t_1}^{t_2}dt \left(\delta p\cdot\dot{q}+p\cdot\delta\dot{q}-\delta p \frac{\partial H(q,p,t)}{\partial p}-\delta q \frac{\partial H(q,p,t)}{\partial q}\right) \\ =&p\delta q\left. \right |_{t_1}^{t_2}+\int_{t_1}^{t_2}dt \left\{\delta p\cdot\left[\dot{q}-\frac{\partial H(q,p,t)}{\partial p}\right]-\delta q\cdot\left[\dot{p}+ \frac{\partial H(q,p,t)}{\partial q}\right]\right\}\end{split}\]

which gives us the Hamilton’s equations:

\[\dot{p}=-\frac{\partial H}{\partial q} \qquad \dot{q}=\frac{\partial H}{\partial p}\]

Note

More generally, the phase-space Lagrangian can depend on \(\dot{p}\) as well, and makes the Lagrange’s equations become:

\[\begin{split}\frac{\delta L_{ph}(q,\dot{q},p,\dot{p})}{\delta q}=&\frac{\partial L_{ph}}{\partial q}-\frac{d}{dt}\frac{\partial L_{ph}}{\partial \dot{q}}=0 \\ \frac{\delta L_{ph}(q,\dot{q},p,\dot{p})}{\delta p}=&\frac{\partial L_{ph}}{\partial p}-\frac{d}{dt}\frac{\partial L_{ph}}{\partial \dot{p}}=0\end{split}\]

References

  1. L.D. Landau & E.M. Lifshitz, Mechanics.
  2. Lecture notes by Bob Dewar on website Classical Mechanics.

Topological Insulator:

Lecture notes

Here is the lecture notes on TI, which is partly based on the lecture notes given by Janos Asboth, Laszlo Oroszlany, and Andras Palyi, Here is the resource, and you can also download a copy here.

Lecture 1 : 1-d SSH model

The Su-Schrieffer–Heeger (SSH) model

The simplest non-trivial topology : 1-d lattice.

Peierls instability makes the atoms dimerize.

_images/1.png

Polyacetylene Structure:

_images/2.jpg
Tight-binding method: first quantization
1-d atom chain

1-d atom chain

Tight-binding method: Single electron total Hamiltonian in atom chain:

\[H=\frac{p^2}{2m}+U(x)\]

with periodical potential: \(U(x+na)=U(x)\)

Assume single atom potential \(V(x)\) with Hamiltonian

\[H_0=\frac{p^2}{2m}+V(x)\]

and well solved eigen-value and eigen-wave-function:(consider only one state)

\[H_0\phi(x)=E_0\phi(x)\]

Do the combination

\[\psi(x)=\sum\limits_n a_n\phi_n\]

with \(\phi_n=\phi(x-x_n), x_n=na\). Define \(\Delta U(x)=U(x)-V(x)\), and substitute following equation

\[H\psi=\left(H_0+\Delta U\right)\psi=E\psi\]

we get

\[\sum_n \langle \phi_m|\Delta U(x-x_n)|\phi_n\rangle a_n=(E-E_0)a_m\]

Define: \(\langle \phi(x-ma+na)|\Delta U(x) |\phi(x)\rangle=-J(x_m-x_n)\)

We get:

\[-\sum_n J(x_m-x_n) a_n=(E-E_0) a_m\]

Because of the Tranformation symmetry of the Hamiltonian, the resulting wavefunction \(\psi(x)\) should take Bloch form, which means we should have the solution \(a_m=e^{ikx_m}\), then, we get

\[E-E_0=-\sum_n J(x_n)e^{-ikx_n}\]

Consider only the nearest-hopping interaction, define \(J_0=J(0), J=J(\pm a)\), then we have:

\[\begin{split}E-E_0+J_0=&-J(e^{ika}+e^{-ika}) \\ =&-2Jcoska\end{split}\]
Second quantization

In the second quantization language, the expectation value of energy becomes a operator, set \(\mathscr{H}=\frac{p^2}{2m}+U(x)\), we have

\[H=\langle \psi|\mathscr{H}|\psi \rangle \Rightarrow \hat{H}=\sum_{m,n}\hat{c}_m^\dagger H_{mn}\hat{c}_n\]

with \(\psi \to \hat{\psi}=\sum\limits_n \hat{c}_n \phi_n, H_{mn}=\langle \phi|\mathscr{H}|\phi \rangle\) \(\phi_n\) is a orthonormal and complete basis in Hilbert space, like plane-waves \(e^{ikx}\) or energy eigen-states of \(H_0\), \(\mathscr{H}\) is the energy operator in single particle first quantization picture, which can only act on Hilbert space, while the second quantization energy operator \(\hat{H}\) acts on Fock space. Here, in tight-binding method, \(\phi_n\) is the wave-function of site \(n\) of the energy eigen-state \(H_0\).

Consider only the nearest interaction, we have:

\[\hat{H}=\sum_{n=1}^M \hat{c}_n^\dagger \hat{c}_{n+1}t_n+h.c.\]

\(h.c.\) means hermitian conjugation. Rewrite it in matrix form: \(\hat{H}=\sum\limits_{mn}\hat{c}_m \tilde{H}_{mn}c_n\), we have

\[\begin{split}\tilde{H}_{mn}=\begin{pmatrix} 0 & t_1 &0 & \cdots&t_M^*\\ t_1^* &0&t_2&\cdots&0\\ 0&t_2^*&0&\cdots&0\\ \vdots&\vdots&\vdots&\vdots&\vdots\\ t_M&0&0&t_{M-1}^*&0 \end{pmatrix}\end{split}\]

In the case when \(t=t_n\), \(\hat{c}_n\) satisfy the Bloch condition, we can transform it into momentum space, with \(\hat{c}_n=\frac{1}{\sqrt{M}}\sum\limits_k \hat{c}_k e^{ikx_n}\),

we can easily get

\[\hat{H}=\sum\limits_k \hat{c}_k^\dagger \hat{c}_k(te^{ika}+t^*e^{-ika})=\sum\limits_k \hat{c}_k^\dagger \hat{c}_k E(k)\]

which gives us the dispersion relation:

\[\begin{split}E(k)=&te^{ika}+t^*e^{-ika}\\ =&2tcoska \qquad \textit{for t is real}\end{split}\]

On the other hand, keep in mind that we would get \(c_n=e^{ik(n-1)a}c_1\), that is

\[\begin{split}\hat{H}=\begin{pmatrix}c_1^\dagger&c_2^\dagger&\cdots&c_M^\dagger \end{pmatrix} \begin{pmatrix} 0 & t &0 & \cdots&t^*\\ t^* &0&t&\cdots&0\\ 0&t^*&0&\cdots&0\\ \vdots&\vdots&\vdots&\vdots&\vdots\\ t&0&0&t^*&0 \end{pmatrix} \begin{pmatrix}c_1\\c_2\\\vdots\\c_M\end{pmatrix}\end{split}\]
\[\begin{split}\Rightarrow &\begin{pmatrix}c_1^\dagger&c_1^\dagger e^{-ika}&\cdots&c_1^\dagger e^{-ik(M-1)a} \end{pmatrix} \begin{pmatrix} 0 & t &0 & \cdots&t^*\\ t^* &0&t&\cdots&0\\ 0&t^*&0&\cdots&0\\ \vdots&\vdots&\vdots&\vdots&\vdots\\ t&0&0&t^*&0 \end{pmatrix} \begin{pmatrix}c_1\\c_1 e^{ika}\\\vdots\\c_1e^{ik(M-1)a}\end{pmatrix}\\ =&\begin{pmatrix}c_1^\dagger&\cdots&c_1^\dagger \end{pmatrix} \begin{pmatrix} te^{ika}+t^*e^{-ika} & 0&0\\ 0&\ddots&0\\ 0&0&te^{ika}+t^*e^{-ika} \end{pmatrix} \begin{pmatrix}c_1\\\vdots\\c_1\end{pmatrix}\end{split}\]

which also gives us \(E(k)=te^{ika}+t^*e^{-ika}\).

More generally, \(t_n\) can be different from each other, for example, if they are all different up to \(4\), but have a super-periodicity with \(t_{5}=t_1\), then there will have 4 sub-bands, in the example we will consider below, we have two \(t\), \(t_1\neq t_2\), and we have two sub-bands.

If each atom have a valance electron, then the above mentioned energy band structure \(E(k)=2tcoska\) is not the stable fundamental mode, it will dimerizes to lower the total energy, that means we’ll get following coupling case:

_images/2.jpg

with the Hamiltonian:

\[H=\sum_{n=1}^N(v_n c_{n,1}^\dagger c_{n,2}+w_nc_{n,2}^\dagger c_{n+1,1}+h.c.)\]

with \(M=2N\).

Review

In 1-d atom chain, we have got the Hamiltonian in second-quantization frame as:

\[\hat{H}=\sum_{n=1}^M \hat{c}_n^\dagger \hat{c}_{n+1}t_n+h.c.\]

\(h.c.\) means hermitian conjugation. Rewrite it in matrix form: \(\hat{H}=\sum\limits_{mn}\hat{c}_m {H}_{mn}\hat{c}_n\), we have

\[\begin{split}\hat{H}=\begin{pmatrix}c_1^\dagger&c_2^\dagger&\cdots&c_M^\dagger \end{pmatrix}\begin{pmatrix} 0 & t_1 &0 & \cdots&t_M^*\\ t_1^* &0&t_2&\cdots&0\\ 0&t_2^*&0&\cdots&0\\ \vdots&\vdots&\vdots&\vdots&\vdots\\ t_M&0&0&t_{M-1}^*&0 \end{pmatrix}\begin{pmatrix}c_1\\c_2\\\vdots\\c_M\end{pmatrix}\end{split}\]
Energy band and topology
\[\begin{split}\hat{H}=\begin{pmatrix}c_1^\dagger&c_2^\dagger&\cdots&c_M^\dagger \end{pmatrix}\begin{pmatrix} 0 & t_1 &0 & \cdots&t_M^*\\ t_1^* &0&t_2&\cdots&0\\ 0&t_2^*&0&\cdots&0\\ \vdots&\vdots&\vdots&\vdots&\vdots\\ t_M&0&0&t_{M-1}^*&0 \end{pmatrix}\begin{pmatrix}c_1\\c_2\\\vdots\\c_M\end{pmatrix}\end{split}\]

\(t_n\) can differ from each other.

  1. For a open chain with \(M\) atoms, we have \(t_M=0\), and this matrix will give us \(M\) eigen-values and eigen-functions.
  2. Possess translational invariance with \(c_{n+1}=c_1e^{ikna}\), it will be diagonalized with \(H(k)=te^{ika}+t^*e^{-ika}\).
  3. Staggered hopping parameters with \(t_1\neq t_2\), but have property \(c_{2n+1}=c_1e^{iknb},c_{2n+2}=c_2e^{iknb}\). We can block the Hamiltonian up in \(2\times 2\) blocks and also pair up \(c_{2n+1},c_{2n+2}\).

The Hamiltonian:

\[H=\sum_{n=1}^N(v_n c_{n,1}^\dagger c_{n,2}+w_nc_{n,2}^\dagger c_{n+1,1}+h.c.)\]

with \(M=2N\).

_images/2.jpg

For a more beautiful notation, define \(\mathbf{c}_n^\dagger=(c_{n,1}^\dagger,c_{n,2}^\dagger)=(c_{2n-1}^\dagger,c_{2n}^\dagger)\), then we have

\[H=\sum_{m,n=1}^N \mathbf{c}_m^\dagger H_{mn}\mathbf{c}_n\]

with

\[\begin{split}\mathbf{c}_n^\dagger H_{nn}\mathbf{c}_n=\begin{pmatrix}c_{n,1}^\dagger&c_{n,2}^\dagger\end{pmatrix}\begin{pmatrix}0&v_n\\v_n^*&0\end{pmatrix}\begin{pmatrix}c_{n,1}\\c_{n,2}\end{pmatrix}=\begin{pmatrix}c_{n,1}^\dagger&c_{n,2}^\dagger\end{pmatrix}U_n\begin{pmatrix}c_{n,1}\\c_{n,2}\end{pmatrix}\end{split}\]

and

\[\begin{split}\mathbf{c}_n^\dagger H_{nn+1}\mathbf{c}_{n+1}=\begin{pmatrix}c_{n,1}^\dagger&c_{n,2}^\dagger\end{pmatrix}\begin{pmatrix}0&0\\w_n&0\end{pmatrix}\begin{pmatrix}c_{n+1,1}\\c_{n+1,2}\end{pmatrix}=\begin{pmatrix}c_{n,1}^\dagger&c_{n,2}^\dagger\end{pmatrix}T_n\begin{pmatrix}c_{n+1,1}\\c_{n+1,2}\end{pmatrix}\end{split}\]
\[\begin{split}\mathbf{c}_{n+1}^\dagger H_{n+1n}\mathbf{c}_{n}=\begin{pmatrix}c_{n+1,1}^\dagger&c_{n+1,2}^\dagger\end{pmatrix}T_n^\dagger\begin{pmatrix}c_{n,1}\\c_{n,2}\end{pmatrix}\end{split}\]

when \(|m-n|>1\), we have \(H_{mn}=0\).

For example, for 6 cells (12 sites), we have

\[\begin{split}H=\begin{pmatrix}U_1&T_1&0&\cdots&T_6^\dagger\\ T_1^\dagger&U_2&T_2&\cdots&0\\ 0&T_2^\dagger&U_3&\cdots&0\\ \vdots&\vdots&\vdots&\vdots&\vdots\\ T_6&0&0&\cdots&U_6 \end{pmatrix}\end{split}\]
\[\begin{split}H_{mn}=\begin{pmatrix}U_1&T_1&0&\cdots&T_6^\dagger\\ T_1^\dagger&U_2&T_2&\cdots&0\\ 0&T_2^\dagger&U_3&\cdots&0\\ \vdots&\vdots&\vdots&\vdots&\vdots\\ T_6&0&0&\cdots&U_6 \end{pmatrix}\end{split}\]

  • Open chain: \(T_6=0\).

  • Closed chain with translational symmetry, \(T_n=T,U_n=U\), with

    \[\begin{split}U=\begin{pmatrix}0&v\\v^*&0\end{pmatrix},T=\begin{pmatrix}0&0\\w&0\end{pmatrix}\end{split}\]

    Using three Pauli matrices

    \[\begin{split}\sigma_x=\begin{pmatrix}0&1\\1&0\end{pmatrix},\sigma_y=\begin{pmatrix}0&-i\\i&0\end{pmatrix},\sigma_z=\begin{pmatrix}1&0\\0&-1\end{pmatrix}\end{split}\]

We get

\[U=Re(v)\sigma_x-Im(v)\sigma_y,T=\frac{1}{2}w(\sigma_x-i\sigma_y)\]

Using \(\mathbf{c}_n=e^{ik(n-1)b}\mathbf{c}_1\), we have

\[\begin{split}\hat{H}=\begin{pmatrix}\mathbf{c}_1^\dagger&\mathbf{c}_2^\dagger&\cdots&\mathbf{c}_N^\dagger \end{pmatrix} \begin{pmatrix} U & T &0 & \cdots&T^\dagger\\ T^\dagger &U&T&\cdots&0\\ 0&T^\dagger&U&\cdots&0\\ \vdots&\vdots&\vdots&\vdots&\vdots\\ T&0&0&T^\dagger&U \end{pmatrix} \begin{pmatrix}\mathbf{c}_1\\\mathbf{c}_2\\\vdots\\\mathbf{c}_N\end{pmatrix}\end{split}\]
\[\begin{split}\Rightarrow \begin{pmatrix}\mathbf{c}_1^\dagger&\mathbf{c}_1^\dagger e^{-ikb}&\cdots&\mathbf{c}_1^\dagger e^{-ik(N-1)b} \end{pmatrix} \begin{pmatrix} U & T &0 & \cdots&T^\dagger\\ T^\dagger &U&T&\cdots&0\\ 0&T^\dagger&U&\cdots&0\\ \vdots&\vdots&\vdots&\vdots&\vdots\\ T&0&0&T^\dagger&U \end{pmatrix} \begin{pmatrix}\mathbf{c}_1\\\mathbf{c}_1 e^{ikb}\\\vdots\\\mathbf{c}_1e^{ik(N-1)b}\end{pmatrix}\end{split}\]
\[\begin{split}=\begin{pmatrix}\mathbf{c}_1^\dagger&\cdots&\mathbf{c}_1^\dagger \end{pmatrix} \begin{pmatrix} U+Te^{ikb}+T^\dagger e^{-ikb} & &\\ &\ddots&\\ &&U+Te^{ikb}+T^\dagger e^{-ikb} \end{pmatrix} \begin{pmatrix}\mathbf{c}_1\\\vdots\\\mathbf{c}_1\end{pmatrix}\end{split}\]

which gives us \(H=H(k)\oplus H(k)\cdots \oplus H(k)=\oplus_{n=1}^N H(k)\) with

\[H(k)=U+Te^{ikb}+T^\dagger e^{-ikb}=\mathbf{h}(k)\cdot \mathbf{\sigma}\]

with

\[\begin{split}h_x(k)=&Re(v)+|w|cos(kb+arg(w))\\ h_y(k)=&-Im(v)+|w|sin(kb+arg(w))\\ h_z(k)=&0\end{split}\]

with \(w=|w|e^{i arg(w)}\).

\[H(k)=\mathbf{h}(k)\cdot \mathbf{\sigma}\]

with

with eigen-energy

\[E(k)=|\mathbf(h)(k)=\pm \sqrt{h_x^2+h_y^2+h_z^2}=\pm \sqrt{|v|^2+|w|^2+2|v||w|cos(kb+arg(v)+arg(w))}\]

and eigen-wavefunctions

\[\begin{split}|\pm\rangle=\begin{pmatrix}\pm e^{-i\phi(k)}\\ 1\end{pmatrix}\end{split}\]

with \(tan\phi=h_y/h_x\).

For example, set \(arg(v)=arg(w)=0\), we have

_images/energy.png

Can not tell the difference \(|v|-|w|=\pm\delta\).

Energy-band description is not completed, it can give us many information, but not the whole, others are hidden in the wave-function. Alternatively, recalling \(H(k)=\mathbf{h}(k)\cdot \mathbf{\sigma}\), the Hamiltonian should contain the whole information, but we have only used \(|h|\), in topological aspect, \(\mathbf{h}(k)\) will suffices.

Set \(arg(v)=0\),\(kb=[0,2\pi]\), we have two cases

  • \(|w|<|v|, \mathbf{inter}<\mathbf{intra}\)
  • \(|w|>|v|, \mathbf{inter}>\mathbf{intra}\)
two cases

two cases

Winding number

\(H(k)=\mathbf{h}(k)\cdot \mathbf{\sigma}\), \(\mathbf{h}(k)=0\) is a degenerate point with \(|v|=|w|\), two bands cross, define \(h(k)=h_x(k)+ih_y(k)\), we have

\[\begin{split}H(k)=\begin{pmatrix}0&h^*(k)\\h(k)&0\end{pmatrix}\end{split}\]
\[ln(h)=ln(|h|)e^{iarg(h)}=ln(|h|)+iarg(h)\]

define

\[\nu=\frac{1}{2\pi i}\int_{-\pi}^{\pi}dk\frac{d}{dk}ln(h(k))\]

When

  • \(|w|>|v|, \nu=1, \mathbf{inter}>\mathbf{intra}\)
  • \(|w|<|v|, \nu=0, \mathbf{inter}<\mathbf{intra}\)

A example, \(N=20, M=2N=40, w=1, v=0.5\), we get eigen-energys:

eigen-energy

eigen-energy

_images/a-.png
_images/a+.png
_images/b-.png
_images/b+.png

Edge-states:

_images/c1.png
_images/c2.png
Chiral symmetry
_images/2.jpg

Recalling the Hamiltonian:

\[H=\sum_{n=1}^N(v_n c_{n,1}^\dagger c_{n,2}+w_nc_{n,2}^\dagger c_{n+1,1}+h.c.)\]

Define projector operators:

\[P_A=\sum_n c_{n,1}^\dagger c_{n,1}, P_B=\sum_n c_{n,2}^\dagger c_{n,2}\]

and the chiral operator \(\Sigma_z=P_A-P_B\), The matrix elements of \(\Sigma_z\) vanish, \(\langle 0|c_r \Sigma c^\dagger_s|0\rangle=0\) if sites \(r\) and \(s\) are in different unit cells.

In first-quantization, we have

\[H=\sum_{n=1}^N(v_n |n,1\rangle\langle n,2|+w_n|n,2\rangle\langle n+1,1|+h.c.)\]

and

\[P_A=\sum_n |n,1\rangle\langle n,1|, P_B=\sum_n |n,2\rangle\langle n,2|\]

In matrix form, we have

\[\Sigma_z=\sigma_z\oplus\sigma_z\oplus\cdots\oplus\sigma=\oplus_{n=1}^N \sigma_z\]

\(\Sigma_z\) is local, it does not mix site between unit cells, and inherits the algebra from \(\sigma_z\):

\[\Sigma_z^\dagger\Sigma_z=1\]
\[\Sigma_z^2=1\]

Recalling

\[\begin{split}\begin{pmatrix} U_1 & T_1 &0 & \cdots&T_N^\dagger\\ T_1^\dagger &U_2&T_2&\cdots&0\\ 0&T_2^\dagger&U_3&\cdots&0\\ \vdots&\vdots&\vdots&\vdots&\vdots\\ T_N&0&0&T_{N-1}^\dagger&U_N \end{pmatrix}\end{split}\]

There are no onsite terms in the Hamiltonian, so we have

\[\Sigma_z H \Sigma_z=-H\]

This is the chiral symmetry.

Note

Actually, here, \(H\) is defined in momentum space, but \(\Sigma_z\) in real space, we should write \(\tilde{H}=U^\dagger H U\) for some unitary matrix \(U\), but the property survives!

Consequences: For eigenstates \(|\psi_n\rangle\) of \(H\), we have

\[H|\psi_n\rangle=E_n|\psi_n\rangle\]
\[H\Sigma_z|\psi_n\rangle=-\Sigma_zH|\psi_n\rangle=-\Sigma_zE_n|\psi_n\rangle=-E_n\Sigma_z|\psi_n\rangle\]\[If\]

  • \(E_n\neq 0\), two orthonormal states \(|\psi_n\rangle, \Sigma_z|\psi_n\rangle\), which gives

    \[\begin{split}\begin{pmatrix}\alpha^* &\beta^*\end{pmatrix}\begin{pmatrix}\alpha\\-\beta\end{pmatrix}=0\end{split}\]
    \[\Rightarrow |\alpha|^2=|\beta|^2\]

  • \(E_n=0\), we particularly have \(\Sigma_z=\Sigma_zH\), \(|\psi_n\rangle\) can be eigen-states of \(H\) and \(\Sigma\) simultaneously, \(\Sigma_z|\psi_n\rangle=\pm|\psi_n\rangle\), gives \(|\psi_n\rangle=\begin{pmatrix}1 \\0\end{pmatrix}\) or \(|\psi_n\rangle=\begin{pmatrix}0 \\1\end{pmatrix}\)

Review 2

In the last class, we have solved 1-d atom chain with staggered hopping parameters and got the Hamiltonian:

\[\begin{split}H(k)=\mathbf{h}(k)\cdot \mathbf{\sigma}=\begin{pmatrix}0&h_x-ih_y\\h_x+ih_y&0\end{pmatrix}=|h(k)|\begin{pmatrix}0&e^{-i\phi(k)}\\e^{i\phi(k)}&0\end{pmatrix}\end{split}\]

with \(tan\phi=h_y/h_x\).

Eigen-values are \(E(k)=\pm|h(k)|\), with eigen-functions: \(|\pm\rangle=\begin{pmatrix}\pm e^{-i\phi(k)}\\ 1\end{pmatrix}\).

\(\mathbf{h}(k)\) depicted as follow:

Topological transition

Topological transition

Topological transition

We have already seen, for \(|w|>|v|\) or \(|w|<|v|\), we got different winding number, means there is a topological transition at \(|w|=|v|\). In the energy-band point of view, it means the gap between two energy bands closes (across each other) and reopens. There are two ways to change the winding number and get a topological transition:

  1. Pull the path through the origin in the \(h_x-h_y\) plane.
  2. Lift it out of the plane (breaking the chiral symmetry).

Case 1(a): \(v=0.5, w=0\to 1\)

_images/a.jpg
_images/figure_1a.png

Case 1(b): \(w=1, v=2.5\to 0\)

_images/c.jpg
_images/figure_3.png

Case 2: Sublattice potential \(H_{sublattice}=u\sigma_z\). Recalling

\[H(k)=U+Te^{ikb}+T^\dagger e^{-ikb}\]

with \(U=\begin{pmatrix}0&v\\v^*&0\end{pmatrix},U=\begin{pmatrix}0&0\\w&0\end{pmatrix}\), now \(U\) changes to \(U=\begin{pmatrix}u&v\\v^*&-u\end{pmatrix}\), and \(H(k)=\mathbf{h}(k)\cdot\mathbf{\sigma}\) with \(h_z(k)=u\).

Set \(w=1, v=1.5-cos\theta, u=sin\theta, \theta=0\to \pi\),

_images/b.jpg
_images/figure_2.png

Case 3: Charge Pump: Using the Hamiltonian \(H=u(t)\sum\limits_{i=1}^N(-1)^{i-1}c_i^{\dagger}c_i+\sum\limits_{i=1}^N[t_0+\delta(t)(-1)^i](c_i^{\dagger}c_{i+1}+h.c.)\), with \(u(t)=u_0 sin\theta, \delta(t)=\delta_0 cos\theta\), add time variation with \(\theta=2\pi t/T\), in previous notation, \(v=t_0-\delta(t), w=t_0+\delta(t)\). set \(u=sin\theta, v=0.75-0.25cos\theta, w=0.75+0.25cos\theta\), the vector \(\vec{h(k)}\) is

_images/5a.png
_images/5b.png
_images/5c.png
_images/5d.png

The energy band evolution is

energy4

energy4

_images/1a.png
_images/2a.png
Winding number v.s. Zak phase

From

\[\begin{split}H(k)=\begin{pmatrix}0&h^*(k)\\h(k)&0\end{pmatrix}\end{split}\]
\[ln(h)=ln(|h|)e^{iarg(h)}=ln(|h|)+iarg(h)\]

We have defined the winding number

\[\nu=\frac{1}{2\pi i}\int_{-\pi}^{\pi}dk\frac{d}{dk}ln(h(k))=\frac{1}{2\pi}arg(h)\left.\right|_{-\pi}^{\pi}=1 \ \textit{or}\ 0\]

From the Zak phase definition:

\[\gamma=i\oint\langle\psi|\nabla_k|\psi\rangle\]

Recalling \(|\psi\rangle=\frac{1}{\sqrt{2}}\begin{pmatrix}\pm e^{-i\phi(k)}\\ 1\end{pmatrix}\Rightarrow\gamma=\frac{1}{{2}}\oint dk\nabla_k\phi=\pm\pi \ \textit{or}\ 0\).

Lecture 2 : Berry Phase and Chern number

Berry Phase review

Assuming a physical system is depended on some parameters \(\mathbf{R}=(R_1,R_2,\cdots,R_N)\), we have the snapshot Hamiltonian \(H(\mathbf{R})\), its eigen-values and eigen-states:

\[H(\mathbf{R})|n(\mathbf{R})\rangle=E_n(\mathbf{R})|n(\mathbf{R})\rangle\]

where \(|n(\mathbf{R})\rangle\) can have an arbitrary phase prefactor.

The parameters \(\mathbf{R(t)}\) are slowly changed with time \(t\), then the adiabatic evolution of time-dependent Schrodinger equation:

\[i\frac{d}{dt}|\psi(t)\rangle=H(\mathbf{R(t)})|\psi(t)\rangle\]

Take the Ansatz: \(|\psi(t)\rangle=e^{i\gamma_n(t)}e^{-i\int_0^tE_n(\mathbf{R(t')})dt'}|n(\mathbf{R(t)})\rangle\), we have

\[-\left(\frac{d}{dt}\gamma_n\right)|n\rangle+i\left|\frac{d}{dt}n\right\rangle=0\]

That gives the Berry phase expression:

\[\gamma_n(\mathcal{C})=\int_\mathcal{C}i\langle n(\mathbf{R})|\nabla_{\mathbf{R}}n(\mathbf{R})\rangle d\mathbf{R}\]

Define Berry connection:

\[\mathbf{A}^{(n)}(\mathbf{R})=i\langle n(\mathbf{R})|\nabla_{\mathbf{R}}n(\mathbf{R})\rangle=-Im\langle n(\mathbf{R})|\nabla_{\mathbf{R}}n(\mathbf{R})\rangle\]

Gauge transformation:

\[|n(\mathbf{R})\rangle \to e^{i\alpha(\mathbf{R})}|n(\mathbf{R})\rangle\]
\[\mathbf{A}^{(n)}(\mathbf{R}) \to \mathbf{A}^{(n)}(\mathbf{R})-\nabla_{\mathbf{R}}\alpha(\mathbf{R})\]

\(\gamma=\oint\mathbf{A}(\mathbf{R})d\mathbf{R}\) is gauge invariant.

Gauge and Parallel transportation: recalling the arbitrary phase

\[|n(\mathbf{R})\rangle \to e^{i\alpha(\mathbf{R})}|n(\mathbf{R})\rangle\]

why shouldn’t we choose one which makes

\[\frac{d}{dt}|n\rangle\equiv 0\]

from

\[-\left(\frac{d}{dt}\gamma_n\right)|n\rangle+i\left|\frac{d}{dt}n\right\rangle=0\]

then we have

\[\gamma_n=0\]

There is no Berry Phase in this frame, which is called inertial frame, the condition \(\frac{d}{dt}|n\rangle\equiv 0\) is called parallel transportation. All the information resorted to \(|n(\mathbf{R})\rangle\), similar to a transformation from active frame to passive frame.

Berry curvature

Define the Berry curvature:

\[\mathbf{B}(\mathbf{R})=\nabla_{\mathbf{R}}\times \mathbf{A}^{(n)}(\mathbf{R})\]

Using Stokes theorem, we have for the Berry Phase:

\[\gamma_n(\mathcal{C})=\int_\mathcal{S}\mathbf{B}^{(n)}(\mathbf{R})d\mathcal{S}\]

where \(\mathcal{S}\) is any surface whose boundary is the loop \(\mathcal{C}\).

Two useful formula:

  • \(B_j=\epsilon_{jkl}\partial_kA_l=-Im\epsilon_{jkl}\partial_k\langle n|\partial_ln\rangle=-Im\epsilon_{jkl}\langle\partial_kn|\partial_ln\rangle\), that is \(\mathbf{B}^{(n)}=-Im\sum\limits_{n'\neq n}\langle\nabla n|n'\rangle\times\langle n'|\nabla n\rangle\).

  • \(\mathbf{B}^{(n)}=-Im\sum\limits_{n'\neq n}\langle\nabla n|n'\rangle\times\langle n'|\nabla n\rangle\), to calculate \(\langle n'|\nabla n\rangle\), start from:

    \[H(\mathbf{R})|n\rangle=E_n|n\rangle\]
    \[\Rightarrow (\nabla H)|n\rangle+H|\nabla n\rangle=(\nabla E_n)|n\rangle+E_n|\nabla n\rangle\]
    \[\Rightarrow \langle n'|\nabla H|n\rangle+\langle n'|H|\nabla n\rangle=E_n\langle n'|\nabla n\rangle\]
    \[\Rightarrow \langle n'|\nabla n\rangle=\frac{\langle n'|\nabla H|n\rangle}{E_n-E_{n'}}\]

then we get:

\[\mathbf{B}^{(n)}=-Im\sum\limits_{n'\neq n}\langle\nabla n|n'\rangle\times\langle n'|\nabla n\rangle=-Im\sum\limits_{n'\neq n}\frac{\langle n|\nabla H|n'\rangle\times\langle n'|\nabla H|n\rangle}{(E_n-E_{n'})^2}\]

which is gauge invariant!

Berry curvature from perturbation theory

We can use time-independent perturbation theory to derive the changes of instant snapshot basis:

\[H(\mathbf{R})|n(\mathbf{R})\rangle=E_n(\mathbf{R})|n(\mathbf{R})\rangle\]

we have

\[|n(\mathbf{R+\Delta R})\rangle=|n(\mathbf{R})\rangle+\sum_{m\neq n}\frac{\langle m|H(\mathbf{R+\Delta R})-H(\mathbf{R})|n\rangle}{E_n-E_m}|m(\mathbf{R})\rangle\]

We see that \(\langle n|\Delta n(\mathbf{R})\rangle=0\), which means we have used parallel transport gauge, more general, we should add a arbitrary phase factor in the above equation for \(|n(\mathbf{R+\Delta R})\rangle\).

\[\nabla_{\mathbf{R}}|n\rangle=\sum_{m\neq n}\frac{\langle m|\nabla_{\mathbf{R}}H|n\rangle}{E_n-E_m}|m\rangle\]

From \(\mathbf{B}^{(n)}=-Im\sum\limits_{n'\neq n}\langle\nabla n|n'\rangle\times\langle n'|\nabla n\rangle\), we also get:

\[\mathbf{B}^{(n)}=-Im\sum_{n'\neq n}\frac{\langle n|\nabla H|n'\rangle\times\langle n'|\nabla H|n\rangle}{(E_n-E_{n'})^2}\]

Also notice:

\[\begin{split}\sum_n\mathbf{B}^{(n)}=&-Im\sum_n\sum_{n'\neq n}\frac{\langle n|\nabla H|n'\rangle\times\langle n'|\nabla H|n\rangle}{(E_n-E_{n'})^2}\\ =&-Im\sum_n\sum_{n'< n}\frac{\langle n|\nabla H|n'\rangle\times\langle n'|\nabla H|n\rangle+\langle n'|\nabla H|n\rangle\times\langle n|\nabla H|n'\rangle}{(E_n-E_{n'})^2}\\ =&0\end{split}\]

Which gives:

\[\sum_n\gamma_n(\mathcal{C})=\int_{\mathcal{S}}\sum_n\mathbf{B}^{(n)}(\mathbf{R})d\mathcal{S}=0\]
Benchmark: Spin-1/2

Gauge!Gauge!Gauge!

2-level Hamiltonian \(H(\mathbf{R})=h_0(\mathbf{R})\sigma_0+\mathbf{h}(\mathbf{R})\cdot\mathbf{\sigma}\), we can set \(h_0=0\), because it does not affect the eigenstates, eigen-energy are \(\pm|\mathbf{h}|\), introduce the unit vector:\(\hat{\mathbf{h}}=\mathbf{h}/|\mathbf{h}|\), the endpoints of \(\hat{\mathbf{h}}\) map out the surface of a unit sphere, called the Bloch sphere shows below:

Bloch sphere

Bloch sphere

Lecture 3 : Chern Insulator

Quantum Hall Insulator

The first Topological Insulator is shown in Integer quantum Hall effect. For 2D electron gas (2DEG), by applying a B field perpendicular to the 2D plane, we get plateaus of Hall conductance \(\sigma_{xy}=1/\rho_{xy}\) which is quantized in \(\nu e^2/h\). In the same time, \(\rho_{xx}=\sigma_{xx}=0\) (which may seems counterintuitive, but noticing the tensor nature of \(\rho\) and \(\sigma\)). The quantization of \(\sigma_{xy}\) is very accurate, it is the second best way to measure the fine structure constant, and it is inert to interactions and impurities. For a charge neutral particle in 2D:

\[i\partial_t\psi(x,y)=H_0\psi(x,y)\]

with \(H_0=\frac{\hat{p}^2}{2m}\).

In a magnetic field with a vector potential \(\vec{A}\), we have \(H=\frac{(\hat{p}-e\vec{A})^2}{2m}\).

For a constant uniform magnetic field, choose Landau gauge: \(A_x=0, A_y=Bx\), we have

\[H=\frac{\hbar^2}{2m}(-i\partial_x)^2+\frac{\hbar^2}{2m}(-i\partial_y-eBx)^2\]

\(H\) is invariant under translations along the y axis, so \([p_y,H]=0\), we can write the eigen states of \(H\): \(\psi=f(x)exp(ik_yy)\) and \(H\psi=\epsilon\psi\). which gives:

\[\psi_{n,k_y}=\phi_n(x-x_0)exp(ik_yy)\]

\(\phi_n(x)\) satisfies the harmonic oscillator potential \(V(x)=\frac{k}{2}x^2\) with \(k=e^2B^2/m, x_0=\frac{\hbar k_y}{eB}\). The eigen energies are the famous harmonic one:

\[\epsilon_{n,k_y}=(n+1/2)\hbar \omega_c =(n+1/2)\hbar \sqrt{\frac{k}{m}} =(n+1/2)\frac{eB\hbar}{m}\]

Some comments

  1. Landau levels are highly degenerated, they form energy bands with \(n\) denotes band index and \(k_y\) the momentum, all the states with the same \(n\) (but different \(k_y\)) have the same energy (flat bands).
  2. The degeneracy of each Landau level is \(D=(2S+1)\frac{\Phi}{\Phi_0}\), where \(S\) denotes spin, and \(\Phi_0\) is the magnetic flux quanta with \(\Phi_0=h/e\).
  3. We can also use symmetry gauge: \(A_x=-\frac{By}{2}, A_y=\frac{Bx}{2}\), though it is more cumbersome, the physics is quite insightful. More on symmetry gauge, see[1].
Edge states with Quantization of the Hall conductivity

Electrons at the edge can move along the edge in one direction, this edge states gives quantized Hall conductivity.

At top edge: \(N=\int \frac{dpdr}{2\pi\hbar}=L\int \frac{dp}{2\pi\hbar}\), we get

\[n=\frac{N}{L}=\int\frac{dp}{2\pi\hbar}\]
\[\delta n_{top}=\int_{k_F}^{k_{Ftop}}\frac{dp}{2\pi\hbar}=\frac{k_{Ftop}-k_F}{2\pi\hbar}=\frac{\mu_{top}-\mu}{2\pi\hbar v_{Ftop}}\]

We have use the relation \(v_F=\frac{\partial E(k)}{\hbar \partial k}=\frac{\mu_{top}-\mu}{\hbar(k_{Ftop}-k_F)}\)

Net current for top edge:

\[I_{top}=ev_{Ftop}\delta_{top}=ev_{Ftop}\frac{\mu_{top}-\mu}{2\pi\hbar v_{Ftop}}=e\frac{eV/2}{h}=\frac{e^2}{h}\frac{V}{2}\]

We have used \(\mu_1-\mu_2=e(V_1-V_2)\). And, similarly, we have \(I_{bottom}=I_{top}\), that gives

\[I=\frac{e^2}{h}V\]
\[\sigma=j/E=\frac{I}{L_y}/\frac{V}{L_y}=I/V=\frac{e^2}{h}\]

If one has \(n\) copies of edge states, then total Hall conductivity is:

\[\sigma=n\frac{e^2}{h}\]
Half BHZ model

See the lattice below: Half-BHZ Consider at each site, there are two orbital modes coupled together, in the tight-binding approximation, we can write the Hamiltonian:

\[H=\sum_{n_x,n_y}c^{\dagger}_{n_x,n_y}\frac{U(n)}{2}c_{n_x,n_y}+c^{\dagger}_{n_x+1,n_y}T^{\dagger}_x(n)c_{n_x,n_y}+c^{\dagger}_{n_x,n_y+1}T^{\dagger}_y(n)c_{n_x,n_y}+\textit{h.c.}\]

Assuming translational invariance along x and y, i.e. \(U(n)=U,T_x(n)=T_x,T_y(n)=T_y\), after Fourier transform: \(c_{n_x,n_y}=\frac{1}{\sqrt{N}}\sum_{k_x,k_y}c_{k_x,k_y}e^{ik_xn_x+ik_yn_y}\) with \(N=N_xN_y, k_xN_x=2\pi m_1,k_yN_y=2\pi m_2\), we get

\[H=\sum_{k_x,k_y}c^{\dagger}_{k_x,k_y}H(k)c_{k_x,k_y}\]

with

\[H(k)=\frac{1}{2}U+T_xe^{ik_x}+T_ye^{ik_y}+\textit{h.c.}\]\[with\]

\(U=\Delta\sigma_z, T_x=\frac{1}{2}\sigma_z-\frac{iA}{2}\sigma_x, T_y=\frac{1}{2}\sigma_z-\frac{iA}{2}\sigma_y\), we get

\[H(k)=(\Delta+cosk_x+cosk_y)\sigma_z+A(sink_x\sigma_x+sink_y\sigma_y)\]
Chern number using \(\mathbf{h}\)

Written the Hamiltonian as: \(H(k)=\mathbf{h}\cdot\mathbf{\sigma}\), we get

\[\begin{split}\mathbf{h}(k_x,k_y)=\begin{pmatrix}Asink_x\\Asink_y\\\Delta+cosk_x+cosk_y\end{pmatrix}\end{split}\]

and we get the eigen-energies:

\[E_{\pm}(k_x,k_y)=\pm |\mathbf{h}(k_x,k_y)|=\pm\sqrt{A^2(sin^2k_x+sin^2k_y)+(\Delta+cosk_x+cosk_y)^2}\]

Set \(A=1\), for some special parameter \(\Delta\), the dispersion relation shows below: dispersion1 dispersion2 dispersion3 dispersion4

From the dispersions also the analysis we can see that Dirac cone appears at \(k_x,k_y=0,\Delta=-2; k_x,k_y=0,\pm\pi,\Delta=0; k_x,k_y=\pm\pi,\pm\pi,\Delta=2\). It means that the monopole seats at \(\mathbf{h}=0\), at each region, we can calculate the Chern number of the valence band in investigating how many times does the torus formed by the image of the Brillouin zone in the space of \(\mathbf{h}\) contail the origin. The Torus for different \(\Delta=-2.5,-1,1,2.5\) shown below (for clarity, only half of the torus is shown): torus1 torus2 torus3 torus4

From \(\mathbf{h}\), we get the Chern number as follow: (Set \(A=1\))

  • \(|\Delta|>2, Q=0\)
  • \(-2<\Delta<0, Q=-1\)
  • \(0<\Delta<2, Q=1\)

As long as the whole band is gapped, we have the same topology and the same Chern number. In the region \(|\Delta|>2\), reminding the Hamiltonian (In what follows, we always consider \(A=1\))

\[H(k)=A(sink_x\sigma_x+sink_y\sigma_y)+(\Delta+cosk_x+cosk_y)\sigma_z\]

It represent Hamiltonians topologically equivalent to the atomic limit, the limit in which all hoppings are set to zero, as if the lattice constant were infinity. The \(|\Delta|>2\) region is topologically the same phase as the phase \(|\Delta|\to\infty\), which has trivial (momentum-independent) eigenstates and zero Chern number.

Chern number as a Skyrmion number

Looking at the vector \(\mathbf{h}\), the \((h_x(k_x,k_y),h_y(k_x,k_y))\) shows below kxy

To understand the Chern number, we can see if the \(h_z\) component change sign or not in the BZ. If it changes sign, it corresponds to a skyrmion, and the Chern number is non-zero, it is the same as Skyrmion number. If it does not change sign, then the topology is trivial. The value of \(h_z(\mathbf{k})\) at high-symmetry points in the BZ are shown below: skyr From which we can get the Chern number.

Low-energy Hamiltonian

We can also calculate the Chern number using the low-energy Hamiltonian. At \(\Delta=-2\), the energy gap collapses at the \(\Gamma=(0,0)\) point, near this point, we have

\[H_{\Gamma+k}=k_x\sigma_x+k_y\sigma_y+(\Delta+2)\sigma_z\]

For the Hamiltonian \(H(k)=k_x\sigma_x+k_y\sigma_y+m\sigma_z\), we can get the monopole field for \(E_-\)state is

\[\Omega_{ij}=\frac{1}{2}\hat{\mathbf{h}}\cdot(\partial_i\hat{\mathbf{h}}\times\partial_j\hat{\mathbf{h}})\]

Then we get

\[\Omega_{ij}=\frac{m}{2(k_x^2+k_y^2+m^2)^{3/2}}\]

Which gives:

\[C=-\frac{1}{2\pi}\int\Omega_{ij}dk_xdk_y=-\frac{sign(m)}{2}\]

Alternatively, from \(\mathbf{B}=\frac{1}{2}\frac{(k_x,k_y,m)}{(k_x^2+k_y^2+m^2)^{3/2}}\),

\[C=-\frac{1}{2\pi}\int\mathbf{B}\cdot d\mathbf{\mathcal{S}}\]

For \(m>0\) and \(m<0\), the integral region and direction shows below:

Bfield

Bfield

Note here the Chern number is not an integer, because we used continuous model, and \(k_x,k_y\) can range from \(-\infty\) to \(\infty\), so the band structure \(\pm\sqrt{k_x^2+k_y^2+m^2}\) run riot and never turns back, which should give us another half of the Chern number (though we have no idea is we should add or subtract half-values). But as we can see below, the differences of Chern number between two states gives the right answer.

So from \(\Delta<-2\to \Delta>-2\), \(m<0\to m>0\), we get

\[\begin{split}C=-\frac{1}{2}sign(m>0)+\frac{1}{2}sign(m<0)=-1\end{split}\]

Also we know \(\Delta<-2\), \(C=0\), so for \(-2<\Delta<0\), \(C=-1\).

Chern number as an obstruction

Write \(\mathbf{h}=|\mathbf{h}|(sin\theta cos\phi, sin\theta sin\phi, cos\theta)\). In principle, for the whole Bloch sphere, we must at least use two different bases which are well defined. If for the whole band, we have \(h_z<0\) or \(h_z>0A\), then, we can use one set of eigen base, which gives us zero Chern number. But for the region \(-2<\Delta<0\), at the point \(\Gamma\), we have \(h_z(0,0)=\Delta+2>0\), \(h_z/h=1=cos\theta\), \(\theta=0\); at the point \((0,\pm\pi)\) or \((\pm\pi,\pm\pi)\), we have \(h_z<0\), \(\theta=\pi\), means we must use two set of eigen bases, denotes no-zero Chern number. The resulting \(\hat{h}=\mathbf{h}/|\mathbf{h}|\) for different\(\Delta=-2.5\to 2.5\) shown below:

h h h h

Haldane Model

Haldane model describe the model of Graphene with real nearest-neighbor-hopping parameters but complex next-nearest-neighbor-hopping parameters resulting from a nonzero magnetic field. But the field is periodic and have zero net flux per plaquette. The real space and reciprocal space shows below (with unit length 1):

lattice

lattice

The vectors are:

\[\begin{split}\mathbf{a_1}=\begin{pmatrix}\sqrt{3}/2\\1/2\end{pmatrix}, \mathbf{a_2}=\begin{pmatrix}-\sqrt{3}/2\\1/2\end{pmatrix}, \mathbf{a_3}=\begin{pmatrix}0\\-1\end{pmatrix}=-(a_1+a_2)\end{split}\]
\[\begin{split}\mathbf{b_1}=\mathbf{a_2}-\mathbf{a_3}=\begin{pmatrix}-\sqrt{3}/2\\3/2\end{pmatrix}, \mathbf{b_2}=\mathbf{a_3}-\mathbf{a_1}=\begin{pmatrix}-\sqrt{3}/2\\-3/2\end{pmatrix}, \mathbf{b_3}=\mathbf{a_1}-\mathbf{a_2}=\begin{pmatrix}-\sqrt{3}\\0\end{pmatrix}\end{split}\]

Set \(\mathbf{b_1},\mathbf{b_2}\) as two base vectors, then the reciprocal lattice spanned by \(\mathbf{b_1^*},\mathbf{b_2^*}\) with \(\mathbf{b_i}\cdot\mathbf{b_j^*}=2\pi\delta_{ij}\), that is

\[\begin{split}\mathbf{b_1^*}=2\pi\begin{pmatrix}-1/\sqrt{3}\\1/3\end{pmatrix}, \mathbf{b_2^*}=2\pi\begin{pmatrix}-1/\sqrt{3}\\-1/3\end{pmatrix}\end{split}\]

Define \(\mathbf{K}=\frac{1}{3}(\mathbf{b_1^*}+\mathbf{b_2^*})=\begin{pmatrix}-4\sqrt{3}\pi/9\\0\end{pmatrix}, \mathbf{K'}=-\mathbf{K}\).

We get the Hamiltonian:

\[\begin{split}H=M\sum_i(a_{iA}^{\dagger}a_{iA}-a_{iB}^{\dagger}a_{iB})+t\sum_{<i,j>}(a_{iA}^{\dagger}a_{jB}+\textit{h.c.})+t_2\sum_{<<i,j>>}(a_{iA}^{\dagger}a_{jA}+a_{iB}^{\dagger}a_{jB}+\textit{h.c.})\end{split}\]

\(M\) breaks inversion symmetry of \(A\) and \(B\) sublattice, and \(t_2\) is the next-nearest-neighbor coupling, for a staggered magnetic field shows below, it changes \(t_{ij}\to t_{ij}e^{i\frac{e}{\hbar}\int_{j\to i}\mathbf{\mathcal{A}}\cdot d\mathbf{l}}=t_{ij}e^{i\frac{e}{\hbar}\chi\Phi_0}=t_{ij}e^{i\frac{e}{\hbar}\chi\frac{h}{e}}=t_{ij}e^{i\phi}\).

mag

mag

Use \(a_{iA}=\frac{1}{\sqrt{N}}\sum_ka_{kA}e^{i\mathbf{k}\cdot\mathbf{R}_{iA}}, a_{iB}=\frac{1}{\sqrt{N}}\sum_ka_{kB}e^{i\mathbf{k}\cdot\mathbf{R}_{iB}}\), \(N\) is the number of unit cell. Then we can get:

\[\begin{split}H=\sum_k\begin{pmatrix}a_{kA}^{\dagger}&a_{kB}^{\dagger}\end{pmatrix}H(k)\begin{pmatrix}a_{kA}\\a_{kB}\end{pmatrix}\end{split}\]

We arrived at

\[H(k)=h_0\sigma_0+\mathbf{h}\cdot\mathbf{\sigma}\]

with

\[h_0=2t_2cos\phi\sum_icos\mathbf{k}\cdot\mathbf{b_i}\]
\[h_x=t\sum_icos\mathbf{k}\cdot\mathbf{a_i}\]
\[h_y=-t\sum_isin\mathbf{k}\cdot\mathbf{a_i}\]
\[h_z=M+2t_2sin\phi\sum_isin\mathbf{k}\cdot\mathbf{b_i}\]

Because of the \(C_3\) symmetry, the Dirac cone can only happen at \(K,K'\). At the vicinity of \(K\), we have

\[H(K+q)=-3t_2cos\phi+\frac{3}{2}t_1(q_x\sigma_x-q_y\sigma_y)+(M+3\sqrt{3}t_2sin\phi)\sigma_z\]

At the vicinity of \(K'\), we have

\[H(K'+q)=-3t_2cos\phi-\frac{3}{2}t_1(q_x\sigma_x-q_y\sigma_y)+(M-3\sqrt{3}t_2sin\phi)\sigma_z\]

where \(-\pi<\phi<\pi\).

Start from \(M\to \pm\infty\), we have zero Chern number, assuming \(t_2>0, \phi>0\), from the previous results, when \(M\) from \(+\infty\) decreases to \(M=3\sqrt{3}t_2sin\phi\), \(K'\) band close and reopen, the Chern number changes to

\[\begin{split}C=-\frac{1}{2}sign((M-3\sqrt{3}t_2sin\phi)<0)+\frac{1}{2}sign((M-3\sqrt{3}t_2sin\phi)<0)=1\end{split}\]

Then when \(M>-3\sqrt{3}t_2sin\phi\) to \(M<-3\sqrt{3}t_2sin\phi\), we have

\[\begin{split}C-1=\frac{1}{2}sign((M+3\sqrt{3}t_2sin\phi)<0)-\frac{1}{2}sign((M+3\sqrt{3}t_2sin\phi)<0) \Rightarrow C=0\end{split}\]

We can do the same calculation for \(\phi<0\), we get the phase chart below:

phase chart

phase chart

Because \(k_x,k_y\) have no dependence on \(M,t_2,\phi\), \(h_z\) in the range of \((M-3\sqrt{3}t_2sin\phi\to M+3\sqrt{3}t_2sin\phi)\), we can draw a spheroid \(\Sigma'\) with height \(6\sqrt{3}sin\phi\) and center \((0,0,M/t_2)\). We can get the Chern number using following pictures:

spheroid

spheroid

Also, we can get the Chern number using the whole 3-d picture (set \(\phi=\pi/2, M/t_2=6,0,-6\)):

haldane1 haldane2 haldane3

Other good Books and review papers on this subject:

More geometrical staffs at this one Geometry, Topology and Physics, by Nakahara.

Berry’s Phase

Preliminary

some topics

  • fiber bundle
  • gauge/parallel transport
  • symmetry

Weyl Semi-metal

Graphene

In the tight-binding approximation, when only nearest neighborhood couplings are considered, the Hamiltonian of Graphene can be written as:

\[\begin{split}\hat{H}(\vec{k})=-t\begin{pmatrix}0 &h(\vec{k})\\h^*(\vec{k}) &0\end{pmatrix}\end{split}\]

where \(h(\vec{k})=\sum\limits_{\vec{\delta}_i}e^{i\vec{k}\cdot\vec{\delta}_i}=|h(\vec{k})|e^{i\phi(\vec{k})},\vec{\delta}_i\) are three position vectors shown in the following diagram.

_images/13.png

Fig. 1 Crystal Structure of Graphene: \(\vec{a}_{1}\) and \(\vec{a}_{2}\) are Bravais crystal vectors for a Graphene unit cell. Each primitive unit cell has two atomic sties, A and B. \(\vec{\delta}_{i}\) specifies B-s’ position around A site. (b) Brillouin Zone for Graphene \(\vec{b}_{1}\) and \(\vec{b}_{2}\) are reciprocal vector basis for intrinsic Graphene. Its corners are known as K and K’ points.

Then the Hamiltonian is:

\[\begin{split}\hat{H}(\vec{k})=-t|h(\vec{k})|\begin{pmatrix}0 &e^{i\phi(\vec{k})}\\e^{-i\phi(\vec{k})} &0\end{pmatrix}\end{split}\]

with the eigen-value: \(E(\vec{k})=-t|h(\vec{k})|\) and eigen-function (only show one of the two):

\[\begin{split}u(\vec{k})=\frac{1}{\sqrt{2}}\begin{pmatrix}e^{i\phi(\vec{k})/2}\\e^{-i\phi(\vec{k})/2} \end{pmatrix}e^{i\psi(\vec{k})}\end{split}\]

We should notice that the \(1/2\) factor is quite important here, when \(\phi\) changes \(2\pi\), the wave-function does not return to its original value, but with a minus sign. If instead, we want the wave-function to be single valued, the function :math:`psi(vec{k})`should change accordingly.

At the vicinity of Dirac point (K or K’, here we expand the things near K), we have:

\[\begin{split}\hat{H}(\vec{K}+\vec{q})=&\alpha\begin{pmatrix}0&q_x+iq_y\\q_x-iq_y&0\end{pmatrix}=\alpha(q_x\sigma_x-q_y\sigma_y) \\ =&\alpha|q|\begin{pmatrix}0 &e^{i\phi(\vec{q})}\\e^{-i\phi(\vec{q})} &0\end{pmatrix}\end{split}\]

We can see that the general \(\phi\) turn out to be the angle of \(\vec{q}\) with the x axis. Then, wind a circle around the Dirac point K at some energy in the band structure shown below (Fig. 2(a)), the corresponding \(\phi\) (Fig. 2(c))winds one round too.

_images/2.png

Fig. 2 Dirac cone and the winding of \(\vec{q}\) and \(\phi\)

So, if \(\vec{q}\) circles around the Dirac point one turn, \(\phi\) changes from 0 to \(2\pi\), in order to keep the basic wave-function \(u(\vec{q})\) single-valued, \(\psi\) must changes \(\pi\).

More explicitly, we calculate the vector potential in momentum space with:

\[\vec{A}(\vec{k})=i\langle u^\dagger |\nabla_{\vec{k}} u(\vec{k})\rangle=-\nabla \psi(\vec{k})\]

Then we get:

\[\gamma=\oint{\vec{A}\cdot d\vec{k}} =-\psi(\phi=2\pi)+\psi(\phi(0))=-\pi\]

We got the Berry’s phase \(\gamma=\pm\pi\). It is this non-trivial phase of the equal-frequency surface that makes us call it Weyl semi-metal, and the Dirac points called Weyl nodes.

Three dimension: Weyl semi-metal and Chern number

In three dimension, things can goes the same way. Using a simple model with Hamiltonian:

\[H(\vec{k})=\left[-2t_x\left(cosk_x-cosk_0\right)+m\left(2-cosk_y-cosk_z\right)\right]\sigma_x+2t_ysink_y\sigma_y+2t_zsink_z\sigma_z\]

It has two Weyl nodes: \(\vec{K}=\pm\left(k_0,0,0\right)\), which means if we treat \(k_x\) as a variable, only when \(k_x=\pm k_0\), the corresponding energy band \(E_{k_x}(k_y,k_z)\) is crossing at the point \(k_y,k_z=(0,0)\).

Also, at the Weyl node (say \(k_x=k_0\)), we have:

\[H(\vec{K}+\vec{q})=v_xq_x\sigma_x+v_yq_y\sigma_y+v_zq_z\sigma_z\]

with \(v_x=2t_xsink_0,v_y=2t_y,v_z=2t_z\). Without loss of generality, we can set \(v_x=v_y=v_z\) (the only effect is the shape changing from sphere to ellipsoid, which has no effect on the topology), then we get:

\[H(\vec{K}+\vec{q})=v\vec{q}\cdot\vec{\sigma}\]

with eigen-value: \(E(\vec{k})=v|\vec{q}|\) and eigen-function (only show one of the two):

\[\begin{split}u(\vec{k})=\begin{pmatrix}sin\frac{\theta}{2}\\-cos\frac{\theta}{2}e^{i\phi}\end{pmatrix}e^{i\chi}\end{split}\]

It is easy to find that this wave-function will give us a magnetic field with a monopole at \(\vec{K}\), which will give us non-trivial equal-frequency surface Chern number \(C=1\).

Bulk-boundary corresponding

In order to see things clearer, also, to see the connection of Weyl semi-metal with Topological insulator, we treat \(k_x\) as a variable, the Hamiltonian is:

\[H_{k_x}(k_y,k_z)=\vec{h}(\vec{k})\cdot\sigma=\left[-2t_x\left(cosk_x-cosk_0\right)+m\left(2-cosk_y-cosk_z\right)\right]\sigma_x+2t_ysink_y\sigma_y+2t_zsink_z\sigma_z\]

For example, we set \(t_x=t_y=t_z=1,m=2,k_0=1\), three typical energy band dispersions shown below:

_images/31.png

Fig. 3 Typical energy band dispersion with (a) \(k_x=0\), (b) \(k_x=k_0=1\), (c) \(k_x=2\).

To see if the system with \(k_x\neq k_0\) is a topological insulator or not, we can plot the diagram of \(\vec{h}(k_y,k_z)\), and see how many times the resulting torus incorporates the origin point. Typical shape of the torus shows below:

_images/41.jpg

Fig. 4 Typical torus of \(\vec{h}(k_y,k_z)\) with \(k_x=0\).

We found that at the region \(k_x=[-k_0,k_0]\), the origin point is in the torus once with Chern number \(C=1\), outside of that, we got \(C=1\) (Noticing we have \(k_x=[-\pi,\pi]\)) . This is why we plot the edge state in Fig. 4(a), but not in Fig. 4(c). In the non-trivial case, for any energy inside the gap, we get a edge state, so different \(k_x\) will give us a edge-state line, which is called Fermi-arc, especially, when we look at the case of Fermi surface with energy \(E=0\), the Fermi-arc stretch from one Weyl node to another, like the picture shown below:

_images/51.png

Fig. 5 Fermi arc.

Alternatively, we can look at the Bulk-boundary corresponding another way. The Weyl points behave like “magnetic” monopoles in momentum space whose charge is given by the chirality; they are actually a source of “Berry flux” rather than magnetic flux.

Consider a curve in the surface Brillouin zone encircling the projection of the bulk Weyl point, which is traversed counterclockwise as we vary the parameter \(\lambda: 0 \to 2\pi; \mathbf{k}_{\lambda} = [k_x (\lambda),k_y (\lambda)]\) [see Fig. 6(a)].We show that the energy \(\lambda\) of a surface state at momentum \(\mathbf{k}_{\lambda}\) crosses \(E = 0\) at some value of \(\lambda\). Consider \(H(\lambda,k_z) = H(\mathbf{k}_{\lambda},k_z)\), which can be interpreted as the gapped Hamiltonian of a two-dimensional system (with \(\lambda\) and \(k_z\) as the two momenta). The two periodic parameters \(\lambda\) , \(k_z\) define the surface of a torus in momentum space. The Chern number of this two-dimensional band structure is given by the Berry curvature integration: \(\frac{1}{2\pi}\int\mathscr{F}dk_zd\lambda\), which, by the Stokes theorem, simply corresponds to the net monopole density enclosed within the torus. This is obtained by summing the chiralities of the enclosed Weyl nodes. Consider the case when the net chirality is unity, corresponding to a single enclosed Dirac node. Then, the two-dimensional subsystem is a quantum Hall insulator with unit Chern number. When the system is given a boundary at \(z = 0\), we expect a chiral edge state for this subsystem [see Fig. 6(b)]. Hence, this surface state crosses zero energy somewhere on the surface Brillouin zone \(\mathbf{k}_{\lambda_0}\) . Such a state can be obtained for every curve enclosing the Weyl point. Thus, at zero energy, there is a Fermi line in the surface Brillouin zone, that terminates at the Weyl point momenta [see Fig. 6(c)]. An arc beginning on a Weyl point of chirality c has to terminate on a Weyl point of the opposite chirality. Clearly, the net chirality of the Weyl points within the \((\lambda,k_z)\) torus was a key input in determining the number of these states. If Weyl points of opposite chirality line up along the \(k_z\) direction, then there is a cancellation and no surface states are expected.

_images/6.png

Fig. 6 Illustration of surface states arising from bulk Weyl points.

References:

    1. Wan, A.M. Turner, A. Vishwanath, and S.Y. Savrasov, Physical Review B 83, 205101 (2011).
  2. A.M. Turner, A. Vishwanath, and C.O. Head, Topological Insulators 6, 293 (2013).
    1. Haldane, Physical Review Letters 93, 206602 (2004).

Condensed Matter Physics:

Linear response theory

Kubo formula

Many times, the world shows us as a black box, we can only get limited knowledge about it and the rest are filled with our theories, in this way, theories can always changing and are never completed. we always use a probe to probe a system and check its response to this probe, and naturally, we expect when the perturbation of the probe is ignorable to the system, the response should be linear to the probe, which is called linear response theory. The most important result of linear response theory is consummated by Kubo formula, we are now going to derive it.

Zero temperature

Let \(\hat{H}_0\) be the full many-body Hamiltonian for some isolated system that we are interested in, and assume the existence of a set of eigenkets \(\{|n\rangle\}\) that diagonalize \(\hat{H}_0\) with associated eigenvalues (energies) \(\varepsilon_n\).

In addition to \(\hat{H}_0\), we now turn on an external probe potential \(\hat{V}(t)\), such that the total Hamiltonian \(\hat{H}(t)\) satisfies:

\[\hat{H}(t)=\hat{H}_0+e^{\eta t}\hat{V}(t)\]

Note

The additional factor \(e^{\eta t}\) means we switch on the external potential adiabatically from \(t\to -\infty\), we’ll see later that it is this factor gives us the way to detour the singular points, it is an analytical continuation which is a reflection of causality.

The Schrödinger equation of the system now reads:

\[i\hbar\frac{\partial}{\partial t}|\psi(t)\rangle=(\hat{H}_0+e^{\eta t}\hat{V}(t))|\psi(t)\rangle\]

Warning

us the way to detour the singular points, it is an analytical continuation which is a reflection of causality.

Important

us the way to detour the singular points, it is an analytical continuation which is a reflection of causality.

Hint

us the way to detour the singular points, it is an analytical continuation which is a reflection of causality.

Fermi’s Golden rule

Some topics want to explore

  • Green’s Function
  • Dirac,Majorana,Weyl fermions

Something about Python

Python 学习

  1. 数组初始化
#初始化一维数组
p1=[1]*100

#初始化二维数组100*10
p2=[[1] * 100 for i in xrange(10)]
#Below does not work!
p2=[[1] * 100]*10
#Or we can use:
p2=[[None for col in range(100)] for row in range(10)]

#初始化三维维数组2*3*4
p3=[[[1] * 2 for i in xrange(3)] for j in xrange(4)]
#Or
p3=[[[1 for i in xrange(2)] for j in xrange(3)] for k in xrange(4)]
#Or
p3=[]
for i in range(2):
    p3.append([])
    for j in range(3):
        p3[i].append([])
        for k in range(4):
            p3[i][j].append(0)
  1. python 中的内积 dot 是没有复共轭的，vdot 有，而matlab dot(a,b) 是对a取复共轭的。
_images/cc_byncsa.png

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