The general strategy here is to convert this potential into a matching condition between left and right parts of space. In particular, if we integrate \(x\) from \(0^-\) to \(0^+\), we get
\[\begin{equation} -\frac1{2m}[\psi'(0^+) - \psi'(0^-)] = \lambda \psi(0). \label{eq:matching} \end{equation}\]The bound state solution to this problem is a simple exponential
\[\begin{equation} \psi_0(x) = \sqrt{m\lambda}\, e^{- m\lambda |x|}, \quad E= -\tfrac12 m\lambda^2 \end{equation}\]We require the bound state to be square integrable, so \(\psi_0(x) = \mathcal N e^{-\kappa x}\) for \(x>0\) and \(\psi_0(x) = \mathcal N e^{\kappa x}\) for \(x<0\). In both cases, we can evaluate \eqref{eq:schro}
\[\begin{equation} -\frac{\kappa^2}{2m} \psi_0(x) = E\psi_0(x), \quad \implies \kappa = \sqrt{-2mE}. \label{eq:bs_energy} \end{equation}\]Using our forms of \(\psi_0(x)\) on either side of zero, we have \(\psi_0'(0^+) = - \kappa \psi_0(0)\) and \(\psi_0(0^-) = +\kappa \psi_0(0)\). Putting these into \eqref{eq:matching}, we have
\[\begin{equation} \frac{\kappa}{m} \psi_0(0) = \lambda \psi_0(0). \end{equation}\]This means that \(\kappa = \lambda m\), and matching this to \eqref{eq:bs_energy} we find \(E = -\frac12 m \lambda^2\). All that is left is normalization and simply
\[\begin{equation} \begin{aligned} 1 & = \int dx |\psi_0(x)|^2 \\ & = 2 \int_0^\infty dx \mathcal N^2 e^{-2\kappa x} \\ & = \frac{\mathcal N^2}{\kappa}. \end{aligned} \end{equation}\]This gives us \(\mathcal N = \sqrt{\kappa} = \sqrt{\lambda m}\), completing our derivation.
But what about the continuum? First, sine waves automatically satisfy \eqref{eq:schro}, and we have
\[\begin{equation} \psi^s_k(x) = \sin(k x), \quad E = \frac{k^2}{2m}. \end{equation}\]It is worth noting that these wave functions are normalized such that
\[\begin{equation} \int dx\; \psi^s_k(x)\psi^s_{k'}(x) = \pi \delta(k - k') \end{equation}\]The continuum states that are modified are the cosine waves
\[\begin{equation} \psi_k^c(x) = \frac{1}{\sqrt{1+\lambda^2m^2/k^2}} \left[\cos(k x) - \frac{\lambda m}{k} \sin | kx|\right], \quad E = \frac{k^2}{2m}. \end{equation}\]This is again chosen such that \(\int dx \; \psi_k^c(x) \psi_{k'}^c(x) = \pi \delta(k - k')\).
First, \(\psi_k^s(x) = \sin(k x)\) is a complete basis for antisymmetric wavefunctions, and therefore, we know that \(\psi_k^c(-x) = \psi_k^c(x)\) in order to maintain orthogonality with \(\psi_k^s(x)\). This implies that if for \(x>0\) we have \(\psi_k^c(x) = A \cos(kx) + B \sin(kx)\), then for \(x<0\), \(\psi_k^c(x) = A \cos(kx) - B \sin(kx)\) justifying our functional form.
We have already guaranteed continuity across \(x=0\), so now we just need to satisfy the matching conditions \eqref{eq:matching} for which we have \({\psi_k^{c}}'(0^+) = B k\) and \({\psi_k^{c}}'(0^-) = -B k\) and therefore by \eqref{eq:matching}
\[\begin{equation} - \frac{B k}m = A \lambda \quad \implies \quad B = -\frac{\lambda m}{k} A \end{equation}\]All that is left is normalization (fixing \(A\)). For this, we introduce a regulator \(e^{-\delta \lvert x \rvert}\) and take it to be small
\[\begin{equation} \begin{aligned} \int dx \, \psi_k^c(x) \psi_{p}^c(x) e^{-\delta |x|} & = 2 A^2 \int_0^\infty dx \, \left(\cos k x - \frac{\lambda m}{k} \sin kx \right)\left(\cos p x - \frac{\lambda m}{p} \sin px \right)e^{-\delta x} \\ & = 2A^2 \frac{\delta (k^2 + p^2 + \delta^2 - 2\delta \lambda^2 m^2 + 2 \lambda^4 m^4)}{(k^2 - p^2)^2 + 2 (k^2 + p^2)\delta^2 + \delta^4} \end{aligned} \end{equation}\]If \(k \neq p\), then this expression goes to zero as \(\delta \rightarrow 0\), and it appears to diverge with \(1/\delta\) if \(k = p\), a hallmark of a Dirac delta function. In fact, making sure this integrates to \(\pi\) when we integrate with respect to \(k\) or \(p\) gives us exactly
\[\begin{equation} \int_0^\infty dp \, 2A^2 \frac{\delta (k^2 + p^2 + \delta^2 - 2\delta \lambda^2 m^2 + 2 \lambda^4 m^4)}{(k^2 - p^2)^2 + 2 (k^2 + p^2)\delta^2 + \delta^4} = A^2 \pi \frac{(\lambda^2 m^2 + k^2 - \lambda m \delta + \delta^2)}{k^2 + \delta^2} \end{equation}\]As \(\delta \rightarrow 0\) then, we obtain
\[\begin{equation} A = \frac1{\sqrt{1 + \lambda^2 m^2/k^2}}, \end{equation}\]completing this derivation.
Taken together, we can now define the (local) density of states for the continuum
\[\begin{equation} \rho(E; x) = \int_0^\infty \frac{dk}{\pi} (|\psi_k^c(x)|^2 + |\psi_k^s(x)|^2) \delta(E - k^2/(2m)) \end{equation}\]To address this, we want to find the difference from the continuum which we call
\[\begin{equation} \rho_0(E;x) = \int_0^\infty\frac{dk}{\pi} (\cos^2(kx) + \sin^2(kx)) \delta(E - k^2/(2m)) \end{equation}\]Subtracting these
\[\begin{equation} \Delta \rho(E; x) = \rho(E;x) - \rho_0(E; x), \end{equation}\]With a bit of work, we can find
\[\begin{equation} \Delta \rho(\tfrac{k^2}{2m}; x) = - \frac{m}{k\pi} \frac{(m \lambda )^2 \cos (2kx) + m \lambda k \sin|2kx|}{(m \lambda)^2 + k^2}. \end{equation}\]We can finally, integrate this over all energies
\[\begin{equation} \int_0^\infty dE \; \Delta \rho(E; x) = - m\lambda e^{-2 m\lambda |x|} = - |\psi_0(x)|^2. \end{equation}\]As you might suspect: the continuum is depleted in the exact way to compensate the bound state. This is known as the Friedel sum rule. But which continuum states are playing the largest roles? For this, we can take
\[\Delta \rho(E) \equiv \lim_{\delta\rightarrow0^+}\int dx \, \Delta \rho(E,x) e^{-\delta |x|},\]from which we find
\[\begin{equation} \Delta \rho(E)dE = \left[-\frac12\delta(k - 0^+) - \frac{1}{\pi} \frac{m\lambda }{(m\lambda)^2 + k^2} \right] dk, \quad E = \frac{k^2}{2m} \end{equation}\]We can tell from this that half of the weight comes from the \(k=0\) (constant) mode. The rest of the weight comes from higher \(k\)’s distributed like a Lorentzian. If we include the ground state as a positive \(\delta\)-function, we get:
This is a simple example of how charge can be pulled from the continuum to make a bound state: leaving a depletion of density in the continuum.
]]>Quasicrystals, a beautiful manifestation of something without a strict crystalline symmetry but nonetheless shows order, have won a Nobel prize and have recently interested my own work with a dodecagonal graphene quasicrystal making its way into Science^{1}.
This led to this beautiful cover in Science^{2}
This phenomena is a perfect example of the kind of research I’ve been doing a lot with these days, and so it inspired the new logo for this website
One can tell how this is done: You find the points where two hexagons are on top of each other, put down a point, and connect. There are three shapes: a rhombus, an equilateral triangle, and a square. This can be done along the entire sheet to create an amazing looking pattern. For completeness, we can fill in the rest of the pictured grid to obtain:
The pattern starts to look even more intriguing the further out in the tiling you go. There is much to learn about such physical systems and their quasiperiodic cousins.
We have been studying how quasiperiodicity interplays with materials that have Dirac nodes, including twisted bilayer graphene. While we have not studied graphene at 30-degrees like the work in Science, that is an extreme where all crystalline periodicity is lost. ↩
This allows us to derive an expression for the DC-conductivity
\[\begin{align} \sigma_{yx} = \int dt \, \sigma_{yx}(t) = \frac1{A_x} \int dt \, j_y(t). \end{align}\]Geometrically, there is a lot going on with \(j_y(t)\) when we have a system with spin-orbit coupling. In particular, take the two band model
\[\begin{align} h(\mathbf p) = \mathbf d(\mathbf p) \cdot \sigma, \end{align}\]where \(\mathbf d\) is 3D, \(\mathbf p\) is 2D, and \(\sigma = (\sigma_x, \sigma_y, \sigma_z)\), the vector of Pauli matrices. The initial states of the system can be represented by where they are on the Bloch sphere \(-\hat{\mathbf d}(\mathbf p)\). But once a pulse is supplied, this state will begin to rotate about a different vector \(\mathbf d(\mathbf p - e \mathbf A)\). Thus, if we add time dependence to \(\hat{\mathbf d}(\mathbf p, t)\equiv -\langle \sigma(t) \rangle\) to represent the state’s location, we can use Heisenberg’s equations of motion to obtain
\[\begin{align} \hbar \frac{\partial \hat{\mathbf d}(\mathbf p, t)}{\partial t} = 2\mathbf{d}(\mathbf p - e\mathbf A) \times \hat{\mathbf{d}}(p, t). \end{align}\]We can rewrite this equation as
\[\begin{align} \hat{\mathbf d}(\mathbf p,t) = \hat{\mathbf d}(\mathbf p - e \mathbf A) [\hat{\mathbf d}(\mathbf p - e \mathbf A) \cdot \hat{\mathbf d}(\mathbf p)] - \hbar \frac{\hat{\mathbf d}(\mathbf p-e \mathbf A) \times \frac{\partial \hat{\mathbf d}(\mathbf p, t)}{\partial t}}{2d(\mathbf p - e A)} \end{align}\]However, this state has an associated current with it, and that can be represented by the operator \(j_\mu = -e \partial_\mu \mathbf d(\mathbf p- e \mathbf A) \cdot \sigma\). And the vector \(\langle\sigma\rangle = -\hat{\mathbf{d}}(p,t)\). Therefore,
\[\begin{align} \langle j_\mu \rangle = e^2 \partial_\mu \mathbf d(\mathbf p - e\mathbf A ) \cdot \hat{\mathbf d}(\mathbf p, t) \end{align}\]Combining these expressions, we have \(\begin{multline} \langle j_\mu \rangle = e^2 \partial_\mu \mathbf d(\mathbf p - e\mathbf A ) \cdot \Bigg[ \hat{\mathbf d}(\mathbf p - e \mathbf A) [\hat{\mathbf d}(\mathbf p - e \mathbf A) \cdot \hat{\mathbf d}(\mathbf p)] \\ - \hbar \frac{\hat{\mathbf d}(\mathbf p-e \mathbf A) \times \frac{\partial \hat{\mathbf d}(\mathbf p, t)}{\partial t}}{2d(\mathbf p - e A)} \Bigg]. \end{multline}\)
Or simplified \(\begin{multline} \langle j_\mu \rangle = e^2 \partial_\mu d(\mathbf p - e\mathbf A )[\hat{\mathbf d}(\mathbf p - e \mathbf A) \cdot \hat{\mathbf d}(\mathbf p)] \\ - \frac{e^2}{2} \partial_\mu \hat{\mathbf{d}}(\mathbf p - e \mathbf A) \cdot \left[\hat{\mathbf d}(\mathbf p-e \mathbf A) \times \frac{\partial \hat{\mathbf d}(\mathbf p, t)}{\partial t} \right]. \end{multline}\)
This is exact. At this point, we make a couple of approximations. First of all, the first term is independent of \(t\) so it cannot contribute to the total current if we have a finite DC conductivity. This leaves the second term. We can do the integral over time — there is an order of limits problem but we can get around this by noting that we do not expect the infinite time state to contribute to the energy (or: it averages to something proportional to \(\mathbf d(\mathbf p - e \mathbf A)\) anway and so the cross product vanishes), so we discard it and therefore, \(\int_0^\infty dt' \, \mathbf d(\mathbf p, t') = - \mathbf d(\mathbf p)\).
Hence, we get the Hall conductivity
\[\begin{align} \sigma_{yx} = \frac{e^2\hbar}{2 A_x} \int \frac{d^2 p}{h^2} \partial_y \hat{\mathbf{d}}(\mathbf p - e \mathbf A) \cdot \left[\hat{\mathbf d}(\mathbf p-e \mathbf A) \times \hat{\mathbf d}(\mathbf p) \right]. \end{align}\]At this point, we actually have not expanded in terms of \(A_x\) yet. The first term will produce a term that is symmetric in \(x\) and \(y\)—however it drops out due to the cross product \(\hat{\mathbf d}(\mathbf p) \times \hat{\mathbf d}(\mathbf p) = 0\) vanishes. Thus, only the second term persists and we see directly that \(x\) and \(y\) must be different and in fact, we get the well-known formula
\[\begin{align} \sigma_{yx}^{\mathrm{Hall}} =\frac{e^2}{h} \int \frac{d^2 p}{4\pi} \hat{\mathbf{d}}(\mathbf p) \cdot \left[\partial_y \hat{\mathbf d}(\mathbf p) \times \partial_x \hat{\mathbf d}(\mathbf p) \right]. \end{align}\]This describes the Chern number of some manifold parametrized by \(\mathbf p\) (usually the Brillioun zone). It is the number of times the vector \(\mathbf d\) wraps the sphere.
To understand why it’s a topological invariant, note that the quantity in the integral looks very much like a Jacobian. In fact, it is; it describes a coordinate transformation from the \(\hat {\mathbf d}\) to \((p_x,p_y)\). In this way, the integrand represents an area element on the sphere, and in general \(\mathbf p\) is a closed manifold. So \(\mathbf d(\mathbf p)\) maps that closed manifold to the sphere, and without any edges or boundaries the area it maps out must be \(4 \pi\) times an integer.
This formula is well-known, but this dynamical way of obtaining it is slightly less well-known. We have extended this idea in a paper published last year to handle the out-of-equilibrium case of quenches. In that situation, new phenonmena appear that are quite different from the equilibrium case—terms that we discarded in this calculation become quite relevant.
]]>There’s a problem though, an electric field accelerates a charge. Consider a classical electron for the time being, then
\[\begin{align} m \ddot x(t) = - e \mathbf{E}. \label{eq:Newton} \end{align}\]Or in terms of current \(j(t) = -e \dot x(t)\), \(\frac{d j}{dt} = \frac{e^2}{m} \mathbf{E}\). From here we can quickly and naively go to frequency space to find \(j(\omega) = i \frac{e^2/m}{\omega} \mathbf{E}(\omega)\). Then one might remember that another way to define \(\mathbf{E}\) is in terms of a vector potential that is purely time-dependent, so \(\mathbf{E}(\omega) = i \omega \mathbf A(\omega)\). Now, if we just plug this into our linear response for the current, we get
\[j(\omega) = - \frac{ e^2}{m} \mathbf A(\omega). \label{eq:jA}\]All is well and good, right? Well, not quite. In electromagnetism, the constant part of \(\mathbf A(t)\) corresponds to the \(\omega = 0\) term of \(A(\omega)\). This represents what is known as a “pure gauge”. These gauges are physically equivalent to the null field \(\mathbf A(\omega=0) = 0\). Thus, whatever linear response is represented above at \(\omega = 0\) must be unphysical, right?
Wrong.
Before explainging why this is wrong, let’s give some further context to this linear response theory. The term \(- e^2/m\) is actually the single particle term of what is known as the “diamagnetic” response to the conductivity when you add in more electrons (usually distributed in a Fermi distribution). This term persists in quantum mechanics, and no other terms appear to cancel it in the simplest case of \(H = \frac{p^2}{2m}\). In fact, while the math becomes more cumbersome, the solution we shall illustate below holds perfectly well for even the non-interacting multi-electron system.
Now, at this point you may have guessed that there’s something strange going on at \(\omega = 0\) due to the fact that the electric field accelerates the particle and doesn’t just have a velocity response. At the \(\omega = 0\) point, the physical field \(E(\omega)\) seems to necessarily be equal to zero in the gauge we have prescribed unless \(A(\omega) \sim 1/\omega\) for small \(\omega\). This would lead to a divergent \(j(\omega)\), restoring our faith that the system is accelerating out of control.
But what about when \(\mathbf{A}(\omega=0) = \mathbf{A}_0\)? It seems like then we have a true velocity response to an unphysical object. The solution is subtle: At some point in the quick derivation we made an assumption that implied \(\mathbf A(t) \rightarrow 0\) at \(t \rightarrow -\infty\). This implies that if \(\mathbf A(t) = \mathbf A_0\) at any finite time, there had to be some time in between where \(d\mathbf A/ dt \neq 0\). Thus, during that “ramp up” time, an electric field was on and it accelerated the charge to a specific velocity resulting in the current \(j(\omega) = - \frac{ e^2}{m} \mathbf A(\omega)\).
The assumption is subtle, but the result is rather simple. For now, just assume that \(\mathbf A(-\infty) = 0\) and at some \(t_0\), \(\mathbf A(t_0) = \mathbf A_0\), then we can integrate Eq. \eqref{eq:Newton} to obtain the velocity:
\[m \dot x (t_0) = e \int_{-\infty}^{t_0} \frac{d \mathbf A}{d t} d t = e \mathbf A_0.\]Or, in other words, \(\mathbf j(t_0) = -\frac{e^2}{m} \mathbf A_0\), the same as before! This is how a constant \(\mathbf A_0\) can be physical: When it represents the change from a different constant vector potential.
Now, to isolate the assumption, let us run through what they were
Now, let us get the last equation (in #4 above) by a simpler route.
Two perfectly legitimate calculations resulting in different results. Firstly, this highlights that the first procedure does actually assume \(A(-\infty) = 0\). Secondly, the only assumptions that could have given \(j(-\infty) = 0\) (an assumption we probably wanted anyway) and \(A(-\infty) = 0\) are that they could be given in terms of Fourier transforms. In order for a function to have a Fourier transform it needs to be absolutely integrable—i.e. \(\int_{-\infty}^\infty \lvert A(t) \rvert dt \lt \infty\). Given \(A(t)\) as a continuous, piece-wise differentiable function, we need \(A(t) \rightarrow 0\) for \(t \rightarrow \pm \infty\). This imposes our gauge, and since we are not interested in future times let alone \(t \rightarrow +\infty\), we can artificially modify the function as we see fit to accomodate that. But how the function began at \(-\infty\) is important, and we must impose that. Hence, we have chosen, at least partially, a gauge.
We are left with a dilemma then about pure plain waves \(A(t) = A(\omega) e^{-i \omega t}\). How do those function?
Technically, they are outside of the bounds of the Fourier analysis and we can see that simply by the fact that if we tried to do the above procedure, we couldn’t have a well defined answer as \(t \rightarrow -\infty\) (too oscillatory). However, we can approximate the plain wave in terms of an absolutely integrable function \(A_\delta(t) = A(\omega) e^{-i (\omega + i \delta) t}\) for any \(\delta \gt 0\), and everything works. This shows us explicitly that \(t = - \infty\) does have \(A_\delta \rightarrow 0\) for all \(\delta \gt 0\). And this is the origin of the well known substitution \(\omega \rightarrow \omega + i\delta\).
The natural question to ask now is how this works for a real system (with dissipation). Why does such a term not exist at zero frequency?
Unless your system is a superconductor, there is some dissipation in the system. The simplest way to include this is classically: When an electron is going at velocity \(\dot x(t)\) it experiences a “drag” that tends to slow it down. Thus, our Newton’s equations become
\[m \ddot x(t) = - m \gamma \dot x(t) - e \mathbf E\]where \(\gamma\) describes how much drag the electron experiences. For more disorder, this would be a larger number. Playing the same Fourier transform game, we can obtain rather quickly that
\[j(\omega) = - \frac{e^2}{m} \frac{\omega}{\omega +i \gamma} A(\omega).\]This is just one step away from the well-known Drude model. We see that if \(A(t) = A_0\), then \(j(\omega) = 0\). But our gauge choice that we described before is still in place, the only difference is that our “kick” at \(t=-\infty\) has an infinite time to dissipate back to rest (the inclusion of \(\gamma\) above is critical for \(j(\omega=0) =0\)). This also suggests a steady state current when \(\mathbf E\) is constant: \(m\ddot x=0\) implies \(j = \frac{e^2}{m \gamma} \mathbf E\). Our current relaxes to zero when there’s nothing around (\(\mathbf E = 0\)), as we would expect.
When a quantum mechanical description is done—by taking a random disorder potential and averaging over disorder configurations—one obtains similar results. The diamagnetic term for a clean system is real and has a physically well defined explanation.
One may not be surprised that this curious “diamagnetic term” occurs for superconductors, however it is sometimes explained that “gauge symmetry is broken” and that is why such a term exists. This is a misleading statement, but one I will explore in a future post.
]]>Many textbooks on Quantum Mechanics mention current density can be derived from the continuity equation and probability. The usual method for figuring this out is to assume you have some Hamiltonian \(H = p^2/2m + V(x)\) where \(p\) is the momentum and \(x\) is the position. In this way the current density is written in terms of the wave function \(\psi(x,t)\) as
\begin{equation} j(x,t) = \frac1{2m i}\left[ \psi^* \overrightarrow{\partial_x} \psi - \psi^* \overleftarrow{\partial_x} \psi \right]. \end{equation}
This then satisfies the continuity equation
\begin{equation} \partial_t \rho(x,t) + \partial_x j(x,t) = 0, \end{equation}
with density \(\rho(x,t)=\psi^*(x,t)\psi(x,t)\). It should be noted that if you write the wave function as \(\psi(x,t) = \sqrt{\rho(x,t)} e^{i \theta(x,t)}\), then the current is just proportional to the gradient of the phase \(j(x,t)= \rho(x,t) \partial_x \theta(x,t)/m\), giving the spatial change in phase a physical significance.
However, there are two lingering questions:
To answer these questions, we consider the more arbitrary Hamiltonian
\begin{equation} H = T(p) + V(x), \end{equation}
where \(V(x)\) is some potential and the kinetic energy is some polynomial
\begin{equation}T(p) = \sum_{n=1} a_n \frac{p^n}{n!}.\end{equation}
We are unworried about bounding the energy, so odd-order Kinetic energy terms are allowed (in the higher dimensional case, the Dirac-like Hamiltonians have linear terms in \(p\)). At this point, we can take our Heisenberg operator \(\dot x(t)\) and find
\[\begin{align} \dot x(t) & = i [H, x(t)] \\ & = T'(p(t)). \end{align}\]where \(T'\) is the derivative of \(T\) with respect to its argument. Now, we would like to obtain a current density from this quantity. We can certainly define the total current at a specific time as
\[\begin{align} I & = \langle\psi_0\lvert \dot x(t) \rvert \psi_0\rangle = \langle\psi_0\lvert T'(p(t)) \rvert \psi_0\rangle \\ & = \langle\psi(t)\lvert T'(p) \rvert \psi(t)\rangle, \end{align}\]where in the last line we go from the Heisenberg to Schroedinger picture. Now to get density, we need to use a complete set position states, so that
\[\begin{align} \label{eq:total-current} I & = \int dx \, dy \, \langle\psi(t)\lvert x\rangle \langle x \lvert T'(p) \rvert y \rangle \langle y \lvert \psi(t)\rangle. \end{align}\]Now, \(p\) acts as a derivative on position kets, so that one can verify that
\[\begin{equation} \langle x \lvert T'(p) \rvert y \rangle = T'(-i \partial_x ) \delta(x-y). \end{equation}\]However, there is an ambiguity here since we can write
\[\begin{align} T'(-i \partial_x)\delta(x-y) & = \sum_{n=0} a_{n+1} \frac{(-i\partial_x)^n}{n!} \delta(x-y) \nonumber \\ & = \sum_{n=0} a_{n+1} \frac{(-i\partial_x)^{n-m} (i \partial_y)^m}{n!} \delta(x-y). \label{eq:T-delta} \end{align}\]This ambiguiuty in how to choose the derivatives leaves us with many way to define the current density. Fortunately, only one of these combinations satisfies the continuity equation. To figure out which one that is, let us reverse engineer the continuity equation to obtain a solution. The density is \(\rho(x,t) = \psi^*(x,t) \psi(x,t)\), and so using the Schroedinger’s equation, we have
\[\begin{align} i \partial_t \rho(x,t) & = i(\psi^*(x,t) \overleftarrow {\partial_t} \psi(x,t) + \psi^*(x,t) \overrightarrow {\partial_t} \psi(x,t) ) \\ & = - [\psi^*(x,t)( T(i \overleftarrow{\partial_x} ) - T(-i \overrightarrow{\partial_x} ) )\psi(x,t) ]. \end{align}\]Thus, the continuity equation must become
\begin{align} \partial_t \rho(x,t) - i [\psi^*(x,t)( T(i \overleftarrow{\partial_x} ) - T(-i \overrightarrow{\partial_x} ) )\psi(x,t) ] = 0. \end{align}
If we now assume that we have a current density that takes the form
\begin{equation} j(x,t) = \psi^*(x,t) \vartheta(\overleftarrow{\partial_x},\overrightarrow{\partial_x}) \psi(x,t), \end{equation}
and satisfies the continuity equation, \(\partial_t \rho + \partial_x j = 0\), then we can equate operators to obtain
\begin{align} \label{eq:diff-ops-cty}
\overleftarrow{\partial_x} \vartheta + \vartheta \overrightarrow{\partial_x} = -i [T(i \overleftarrow{\partial_x} ) - T(-i \overrightarrow{\partial_x} ) ].
\end{align}
Anticipating the answer, we write the general form of \(\vartheta\) as
\begin{equation} \vartheta = \sum_{n=0} \sum_{m=0}^n (-1)^{m} i^n b_{n,m} \overleftarrow{\partial}{}_x^{n-m} \overrightarrow{\partial}{}_x^m. \end{equation}
Then we can take the left hand side Eq. \eqref{eq:diff-ops-cty} and write
\[\begin{multline} \overleftarrow{\partial_x} \vartheta + \vartheta \overrightarrow{\partial_x} = -i \sum_{n=1} i^n b_{n-1,0} \overleftarrow{\partial}{}_x^{n} - \sum_{n=0} \sum_{m=0}^{n-1} (-1)^m i^n \left[ b_{n,m+1} - b_{n,m} \right] \\ \times \overleftarrow{\partial}{}_x^{n-m} \overrightarrow{\partial}{}_x^{m+1} + i \sum_{n=1} (-i)^{n} b_{n-1,n-1} \overrightarrow{\partial}{}_x^{n} . \end{multline}\]On the other hand, we can calculate the right hand side of Eq. \eqref{eq:diff-ops-cty} to be
\[\begin{equation} -i [T(i \overleftarrow{\partial_x} ) - T(-i \overrightarrow{\partial_x} ) ] = -i \sum_{n=1} a_n i^n \frac{\overleftarrow{\partial}{}_x^n}{n!} + i \sum_{n=1} a_n (-i)^n \frac{\overrightarrow{\partial}{}_x^n}{n!}. \end{equation}\]Equating the left and right sides, we can just read off that \(b_{n-1,0} = a_n/n!\), \(b_{n-1,n-1}= a_n/n!\) and \(b_{n,m+1} = b_{n,m}\), so that \(b_{n,m} = a_{n+1}/(n+1)!\).
Thus, we have
\begin{equation} \vartheta = - \sum_{n=0} \sum_{m=0}^n (-1)^{m} i^n \frac{a_{n+1}}{(n+1)!} \overleftarrow{\partial}{}_x^{n-m} \overrightarrow{\partial}{}_x^m. \end{equation}
Returning all the way to when we were considering \(\dot x(t)\) as an integral over position, this suggests that in Eq. \eqref{eq:T-delta}, we want to consider
\begin{equation} T’(-i \partial_x)\delta(x-y) \ = \sum_{n=0} \frac{a_{n+1}}{n!} \frac1{n+1}\sum_{m=0}^n (-i\partial_x)^{n-m} (i \partial_y)^m \delta(x-y). \end{equation}
Given the expression for total current Eq. \eqref{eq:total-current} and integrating the delta function by parts numerous times, we can replace \(\partial_x\) with \(-\overleftarrow \partial_x\) and \(\partial_y\) with \(- \overrightarrow\partial_x\), and then the total current is just
\begin{equation} I = \int dx \, \psi^*(x,t) \sum_{n=0} \frac{a_{n+1}}{(n+1)!} \sum_{m=0}^n (i \overleftarrow \partial_x)^{n-m} (-i \overrightarrow \partial_x)^m \psi(x,t). \end{equation}
which actually integrates the current density! Thus, we have shown that
\begin{equation} j(x,t) = \psi^*(x,t) \sum_{n=0} \frac{a_{n+1}}{(n+1)!} \sum_{m=0}^n (i \overleftarrow \partial_x)^{n-m} (-i \overrightarrow \partial_x)^m \psi(x,t), \end{equation}
and that
\begin{equation}\langle \dot x(t) \rangle = \int dx \, j(x,t). \end{equation}
Indeed, \(\dot x(t)\) does track the current of the problem and can even be written as the integral of a current density. Even for the more arbitrary Hamiltonian \(H = T(p) + V(x)\).
]]>We look at the relatively simple problem^{1} of finding the energy spectrum for a particle in the lattice potential
\begin{equation} U(x) = \alpha\sum_{n=-\infty}^\infty \delta(x - n a)\end{equation}.
The time-independent Schrödinger equation takes the form
\begin{equation} \left[ - \frac{\partial_x^2}{ 2 m} + U(x) \right] \psi(x) = E \psi(x),\end{equation}
where \(E\) is the energy.
Since we can solve the problem between the delta-functions quite simply (\(U(x) =0\) there), let us restrict our focus to \(na \lt x \lt (n+1)a\). Here, the wave function takes on the form
\begin{equation} \psi(x) = A_n e^{ik(x-na)} + B_n e^{-ik(x-na)},\end{equation}
where we have \(k = \sqrt{2m E}\). Now, we can find an operator that commutes with the Hamiltonian so that we can diagonlize it to help solve the problem — this will be the operator that translates us by \(a\).
To make this clear, let us abstract things to operators so that we have a momentum operator \(p\) and a position operator \(x\), then we have the commutator \([ x, p] = i\). The operator \(p\) commutes with functions of \(x\) as though it were a derivative \([ p, f( x)] = -i f'( x)\), so considering the translation operator \(e^{i a p }\), we can write
\begin{equation} e^{i a p} U( x) e^{ - i a p} = \sum_{n=0}^\infty \frac1{n!} U^{(n)}( x) a^n = U( x + a),\end{equation}
and since \(U(x)\) is periodic in \(a\), we have that \(e^{i a p} U( x) e^{ - i a p} = U( x)\). Thus, the operator \(T_a = e^{i a p}\) commutes with the Hamiltonian and we can simulataneously diagonalize both it and the Hamiltonian. We say \(T_a \lvert \psi\rangle = e^{i a q} \lvert \psi \rangle\) has quasi-momentum \(q\). It is important to note that this not the same as real momentum which is not a well-defined quantum number in this problem (that needs translation symmetry).
In other words, we can write our eigenfunctions such that \(\psi(x+a) = e^{i q a} \psi(x)\), and this naturally leads us to relate the coefficients for our eigenfunctions above as
\begin{equation} A_{n-1} = e^{-i q a} A_n, \quad B_{n-1} = e^{-i q a} B_n.\end{equation}
Now, we can apply matching conditions at \(x = na\) remembering that our wavefunction should be continuous, but that the delta-function will cause the first derivatives to be discontinuous. The equations for the coefficients are
\begin{align}
A_n + B_n & = e^{i k a} A_{n-1} + e^{-i k a} B_{n-1}, \\
\left( 1 + \frac{2 i m \alpha}{k} \right) A_n - \left( 1 - \frac{2 i m \alpha}k \right) B_n & = e^{i k a} A_{n-1} - e^{-ik a} B_{n-1},
\end{align}
and if we insert the relation between the coefficients at \(n-1\) and \(n\), we get a matrix equation
\begin{align}
\begin{pmatrix}
e^{i q a} - e^{i k a} & e^{i q a} - e^{- i k a} \\
e^{i q a} \left(1 + \tfrac{2 i m \alpha}k \right) - e^{i k a} & - e^{i q a } \left( 1 - \tfrac{2 i m \alpha}k \right) + e^{-ik a}
\end{pmatrix}
\begin{pmatrix}
A_n \ B_n
\end{pmatrix} = 0.
\end{align}
This equation has a non-zero solution only if the determinant of the matrix is zero which can be written as
\begin{equation} \cos q a - f(E) = 0 , \quad f(E) = \cos ka + \frac{m \alpha}{k} \sin ka. \end{equation}
This is an equation which relates the energy to the quasi-momentum. Since \(\cos q a\) can only be between -1 and 1, this equation only has a solution when \(f(E)\) is between -1 and 1. The oscillatory nature of \(f(E)\) means that it should pass in this range multiple (in fact, a countable infinite) number of times.
Armed with this equation, we can use \(q\) and an integer to label our energies and we obtain the following energy bands by just solving for \(E\) (setting \(m = 10\), \(a = 1\), and \(\alpha = 0.3\))
The dotted lines in this plot represent the spectrum if there were no delta-function potentials (displaced in energy by \(\alpha / a\) for clarity). Notice how the introduction of the delta-functions opens up gaps in this energy spectrum, so that there are some energies that are inaccessible. The gaps actually open up when \(\lvert f(E)\rvert \geq 1\) since there is no solution to our equation there. This energy gap for small \(\alpha\) just goes like \(\alpha/a\), vanishing as we’d expect when \(\alpha = 0\).
Notice that in this energy spectrum, we see that if \(q \rightarrow -q\) we get the same energy.
Additionally, if the energy ever goes negative the solutions turn from plane waves \(e^{\pm i k (x-na)}\) into functions localized around the delta functions \(e^{\pm\kappa(x-na)}\). In fact, the wave functions look like this for a \(q=0\) state:
These only appear when \(\alpha \lt 0\), and are related to the fact that the delta-function potential has a bound state. Additionally, only one band can ever have this state. This is due to the fact that the oscillatory sine and cosine change to their non-oscillatory hyperbolic counterparts. However, these states in the delta-function lattice are spread out throughout the crystal, and can not be said to be “localized” to a specific site — they still all have a definite quasi-momentum.
As with other single particle problems, upon considering the many particle picture, these energy bands get filled up to a set energy level (if we are considering fermions).
This is problem 2.53 in Exploring Quantum Mechanics by Galitski, Karnakov, Kogan, and Galitski. ↩
To address this, there is an extension to gmail called GmailTeX available on Chrome, Firefox, Safari, and Opera (also it has a bookmarklet for any other browser).
If someone sends you an email in pseudo-LaTeX it can try to parse what the math with simple math. The resulting output is kind of what you’d get by using Unicodeit. Or, if someone sends you LaTeX inside dollar signs (i.e., $ […] $), then it can just compile that to LaTeX for you with its rich math function.
But best of all, when you compose emails and use the rich math function, it creates an image of your math hosted on a remote server. They don’t need to have GmailTeX installed unless you decide to send them the code itself.
The best collaborative apps require little to no commitment for collaborators to use, and this is one of those cases. Collaborators will just receive email with LaTeX images without needing to install anything.
The only issue that I’ve found from playing around with this extension is that the receiver may have to flag your email as “trustworthy” and/or allow images from remote servers to be viewed in their email client. Otherwise, they may not see any mathematics. Additionally, as you might have guessed, the emails won’t be viewable in an offline mode. Unless you’re going through these emails on an airplane though, I don’t think that should be too big of an issue.
(h/t Brian Danielak)
]]>\begin{equation} H = \frac{p^2}{2m} + \alpha (\boldsymbol{\sigma} \times \mathbf{p})\cdot \hat{\mathbf{z}} + \Delta \sigma_z. \end{equation}
\(m\) is the mass, \(\alpha\) is the spin-orbit coupling strength, and \(\Delta\) is some Zeeman field (it acts as magnetic field on the spin).
In this post, we go through the calculation of the energy spectrum and eigenvectors – a straight forward exercise in undergraduate linear algebra.
First of all, instead of the normal method of finding eigenvectors, we note that we can rewrite this Hamiltonian in the form
\begin{equation} H = \frac{p^2}{2m} + \mathbf{b}(p) \cdot \boldsymbol{\sigma} \end{equation}
where \(\mathbf{b}(p) = (\alpha p_y, -\alpha p_x, \Delta)\). Now, \(\mathbf{b}(p)\) represents a point on the Bloch sphere, and so we expect the eigenvectors to be parallel and anti-parallel to this vector. The energies in this case are very straight forward and amount to the positive and negative of \(\lvert\mathbf{b}(p)\rvert\):
\begin{equation} \epsilon_\pm(p) = \frac{p^2}{2m} \pm \sqrt{ \alpha^2 p^2 + \Delta^2}. \end{equation}
With these eigenvalues, it is a straight forward exercise in linear algebra to find the eigenvectors. After a bit of algebra, the eigenvectors of \(H\) in terms of the eigenvectors of \(\sigma_z\) ( \(\sigma_z\left\lvert\uparrow\right\rangle = \left\lvert\uparrow\right\rangle\) and \(\sigma_z\left\lvert\uparrow\right\rangle = -\left\lvert\uparrow\right\rangle\) ) are
\begin{equation}\left\lvert\pm\right\rangle = \frac1{\sqrt2}\left[\sqrt{1 \pm \frac{\Delta}{\sqrt{\Delta^2+\alpha^2 p^2}}}\left\lvert\uparrow\right\rangle + e^{-i\phi} \sqrt{1 \mp \frac{\Delta}{\sqrt{\Delta^2+\alpha^2 p^2}}}\left\lvert\downarrow\right\rangle \right]\end{equation}
where we have defined \(\phi\) by \(p_y+ip_x = p e^{i\phi}\). Note that when \(p_{x,y} \rightarrow -p_{x,y}\), the occupations stay the same. However, if we just look at one energy, \(\epsilon_-(p)\) the ground state energy, we see that the state we get when \(p_{x,y} \rightarrow -p_{x,y}\) is almost orthogonal to the original state.
The energy bands themselves look like the figure on the right where the vertical axis is energy (and for this particular example, \(m=1\), \(\alpha = 3\), and \(\Delta=2\)). Interestingly, the introduction of \(\Delta\) actually causes the gap to open up – the dotted lines are for when \(\Delta=0\).
Now, if we have a bunch of fermions filling up these energies, if we set the chemical potential to be in the gap, we would find that the only excitations would states that are spin-locked to the momentum.
Many things can be done with this Hamiltonian to interesting effect. It finds its way into cold atom physics as well as condensed matter.
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