--- jupytext: cell_metadata_filter: -all formats: ipynb,md:myst text_representation: extension: .md format_name: myst format_version: 0.13 jupytext_version: 1.14.0 kernelspec: display_name: Python 3 (ipykernel) language: python name: python3 --- ```{code-cell} ipython3 import mmf_setup;mmf_setup.nbinit() import logging;logging.getLogger('matplotlib').setLevel(logging.CRITICAL) %matplotlib inline import numpy as np, matplotlib.pyplot as plt ``` # Landau-Zener Transitions $\newcommand{\sa}{\ket{\downarrow}}\newcommand{\sb}{\ket{\uparrow}}$ The [Landau-Zener formula] for nonadiabatic transitions is a non-trivial example of the type of manipulation and level of mathematical sophistication expected in this course. This example serves several purposes. In particular, it: 1. contains the most general dynamics of a two-state system (qubit), providing a connection between dynamics on the {ref}`sec:BlochSphere` and the analytic formulation of quantum mechanics. 2. represents the essence [adiabatic quantum computing]. 3. demonstrates a technique of analytically studying systems that are not analytically solvable. ## General Qubit Dynamics :::{margin} If the qubit is implemented as a spin-½ particle, then this corresponds to placing the particle in a time-dependent magnetic field \begin{gather*} \vec{B}(t) = \frac{-2}{\hbar \mu_s}\vec{b}(t) \end{gather*} where $\mu_s$ is the [spin magnetic dipole moment]. ::: The most general dynamics for a single qubit can be described by the following Hamiltonian: \begin{align*} \frac{\mat{H}(t)}{\hbar} &= \vec{b}(t) \cdot \vec{\mat{\sigma}} = b_x(t)\mat{\sigma_x} + b_y(t)\mat{\sigma_y} b_z(t)\mat{\sigma_z}\\ &= \begin{pmatrix} b_z(t) & b_x(t) - \I b_y(t)\\ b_x(t) + \I b_y(t) & -b_z(t) \end{pmatrix}. \end{align*} The general dynamics of a qubit state $\ket{\psi(t)}$ follow the time-dependent Schrödinger equation \begin{gather*} \I\hbar \diff{}{t}\ket{\psi(t)} = \I\hbar \ket{\smash{\dot{\psi}}(t)} = \mat{H}(t)\ket{\psi(t)}. \end{gather*} :::{margin} One will often see the following notation for **the commutator** of two matrices: \begin{gather*} [\mat{A}, \mat{B}] = \mat{A}\mat{B} - \mat{B}\mat{A}. \end{gather*} ::: What makes this problem tricky *(and quantum dynamics tricky in general)* is that the matrices $\mat{H}(t)$ at different times may not commute: i.e. there exists times $t$ and $t'$ such that \begin{gather*} \mat{H}(t)\mat{H}(t') \neq \mat{H}(t')\mat{H}(t). \end{gather*} :::{admonition} Time Ordering: why it's tricky. :class: dropdown If the Hamiltonian commutes at all times $[\mat{H}(t), \mat{H}(t')] = 0$ -- e.g., if $\mat{H}(t) = \mat{H}$ is constant -- then the formal solution to the problem is simply: \begin{gather*} \ket{\psi(t)} = \underbrace{ \exp\Biggl( \overbrace{\int_0^{t}\d{\tau}\; \frac{\op{H}(\tau)}{\I\hbar}}^{\mat{Q}(t)} \Biggr) }_{\mat{U}(t) = e^{\mat{Q}(t)}}\ket{\psi(0)}, \end{gather*} where the unitary matrix $\mat{U}(t)$ is **the propagator**. Recall from {ref}`sec:MatrixExponential` that this matrix exponential can be defined in terms of the Taylor series: \begin{gather*} \mat{U}(t) = \mat{1} + \mat{Q}(t) + \frac{\mat{Q}^2(t)}{2!} + \cdots. \end{gather*} Even in the second term, we see the problem: \begin{gather*} \dot{\mat{Q}}(t) = \frac{\mat{H}(t)}{\I\hbar}, \\ \I \hbar \dot{\mat{U}}(t) = \op{H}(t) + \frac{\mat{H}(t)\mat{Q}(t) + \mat{Q}(t)\mat{H}(t)}{2} + \cdots. \end{gather*} To recover the Schrödinger equation, we must pull all factors of $\mat{H}(t)$ to the left so we have: \begin{gather*} \I \hbar \dot{\mat{U}}(t) = \op{H}(t)\underbrace{ \Biggl( \mat{1} + \mat{Q}(t) + \frac{\mat{Q}^2(t)}{2!} + \cdots \Biggr)}_{\mat{U}(t)}, \end{gather*} but we cannot do this if $\mat{Q}(t)$ and $\mat{H}(t)$ do not commute, which will generally be the case if the Hamiltonian does not commute at different times. The solution is to work through this expansion, manually ordering all products of $\mat{H}(t)$ so that later times appear to the left while earlier times appear to the left. This is done with the [time ordering] operator $\mathcal{T}$: \begin{gather*} \mathcal{T}\Bigl\{\mat{A}(t_a)\mat{B}(t_b)\Bigr\} = \begin{cases} \mat{A}(t_a)\mat{B}(t_b) & t_a > t_b\\ \mat{B}(t_b)\mat{A}(t_a) & t_a < t_b. \end{cases} \end{gather*} Thus, once sometimes the solution written as \begin{gather*} \mat{U}(t) = \mathcal{T}\left\{ \exp\Biggl( \int_0^{t}\d{\tau}\; \frac{\op{H}(\tau)}{\I\hbar} \Biggr) \right\}, \end{gather*} but this does not really help solve the equation. Solving the differential equation numerically is usually the easiest approach, but this time-ordering can be useful when the time-dependence is perturbative. ::: Edwin Barns presents an interesting solution in {cite:p}`Barnes:2013` that turns the problem around. Instead of specifying $\vec{b}(t)$ and trying to find a solution, he shows that one can directly parameterize the propagator $\mat{U}(t)$, then determine what magnetic field $\vec{b}(t)$ gives this behavior. :::{admonition} Barns's original presentation. Barns introduces the functions $β(t)$, $φ(t)$, and $χ(t)$, subject to the constraint \begin{gather*} \abs{\dot{χ}(t)} < \abs{β(t)}, \\ χ(t) = 0, \qquad \dot{χ}(0) = -sβ(0). \end{gather*} In the following expressions, we suppress the argument $t$. All quantities except $s = \pm 1$ are time-dependent: \begin{gather*} \mat{U} = \begin{pmatrix} u_{11} & -u_{21}^*\\ u_{21} & u_{11}^* \end{pmatrix} = \begin{pmatrix} \vec{u} & -\I\mat{Y}\vec{u}^* \end{pmatrix},\\ \vec{u} = \begin{pmatrix} \cos(χ) e^{\I (ξ_{-} - φ/2)}\\ \I s \sin(χ) e^{\I (ξ_{+} + φ/2)} \end{pmatrix}, \\ ξ_{±} = \int_0^{t}β\sqrt{1-\frac{\dot{χ}^2}{β^2}}\csc(2χ)\d{\tau}\; \pm \frac{1}{2}\sin^{-1}\frac{\dot{χ}}{β} \pm s\frac{π}{4},\\ \vec{b} = \begin{pmatrix} β\cos φ\\ β\sin φ\\ \frac{\ddot{χ} - \dot{χ}\dot{β}/β}{2β\sqrt{1-\frac{\dot{χ}^2}{β^2}}} -β\sqrt{1-\frac{\dot{χ}^2}{β^2}}\cot(2χ) + \frac{\dot{φ}}{2} \end{pmatrix}. \end{gather*} We replace $\beta(t)$ with the angle $\phi(t)$ where $\cos \phi = \dot{\chi}/\beta$. This automatically satisfies the constraints if start with $\phi(0) \in \{0, \pi\}$ (corresponding to $s=-1$ and $s=+1$ respectively): \begin{gather*} \vec{u} = \I^{(1-s)/2} e^{\I\eta}\begin{pmatrix} \cos(χ)\\ -\I\sin(χ) e^{\I(φ + \phi)} \end{pmatrix}, \\ \eta = \int_0^{t}β\sin \phi\csc(2χ)\d{\tau}\; - \frac{\phi}{2},\\ \beta = \frac{\dot{χ}}{\cos\phi},\\ \vec{b} = \begin{pmatrix} β\cos φ\\ β\sin φ\\ -β\sin \phi \cot(2χ) + \frac{\dot{φ} - \dot{\phi}}{2} \end{pmatrix}. \end{gather*} *(This still needs the phases checked...)* ::: ```{code-cell} ipython3 #:tags: [hide-cell] # Numerical checks of these equations from scipy.integrate import solve_ivp T = 5.0 hbar = 1 from phys_555.utils import sigmas # Pauli matrices # "Random" functions import sympy t_, T_ = sympy.var('t_, T_') beta_ = sympy.exp(2*t_ / T_) / T_ varphi_ = sympy.cos(2*np.pi * t_ / T_)**2 + 1 chi_ = (t_ / T_) # Differentiate and make functions get_beta, get_varphi, get_chi = [ sympy.lambdify([t_, T_], [_x, _x.diff(t_), _x.diff(t_, t_)], "numpy") for _x in (beta_, varphi_, chi_)] def b(t, T=T): varphi, dvarphi, _ = get_varphi(t, T) chi, dchi, ddchi = get_chi(t, T) beta, dbeta, _ = get_beta(t, T) eta = np.sqrt(1 - (dchi/beta)**2) return [ beta * np.cos(varphi), beta * np.sin(varphi), (ddchi - dchi*dbeta/beta) / 2 / beta / eta - beta*eta/np.tan(2*chi) + dvarphi/2, ] def get_H(t): return np.einsum('i,iab->ab', b(t), sigmas) def rhs(t, psi): dpsi = get_H(t) @ psi / 1j / hbar return dpsi psi0 = np.array([1, 0j]) print(get_H(1e-5)) res = solve_ivp(rhs, t_span=(1e-5, T), y0=psi0) t = res.t res.y.shape plt.plot(t, abs(res.y.T)) chi, dchi, ddchi = get_chi(t, T) varphi, dvarphi, ddvarphi = get_varphi(t, T) beta, dbeta, ddnbta = get_beta(t, T) plt.plot(t, abs(np.cos(chi)), ':') plt.plot(t, abs(np.sin(chi)), ':'); ``` ```{code-cell} ipython3 # New formulation phi_ = sympy.acos(chi_.diff(t_)/beta_) get_phi, = [ sympy.lambdify([t_, T_], [_x, _x.diff(t_), _x.diff(t_, t_)], "numpy") for _x in (phi_,)] def bnew(t, T=T): varphi, dvarphi, _ = get_varphi(t, T) phi, dphi, _ = get_phi(t, T) chi, dchi, _ = get_chi(t, T) beta = dchi / np.cos(phi) return [ beta * np.cos(varphi), beta * np.sin(varphi), -dchi * np.tan(phi)/np.tan(2*chi) + (dvarphi-dphi)/2, ] ts = np.array([0.0001, 1.0]) np.allclose(b(ts), bnew(ts)) ``` ```{code-cell} ipython3 #:tags: [hide-cell] # Complete new solution from scipy.integrate import solve_ivp T = 5.0 hbar = 1 from phys_555.utils import sigmas # Pauli matrices # "Random" functions import sympy t_, T_ = sympy.var('t_, T_') varphi_ = sympy.cos(2*np.pi * t_ / T_)**2 chi_ = (t_ / T_)**2 phi_ = sympy.sin(2*np.pi * t_ / T_) # Differentiate and make functions get_phi, get_varphi, get_chi = [ sympy.lambdify([t_, T_], [_x, _x.diff(t_)], "numpy") for _x in (phi_, varphi_, chi_)] def b(t, T=T): varphi, dvarphi = get_varphi(t, T) chi, dchi = get_chi(t, T) phi, dphi = get_phi(t, T) beta = dchi / np.cos(phi) return [ beta * np.cos(varphi), beta * np.sin(varphi), -beta*np.sin(phi)/np.tan(2*chi) + (dvarphi-dphi)/2, ] def get_H(t): return np.einsum('i,iab->ab', b(t), sigmas) def rhs(t, psi): dpsi = get_H(t) @ psi / 1j / hbar return dpsi psi0 = np.array([1, 0j]) print(get_H(1e-5)) res = solve_ivp(rhs, t_span=(1e-5, T), y0=psi0) t = res.t res.y.shape plt.plot(t, abs(res.y.T)) chi, dchi = get_chi(t, T) varphi, dvarphi = get_varphi(t, T) phi, dphi = get_phi(t, T) plt.plot(t, abs(np.cos(chi)), ':') plt.plot(t, abs(np.sin(chi)), ':'); ``` ```{code-cell} ipython3 a = beta/dchi da = dbeta/dchi - beta*ddchi/dchi**2 w = np.arctanh(1/a) #plt.plot(t, a); gamma = 1/np.sqrt(1-(dchi/beta)**2) plt.plot(t, 1/((ddchi - dchi*dbeta/beta)/2/beta*gamma), '-'); plt.plot(t, -1/(da/2*np.sinh(w)/a), ':'); plt.plot(t, -1/(da/2/a/np.sqrt(a**2-1)), ':'); #plt.plot(t, -beta/gamma, '-'); #plt.plot(t, -dchi/np.sinh(w), ':'); #plt.plot(t, -dchi*a/np.cosh(w), ':'); ``` ``` {code-cell} ipython3 #:tags: [hide-cell] # Numerical checks of these equations from scipy.integrate import solve_ivp T = 1.0 hbar = 1 from phys_555.utils import sigmas # Pauli matrices # "Random" functions import sympy t_, T_ = sympy.var('t_, T_') w_ = 1+sympy.sin(2*np.pi * t_ / T_)**2 phi_ = sympy.cos(2*np.pi * t_ / T_)**2 + 1 chi_ = (t_ / T_)**2 # Differentiate and make functions get_w, get_phi, get_chi = [ sympy.lambdify([t_, T_], sympy.Array([_x, _x.diff(t_)]), "numpy") for _x in (w_, phi_, chi_)] def b(t, T=T): phi, dphi = get_phi(t, T) chi, dchi = get_chi(t, T) w, dw = get_w(t, T) beta = dchi / np.tanh(w) return [ beta * np.cos(phi), beta * np.sin(phi), dw / 2 /np.cosh(w) - dchi/np.tan(2*chi)/np.sinh(w) + dphi/2, ] def get_H(t): return np.einsum('i,iab->ab', b(t), sigmas) def rhs(t, psi): dpsi = get_H(t) @ psi / 1j / hbar return dpsi psi0 = np.array([1, 0j]) res = solve_ivp(rhs, t_span=(1e-10,T), y0=psi0) t = res.t res.y.shape plt.plot(t, abs(res.y.T)) w, dw = get_w(t, T) chi, dchi = get_chi(t, T) phi, dphi = get_phi(t, T) beta = dchi/np.tanh(w) plt.plot(t, abs(np.cos(chi)), ':') plt.plot(t, abs(np.sin(chi)), ':') ``` ## The Landau-Zener Formula :::{margin} These energies $E_n(\lambda)$ are sometimes called energy "bands", especially if the parameter $\lambda = k$ is a (quasi-)momentum in some electronic material. ::: The general idea is to consider the eigenstates of a Hamiltonian $\op{H}(\lambda)\ket{n(\lambda)} = \ket{n(\lambda)}E_n(\lambda)$ that depends on some parameter $\lambda(t)$ which varies in time. As two bands cross, say $E_0(\lambda_*) \approx E_1(\lambda_*)$, the [Landau-Zener formula] gives the transition probability as \begin{gather*} P \approx e^{-2\pi \Gamma}, \qquad \Gamma = \frac{\norm{\braket{0|\op{H}|1}}^2} {\hbar\left\lvert\pdiff{(E_0 - E_1)}{t}\right\rvert} \end{gather*} where everything is evaluated at the transition $\lambda(t) = \lambda_*$ when the levels approach. *(This formula is exact if several assumptions are made: see [Landau-Zener formula] for a discussion.)* :::{error} The numerator is not correct! ::: Note that the transition probability can be suppressed (i.e. large $\Gamma$) by: * Changing the system slowly: This is the [adiabatic theorem] that ensures that a system will remain in its instantaneous eigenstate if varied slowly enough. ## Alternative Formulation: Induced Transition An alternate formulation is to consider a time-dependent Hamiltonian of the form: \begin{gather*} \op{H}(t) = \op{A} + \lambda(t)\op{B}, \qquad \lambda(t\rightarrow \pm \infty) = 0. \end{gather*} The question is: what is the probability of a state remaining in an specified eigenstate $\ket{0}$ of $\op{A}$ (typically the ground state, hence our notation) far in the future? I.e., if we start the system in the state $\ket{\psi} = \ket{0}$ at time $t=-\infty$, what is: :::{margin} Convince yourself that $P(\infty)$ is well defined. Hint: how do we know that once $\lambda(t) \rightarrow 0$, $P(t)$ does not oscillate? ::: \begin{gather*} P(\infty) = \lim_{t\rightarrow \infty} \overbrace{\norm{\braket{0|\psi(t)}}^2}^{P(t)}. \end{gather*} Here we will consider the two-state problem where $\op{A}$ has two eigenstates $\ket{0}$ and $\ket{1}$ with energies $E_0=-E$ and $E_1=E$ respectively, and the time-dependence is expressed through a coupling term: \begin{gather*} \op{H}(t) = \ket{1}E\bra{1} - \ket{0}E\bra{0} + \lambda(t)\Delta\frac{\ket{0}\bra{1} + \ket{1}\bra{0}}{2}. \end{gather*} Expressed in the $\{\ket{0}, \ket{1}\}$ basis, the Hamltonian has the following matrix elements: \begin{gather*} \mat{H}(t) = \begin{pmatrix} -E & \lambda(t)\frac{\Delta}{2}\\ \lambda(t)\frac{\Delta}{2} & E \end{pmatrix} = \tfrac{\Delta}{2}\lambda(t)\mat{\sigma}_x - E\mat{\sigma}_z. \end{gather*} Appealing to our previous formulation, this is implemented by magnetic field: \begin{gather*} \vec{u} \propto \begin{pmatrix} \cos(χ)\\ -\I\sin(χ) e^{\I(φ + \phi)} \end{pmatrix}, \qquad \beta = \frac{\dot{χ}}{\cos\phi},\\ \vec{b} = \begin{pmatrix} β\cos φ \\ β\sin φ \\ -β\sin \phi \cot(2χ) + \frac{\dot{φ} - \dot{\phi}}{2} \end{pmatrix} = \begin{pmatrix} \lambda(t)\tfrac{\Delta}{2\hbar}\\ 0\\ E/\hbar \end{pmatrix}. \end{gather*} Thus, we are free to choose functions $\chi(t)$ and $\phi(t)$ such that \begin{gather*} \chi(-\infty) = 0, \qquad \dot{\chi}(\infty) = 0, \qquad \chi(\infty) = \sin^{-1}\sqrt{P(\infty)}. \end{gather*} Hence, \begin{gather*} \varphi(t) = 0, \qquad \beta(t) = \lambda(t)\tfrac{\Delta}{2\hbar},\\ -\lambda(t)\tfrac{\Delta}{2}\sin \phi \cot(2χ) - \frac{\dot{\phi}}{2} = E/\hbar. \end{gather*} Suppose we would like the transition ## Analytic Solutions The general Landau-Zener problem is not analytically solvable, but we can use the results from {cite:p}`Barnes:2013`: \begin{gather*} \vec{u} \propto\begin{pmatrix} \cos(χ)\\ -\I\sin(χ) e^{\I(φ + \phi)} \end{pmatrix}, \\ \vec{b} = \begin{pmatrix} β\cos φ\\ β\sin φ\\ -β\sin \phi \cot(2χ) + \frac{\dot{φ} - \dot{\phi}}{2} \end{pmatrix} = \begin{pmatrix} \tfrac{\Delta}{2\hbar}\\ 0\\ \frac{E}{\hbar}\lambda(t) \end{pmatrix}, \end{gather*} hence \begin{gather*} \varphi(t) = 0, \qquad \beta(t) = \tfrac{\Delta}{2\hbar} = \frac{\dot{χ}(t)}{\cos\phi(t)}, \\ \phi(t) = \cos^{-1}\Bigl(\frac{2\hbar\dot{χ}(t)}{\Delta}\Bigr),\\ \tfrac{E}{\hbar}\lambda(t) = -\tfrac{\Delta}{2\hbar}\sin \phi(t) \cot\bigl(2χ(t)\bigr) - \frac{\dot{\phi}(t)}{2} \end{gather*} Note that if we start in state $\ket{0}$ at time $t=0$ with $\chi(0)=0$, then the transition probability at time $t$ is $P_1(t) = \sin^2\chi(t)$. How fast can we effect such a transition? Well, we must keep $\abs{\cos(\phi)} \leq 1$, so we have the so-called [quantum speed limit]: \begin{gather*} \abs{\dot{\chi}} \leq \frac{\abs{\Delta}}{2\hbar}. \end{gather*} The fastest transition can be implemented with: \begin{gather*} \chi(t) = \frac{\Delta}{2\hbar}t, \qquad \phi(t) = 0, \qquad \lambda(t) = 0, \end{gather*} effecting a complete transition in time $t = 2\hbar/\abs{\Delta}$. This should make intuitive sense: we just let the magnetic field $b_x$ effect the rotation. Any interference from $b_z$ will rotate the spin towards $b_x$, reducing its efficiency. For the Landau-Zener problem we have \begin{gather*} \vec{b} = \begin{pmatrix} \frac{\Delta}{2}\\ 0\\ E\lambda(t) \end{pmatrix}, \qquad \varphi = 0, \qquad \beta = \frac{\Delta}{2}, \\ E\lambda(t) = \frac{\hbar^2\ddot{\theta}(t)}{\sqrt{\Delta^2-4\hbar^2\dot{\theta}^2(t)}} -\frac{\cot 2\theta(t)}{2}\sqrt{\Delta^2 - 4\hbar^2\dot{\theta}^2(t)}. \end{gather*} Note that if we start in state $\ket{1}$ at time $t=0$, then the transition probability at time $t$ is: \begin{gather*} P(t) = \abs{s(t)}^2 = \sin^2 \theta(t). \end{gather*} We, if we start with $\theta(0) = 0$, we can thus effect a perfect conversion by taking $\theta(t) = \pi/2$. Note, however, that we must keep \begin{gather*} \abs{\dot\theta} < \frac{\abs{\Delta}}{2\hbar}, \qquad t > \frac{\pi \hbar}{\abs{\Delta}}. \end{gather*} This is sometimes called the [quantum speed limit]. :::{admonition} Alternative Formulation In the alternative formulation, we have \begin{gather*} \vec{b} = \begin{pmatrix} \frac{\Delta}{2}\lambda(t)\\ 0\\ E \end{pmatrix}, \qquad \varphi = 0, \qquad \beta = \frac{\Delta}{2}\lambda(t), \\ \diff{}{t}\ln \lambda(t) = \frac{ \hbar^2\ddot{\theta}(t) - E\sqrt{\Delta^2-4\hbar^2\dot{\theta}^2(t)} + \frac{\cot 2\theta(t)}{2}(\Delta^2 - 4\hbar^2\dot{\theta}^2(t))}{\dot{\theta}}. \end{gather*} If we let $\epsilon(t) = \sqrt{\Delta^2-4\hbar^2\dot{\theta}^2(t)}$, then we have \begin{gather*} \dot{\theta}(t) = \frac{\sqrt{\Delta^2 - \epsilon^2(t)}}{2\hbar},\\ \ddot{\theta}(t) = \frac{-\epsilon(t)\dot{\epsilon}(t)}{2\hbar\sqrt{\Delta^2 - \epsilon^2(t)}},\\ \frac{\ddot{\theta}(t)}{\dot{\theta}(t)} = \frac{-\epsilon(t)\dot{\epsilon}(t)}{\Delta^2 - \epsilon^2(t)},\\ \diff{}{t}\ln \lambda(t) = \frac{ \hbar^2\ddot{\theta}(t) - E\epsilon(t) + \frac{\cot 2\theta(t)}{2}\epsilon^2(t)}{\dot{\theta}}. \end{gather*} ::: The idea is as follows. Consider a time-independent Hamiltonian $\op{H}_0$ with two eigenstates $\ket{0}$ and $\ket{1}$ with energies $E_0$ and $E_1$ respectively. To this, we add a time-dependent piece which mixes these: The model considers the dynamics of the following time-dependent Hamiltonian that couples two states as expressed in the $\op{S}_z$ basis $\{\sa, \sb\}$: :::{margin} In the notation of [Barnes:2012](http://dx.doi.org/10.1103/PhysRevA.88.013818), $\beta = \Delta/2$, $\alpha = E \omega t^2$. ::: \begin{gather*} \op{H} = \begin{pmatrix} E \omega t & \frac{\Delta}{2}\\ \frac{\Delta}{2} & -E \omega t \end{pmatrix} = \frac{\Delta}{2}\mat{\sigma}_x + E \omega t\mat{\sigma}_z. \end{gather*} The question is: if we start in the state $\sb$ at time $t \ll 0$, what is the probability that the system will eventually transition to the state $\sa$ far in the future $t\gg 0$? ```{code-cell} ipython3 from scipy.integrate import solve_ivp from functools import partial def get_H(t, w, delta, E=1.0): return np.array([ [E*w*t, delta/2], [delta/2, -E*w*t]]) wts = np.linspace(-3, 3) hbar = 1.0 w = 1.0 delta = 1.0 E = 1.0 Es = [np.linalg.eigvalsh(get_H(_t, w=w, delta=delta, E=E)) for _t in wts/w] fig, ax = plt.subplots() ax.plot(wts, Es) ax.set(xlabel=r"$\omega t$", ylabel="$E_n/E$"); def dpsi_dt(t, psi, w): Hpsi = get_H(t, w, delta=delta, E=E) @ psi return Hpsi / (2j * hbar) wT = 10.0 psi0 = np.array([1, 0]) + 0j ws = [0.1, 0.5, 1.0, 5.0, 10.0] for w in ws: res = solve_ivp(partial(dpsi_dt, w=w), t_span=[-wT/w, wT/w], y0=psi0, t_eval=wts/w) Es = np.array( [(psi.T.conj() @ get_H(t, w=w, delta=delta, E=E) @ psi).real for (t, psi) in zip(res.t, res.y.T)]) ax.plot(res.t*w, Es/E, ls="--", label="$\hbar\omega={:.1f}E$".format(hbar*w/E)) ax.legend(); ``` Our expectation is that, if the gap $\Delta$ is large, and the rate of change is slow (large $\tau$), then we should remain in the ground state, otherwise we will transition. ## Solution We must solve the following differential equation \begin{gather*} \I\hbar \ket{\dot{\psi}(t)} = \mat{H}(t)\ket{\psi(t)}, \qquad \ket{\psi(t\rightarrow - \infty)} = \begin{pmatrix} 1\\ 0\end{pmatrix} = \sb. \end{gather*} ## References * {cite:p}`Barnes:2013`: An analytic reformulation of the problem that gives formulae for the magnetic field $\vect{B}(t)$ required to effect transitions in the two-state system. * {cite:p}`Jaffe:2010`: A clean derivation of the classical Landau-Zener result using a semiclassical formalism that derives reflection probabilities using the duality between momentum and position. [Landau-Zener formula]: [adiabatic theorem]: [quantum speed limit]: [adiabatic quantum computing]: [spin magnetic dipole moment]: [time ordering]: