Likewise, any operator, including the most generic Hamiltonian, can be written in this basis as H11 H

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1 Finite Dimensional systems/ilbert space Finite dimensional systems form an important sub-class of degrees of freedom in the physical world To begin with, they describe angular momenta with fixed modulus and particle spins More generally, any set of closely spaced levels in the quantum system weakly connected to the rest of the spectrum can be dealt with approximately as a finite-dimensional system Any pair of levels in the spectrum addressed by near-resonant radiation also forms an effective two level system Neutrino flavors, isospins, tunneling states in the double-well potential, etc etc are all example of exact or well isolated finite-dimensional systems For simplicity, I will focus here on the two-level system (the larger the dimension of space the more complex is the general solution but most principles remain essentially the same I would rank it as number two on the list of most important QM problems after harmonic oscillator It does not really matter whether we are thinking of the spin-/, tunneling states in the double well potential, or isospin The formal description is identical for all of them, though the language used may depend on the context To make the point I will try to deliberately avoid the magneticfield/spin-/ projection which is the most common one in textbooks So, assume that we are dealing with the two-dimensional ilbert space characterized by two basis states e and e of whatever origin, eg they may be eigenstates of some specific amiltonian Then any state in this basis can be represented as ( α ψ = = α e β + β e Likewise, any operator, including the most generic amiltonian, can be written in this basis as ( Before we start solving for eigenstates and eigenvalues of this amiltonian, let us simplify it a little bit and introduce better notations Since Ĥ is ermitian, we have = I will call this non-diagonal (in a given basis matrix element a mixing matrix element and denote it as e iϕ If = 0 then our basis and amiltonian eigenstates coincide because ( 0 0 ( 0 0 ( 0 ( 0 ( = = E 0 e ( 0 = = E e E ( E E / * E It make perfect sense to count energy from the middle point (E + E /, so in the rest I will consider it as energy zero This results in the following generic amiltonian ( ξ e iϕ e iϕ ( ξ

2 where ξ = (E E / is often called an energy bias The dynamics of the system is determined from conventional i ( ( t α β ψ = Ĥ ψ i ξ = e iϕ e iϕ ξ ( α β and the best way to solve it to find the eigenvalue basis of Ĥ from ( ( ξ E e iϕ α e iϕ = 0 ( ξ E β The eigenvalue Equation is found from the zero determinant condition ξ E e iϕ e iϕ ξ E = 0 E ξ = 0 (3 and produces two eigenenergies Substituting this back to Eq ( we find the eigenvectors α = E ± = ± ξ + ±ɛ (4 ɛ + ξ eiϕ β, α + = ɛ ξ eiϕ β + (5 To complete the job we have to normalize them using α + β = This is best obtained through the parameterization (which satisfies the normalization condition by construction ( ( ( ( α sin θ e iϕ α+ cos θ e iϕ = =, + = =, (6 cos θ sin θ β with the mixing angle θ = tan, with cos θ = ξ + ɛ Now time evolution of an arbitrary state is a piece of cake: β + + ξ ɛ, sin θ = ξ ɛ (7 ψ(t = + ψ(t = 0 e iɛt + + ψ(t = 0 e +iɛt, all one has to do is to project an initial state on eigenvectors and write down the sum for the final answer Let us do it for two initial states: ψ(t = 0 = e and ψ(t = 0 = e In the first case we find ψ(t = α + e iɛt ( α+ β + + α e iɛt ( α β = e iɛt ( α+ α +β + + e iɛt ( α α β This defines the probability amplitude of finding the system in state at time t if it was in state at t = 0: U = e iɛt α + + e iɛt α = e iɛt cos θ + e iɛt sin θ = cos(ɛt i ξ ɛ sin(ɛt The probability itself is P = cos (ɛt + ξ ɛ sin (ɛt ɛ sin (ɛt

3 Likewise, for ψ(t = 0 = e we find and U = e iɛt β + + e iɛt β = e iɛt sin θ + e iɛt cos θ = cos(ɛt + i ξ ɛ sin(ɛt P == cos (ɛt + ξ ɛ sin (ɛt ɛ sin (ɛt The probabilities P and P are complementary to P and P, respectively, because j=, P ji = for i =, We see that oscillations have the largest amplitude of unity and the smallest frequency [sin (-squared oscillates twice as fast as sin(] for the symmetric two-level system when ξ = 0 As ξ is increased, the amplitude of oscillations decreases while the frequency increases as ɛ, ie the particle probes the other state at a much faster rate but only a tiny bit P ( t 0 ( P ( t ( / t / t It is time for homework and physical examples Problem 3 Two level system thermodynamics Use amplitudes U ii to compute the partition function of the two-level system from Z = Tr e β i e βĥ i, (recall the relation between the statistics and evolution in imaginary time discussed in the pathintegralpdf part of notes Check that your answer agrees with the one obtained from the sum of Gibbs exponentials based on known spectrum i=, Problem 33 Rabi oscillations Consider a two level system with the following time-dependent amiltonian: ( ( ĤT LS + ˆV ξ 0 0 δ e iωt rad = δ e iωt, 0 where the first term represent the two level system with the energy splitting ξ, and the second one describes coupling to the radiation field which induces transitions between the levels and is oscillating with frequency ω 3

4 a Consider first the case of exact resonance ω = ξ, and determine the probability of return P [int: use the so-called rotating frame basis which formally accounts for seeking the solution in the form ( α(te iωt ψ(t = β(t You will notice that for α(t and β(t the problem is reduced to the time-independent two-level system described in the notes] b Your solution in part a is often referred to as resonant Rabi oscillations More generally, Rabi oscillations are present in any two-level system selected by the near resonant radiation, so consider now the case ξ ω and assume that δ/ ξ ω is arbitrary Use the same approach to the solution and formulate an effective two-level amiltonian for the system in rotated frame, ie acting on α(t and β(t coefficients so that i d ( ( α α = dt β eff β c What is the frequency and amplitude of Rabi oscillations in the general near-resonant case discussed in part b? The general approach to an arbitrary n-dimensional system is along the same lines The amiltonian operator is represented by the n n matrix with ij = ji in some basis To solve it one has to diagonalize this matrix, ie to find its eigenvalues and eigenvectors, Ĥ ψ k = E k ψ k The dynamics immediately follows from χ(t = n k= ψ k χ(t = 0 e iekt ψ k Diagonalizing large matrix quickly becomes an impossible task analytically, except in special cases Numerically, nowadays one can diagonalize generic matrices for n as large as n For the lowest energy eigenstates more refined techniques are available which can handle ilbert spaces with up to several billion levels! Problem 34 Three level system TLS Consider the following amiltonian for the three level system (representing particle in the threewell potential with weak tunneling between the nearest-neighbor wells and the middle well having its energy shifted upwards 0 0 U, 0 0 a Diagonalize this amiltonian, ie find the spectrum of eigenenergies for this system Do not solve for eigenfunctions b Consider the lowest two states in the limit of U (keep terms proportional to /U only Given that the original model is symmetric with respect to left and right wells, determine an effective two-level amiltonian which has exactly the same lowest energy eigenvalues ( h h Ĥ eff = h h 4

5 It is representing an effective model for tunneling between left and right wells through a virtual intermediate state You may not be able to determine the sign of h Take it as negative for now The ammonia molecule Ammonia has the chemical form N 3 If we consider a simple picture with the molecule spinning about its symmetry axis, as shown in the figure (Note: N 3 has a tetrahedral structure with the hydrogen atoms forming a plane and the nitrogen being either above or below then there are two possible locations for nitrogen atom relative to the direction of spin, as shown (if the molecule was not spinning, then the two positions would be indistinguishable up to rotating the molecule as a rigid body upside down z N Center of mass N Of course from symmetry the energies of these two configurations should be the same ie taking these as basis states we have = = E 0 As usual we will consider E 0 as an energy zero In quantum mechanics particles have exponentially small but non-zero probability amplitudes to tunnel through the potential barrier and thus covert state e to state e We can represent this in terms of = = Thus our amiltonian becomes ( 0 Ĥ eff = 0 which has two eigenstates with energies and eigenfunctions ( E =, = E + = +, + = ( That it, the eigenstates are the simple even/odd mixtures of basis states and The tunneling (or mixing matrix element is generically negative when tunneling occurs between the lowest energy states in the potential The level splitting E + E = can be readily seen my the microwave 5

6 spectroscopy by radiating the molecule (see the Rabi oscillations problem Next, one can apply an electric field to couple E ( E to the molecules dipole moment µ oriented along the molecule s ẑ-axis having opposite directions in states and If the field is applied along the ẑ-axis then it adds a bias to our two-level system ξ = µe The resulting spectrum as a function of electric field strength is shown in the figure In E the absence of tunneling one would have an exact crossing of levels in zero field Mixing between the levels leads to the so-called avoided crossing and ( E opens a gap between the ground and first excited states One can easily see that in the strong-field limit µe the eigenstates are very well represented by the original and states Indeed, the mixing angle in this limit is small θ / µe At the avoided crossing point the eigenstates are equal superpositions of the original states, and as the electric field is varied, the nature of the ground state continuously changes from to An interesting problem (not discussed in this course is to look at transition probabilities between the states when the amiltonian parameter ξ is changed at a finite rate from large negative to large positive values the so-called Majorana, or Landau-Zener, transition (the attachment of names depends on who set the problem first, or solved it semiclassically, or exactly K 0 K0 and neutrino oscillations There are many other examples which are accurately described by the two-level, or n-level, system One is the K 0 K0 system ere the relevant degrees of freedom are particle/antiparticle pairs, one having positive strangeness quantum number S = +, and the other having negative strangeness S = (we are not really concerned here what this word means : Because we are talking about particle/antiparticle pair the two have identical masses Due to weak interaction these prime particle states are actually mixed, ie they can convert one into another This leads to the description of the K 0 K0 system in the form ( m + M δ δ m + M In the real world then a state which starts as a K 0 becomes a K 0 later on then back K 0, etc One slight complication is that these states are unstable on a longer timescale and decay into pions (the decay is dominated by the two-pion decay of the symmetric superposition state This results in damped oscillations but they reamain clearly visible, as in the figure below, which shows the 6

7 probability for the particle which starts out at time t = 0 as K 0 to be detected at time t as K 0 P 0 0 K K P e t 0 (0 sec 3 t 0 (0 sec Neutrino oscillations are similar One of the mysteries was that the flux of electron-neutrinos coming from the sun is about a factor of three smaller then expected from very robust calculations, according to both the omestake gold mine experiment in South Dakota and to the Kamiokande experiment in Japan An explanation for this phenomenon is that neutrinos have a (small mass and that there are mixing terms between neutrinos of different kind Taking the basis states as electron-, ν e, and muon-neutrinos, ν µ, the amiltonian then reads It leads to the superposition eigenstates ( m δm δm m ν + = cos θ ν e + sin θ ν µ, ν = sin θ ν e + cos θ ν µ Now the low energy solar neutrinos can be detected in the state ν e since they can create an electron via the weak interaction, and detectors were designed for this purpose owever, the state ν µ is sterile since the energy is not large enough to allow a muon to be created [Note: m µ 0 m e ] Thus the explanation for the solar neutrino deficit is that most of them convert to ν µ state by the time they arrive to Earth In the figure you can see a model plot for conversion from ν e to ν µ state There many more examples from atomic physics, cavity QED, amorphous solids, magnetism, quantum information (qubits you name it The origin of two states is, of course, important in defining the matrix elements in the effective amiltonian, but the rest of the description is absolutely universal I will do spins-/ in magnetic field later in the course, but you probably will be bored 7

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