Physics

Physics · modern-physics

Problem

2. The atoms are initially prepared in the spin up states right after leaving the screen, where μz=γℏ=∣μx∣\mu_{z}=\gamma \hbar=\left|\mu_{x}\right|. This means the atoms will precess at rates covering a range of values Δω\Delta \omega with respect to the xx component of B2\boldsymbol{B}_{2}, specifically B2x=B0+CxB_{2 x}=B_{0}+C x. Prove that the spread in the precession angle Δωt\Delta \omega t is so large and hence we cannot measure both μx\mu_{x} and μz\mu_{z} simultaneously. In other words, the measurement of μx\mu_{x} destroys the information on μz\mu_{z}. Given constants: All matters in the universe have fundamental properties called spin, besides their mass and charge. Spin is an intrinsic form of angular momentum carried by particles. Despite the fact that quantum mechanics is needed for a full treatment of spin, we can still study the physics of spin using the usual classical formalism. In this problem, we are investigating the influence of magnetic field on spin using its classical analogue. The classical torque equation of spin is given by τ=dLdt=μ×B\boldsymbol{\tau}=\frac{d \boldsymbol{L}}{d t}=\boldsymbol{\mu} \times \boldsymbol{B} In this case, the angular momentum L\boldsymbol{L} represents the "intrinsic" spin of the particles, μ\boldsymbol{\mu} is the magnetic moment of the particles, and B\boldsymbol{B} is magnetic field. The spin of a particle is associated with a magnetic moment via the equation μ=−γL\boldsymbol{\mu}=-\gamma \boldsymbol{L} where γ\gamma is the gyromagnetic ratio. In this problem, the term "frequency" means angular frequency (rad/s), which is a scalar quantity. All bold letters represent vectors; otherwise they represent scalars. Part A. Larmor precession Context question: 1. Prove that the magnitude of magnetic moment μ\mu is always constant under the influence of a magnetic field B\boldsymbol{B}. For a special case of stationary (constant) magnetic field, also show that the angle between μ\boldsymbol{\mu} and B\boldsymbol{B} is constant. (Hint: You can use properties of vector products.) Context answer: 证明题 Context question: 2. A uniform magnetic field B\boldsymbol{B} exists and it makes an angle ϕ\phi with a particle's magnetic moment μ\boldsymbol{\mu}. Due to the torque by the magnetic field, the magnetic moment μ\boldsymbol{\mu} rotates around the field B\boldsymbol{B}, which is also known as Larmor precession. Determine the Larmor precession frequency ω0\omega_{0} of the magnetic moment with respect to B=B0k\boldsymbol{B}=B_{0} \boldsymbol{k}. Context answer: ω0=γB0\omega_{0}=\gamma B_{0} Extra Supplementary Reading Materials: Part B. Rotating frame In this section, we choose a rotating frame S′S^{\prime} as our frame of reference. The rotating frame S′=(x′,y′,z′)S^{\prime}=\left(x^{\prime}, y^{\prime}, z^{\prime}\right) rotates with an angular velocity ωk\omega \boldsymbol{k} as seen by an observer in the laboratory frame S=(x,y,z)S=(x, y, z), where the axes x′,y′,z′x^{\prime}, y^{\prime}, z^{\prime} intersect with x,y,zx, y, z at time t=0t=0. Any vector A=Axi+Ayj+Azk\boldsymbol{A}=A_{x} \boldsymbol{i}+A_{y} \boldsymbol{j}+A_{z} \boldsymbol{k} in a lab frame can be written as A=Ax′i′+Ay′j′+Az′k′\boldsymbol{A}=A_{x}{ }^{\prime} \boldsymbol{i}^{\prime}+A_{y}{ }^{\prime} \boldsymbol{j}^{\prime}+A_{z}{ }^{\prime} \boldsymbol{k}^{\prime} in the rotating frame S′S^{\prime}. The time derivative of the vector becomes dAdt=(dAx′dti′+dAy′dtj′+dAz′dtk′)+(Ax′di′dt+Ay′dj′dt+Az′dk′dt)\frac{d \boldsymbol{A}}{d t}=\left(\frac{d A_{x}{ }^{\prime}}{d t} \boldsymbol{i}^{\prime}+\frac{d A_{y}{ }^{\prime}}{d t} \boldsymbol{j}^{\prime}+\frac{d A_{z}{ }^{\prime}}{d t} \boldsymbol{k}^{\prime}\right)+\left(A_{x}{ }^{\prime} \frac{d \boldsymbol{i}^{\prime}}{d t}+A_{y}{ }^{\prime} \frac{d \boldsymbol{j}^{\prime}}{d t}+A_{z}{ }^{\prime} \frac{d \boldsymbol{k}^{\prime}}{d t}\right) (dAdt)lab=(dAdt)rot+(ωk×A)\left(\frac{d \boldsymbol{A}}{d t}\right)_{l a b}=\left(\frac{d \boldsymbol{A}}{d t}\right)_{r o t}+(\omega \mathbf{k} \times \boldsymbol{A}) where (dAdt)lab\left(\frac{d \boldsymbol{A}}{d t}\right)_{l a b} is the time derivative of vector A\boldsymbol{A} seen by an observer in the lab frame, and (dAdt)rot \left(\frac{d A}{d t}\right)_{\text {rot }} is the time derivative seen by an observer in the rotating frame. For all the following problems in this part, the answers are referred to the rotating frame S′S^{\prime}. Context question: 1. Show that the time evolution of the magnetic moment follows the equation (dμdt)rot=−γμ×Beff\left(\frac{d \boldsymbol{\mu}}{d t}\right)_{r o t}=-\gamma \boldsymbol{\mu} \times \boldsymbol{B}_{e f f} where Beff =B−ωγk′\boldsymbol{B}_{\text {eff }}=\boldsymbol{B}-\frac{\omega}{\gamma} \boldsymbol{k}^{\prime} is the effective magnetic field. Context answer: 证明题 Context question: 2. For B=B0k\boldsymbol{B}=B_{0} \boldsymbol{k}, what is the new precession frequency Δ\Delta in terms of ω0\omega_{0} and ω\omega ? Context answer: Δ=γB0−ω\Delta =\gamma B_{0}-\omega Context question: 3. Now, let us consider the case of a time-varying magnetic field. Besides a constant magnetic field, we also apply a rotating magnetic field b(t)=b(cos⁡ωti+sin⁡ωtj)\boldsymbol{b}(t)=b(\cos \omega t \boldsymbol{i}+\sin \omega t \boldsymbol{j}), so B=B0k+b(t)\boldsymbol{B}=B_{0} \boldsymbol{k}+\boldsymbol{b}(t). Show that the new Larmor precession frequency of the magnetic moment is Ω=γ(B0−ωγ)2+b2\Omega=\gamma \sqrt{\left(B_{0}-\frac{\omega}{\gamma}\right)^{2}+b^{2}} Context answer: 证明题 Context question: 4. Instead of applying the field b(t)=b(cos⁡ωti+sin⁡ωtj)\boldsymbol{b}(t)=b(\cos \omega t \boldsymbol{i}+\sin \omega t \boldsymbol{j}), now we apply b(t)=b(cos⁡ωti−sin⁡ωtj)\boldsymbol{b}(t)=b(\cos \omega t \boldsymbol{i}-\sin \omega t \boldsymbol{j}), which rotates in the opposite direction and hence B=B0k+b(cos⁡ωti−sin⁡ωtj)\boldsymbol{B}=B_{0} \boldsymbol{k}+b(\cos \omega t \boldsymbol{i}-\sin \omega t \boldsymbol{j}). What is the effective magnetic field Beff \boldsymbol{B}_{\text {eff }} for this case (in terms of the unit vectors i′,j′,k′\boldsymbol{i}^{\prime}, \boldsymbol{j}^{\prime}, \boldsymbol{k}^{\prime} )? What is its time average, Beff ‾\overline{\boldsymbol{B}_{\text {eff }}} (recall that cos⁡2πt/T‾=sin⁡2πt/T‾=0\overline{\cos 2 \pi t / T}=\overline{\sin 2 \pi t / T}=0 )? Context answer: Beff=(B0−ωγ)k′+b(cos⁡2ωti′−sin⁡2ωtj′)\mathbf{B}_{\mathrm{eff}}=\left(B_{0}-\frac{\omega}{\gamma}\right) \mathbf{k}^{\prime}+b\left(\cos 2 \omega t \mathbf{i}^{\prime}-\sin 2 \omega t \mathbf{j}^{\prime}\right) , Beff‾=(B0−ωγ)k′\overline{\mathbf{B}_{\mathrm{eff}}}=\left(B_{0}-\frac{\omega}{\gamma}\right) \mathbf{k}^{\prime} Extra Supplementary Reading Materials: Part C. Rabi oscillation For an ensemble of NN particles under the influence of a large magnetic field, the spin can have two quantum states: "up" and "down". Consequently, the total population of spin up N↑N_{\uparrow} and down N↓N_{\downarrow} obeys the equation N↑+N↓=NN_{\uparrow}+N_{\downarrow}=N The difference of spin up population and spin down population yields the macroscopic magnetization along the zz axis: M=(N↑−N↓)μ=Nμz.M=\left(N_{\uparrow}-N_{\downarrow}\right) \mu=N \mu_{z} . In a real experiment, two magnetic fields are usually applied, a large bias field B0kB_{0} \boldsymbol{k} and an oscillating field with amplitude 2b2 b perpendicular to the bias field (b≪B0)\left(b \ll B_{0}\right). Initially, only the large bias is applied, causing all the particles lie in the spin up states ( μ\boldsymbol{\mu} is oriented in the zz-direction at t=0t=0 ). Then, the oscillating field is turned on, where its frequency ω\omega is chosen to be in resonance with the Larmor precession frequency ω0\omega_{0}, i.e. ω=ω0\omega=\omega_{0}. In other words, the total field after time t=0t=0 is given by B(t)=B0k+2bcos⁡ω0ti.\boldsymbol{B}(t)=B_{0} \boldsymbol{k}+2 b \cos \omega_{0} t \boldsymbol{i} . Context question: 1. In the rotating frame S′S^{\prime}, show that the effective field can be approximated by Beff ≈bi′,\boldsymbol{B}_{\text {eff }} \approx b \boldsymbol{i}^{\prime}, which is commonly known as rotating wave approximation. What is the precession frequency Ω\Omega in frame S′S^{\prime} ? Context answer: Ω=γb\Omega=\gamma b Context question: 2. Determine the angle α\alpha that μ\boldsymbol{\mu} makes with Beff \boldsymbol{B}_{\text {eff }}. Also, prove that the magnetization varies with time as M(t)=Nμ(cos⁡Ωt).M(t)=N \mu(\cos \Omega t) . Context answer: 证明题 Context question: 3. Under the application of magnetic field described above, determine the fractional population of each spin up P↑=N↑/NP_{\uparrow}=N_{\uparrow} / N and spin down P↓=N↓/NP_{\downarrow}=N_{\downarrow} / N as a function of time. Plot P↑(t)P_{\uparrow}(t) and P↓(t)P_{\downarrow}(t) on the same graph vs. time tt. The alternating spin up and spin down population as a function of time is called Rabi oscillation. Context answer: P↓=sin⁡2Ωt2P_{\downarrow}=\sin ^{2} \frac{\Omega t}{2} , P↑=cos⁡2Ωt2P_{\uparrow}=\cos ^{2} \frac{\Omega t}{2} Extra Supplementary Reading Materials: Part D. Measurement incompatibility Spin is in fact a vector quantity; but due to its quantum properties, we cannot measure each of its components simultaneously (i.e. we can know both ∣μ∣|\boldsymbol{\mu}| and μz\mu_{z} as in above problems; but not all ∣μ∣,μx,μy|\boldsymbol{\mu}|, \mu_{x}, \mu_{y}, and μz\mu_{z} simultaneously). In this problem, we will do a calculation based on the Heisenberg uncertainty principle (using the relation ΔpqΔq≥ℏ\Delta p_{q} \Delta q \geq \hbar ) to show how these measurements are incompatible with each other. Context question: 1. Let us consider an oven source of silver atoms, which has a small opening. The atoms stream out of the opening along −y-y direction (see Figure below) and experience a spatial varying field B1\boldsymbol{B}_{1}. The field B1\boldsymbol{B}_{1} has strong bias field component in the zz direction, where the atoms with different magnetic moment μz=±γℏ\mu_{z}= \pm \gamma \hbar are split in the zz direction. At a distance DD from the oven source, a screen SC1S C_{1} is put to allow only spin up atoms to pass (blocking spin down atoms). Thus, at the instant after passing the screen, the atoms are prepared in spin up states. After the screen, the atoms enter a region of nonhomogenous field B2\boldsymbol{B}_{2} where the atoms feel a force Fx=μxCF_{x}=\mu_{x} C The field B2\boldsymbol{B}_{2} has strong bias field component in the xx direction, where the atoms have magnetic moment μx=±γℏ\mu_{x}= \pm \gamma \hbar. <image_1> In order to determine μx\mu_{x} by observing the splitting in xx direction, show that the following condition must be fulfilled: 1ℏ∣μx∣ΔxCt≫1\frac{1}{\hbar}\left|\mu_{x}\right| \Delta x C t \gg 1 where tt is the duration after leaving the screen SC1S C_{1} and Δx\Delta x is the opening width on SC1S C_{1}. Context answer: 证明题

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