The curious properties of spin

Similar documents
The Spin (continued). February 8, 2012

Welcome back to PHYS 3305

Welcome back to PHYS 3305

Lecture 18: 3D Review, Examples

Magnetic Moments and Spin

(n, l, m l ) 3/2/2016. Quantum Numbers (QN) Plots of Energy Level. Roadmap for Exploring Hydrogen Atom

Atkins & de Paula: Atkins Physical Chemistry 9e Checklist of key ideas. Chapter 8: Quantum Theory: Techniques and Applications

4/21/2010. Schrödinger Equation For Hydrogen Atom. Spherical Coordinates CHAPTER 8

L z L L. Think of it as also affecting the angle

Final Exam Tuesday, May 8, 2012 Starting at 8:30 a.m., Hoyt Hall Duration: 2h 30m

Angular Momentum Quantization: Physical Manifestations and Chemical Consequences

2.4. Quantum Mechanical description of hydrogen atom

i = cos 2 0i + ei sin 2 1i

Assignment 2 Solutions. 1. The general state of a spin half particle with spin component S n = S ˆn = 1 2 can be shown to be given by

The experiment consists of studying the deflection of a beam of neutral ground state paramagnetic atoms (silver) in inhomogeneous magnetic field:

The Hydrogen Atom. Dr. Sabry El-Taher 1. e 4. U U r

1.6. Quantum mechanical description of the hydrogen atom

The Postulates of Quantum Mechanics Common operators in QM: Potential Energy. Often depends on position operator: Kinetic Energy 1-D case: 3-D case

The Stern-Gerlach experiment and spin

The Hydrogen Atom. Thornton and Rex, Ch. 7

Atomic Structure. Chapter 8

CHAPTER 7 FORCE ON A CURRENT IN A MAGNETIC FIELD

Properties of Elementary Particles

1 Recall what is Spin

Let us go back to what you knew in high school, or even earlier...

The 3 dimensional Schrödinger Equation

20 The Hydrogen Atom. Ze2 r R (20.1) H( r, R) = h2 2m 2 r h2 2M 2 R

Physics 221A Fall 1996 Notes 13 Spins in Magnetic Fields

St Hugh s 2 nd Year: Quantum Mechanics II. Reading. Topics. The following sources are recommended for this tutorial:

Stern-Gerlach Experiment and Spin

Homework assignment 3: due Thursday, 10/26/2017

Lecture 9. Angular momentum - 2

Uncertainty principle

Vibrational motion. Harmonic oscillator ( 諧諧諧 ) - A particle undergoes harmonic motion. Parabolic ( 拋物線 ) (8.21) d 2 (8.23)

How the Result of a Measurement of a Component of the Spin of a Spin- 1 2 Particle Can Turn Out to be 100

H atom solution. 1 Introduction 2. 2 Coordinate system 2. 3 Variable separation 4

Stern-Gerlach experiment

Rotational & Rigid-Body Mechanics. Lectures 3+4

NYS Master Teacher Program March 2016

C/CS/Phys C191 Particle-in-a-box, Spin 10/02/08 Fall 2008 Lecture 11

CHAPTER 8 The Quantum Theory of Motion

Quantum Field Theory. Chapter Introduction. 8.2 The Many Particle State

1 1D Schrödinger equation: Particle in an infinite box

1 1D Schrödinger equation: Particle in an infinite box

Lecture3 (and part of lecture 4).

PH425 Spins Homework 5 Due 4 pm. particles is prepared in the state: + + i 3 13

Handout 7: Torque, angular momentum, rotational kinetic energy and rolling motion. Torque and angular momentum

Chm 331 Fall 2015, Exercise Set 4 NMR Review Problems

Announcements. Lecture 20 Chapter. 7 QM in 3-dims & Hydrogen Atom. The Radial Part of Schrodinger Equation for Hydrogen Atom

Rotational Kinetic Energy

ECE440 Nanoelectronics. Lecture 07 Atomic Orbitals

Quantum Physics III (8.06) Spring 2007 FINAL EXAMINATION Monday May 21, 9:00 am You have 3 hours.

Lecture II: Rigid-Body Physics

Solutions to chapter 4 problems

Introduction to Quantum Mechanics Prof. Manoj Kumar Harbola Department of Physics Indian Institute of Technology, Kanpur

Reference Texts. Principles of Quantum Mechanics R Shanker Modern Quantum Mechanics J.J. Sakurai

= o + t = ot + ½ t 2 = o + 2

Angular momentum and spin

PHYSICS 250 May 4, Final Exam - Solutions

STERN-GERLACH EXPERIMENTS

Problems with the atomic model?

Experiment 08: Physical Pendulum. 8.01t Nov 10, 2004

Lecture 10 - Moment of Inertia

PAPER No. : 8 (PHYSICAL SPECTROSCOPY) MODULE No. : 8 (ALKALI METAL SPECTRA)

Rotation. Rotational Variables

14. Rotational Kinematics and Moment of Inertia

16. Elastic Scattering Michael Fowler

Modern Physics for Scientists and Engineers International Edition, 4th Edition

4E : The Quantum Universe. Lecture 28, May 19 Vivek Sharma

Electric fields in matter

Time part of the equation can be separated by substituting independent equation

Handout 6: Rotational motion and moment of inertia. Angular velocity and angular acceleration

Integral Theorems. September 14, We begin by recalling the Fundamental Theorem of Calculus, that the integral is the inverse of the derivative,

TRANSFER OF RADIATION

Physics 221A Fall 1996 Notes 12 Orbital Angular Momentum and Spherical Harmonics

Let b be the distance of closest approach between the trajectory of the center of the moving ball and the center of the stationary one.

d 1 µ 2 Θ = 0. (4.1) consider first the case of m = 0 where there is no azimuthal dependence on the angle φ.

Spinning Electrons as Physics Fantasy

Physics of Rotation. Physics 109, Introduction To Physics Fall 2017

Chapter 11. Angular Momentum

Particle collisions and decays

(a) What is the magnitude of the electric force between the proton and the electron?

Magnetism, Angular Momentum, and Spin

Modern Physics. Unit 6: Hydrogen Atom - Radiation Lecture 6.3: Vector Model of Angular Momentum

Maxwell s equations for electrostatics

Total Angular Momentum for Hydrogen

C/CS/Phy191 Problem Set 6 Solutions 3/23/05

Stern-Gerlach Simulations

Lecture 4 (Sep. 18, 2017)

G : Quantum Mechanics II

Lecture 4 Quantum mechanics in more than one-dimension

Rotation of Rigid Objects

Physics 228 Today: Ch 41: 1-3: 3D quantum mechanics, hydrogen atom

Lecture 4 Quantum mechanics in more than one-dimension

5.5. Representations. Phys520.nb Definition N is called the dimensions of the representations The trivial presentation

Applied Statistical Mechanics Lecture Note - 3 Quantum Mechanics Applications and Atomic Structures

Grasp the New Image of the Spin Angular Momentum

Symmetries 2 - Rotations in Space

Does Quantum Measurement Violate the Conservation of Angular Momentum? A Paradox Resolved Rick Bradford, 6/12/16

APPENDIX E SPIN AND POLARIZATION

Transcription:

The curious properties of spin February, 07 The Stern-Gerlach experiment The Schrödinger equation predicts degenerate energy levels for atoms electron states that differ only in the z-component of their angular momentum will have equal energies. Specifically, with the angular momentum quantized to integer multiples of the reduced Planck constant, l, quantization still allows l + different allowed orientations of the direction of the angular momentum vector. In an attempt to verify this prediction experimentally, Stern and Gerlach performed a series of experiments through the 90s. By passing atoms through an inhomogeneous magnetic field, they could separate the degenerate states of an atomic beam into an odd number, l + ), of separate beams. Surprisingly, some of these experiments produced an even number of beams. Uhlenbeck and Goudsmit interpreted this as the presence of intrinsic spin, of magnitude l =, of the electron. The half-integer value, l = gives l + = -fold degeneracy.. The classical prediction.. Magnetic moment of a spinning charged particle The magnetic moment of a distribution of currents is given by integrating dµ = r j where j is the current density. If j is given by a moving charge density, j = ρ q v Then we may write this as dµ = ρ qr v Consider a rotating sphere of radius R, of uniform charge Q, and uniform mass M. Then so that ρ q = 3Q 4πR 3 = Q M ρ mass µ = ˆ ) Q r M ρ mass vd 3 x = Q M S where S = ρ mass r vd 3 x is the total angular momentum vector of the spinning ball. For a nonuniform distribution of charge and to include quantum field theory effects), we may include an additional factor g, expected to be of order one, µ = gq M S

Exercise: How large could g be? Suppose all the charge resided on a ring around the equator of the sphere instead of being distributed throughout the volume. Then the charge density would be ρ q = Q θ πr δ r R) δ π ) Show that, in terms of the angular momentum of a sphere, L = Iω = 5 MR ω, that µ = 5 Exercise: Suppose the intrinsic angular momentum of an electron,, is given by the classical form, 5 mr ω = Q M S where m is the electron mass, r the Compton radius of the electron, mc where c is the speed of light. Let ω = v r where v is the tangential velocity of the surface of the electron. What is v? If a classical particle with magnetic moment µ passes through a magnetic field B, it experiences a torque, τ = µ B. In addition, if the magnetic field is nonuniform, but with vanishing curl) there is a net translational force as well, arising from a potential, V = µ B. For the magnetic fields we discuss, the force may be written in the form F = µ B) Let the magnetic field have its gradient in the z-direction. This means that if the magnetic moment of the spinning, charged particle is directed parallel to B it is deflected upward; antialignment leads to a downward deflection. For a beam of such particles, we expect the directions to be randomly directed, leading to a distribution of deflections... The classical distribution of deflections For a uniform distribution of magnetic moments, µ = µˆn the unit vector ˆn is equally likely to point in any direction. The probability that ˆn points into any element of solid angle, dω = sin θdθdϕ is therefore P θ, ϕ) sin θdθdϕ = If we have a total of N particles, then the number dn in dω is dn = N sin θdθdϕ 4π sin θdθdϕ 4π Since µ B = µb cos θ is independent of θ, we integrate over all ϕ to get dn = N sin θdθ Let B z z) be constant. Then the impulse p z 0 = p = F t delivered to a particle with z-component db of magnetic moment µ z will be approximately µ z dbz z dz t = µ cos θ dz t where t is the short) time spent in the magnetic field. A particle with its magnetic moment oriented at an angle θ will be acquire an upward component of velocity of v = p z m = µ t m db z dz cos θ

so a change in the angle of the magnetic moment of dθ changes the z-component of the velocity by dv = µ t db z sin θdθ m dz dv = µ t db z dθ m dz sin θ Travelling for a time T, a particle with orientation θ strikes a distant screen a height vt above the centerline. The number of particles in an interval dθ is dn = N sin θdθ so the number of particles per unit distance dl = T dv along the screen is dn dl = dn dθ dθ dl = dn dθ dθ T dv = N sin θ T µ t db z m dz sin θ mn = T µ t dbz dz The distribution is therefore constant across the full interval, out to maximum displacements of L = ± µ t db z m dz T. Classically, we expect the the beam to become a uniform vertical smear on the screen.. The experimental result When the experiment is carried out on atoms such as silver or cesium, which have total angular momentum depending only on the spin of a single outer electron, the result is strikingly different from this. Just as the orbital angular momentum within atoms is quantized into discrete total angular momentum and discrete z-components given as integer multiples of, the electron spin is only ever measured to have one of two discrete values, ±. The Stern-Gerlach beam discribed above produces only two sharp spots, above and below the centerline, and corresponding to a magnetic moment of µ electron = ge m e S where the intrinsic spin angular momentum, or just spin, of the electron is S = measured experimentally to be g =.003930499±.5 0 ). and the g factor is Implications of sequential Stern-Gerlach measurements If the initial beam directed into a Stern-Gerlach device, we may easily check that the distribution is random by rotating the orientation of the inhomogeneous magnetic field. No matter what orientation we choose, the beam splits into two bright spots of equal intensity. There is no preferred orientation for a suitably prepared initial beam. We always get two resultant beams, one called spin up and the other, spin down, measured relative to the magnetic field gradient. However, if we follow one such division of the beam by a second splitting applied to only the spin up part, using a Stern-Gerlach device with the same orientation as the first, only a single, spin up beam results. It is as if the use of the first device has fixed the direction of the subsequent beams. If we rotate the second device say, orienting the field in the x-direction instead of the z-direction) we once again get two beams. The behavior is much like polarizing lenses. [http://physics.nist.gov/cgi-bin/cuu/value?eqae%7csearch_{f}or=electron+magnetic+moment] 3

Suppose we introduce two symbols, z; + = + z; = for spin up and spin down, respectively. The z tells us that that we have oriented the magnetic gradient in the z direction. In general, we may write ˆn; +, ˆn; when the field gradient is in the ˆn direction. If no direction is explicitly specified, we always mean the zdirection. We can correctly model the effect of a Stern-Gerlach device if we assume the initial beam is an equal mixture of these, + + ) and that choosing the up beam after the beam separates acts like a projection, + + ) + Think of the pair +, as basis vectors) in a -dimensional vector space, with the vector describing the original beam having components, in this basis. A second projection in the same orientation leaves the + state unchanged, with no down beam emerging. ) ) Introducing a vector norm, the magnitude of + + ) is + =. We will soon introduce additional notation for this, but for now we simply use what we know about vectors. Like the norm of the wave function, ψ ψ, it is the square of the amplitude that is measured. Since the state + + ) ) has amplitude for both the up and down states, we find probability = of measuring either of the two possible spins. Now suppose we split the + state using a second Stern-Gerlach device oriented in the x-direction. Then this weaker beam, all in the z; + direction, breaks equally into two new beams, up and down with respect to the x-direction, + [ ] x; + + x; ) Notice what has happened to the intensities. Since our spin up beam had half the amplitude of the original random beam, each of the x; + and x; beams has 4 the overall amplitude. We may think of the selection of the up beam as a measurement. Measuring the initial beam, we find it to be up half the time; measuring the resulting up beam, it is always up. Once we have measured the spin property, and determined it to be z; + = +, it remains so. The initial state + + ) is a mixture of up and down, essentially a list of what we might possibly measure. After measuring the beam to be all spin up, that list is reduced and we have only + left on the list of possible measurements. ) If we choose the down beam using our Stern-Gerlach device, we find its intensity is also = that of the original beam. We found that the pure up beam + may be split into two x-orientations, so we may write + = x; + + x; ) or simply + = x; + + x; ) In this way, we are treating the states as vectors, and the two possible different orientations as different pairs of basis vectors. Any state may be written as a linear combination of +,, or equally well as a linear combination of x; +, x;. 4

3 Spin vectors Now, we study the relationships between the various pairs of basis vectors more carefully. Contemplating the splitting of beams using Stern-Gerlach devices oriented in the x, y or z directions, we see that we may write a given spin state as a linear combination of any of three pairs, +, x; +, x; y; +, y; Furthermore, the six basis states here are mutually orthogonal, in the sense that a pure state in any of the three directions is found to split into equal measures of the other two. For example where we must have x; + = µ + + + µ + ) x; + = σ + y; + + σ y; ) µ + = µ = σ + = ν = This cannot be accomplished with real coefficients since only ± have norm. Since complex numbers have infinitely many numbers with norm, we take the spin space to be a complex two dimensional space. Replacing the constants µ ± with phases, the x basis may be written as x; + = e iϕ + + e iϕ ) x; = e iϕ + e iϕ ) where the relationship between the spin up and spin down coefficients follows from othogonality, x; + x; = 0. Quantum states include an arbitrary overall factor. The real part of this factor is used to normalize the state, leaving an arbitrary phase with no physical consequence. We therefore may multiply these states by e iϕ. Defining ϕ = ϕ ϕ, x; + = + + e iϕ ) x; = + e iϕ ) Inverting these equalities, we have + = x; + + x; ) = e iϕ x; + x; ) Now consider two ways or writing the spin up state along the y axis. Since it will split equally if measured along either the x- or the z-axis, and we may apply the same phase arguments as for the x-basis, we may write y; + = x; + + e iξ x; ) y; + = + + e iθ ) 5

for phases ξ and θ. Equating these up to a phase, then substituting for the x-basis terms, Equating the phases, e iζ + + e iθ ) = x; + + e iξ x; ) = + + e iϕ ) + e iξ + e iϕ )) = + e iξ e iζ = + eiξ e iζ e iθ = eiξ e iϕ + + ) eiξ e iϕ The first requires +eiξ to have norm, + e iξ + e iξ = + cos ξ + = cos ξ = 0 ξ = n + ) π With e iξ = e in+) π = ) n i, the second becomes + ) n i e iθ = )n i e iϕ e ± iπ 4 e iθ = e iπ 4 e iϕ θ = ϕ π with the upper sign holding for neven and the lower for nodd. Substituting into the expressions for the x- and y-basis, x; + = + + e iϕ ) x; = + e iϕ ) y; + = + ie iϕ ) y; = + ± ie iϕ ) The choices of nand ϕ correspond to a rotation about the z-axis and therefore does not affect the essential orthogonality relationships. Choosing the lower sign, and ϕ = 0 gives our choice of basis, x; + = + + ) x; = + ) 6

y; + = + + i ) y; = + i ) 7