Physics 2203, Fall 2012 Modern Physics

Transcription

1 Physics 2203, Fall 2012 Modern Physics. Wednesday, Oct. 10 th, Finish examples from Ch. 8 Chapter 9 with examples Take home average 85, sum average 63. Announcements: Quiz on Friday on Ch. 8 or Ch. 9 Midterm grades posted early next week conservanve Submit a detailed outline of your paper by Monday, Oct. 29th

2 hvp:// index.php?id=1829 David Wineland (US), Serge Horoche (France)

3 hvps://lasers.llnl.gov/mulnmedia/virtual_tours/ NIF: The Crown Joule of Laser Science

4 This could be worse than the death of the Supercollider for American Science

5 The allowed magnitudes of the angular momentum are L= l ( l +1) in QM. They are l in the Bohr model. (a) Find the ratio of correct/bohr for l=1, 2, 3, 4, 10, and 100 (b) How large does l have to be to have only a 1% error? ( ) The ratio of correct Bohr = l l + 1 l ( ) (b) correct Bohr = l l + 1 l l 50 = 1.01 l Ratio What conclusion do you draw?

6 Hydrogen Atom

7 The picture we have developed for a H like atom fails when you apply a magnenc field. What was a single emission line in zero field become a mulnplet when a field is applied First an electron moving in a circular orbit has a magnenc moment. Second, the electron has an inherent magnenc moment called spin You can think about this in analogy with the total angular momentum of the earth around the sun. J = L + S L is the angular momentum around the sun rxp S is orbital angular momentum Iω

8 An electron moving in a circular orbit has a magnenc moment. If there is a magnenc field B there will be a torque. B If the current i flowing around a small plane loop of area A in a magnetic field B, there will be a torque τ τ = ia xb A is a vector with magnitude of the area, direction according to right hand rule with i But usual notation is µ = ia µ is magnetic moment of the loop. Work is done by torque moving through angle θ W = τ dθ = µb sinθdθ = µbcosθ + constant The potential U is - work

9 We can always define a zero for the potenital U = µbcosθ = µ B Lets calculate the classical µ The current is the charge divided by the period. i = e T : but v= 2πr T µ = ia = e v 2πr πr2 µ = ia = ev 2πr πr 2 = evr 2 But L=mvr µ L = e 2m Gyromagnetic Ratio µ = e 2m L Electrons charged neganve

10 µ = e 2m L µ B = e 2m e = 5.79x10 5 ev / T MagneNc moment of an orbinng charge Bohr Magneton We know that angular momentum will be quannzed figure µ z = e 2m e L z = e 2m e m l = µ B m l

11 τ = µ xb Classical this creates a torque that would align the moment with the B field. µ = e 2m L With quannzed angular moment, we get precession around the field direcnon: Larmor precession Look at the bovom of the figure: L sinθidφ = dl Giving: L sinθidφ = dl = τ dt = e LBsinθ dt 2m e ω L = dφ dt = 1 dl L sinθ dt = e 2m e B Larmor frequency Work done dw=-µbsinθdθ=d(µbcosθ)=d(µ ib ) U = µ ib look back.

12 τ = µ xb µ = e 2m L ω L = e 2m e B U = µ ib = e L ib = eb L z = ω L m l 2m e 2m e The Total Energy of the electron is energy without the field + U E = E 0 +U = µ ib = E 0 + ω L m l Example for l=2 What does this look like for n=1 system?

13 B = B 0 k ml=0

14 Δm l = 0, ± 1 Δl = ±1

15

16 Calculate the magnenc energy and Larmor frequency for an electron in the n=2 state of hydrogen, assuming the atom is in a magnenc field of strength B=1.00T. Take the z axis to be along B U = ω L m l ω L = e 2m e B=µ b = (5.79x10 5 ev / T )(1T ) ω L = 5.79x10 5 ev n = 2, l=0, m l = 0 no U n = 2, l=1, m l = ±1 no U=± ω L allowed Energies, - ω L,0,+ ω L Larmor frequency= ω L ω L = 8.80x rad / s

17 Red allowed Black Forbidden

18 Pieter Zeeman Dutch A magnenc field splits emission lines in the Rydberg series. Proves electrons are moving around the nucleus. Shared 1902 Nobel Prize with Lorentz To make the arguments simple we need a system where we can focus on the orbital monon of the electron without having to worry about the spin. To accomplish this we need a system with two electrons where the spins are opposite. This is called a singlet state. One option would be two electron in the 1s state of He, but that will not work because we need m l 0. What will work is having electrons in two different orbits of He but still as a singlet.

19 Pieter Zeeman Dutch Shared 1902 Nobel Prize with Lorentz It is easy to calculate the energy ΔE ΔE=-µ B e = 2m e L B Pick the z axis to be aligned with B L z = m e ΔE= 2m e L zb = e 2m e m lb µ B = e 2m e BohrMagnetron µ B = e 2m e = 9.27x10 24 A m 2 µ B = e 2m e = 5.79x10 5 ev / T

20 Pieter Zeeman Dutch µ B = e 2m e = 5.79x10 5 ev / T ΔE=mµ B B µ B = e 2m e BohrMagnetron Consider a He atom with two electrons, one is in the 1s state and the excited one is in the 2p state, singlet state. Shared 1902 Nobel Prize with Lorentz Lets see what happens in a 5 T magnenc field. The magnetic field spits the one line (B=0) into three for m l = 1,0,+1 Each line separated by ΔE=µ B B = 3.0x10 4 ev ΔE E = 1x10 5 ev

21 Pieter Zeeman Dutch µ B = e 2m e = 5.79x10 5 ev / T ΔE=mµ B B µ B = e 2m e BohrMagnetron Consider a He atom with two electrons, one is in the 1s state and the excited one is in the d state. Nothing happens to the 1s. Shared 1902 Nobel Prize with Lorentz Now what happens? How many lines will you see? Good Quiz quesnon!

22 Imagine a hydrogen atom in which the electron has no spin (only orbital moment counts). The atom is placed in a magnenc field of 1.5T along the z axis (a) Describe the effect of the B field on the 1s and 2p states of the hypothencal H atom. Sketch the energy levels. (b) When B=0, there is a single spectral line corresponding to the 2p to 1s transinon. How many lines does this become when B 0? (c) What is the fracnonal separanon between the adjacent lines. ΔE=µ B B ΔE=µ B B = 0.9x10 4 ev E(2 p 1s) = = 10.2eV ΔE E = 0.9x x10 5

23 Fact, The Normal Zeeman effect is frequently not observed. Instead you see four, six or even more unequally spaced lines in the emission spectra. This is known as the Anomalous Zeeman Effect. There is also another effect which doubles the number of lines call Fine structure. Everything is a result of the electron spinning around its axis and having a magnenc moment. Stern Gerlach experiment S z = m s m s = ±1 / 2

24 If we create a beam of atoms such that we are in the n=2, l=1 state there are 2l+1=3 different values of m l. If this beam is directed into a region with a constant magnenc field all that happens is the vector precesses around the field. But in an a nonuniform magnenc field there is a force and we should see three images of the slit as shown. 2(l+1) Stern and Gerlach carried out this experiment on Ag atoms where l=0 so they should have seen no splipng but they saw two lines, caused by the intrinsic spin of the electron OVo Stern Nobel Prize 1943

25 Spin Quantum Number s for the electron is ½! Stern Gerlach magnet experiment split the beam into to lines =(2s+1); Spin up spin down. According to quannzanon rules for angular momentum applied to spin. S z = m s m s = ± 1 2 By defininon the magnitude of the spin angular momentum is S = s(s +1) = 3 2 The angular momentum of rotanon cannot be changed in any way, but is an intrinsic property of the electron

26 Spin, as angular momentum and torque, is known as a pseudovector (or an axial vector), Opposed to a true or polar vector such as velocity. The property of a pseudovector is that its mirror image is equal in magnitude but flipped in direcnon because a pseudovector has a chirality. An ordered array of spin in a solid breaks the symmetry of the crystal. How can you see this? PHYSICAL REVIEW B 86, (R) (2012)

27 An LSU team, first authored by graduate student Guorong Li, has found a way to see such a broken symmetry. They reported in late August (Phys. Rev. B 86, (R) (2012)) that, in one of the parent compounds of new Fe based superconductors BaFe2As2, the surface dramancally enhances the spin lapce coupling so that dynamic annphase spin domains in the bulk are frozen out at the surface. This leads to the coexistence of structural and spin annphase domain walls. Due to the large spin electron lapce coupling at the surface, these domain walls have the broken mirror plane symmetry associated with spin, resulnng in a leu and right handed chirality, which can be described as a spin toroidal moment chirality (Fig. 1b).

28 While layer ruthenate materials are well known to display an array of excinng phenomena such as metal insulator transinons (MIT), spin orbital ordering, exonc superconducnvity, and quantum crincality, this neutron scavering work has revealed an unusual E type annferromagnenc (AFM) structure in Mn subsntuted Sr 3 Ru 2 O 7 (x = 0.16) [See Fig.]. They found that this layered ruthenate behaves as a quasi two dimensional (2D) annferromagnet with in plane (ab) long range ordering but only single bilayer (5 6Å) ferromagnenc correlanons along the c direcnon bellow T N = 78 K. Such Mn induced magnenc structure is unusual because the crincal behavior of the staggered magnenzanon (i.e. the AFM order parameter) does not reflect the expected behavior of a 2D magnenc phase transinon. PRB 85, (R) (2012)].

29 We will do many examples of this on Friday.

30 We now have four quantum numbers n = 1,2, 3,... l = 0,1,2,3,...,(n 1) m l = l,l 1,.,0,..., l m s = ± 1 2 What is the degeneracy for n=1? n = 1 l = 0 m l = 0 What is the degeneracy? n = 2, l = 0 and l=1 m l = 0 and 1,0,-1 m s = ± 1 2 Degeneracy is 2n 2 m s = ± 1 2

31 Make a table showing the values of the four quantum numbers, n, l, m, and m s for 10 lowest lying states of the H atom. n=1, we have l=0, m=0, m s =± 1 2 n=2, l=0, m=0, m s =± 1 2 n=2, l=1, m=1, m s =± 1 2 n=2, l=1, m=0, m s =± 1 2 n=2, l=1, m=-1, m s =± 1 2

32 Compute the ranon S/L of the magnitudes of the spin angular momentum to the orbital angular momentum for (a) an s electron, (b) a p electron, (c) a d electron, and (d) an f electron. ( ) ( ) S = s( s + 1) : L = l( l + 1) : S/L= s s + 1 l l + 1 l S / L When will this rano be 10%?