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# INTR MODRN PHYSCS I PHYS 321

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This 31 page Class Notes was uploaded by Maverick Harris on Sunday September 27, 2015. The Class Notes belongs to PHYS 321 at Iowa State University taught by Staff in Fall. Since its upload, it has received 10 views. For similar materials see /class/214478/phys-321-iowa-state-university in Physics 2 at Iowa State University.

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Date Created: 09/27/15

Chapter 8 Outline Magnetic Dipole Moments Spin and Transition Rates Relation between magnetic dipole moment and angular momentum Orbital magnetic dipole moments Bohr magneton orbital g factor Larmor precession magnetic dipole moments in a uniform magnetic field effects of a nonuniform field The StemGerlach experiment and electron spin comparison with Schrodinger predictions Phipps Taylor experiment spin quantum numbers s and ms Zeeman effect Spin Orbit interaction Thomas precession Total Angular Momentum coupling between orbital and spin angular momenta conditions satisfied by the quantum numbers j and mj Spin Orbit interaction energy and the hydrogen energy levels Lamb shift hyperfine structure Hamiltonian operator Transition rates and selection rules electric dipole matrix element QED picture of stimulated and spontaneous emission symmetries HW 8 Chapter 8 due 1118 questions 10 19 problems 2 3 5 6 10 16 17 18 We ll begin with a discussion of experiments which measure the orbital angular momentum L However we don t actually measure L what is the thing we actually measure 7 The magnetic dipole moment ul Shortly we ll develop a relation between L and ul Our methods will use a combination of simple EampM pseudoclassical approaches such as the Bohr model and QM This procedure is justified in that the results agree with a full QM treatment which is beyond the scope of this course Orbital magnetic dipole moments Consider an electron of mass m and charge e moving with velocity v in a circular Bohr orbit of radius r 2757 From a magnetic point of view what does a current loop look like if you are a large distance away from it 7 6 We have charge c1rculat1ng in a loop what is the current 7 z a dipole For a current i in a loop of area A what is the magnetic dipole moment 7 anyone remember u 1A or in our case ul 1A The quantity ul specifies the strength of the magnetic dipole it equals the product of the poles strengths times their separation What is the angular momentum of our electron 7 L mvr ls L parallel or antiparallel to ul 7 Since 6 is negative its magnetic dipole moment Hz is antiparallel to L Now putting our expression for i into our expression for M we nd ev 2 evr ul 1A 7m 7 2m 2 Combining this with our expression for L we see that which is usually written as 1 giftquot L Where m 1 0927x1039 anlp m2 Bohr Magneton m and gl l is called the orbital g factor later on we ll have cases where g i 1 8141 Z h Writing our expression for Ml as a vector equation we have 71 By repeating this calculation for an elliptical orbit it can be shown that ulL is independent of the shape of the orbit Why Also note that ulL is independent of the details of the orbit 7 suggests that it might not even depend on the details of the mechanical theory used to evaluate it Indeed it can also be shown that this relation is quantum mechanically valid so 1 g17 7Lg17 Azltzlgth gnawKPH gzbL gzb h z 72 We now fall back to classical EampM and ask what happens to a dipole in a magnetic field 7 and zz mlh 2 31 me It experiences a torque How do we express this mathematically 7 f l X B What does this do physically 7 Tends to align the dipole with the field What about potential energy for a given orientation between 1 and 3 what is the potential energy 7 AE 17 o E Example 81 Calculate the energy required to turn a moment 1 uh from parallel to antiparallel to a field B 1 T AE Ealp EP AuB chos7r lecos0 2MB 2x0927gtlt103923gtlt1 J 116gtlt10394 eV This AE is very small AEk 134 K but it is the basis for electron paramagnetic resonance EPR also known as electron spin resonance ESR It is 2000 times larger than the corresponding AE s of nuclear magnetic moments which are the basis for nuclear magnetic resonance NMR Consider the case where there is no process by which a magnetic moment y in a magnetic field B can dissipate its energy what do we expect will happen 7 What happens to 6 7 t9 cos391u39BuB constant And what does ul do 7 ul precesses about B since I uxB is perpendicular to L just like the precession of a spinning top which is due to the torque created by the gravitational force So what s the frequency of precession 7 hintl f analog of Newton39slaw F m5 m A R d dt dt K hint2 f7gtltB7 M LxB V 7 dt a W hint 3 From the figure what is dL equal to 7 dL Lsin60dl A T hjm q stin6dirgl bwsin6 A r dt h so a 571le or as a vector Lmour h Frequency QM treatment gives same result Gout in terms of expectation values B I For the case where B is uniform is there a net translational force acting on u 7 No the e39 simply executes a circular motion about an axis II B Suppose however that B is nonuniform converging is there now a net translational force 7 Yes upward Clarification classical dipole in a field I Consider a simple dipole in a field Since AU rill OB We have the lowest energy when m amp B line up And recalling that in general 17 i U shouldn39t there be a force to align m amp B 7 Wht39 V39 h 39al di te a 543 17 7i a s in sp enc coor na s er rag Hinge Hmmm sure look like there39s a force to align m amp B However 7 What do we expect if our magnet is spinning Conservation of angular momentum won t let it simply line up with the eld Instead we get precession KL DD Looking at this another way let s consider the force acting on the poles of a magnetic dipole In analogy with the behavior of an electric dipole in a converging FN electric field what can we say about the the force on the poles 7 N BE 5 FN gt F3 upward In generalcan show that Fz T BBL pg 2 z where z is in the direction of increasing field strength Conclusion a magnetic dipole in a nonuniform magnetic field experiences a torque causing precession and a force causing displacement in the direction of increasing B The Stern Gerlach ex eriment and electron s in 1922 Stem amp Gerlach sent a monoenergetic beam of Ag atoms through a nonuniform field B Since Ag atoms are neutral the only force acting on them will be F z BBdz llz Classically what would we expect 7 Classically we expect that bk can have any Value from yl to 1 and Fz accordingly Quantumrmechanically what do we expect 7 Quantumrmechanically u rglybml where ml 71 71 1 71 2 If 1 1 Hence the positions of the Ag atoms on the screen will be discrete And what did the experiment show 7 Qualitatively the same for other z orientations and for 39fferent atoms ie two or more discrete bands What39s wrong with this picture 7 Observed Classmtu predicted Possible Values of u is 21 1 ie an odd number 7 not 2 Hence the Schrodinger description must be wrong or incomplete 1927 Phipps amp Taylor using the Steranerlach technique observed two lines produced by a beam of H atoms where 1 0 in the ground state simple single electron atom Who hasn t guessed the solution yet 7 We assume that the K has an intrinsic spin magnetic moment uJ due to an intrinsic angular momentum S 7 to first order we can think of this as the electron producing the external magnetic fieldofa 39d39r d L 39 dwith39 39 39 L We further assume S ss1h 7 git 5 where gJ is called the spin g factor 5 mi 5 17th Based on the Phipps amp Taylor experiment how many Values can us assume 7 two plit 39 t t p t what can we say about these And 39 W0 Values 0f H5 7 They are equal in magnitude and opposite in sign If we make the nal assumption that the possible Values of m should differ by one and range from S to s what can we conclude about two possible Values of m 7 m 712 12 and we can also conclude that s l2 BB By analogy With our equation relating Fz and pk we can write Fz a Z And since EZ and uh are known this allows gsm to be experimentally determined z ngm Within experimental uncertainty it was found that gim 1 Since we ve already concluded that m r 2 we must have that gs 2 These Values ofs and g have been confirmed many many times by the Zeeman effect which is the splitting of spectral lines by a B eld For example the Zeeman effect is demonstrated by H atoms in the ground state That energy is split into 2 levels symmetric about the Value at B 0 E eugB 74113 gs bm igs bBZ Measurements by Lamb showed that actually g 200232 This Value of g is accounted for by quantum electrodynamics developed by Feynman Schwinger Dyson and Tomonaga Example 82 A beam of H atoms is emitted from an oven at T 400 K and is sent through a Stem Gerlach magnet of length X 1 m The atoms experience a field gradient of aBzaz 10 Tm Calculate the transverse de ection of a typical atom in each spin component of the beam due to the force exerted on its spin at the point where the beam leaves the magnet 1 What 221 relation will you use to solve this 7 Z 5612 Now what 7 TO find az we need the force and to find t we need vx we already have the magnet length BB 3B What force do they experience 7 Fz a z ybgsms i a z yb Z Z Now we need vx how do we find it 7 From the average kinetic energy which is 7 The average kinetic energy of the atoms in the beam is 2kT rather than 3kT2 why7 va22 2kT gt vx 4kTM 2 and how do we find t 7 t Xvx XM4kT12 2 723 so Zziaztzziixz M iaBzaznbx 2i 10x093gtltf3 gtlt1 lmm 2 2M 4kT 8kT 8x138gtlt10 x400 The StemGerlach experiment was actually motivated by the predictions of Goudsmit amp Uhlenbeck 1925 who postulated the 6 spin of ms 112 amp gs 2 to explain the ne structure of alkali atom spectra The nature of the spin7 It is not due to any spinning motion of the e39 If that were the case the surface of the 639 would spin at speeds gtgt c And that s only one problem with that picture since it now appears that the 639 really is a pointlike particle with an upper bound of 103916 m on its size Finally Dirac showed that the existence of the 639 spin is postulated by the relativistic form of Schrodinger s equation eek ending PRL970308012006 PHYSICAL REVIEW LETTERS ZlVJULYZOOG New Measurement of the Electron Magnetic Moment Using a OneElectron Quantum Cyclotron B Otlom k D Hanneltc B D39Ut so and G ani39ielseY Department af Pliyxicx Harvard Univemiry Cambridge Massarllilserrs 02138 USA Received 17 May 2006 published 17 July 2006 A new measttrement resolves cyclotron and spin levels for a singlerelecti39on quantum cyclotron to obtain an electron magnetic moment given by gZ 1001 15965218085 7s075 ppt The uncei39r tainty is nearly 6 times lower than in the past and g is shifted downward by 17 standard deviations The new I 39k 39 39 39 quotr 39 39 39 detemiines m n wit a O b IlncertaintyilO times smaller than for ntonHecoil determinations Remarkably this 100 mK measurement probes for internal electron structure at 130 GeV DOI l0llOSPhysReVLeu9703080l PACS numbers 062011 lZZOFV 134031 M60Cd a u Wek nd39t39 PRL9703080312000 PHYSKAL REVIEW LE1 FEM 111311135665 New Determination ufthe Fine Structure Constant from the Electron g Value and QED G Guht39iclsu1 l Htiiiickc39 T Kinttshili Lmtnn Lulmrumn Hm JLubumiang I mien quotTitaaraami Pli tRe 39uitl B tltunquot quot 111mm 39lmuem39 03L USA 1r Ply Ilium New York 14351 USA LilInfring RIKEN Wlkll Suitmm Japan 354703 M 2006 published 17 July 2006i Inem 11 t u 39 39 i 39 uttuncut of the electron lg and the ne structure constant a A new measurement of g using a onerelectron quantum cyciatrnu together with u QED uttluiluliuu iumitittg XUI eighthorder Feynman tungrants detennine ctquot 37035 999710096 070 ppb The uncenainu39es are 10 times smaller than those of nearest rivttl Inclllml ilttu mtIutlc tll ml ccnll Illcasllrcmcllts Ctttnpttrisnus ot39ntcasutetl and calculated 1 test QED most stringently anti set a limit on internal eiectmu structure DOl lUllOSPliysRe Leu97030801 PACS numbers 0620Jr lZquotJFv 1340Em ll60Cd 71 S g E 339 a trap cavity electron top endcap electrode quartz Spacer compensation K Ar electrode ring electrode it 7 nickel rings 05 cm1 compensation 1 electrode bottom endcap field emission point SIGCtrOde ll 63 2 77m quotI I 1139 v u m M 0 n0 n I L T vagvc2 vC 7662 nO ms 12 g l 1 1 2 En ms Elwcms n Iwc 511511 E my 2 w m 39 a N 1 1 EZf39c f 352 7732fJ 3139 20 quotcoloun39nnu39o 39 39 39 quoto n39 39 Q f g 10 601 L b gtN E 0 I I I Foo0905quot anvilo I uuiutlou 0 20 4O 60 O 20 4O 60 time 5 time s C 9 E 020 d a 015 gt E 2 010 E 3 005 C Q 000 U 1 O 1 2 0 1 frequency 7a in Hz frequency fC in kHz A new value for the electron magnetic moment gZ 1001 159 652 180 85 76 076 ppt 1a III Di Ifc39i 11quotde Hm Ill IthIlJ IVfD39 g I v39tyc39 I v t 39d 1 V FIG 3 Typical diagrams from each gauge invariant subgroup that contributes to the eighth order electron magnetic moment Solid and wigeg curves represent the electron and photon respectively Solid horizontal lines represent the electron in an external magnetic eld 2 3 4 E 1 C43 C4lt3gt C6lt3gt C83gt 2 7T 7T 7T 77 alm ahadronic aweakr b 109 5 pp How well is at known 10 0 5 10 a 39 electron g Harv rd 2006f tH 39 39 CS E20016 1 0 1 i eiectron g UW 1987 Rb 2006 0 3 5 90 95 100 10 5 b a electron 9 Harvard 2006 Rb2006 H i t H 0312006 i uanturn Hall 2001 neutron 1999 l O Iq muonium his i D i 1999 at Josephson etc 1998 m electron g UW 1987 5 0 5 10 15 20 25 d1 137035 990 105 FIG 1 color The least uncertain at determinations 356 at with older determinations 2 on a 10 times larger scale b Measured g are converted to or using current QED theory aquot 137035 99971090 33 066 ppb024 ppb 137035 999 710 96 070 ppb The SpinrOrbit Interaction As the title suggests this is the inmtaction between the electIon s 52in magnetic dipole moment and its orbital magnetic dipole moment SimilaI intetaction between the spin and otbital magnetic moments of nucleons Lei piotons OI neuttons is so stIong it govetns the petiodic piopetties of nuc eil Recall the Boht atom r in out usual pictme the election ttavels anund nucleus This is ceItainly the View rom the nucleus point of View Howevet in the IefeIence flame of the e the nucleus moves about the e so the e is in effect inside a cunent loop which pIoduces a magnetic field What is the cunent element ptoduced by the chatged nucleus 7 j rZeV And What law rom 222 gives us the B field rom a cunent loop 7 BiotSavaIt law ij 7 47 r 47 r BiotSavart law 3 amp 47 r3 47 r3 Ze 17 Ze 17 Recall classical EampM and ask what 3 is 7 E Field acting on the electron E 3 47580 r 47580 r a a b C 1 B 1 E So B VXE ut so 17gtlt 0 V go o C2 What do you expect the orientational potential energy of the equots spin magnetic dipole moment to be 7 E y uSB gs bhSB However this is the energy in the equots reference frame it can be shown that in the nucleus s reference frame E y 12gSybI zSB see Appendix 0 Thomas precession It will be convenient to express this in terms of SL instead of SB to begin we ask what is the force on the electron due to the electric field E of the nucleus 7 F 2 6E dV A dV 7 And what is the relation between Force F and potential V 7 F d r d r r r a F 1dV 1 lldVau So E and B 2vgtltE 2 e 6 dr r c 66 r dr a a a a a a 1 1dV Nowwhat1sLequalto7 Lrgtltprgtltmv mvgtltr so B emc r dr Note that the B fieldexperienced by the 6 due to its motion about the nucleus is olt to L 8W2 ldV z Zemc h r dr Putting this into our expression for E y we have E a h 1 1 W a a 39 L Recalling that gs 2 and ub 26 we get E 39 Squot L m gsll39lb 1 a E S k 2majc 2 r dr 2emczh r dr Note that this result is correct in the relativistic limit Example 83 Estimate E y for the e in the n 2 l 1 state of the H atom and compare it to the observed finestructure splitting 2 1 IV 62 1 Clearl we need Vr V e y M 472290 r 80 dr 472290 r2 Therefore E 2 62 i g E Since the magnitude of S and L are each 5 3 47Z3960X2m262 r a a 2 assume S 0 L z E and for the n 2 l 1 state lt1r3gt z 13aO3 g h 10 23 J 10394 eV 47r 0gtlt2mzcz 33 47rgo3h6 5447zg0 c2h4 2 36 8 e 1 me 2 me I so the splitting is 2gtlt10 4 eV comparing this to the energy of the e in the n 2 l 1 state E2 34 eV we find a finestructure splitting of 1 part in 10 same as discussed in chap 4 Example 84 Estimate the field B created by the orbital magnetic moment acting on the spin magnetic dipole moment of the equot in the n 2 l 1 state of the H atom Th1s IS big at the Since IE l lugBI ySB and us uh 103923 Amz B IE lyb 1 JAm2 1 T limit of an iron core electromagnet Total Angular Momentum If there were no spinorbit interaction L and S would be independent of each other But since there is a spinorbit interaction ie a strong internal magnetic field is acting on the spin of the e39 it exerts a torque on S This torque couples L and S causing them to precess about their sum with the orientation of each dependent on the orientation ofthe other We find that they precess about their sum instead of lyingon cones symmetric about the z axis LHence the values 011 and S are not xed If the atom is in free space subject to no net external torque the total angular momentum J L S has a constant magnitude and a constant Jzi Similar to L and S the magnitude of J is Note also that clearly J j39 M 2 h and Jz m with m j jl j2 mjlj Jz Lz Sz In the absence of the spinorbit interaction we would have Lz quot117 and Sz mjl We would still be able to de ne J as above and get m 2 ml my Then clearly m1max l 12 This last result is actually valid even with spinorbit interaction because conservation of angular momentum prevents an interaction which is internal in the isolated atom from changing the vector J ie from changing Jz as well 39 differ byime Pr and in p rm1m zpL j z 12 17 12 1732 17 52 but where does it laminate 7 Note that depending on me oriemation of s with respect to L we will have z 12 But since ILS2 IILIVISII we have UIEIILIV ISII S LS s S S LlHS n5 S L L L LS L L LS hLliSH i LS L So jj1h2 ll1h7ss1h sa sfyylhis inequality L r12 z12 Notelhatifl012 Examgle 85 Enumerale 1115 possible Values of and m for l 2 and r 2 The possible Values of j are 52 and 32 A N 3 N u 2 quotLi a a J 32 m N N H IV N u N 3 u 12 12 32 FOIJZSZ m 752r3271212 32 52 FOIJZ32 m 7327121232 Spin Orbit Interaction Energy and the Hydrogen Energy Levels We first obtain an expression for the spinorbit interaction energy in terms of Vr l s and j Then we show how it predicts the detailed structure of the energy levels of the H atom 1 ldV g We saw that Z 2mzc2 r dr SinceJ L s JJ LL 55 2 sL sL JJ LL sS2 12 L2 sz2 sL h2jj 1 1 1 ss 12 SO Of ie 2 h E 2 jjl lll ssl 4m262 In Example 83 we saw that for the n 2 l 1 state of the equot in the H atom E 10394 eV But that is also the order of the relativistic corrections to the energy levels The complete relativistic treatment yields Ue a2 1 i j 1 2 4n E quot 471398022h2n2 n 2 e l is the fine structure constant h a W ere 47reohc 137 Splittings are exaggerated by a factor of 1372 188 x 104 ie as if on 1 ml and the corresponding those energies are degenerate if there is no external field Bohr Summerleld Energy eV r S n3 v12 quot91 1z zo1 4 2 E1 7 Fri 7 Schrodinger quot 4mofzh2n2 n j12 4n 7 ll 284 L 1 ZZZ 2 L7 i Sommerfeld 391 247r802h2 quot2 71 quot9 471 313134 71 1 39 What do you notice about the case of the 181 x 10quot eV Hydrogen atom 7 eqns are ngzl j12yl0 thesame Since the Sommerfeld model is based on the Bohr model it is only a very rough approximation The Dirac theory on the other hand represents our most re ned view of how the world works How is it that these two models give the same results for the Hydrogen atom 7 Coincidence But W 39 quot h f 39 during the 107 L j theories were developing What else do we notice about the Dirac energy level 7 For a given j looks degenerate wrt 1 Actually these levels are not degenerate either Hyper ne Complete QED treatment predicts energy differences in the microwave radio region F j Lamb Shift was measured by Lamb in 1947 Can anyone guess where this shift comes from 7 Where does the spinrorbit coupling come from 7 Interaction of electron spin magnetic moment with internal magnetic field of the atom due to motion of electron Are there any other sources of magnetic fields in an atom 7 Nuclear spin magnetic dipole moment ome H atoms kicked to metastable state n 2 l 0 Waveguide r alternating B field 1 with hv 4 ueV microwaves 7 T0 1 7 ilmpllller sensitive only to metastable atoms 4i Melastahle stale l electron gun heated cathode u 2112 44 x 1 Eevl n2l0 gt n 110highlyinhibitedbyAl selection rule and that all other states lie above except n 2 l 1 j 1 which Dirac said was degenerate 7 2 9 Experiment 0 nd n112 0 Ground siatc Transition Rates and Selection Rules Recall Chap 4 excited atoms may decay to lower states by photon emission but not all conceivable transitions occur and those that do occur at widely different rates Photons are observed only with frequencies corresponding to transitions between energy levels whose quantum numbers satisfy the selection rules AlilAj0il If the wavefunction describing an electron in an atom is an eigenfunction of a single quantum state what can we say about the time dependence of the probability density Constant If however the wavefunction is a linear combination of eigenfunctions with different energies then what can we say about the time dependence Pn lr will contain terms of the form expiE2 E1t 270 which oscillate with t at a frequency 0 E2 E127Eh Note probability density of the e is also its charge distribution function Charge distribution of the e and consequently of the whole atom will oscillate between the states 2 and 1 But the atom is neutral so the simplest quantity related to its charge distribution that can oscillate with time is its electric dipole moment the product of the equots charge and the expectation value of its displacement from theessentiallyfixed nucleus We can actually use the classical formula for the rate of emission of energy by an oscillating electric dipole to obtain some factors for atomic transition rates Appendix B shows that a classical oscillating dipole radiates energy at an average rate 47r3v4 2 R lt 38063 p where p is the amplitude of the oscillating dipole moment and V is the frequency Since the energy is carried off by photons whose energies are of magnitude IN the rate of emission of photons is 47313 hv 38ch3 Note that this is the probability per second that a photon will be emitted and therefore the probability per second that the atom has undergone the transition Relative to the essentiallyfixed nucleus the electric dipole moment p of the 1 639 atom is p er To obtain an expression for the oscillating electric dipole moment of the atom when it s in a state which is a mixture of two eigenstates we need to calculate the expectation value of p To that end we need to evaluate gtllt gtllt gtllt gtllt gtllt gtllt gtllt gtllt gtllt gtllt gtllt 1 1 61 Pl CZ 1 2 cl l 1 CZ I Z c1 cl l l I l CZ CZ PZ 1 2 CZ cl I z I l 66 where cc is the complex conjugate of the previous term Yet how do we determine c1 and CZ The answer is that we can t but since we ll see that the results are independent of c1 and CZ we set them to 1 c1 c2 1 Then 111 1 915 15 ylfkyli yiexpiEi Eft Zm h ac SO f7olt I lle7ldeJ zlfe71Iid7expiEi EfthJ lIfe lldeCC If V is spherically symmetric gift5 and llf lli will have even parity so er and ll eryi will have odd parity so the first two integrals will vanish It can be shown that even is V is not spherically symmetric those two integrals vanish Therefore the amplitude of the oscillating electric dipole moment is proportional to P E Weft4614 This is called the matrix element of the electric dipole moment taken between the initial and final states Thus the transition rate is 167314 2 R lt gt 380hc3 p QED comment

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