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Intrmd Quant Mech I

by: Yvonne McKenzie

Intrmd Quant Mech I PHYSICS 614

Yvonne McKenzie
GPA 3.62

Nikolay Prokofiev

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Nikolay Prokofiev
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This 17 page Class Notes was uploaded by Yvonne McKenzie on Friday October 30, 2015. The Class Notes belongs to PHYSICS 614 at University of Massachusetts taught by Nikolay Prokofiev in Fall. Since its upload, it has received 16 views. For similar materials see /class/232311/physics-614-university-of-massachusetts in Physics 2 at University of Massachusetts.

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Date Created: 10/30/15
Physics 614 As a prelude to the quantum mechanical formalism which we shall explore it is useful to examine the basic tenets as outlined by Feynman in his pedagogical approach to the subject The following material is extracted from Feynman s Lectures in Physics Volume 111 Chapter 1 5 The reader is invited to consult the original source for more details Electron Diffraction Consider electrons passing through a two slit system If we had particles passing through we know the answer of what would occur a probability pattern would emerge as shown in Fig 1 WSW Puz 39 2 Figure 1 Pt as shown is just the sum of the probabilities P1 and P2 which would result if either slit 1 or slit 2 were blocked 011 as indicated in Fig 2 This is equivalent to saying that the particle went through either slit 1 o r slit 2 BACKwa Figure 2 Suppose now however that Sis alight source We know here too what would occur If the slit separation is of the order of the incident wavelength we nd an interference pattern as shown in Fig 3 i 395 ate g 2 wan assuage Im 12Ihhzl Iz lquotzl2 Figure3 The existence of such a pattern is assured since light is a wave and the eld amplitudes E and B add rather than the intensities Of course if we blocked 01f either slit 1 or slit 2 we would nd curves P1 and P2 similar to those indicated in Fig 2 However if both slits are open the light passes through both slits and interferes as seen on the screen We can easily analyze the situation as follows Assume the slit separation a is much less than the slit screen distance L Then w r L Figure 4 2 2 a 2 D1 L Ltan9 i 2 L2sec29 Ltan9a a 1 D1 m Lsech1 mLsecQ iasinf Similarly we nd D2 m L sec 9 a sin 9 The phase difference of the two beams upon arrival at point P is then 27r 27f Avl p D1 7ast The the total electric eld on the screen will be given by Em En coskv wt A En coskv wt ll 1 1 1 1 Encos kw wt inn cos inn sin kw wt inn sin inn 1 1 1 1 Encos kw wt inn cos inn sin kw wt inn sin inn ll 1 1 2E0 cos kw wt A cos A and the intensity is 1 I9olt 4 1 604E cos2 iAz 60133 cos2 737a sin 9 which is the pattern shown in Fig 3 If we perform this experiment now with electrons we nd a similar interference pattern That is the electrons did not go through either slit 1 or slit 2 They were acting as if they were a wave Now we might try to answer the question of which slit any given electron went through by positioning light sources D1 and D2 near slits 1 and 2 Then each time an electron is detected striking the screen we look to see if there was a corresponding flash near slit 1 or 2 These are recorded separately If at the end we total those arrivals correlated with flashes at 1 and 2 respectively we nd curves like P1 and P2 in Fig 2 When we combine them to nd the total probability we nd or course P1 P2 as in Fig 1 The interference has disappeared Once we determine which slit the electron passed through it really does pass through a de nite slit and acts like a particle Another way of looking at this rather peculiar result is to say that the photon has interfered with the phase of the electron amplitude In order to reduce this interference we must go to longer wavelength tie lower energy photons We nd that nally around A w a the pattern returns 7 we have electron electron interference again But now we can no longer localize the electrons to one slit or the other since light can only locate particles to within a wavelength Now let us construct a quantum mechanical formalism which describes this sit uation Let ltP Sgt be the amplitude that an electron is emitted by the source and arrives at the screen at point P There are of course two possible paths If LS is the amplitude for the electron to travel from the source to slit 1 and Pgl is the amplitude to travel from slit 1 to point P on the screen we have ltP Sgt ltP 1gtlt1 SgtltP 2gtlt2 Sgt The probability for the process in question is given in quantum mechanics by the absolute square of the amplitude Note amplitudes are in general complex with real and imaginary parts If A 7quot 219 then Prob A r2 52 Now deBroglie suggested that for particles just a for photons h Ez w and pz kzx Thus we write expza2 251 1131 Ex 1 and ltP Sgt 2 oc egtltpiLsecl9 gsinl 39egtltpigLsec 939gsinm2 2 2 expz Lsec9 expz 2sin9exp z 2sin9 ll 4 cos2 8 9 sin 9 h 2 which thus predicts the observed interference pattern But now what happens when the light sources are turned on Let ltPf Sgt ltP 1gtalt1 3gt represent the amplitude that an electron passes through slit 1 scatters light from the photon source into a detector near 1 and arrives at P Let ltPJESgtltPi1gtblt1iSgt 4 represent the amplitude that an electron passes through slit 1 scatters a photon near detector 2 and arrives at P Then the amplitude that an electron arrive at P with a count in detector 1 is ltP 1gtalt1 3gt ltP 2gtblt2 Sgt while the amplitude that an electron arrive at P with a count in detector 2 is ltP 2gtalt2 3gt ltP 1gtblt1 Sgt An important axiom of quantum mechanics is that probabilities not amplitudes of distinguishable events are added Thus the net probability of nding an electron at point P is Prob altP 1gtlt1Sgt bltPi2gtlt2iSgti2 altP 2gtlt2Sgt bltP 1gtltliSgt 2 Now if A lt a A being the photon wavelength then for a well designed system only electrons passing through slit 1 will scatter photons into detector 1 so that gbg lt gag Then Prob w a 2iltP 1gtlt1 3gt 2 ltP 2gtlt2 3gt 2 oc a 2 ltP 1 2 ltP 2 2 const as that no interference is predicted On the other hand if A 2 a then when a photon is scattered a large blob of light will be seen rather than a pin point ash as that then is a sizable probability that an electron passing through slit 1 can scatter light into detector 2 Thus 6 w a and Prob w 2 a 2 ltP 1gtlt1 3gtltP 2gtlt2 3gt 2 oc SEaEZcosZEEsinQ H2 tie an interferences pattern as before SternGerlach Apparatus Let s see what else we can do with our quantum mechanical formalism We examine what happens when a classical magnetic moment I interacts with an inhomogeneous magnetic eld First consider the interaction of the dipole with a constant eld B Bquot We know that in such a case T X B where T is the torque Then the dipole precesses about the eld direction as shown and AL MB sin 9A2 ILl sin QAqb TB Figure 5 Then w Lf l is the angular frequency of precession Thus for the Bohr atom we recall that u IA 1 being the current ow around a closed loop and A being the area enclosed by the loop where A7rrz Ieuei 27W Figure 6 Then U E e 7 39 x w 2 7 L M 27r7 2m 2m 6 Then the precessinnal frequency is given by e30 T 2112 Now consider what happens when the dipole is placed in a magnetic eld which does not vary with y and is symmetric with respect to the yz plane as drawn below E 5 l 9 M V N Figure 7 Consider the force acting on a dipole placed in this eld The interaction energy is given by U B Thus BB BB BB BB 03 BB F m I v z I w y z z VU 01 61 PM in p 35 1 1 By My By 4quot 81 J L 631 0By 0B it Bz M 6239 M 9239 Since there is no variation with y Fy 0 Now since if a moment which enters the system is not aligned along one of the B lines it will precess n is not xed and must be averaged over For a moment traveling along the symmetry axis we have then It 1 0 Then 7 7 BB BBZIC 7 I By J 0m 39 02 But 36 0 since there is no variation in y and a OB 0 since we are along the symmetry axis Thus 6B A F 12 a k iiei particles are de ected in the z direction in amounts depending oil13 Although classically if an atom with a magnetic moment is passed through the device there would be a continuum of possible values of 1 corresponding to differing angles of alignment with the z axis quantum mechanically only discrete values for Lg are allowed Hence atoms passing through such a device called by the way a StemGerlach apparatus are separated into a discrete number of beams It turns out that there are certain kinds of atoms called spin 1 atoms for which znHo nLOrI Thus when such an atom passes through a Stern Gerlach machine it must take one of three paths as shown depending on its value of n Figure 8 Now imagine we build an improved SternGerlach apparatus as shown below Figure 9 It consists of three high gradient magnets The rst on the left is an ordinary SterneGerlaeh machine and splits the incoming beam into three separate beams for a spin 1 particle The second magnet is twice as long and has its polarity reversed It binds the paths of the particles back toward the axis as shown The third magnet is just like the rst and brings the particles back together again Finally imagine that there is a device at A which accelerates the atoms from rest and sends them through the Stern Gerlach machine and at B there is a device which decelerates them and brings them to rest again We denote this improved Stern Gerlach machine by the shorthand symbol 0 S This is not a quantum mechanical symbol It is only a convenient way to represent the improved Stern Gerlach device Since later we shall use several of these machines at once and with various orientations we identify each by a letter underneath Thus we call this one machine 3 It is interesting to see what happens when we use the Stern Gerlach device as a lter by putting in absorbers which block one or more beams We done these by the shorthand symbols given to the left If N particles are sent now into a Stern Gerlach device which has the 0 and channels blocked in general a fraction cc of the original number of particles will emerge If these particles now enter another Stern Gerlach device with the same orientation as the rst we call this one 3 and with the 0 and channels blocked we nd that all the particles get through the second device On the other hand is S has the and 0 channels or the and channels blocked none of the particles get through N gt 0 I aN 0 I 1N S Squot I Ngt 0 I 1N 0 0 S Squot N gt 0 I G 0 0 Now we want to describe these experiments quantum mechanically We say that an atom is in a 3 state is it has passed through a Stern Gerlach machine 3 which has the 0 and channels blocked it is in the 03 state if it has passed through 3 with the and channels blocked and it is in the 3 state if it has passed through 3 with the and 0 channels blocked Then we let bga be the amplitude that an atom in state go gt will get through an apparatus into the gb gt state The experiments above yield ltS Sgt1 ltUSS0 lt SS0 Similarly if the rst machine produces particles in the 03 state we nd lt3 03gt 0 lt03 03 1 303 0 and if particles are produced in the 3 state we nd ltS Sgt0 ltUS S0 lt S S1 Then if we are only dealing with pure states tie only one channel open at a time we can display this information as a matrix from Sgt 03 i Sgt ltS 1 0 0 to lt03 0 1 0 since there are only nine different experiments we can do Now suppose that after passing through the 3 machine we try to pass the beam through a machine T which is oriented at some angle to the 3 device We can do experiments as above and construct a new matrix composed of elements ltiT j3gt where 239 j are either 0 or Ngt 0 I 1N 0 10T 32N Thus if we do the experiment below and ozN particles leave the 3 machine in state 3 the number of particles which get through the T machine is oaN gtlt lt0T 3 The new matrix is 10 from i S 03 3 ltT ltT S ltT 03gt ltT S to UT UT S UTIJS UT 3 HT lt T 3 lt TUSgt lt T 3 Is there anything we can say about these matrix elements in general Yes because we know that if a particle which comes out of the 3 machine in my state enters the T machine it must come out in one of the three states T GT or T That is there are no particles which are lost This says that ltT 3gt 2 lt0T 3gt 2 lt T Sgti2 ltT 03gt 2 lt0T 03gt 2 ltT 03gt 2 1 ltT 3gt 2 lt0T 3gt 2 lt T 3gt 2 1 I H or in a more convenient notation Z ltjTii3gt 2 1 239 41 j707 Now consider a case wherein we send a beam through a machine 3 with two channels blocked then through a T machine with another two channels blocked and nally through another S machine identical to the rst I N gt OI 0N 0 GBN 0 OWN We nd that the probability of a particle passing through the nal 3 machine after coming out of the T machine is completely independent of the fact that the beam entering the T machine is in a 3 state That is the fraction 6 measured above is the same as would be seen in the experiment below I I N gt 0 7N 0 761V 0 I 7 6va I I S T S 11 Thus the particle forgets completely what sate it was in before entering the T machine Now suppose all the blocks in the T machine are removed What happens We nd a remarkable behavior as shown below that it is as if the T machine were removed N gt 0 I 1N 0 1N 0 I 1N S I 1N 0 0 gt T J The particle remembers completely which state it was produced in In fact we nd that for the case indicated at the top of the page a number 16617 particles get through when there are two baffles on the T apparatus but no particles would get through if the baffles were removed 839 0 O I Q What we nd from the bottom experiments is then that ltSg S ltSg TltT S ltSUTlt0T S ltSg Tlt T S US S US TltT S lt0S0Tlt0Tg S OS Tlt T S or more succinctly ltS Sgt ltSEJTgtltJTE3gt j0 ltOS Sgt Z lt03 jTgtltjTESgt ju In fact it is found that a more general behavior obtains 7 for any two states giR ng with 239 j 0 and R 3 being Stern Gerlach devices of arbitrary orientations 7 we nd ltiR jSgt Z ltiRikTgtltkTijSgt 1pm so that insertion of an open T lter never makes any difference The states ng are called basis states Any atomic system can be separated by a ltering process into a set of basis states such that the future behavior of atoms in a single basis state is completely independent of any past history and depends only on the nature of the state Of course there is no unique set of basis states g T 0T7 g T is one set S US S is another and there are clearly an in nite number of possibilities The requirements which sets of states must ful ll in order to be basis states are 12 i They must all be completely different Or in other words there must be no amplitude for a g T state to go into a 0T or g T state We can write this succinctly as 1 if 239 j 0 if 239 j ii They must be complete If we pass a system of atoms into the ltering appa ratus each atom must be able to pass through via some beam That is for example T7 0T satisfy ltiT jT 674 but do not constitute a complete set since an atom which happened to be in the T state could not pass through a lter which only accepted g T g0T ltiT jTgt 57 Then given our de nition of basis states we can write ltiR jSgt ZltiR kgtltkijSgt where the set of states gk constitute a complete set of basis states For example if the set of states gk are S7 US S we have ltiR jSgt Z ltiR kSgtltkS jSgt k0 Z ltiR kSgt6jkltiR jSgt k0 An identity Thus things seem to hang together We can get one more result rather easily We showed previously that Z ltjT iS 21 ltT iSltT iSlt0T iSlt0T iS j0 lt Tz Slt ng S Similarly from above we have 1ltiS iS Z ltiSkTltkTgiS lt2 STltng 3ltiS 0Tlt0TgiS k0 lt2 S Tlt T iS These two results are consistent for all possible orientations of S and T if and only if ltjT iSgt ltiS jTgt The three results we have from above can be generalized and are valid for all quantum mechanical systems Thus i There is a complete set of basis states gk such that jgk jk ii For any two states 30 we have Zlt igtltiixgt lt xgt where the sum is over the complete set mentioned previously iii For any two states gr gx we have lt X ltX gt These are the basic laws of quantum mechanics and should be memorized Now let us examine the usefulness of this idea of basis states Suppose we have some device A not necessarily a Stern Gerlach machine which operates on our spin 1 particles Suppose we want to know what will happen if we sent into A a beam in the state z z could be 3 0R T etc 7 any state and then try to pass the particles coming out of A into a machine which only accepts particles in the state 1 again as could be 3 0R T etc 7 any state We write the amplitude for this as ltwiAi gt Le Amp 0 I A a URgA S so that if N particles enter A machine from the 3 machine the number coming out of the R machine is glt0R A 3 37V What can we say about the number ltX A It is simplest to analyze this matrix element by imagining an open T machine to be placed before and after A 0 A 0 X Then we have ltx Ailtzgtgt ZltXijTltjT AEiTltiTE Thus provided we know the amplitudes iqu jTgx eg ltT 3 OT S ltT R etc my experiment involving the A machine can be described in terms of the nine complex numbers iTgAng with 239 j 0 We can write this as a matrix A with A iTgAng The matrix is then ltTgA T ltTAUT ltT A T A lt0T A T lt0T A 0T lt0T A T lt T ATgt lt T A0Tgt lt T A Tgt If A is made up of two other machines B and C in series A B c we can imagine this as z 0 B 0 C 0 X T Then ltX Ai gt Zltx iTgtltiTiA jTgtltjTi gt if Zltx iTgtltiT C kTgtltkTiB jTgtltjTi gt ijk That is ltiT AEjTgt ZltiTiCikTgtltkT BijTgt k 01 A51 2 Cz jBka k This is just matrix multiplication Thus all possible experiments are describable in terms of nine complex numbers for each machine plus knowledge of ng3 for arbitrary orientations of the S and T Stern Gerlach devices Now all this machinery we have set up is very similar to vector algebra Consider 132623 to be the basis vectors Then just as ltiS jS 6434 for basis states S7 US7 S 15 we have i j 6434 for the basis vectors just as ltX gt ZltxiiSgtltisi gt t we have B A A 2i BiAi 234 B A i i A A There is one difference having to do with the fact that quantum mechanical amplitudes are complex The condition lt Xgt 0699 in quantum mechanics corresponds to B A A A A B with ordinary vectors Until now we have written all of our quantum mechanical formalism using x which are numbers We could of course do the same with vectors If F ma we could write C A F m0 4 a which is true for any vector C But it is of course silly to do so Just as the vector A is the important thing 7 not its dot product with another vector so in quantum mechanics it is the state gx which is important not its inner product 30 with another state Thus since lt Xgt Zlt ii3gtltiS Xgt is true for any state 15 we can throw away gt and write X Z iSgtltiS Xgt i This corresponds in vector algebra to A Z i i A i In fact since the above is true for any state X why keep X Indeed we can write 1 E was a If we de ne lt63 Ci 16


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