It ˆ+( J ˆ?) raises (lowers) the eigenvalue

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Last updated: May 22, 2019

It is convenient to define the raising and lowering operators (note thesimilarity to the Harmonic oscillator!):Jˆ±? J ˆ± Ji ˆyxWhich satisfy the commutation relations:ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ+z z ± ± J,J?= 2 ZJ J,J±= ± ZJ J,J2= 0 The raising and lowering operators have a peculiar effect on theeigenvalue ofJˆ:zˆ ˆ, , ,Jˆ(Jˆ±??)=( Jˆ,J+ J J ˆ)??=(?± Z) (Jˆ? ?)z z ± ± z ± Thus, J ˆ+( J ˆ?) raises (lowers) the eigenvalue of J ˆbyZ , hence theznames. Since the raising and lowering operators commute withJ ˆ2they do not change the value of ? and so we can writeJˆ±?????± Z, ,and so the eigenvalues of J ˆare evenly spaced.zWhat are the limits on this ladder of eigenvalues? Recall that for theharmonic oscillator, we found that there was a minimum eigenvalueand the eigenstates could be created by successive applications ofthe raising operator to the lowest state.

There is also a minimumeigenvalue in this case. To see this, note that:12 2ˆ2ˆ, ,(???=) ? ?J ? J ? ?= ,? ?( J ˆJˆ+ J ˆ+ ?,J ˆ)? ?z? +21 = ? ?( J ˆJˆ+ J ˆ+ +, ,† †Jˆ)? ?+ +221 1=,? ?Jˆ+2+,J ˆ+? ?? 0 2 2 2?Hence???and therefore ??? ??. Which means that thereare both maximum and minimum values that ?can take on for agiven?. If we denote these values by?maxand?min, respectively,then it is clear that,J ˆ+? ?,= 0 J ˆ?? ?m i n= 0 .ma xWe can then use this knowledge and a trick to determine therelationship between?and?max(or?min):ˆJˆˆJˆ,? J + ?? ?,= 0 J ? +? ?m i n= 0ma x2 2ˆ ˆ2 2ˆˆ(ˆy x?J Jˆ)) ? ?? ( J+Jˆ?J J i,ˆ(ˆy x?J Jˆ)) ? ?= 0 ( J+Jˆ+J J iˆ= 0x y x ymaxx y x y,min2ˆ2 2ˆ ˆˆ2ˆ,? ( J ? J ? ZJ) ? ?=0( J ? J + ZJˆ) ??= 0z zmaxz z,min2 2? (???? Z?)= 0 (???+ Z?min)= 0max max min? ?=? (?+ Z)=? (?? Z)max max min min? ?= ??min? ZjmaxSo we have that? Zj ??? Zj . Further, since we can get from thelowest to the highest eigenvalue in increments ofZ by successiveapplications of the raising operator, it is clear that the differencebetween the highest and lowest values Zj?(? Zj)= 2Zj must be an integer multiple ofZ . Thus,j itself must either be an integer or ahalf-integer.

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Putting a,(,ˆ 23212=+= jmjjjmj Z2JjjjjmmjmmjJz1,,,ˆ?==Z … 2 ,,1, ,0 )1,1 …

+ ? ? ll these facts together, we conclude (Definem ? b / Z ):and2where in the first equation, we have noted that 0 ? J ˆ2= Z j( j + 1) impliesj ? 0 . These are the fundamental eigenvalue equations for allforms of angular momentum.The other matrix elements we might be interested in are those of theraising and lowering operators. As we saw beforeJˆ±j, m ? j, m ± 1 and so±’j , ‘ J m ˆj, m = Cj ,m? ?m m ± 1± j , j , and one only needs to determine the value of C ±j ,m . To this end,2†ˆ2ˆ ˆj, m = j, J J m j, m = j, J J m ˆj, m = j, m J ˆ2? Jˆ_ ZJˆzj, mJˆ± ± ± _ ± z ±22 21)= Z j(j ? + m ± m= C j ,m ±the phase of Cj ,m is undetermined, because the phase of theeigenstate j, m is arbitrary.

We will choose the phase of j, m so±that Cj ,m is real and positive which leads to:

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