Lecture 12 Updated for 2005

Notes for Quantum Mechanics

Richard Seto Lecture 12 Updated for 2005

Date

2005,@D 11, 4, 17, 37, 42.5609024

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Lecture 12 Changing basis (or "representation")

We have decided to use the Sz basis, i.e. + and - . Suppose we want to switch to the Sx basis. Is the a way to do this? Before we answer this, lets see if there is a way to go from one set of eigenkets say + and - , to another set, perhaps ` ` Sx+ and Sx- . Lets think of this more generally as A=Sz with eigenkets ai . Is there a way to find a different set of ` ` † \ † \ ` ` eigenkets of another operator B=Sx with eigenkets bi ? In general we can let A and B be any two operators which represent observables so that the kets a and b are complete, orthogonal eigenkets.† \ Let us† \suppose that we have an ` i i †operator\ U† which\ does this. † \ † \ ` † \ † \ i.e. bi =U ai . (1)

` The operator U will be a special kind of transformation matrix called a unitary operator. † \ † \

` ` † ` ` ` ` † ` ` -1 ` † A unitary operator U is one in which U U=1 and UU =1 This also means U =U (2)

` ` Now we want an operator U such that bi =U ai . By construction this will be k bk ak (3)

` Proof: U ai = k bk ak ai = k bk dki= bi QED † \ † \ ⁄ † \ X » ` † ` ` we can also show that this is unitary: U U= j a j b j k bk ak = j k a j b j bk ak = j a j a j =1 ` Now we can† think\ ⁄ of† U\inX a matrix» \ ⁄notation.† \ The† matrix\ elements are (see lecture 7) ⁄ † \ X »⁄ † \ X » ⁄ ⁄ † \X † \ X » ⁄ † \X » ` ` a1 b1 a1 b2 ai U a j = ai k bk ak a j = ai b j so for the a two component case U= (4) a2 b1 a2 b2

i X † \X † \ y where we haveX used» † (3).\ NowX » ⁄lets† see\ Xhow» we\ canX »use\ this. j z k X † \X † \ { We have typically written things in the Sz basis, but suppose we want to switch to a different basis, say Sx . How do we do

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this?

First lets start out by figuring out how to change the basis of kets. In the Sz basis we write a = + + a + - - a . ` ` ` So that we can write this symbolically as a = a c = a a a where the a are eigenkets of A and A=S . We see ` i ` i i i i i i ` ` z that the coefficients ci = ai a . Now lets let B=Sx and bi are the eigenkets. In general we† can\ † let\X A† and\ † B\X be† any\ two operators which represent observables so that the kets ai and bi are complete, orthogonal eigenkets. Remember ` ` † \ ⁄ † \ ⁄ † \ X † \ † \ ai ai = 1 and i bi bi =1 Then we can write i X † \ † \ ' a = i ai ai a = ij b j b j ai ai a = j b j i b†j a\i ai †a \ So we can now write a = j b j c j where the new ' ` ‚coefficients† \ Y ° b j a =c j⁄= †i \bXj a†i ai a . Thinking of this in matrix notation is particularly nice since originally in the A a1 a basis we write a column vector † \ ⁄ † \ X † \ ⁄ † \ X † \aX2 a† \ ⁄ † \ ⁄ X † \ X † \ † \ ⁄ † \ X † \ ⁄ X † \ X † \ b a b a b a a a X † \ 1 1 1 1 2 1 We can transform toji the newzy basis as follows: = . (5) j z b2 a b2 a1 b2 a2 a2 a k X † \ { Notice how this matrix multiplication does exactly the samei Xthing† \ asy thei X sum.† \XNow notice† \ yi thatX †this\ y matrix is just the ` j z j zj z transpose-complex conjugate of U so we can write k X † \ { k X † \X † \ {k X † \ {

b1 a ` † a1 a ` =U where it is now understood that U is in matrix notation. i.e. (6) b2 a a2 a ` † (7) (newi X †basis)=\ y Ui X(old† basis)\ y j z j z k X † \ { k X † \ { Example:

Lets first recall from Lecture 8: 1 1 Sx; ≤ = ÅÅÅÅÅÅÅÅÅÅ ( + ± - ) Sy; ≤ = ÅÅÅÅÅÅÅÅÅÅ ( + ±i - ) (a) 2 2 1 1 + = ÅÅÅÅÅÅÅÅÅÅ ( Sx + + Sx - ) - = ÅÅÅÅÅÅÅÅÅÅ ( Sx + - Sx - ) (b) 2 è!!!! 2 è!!!! † \ 1 † \ † \ † \ 1 † \ † \ + = ÅÅÅÅÅÅÅÅÅÅ ( Sy + + Sy - ) - = ÅÅÅÅÅÅÅÅÅÅÅÅÅ ( Sy + - Sy - ) (c) è!!!!2 è!!!!2 i † \ † \ † \ † \ † \ † \ Now lets take ± as the old basis and Sx± as the the new basis è!!!! è!!!! † \ † \ † \ † \ † \ † \ 1) find the representation of ± in the Sx± basis. We know what the answer should be just by looking at (b). 1 1 1 1 1 1 It should be + = ÅÅÅÅÅÅÅÅÅÅ† \ ( Sx + + Sx - ) U ÅÅÅÅÅÅÅÅÅÅ and - =† ÅÅÅÅÅÅÅÅÅÅ\ ( Sx + - Sx - ) U ÅÅÅÅÅÅÅÅÅÅ in the Sx± basis. 2 2 1 2 2 -1

Its very important that we recall† \ we are in† the\ Sx± basis! Lets now use what we learned to see if we get the è!!!! è!!!! i y è!!!! è!!!! i y same answer.† \ † \ † \ j z † \ † \ † \ j z † \ ` old1 new1 old1 new2 + Sx +k {+ Sx - + + + - + k+ -{ - 11 UU = == ÅÅÅÅÅÅÅÅÅÅ1 = ÅÅÅÅÅÅÅÅÅÅ1 old2 new1 old2 new2 - Sx + † - S\x - 2 - + + - - + - - 2 1 -1 ` † 11 1 0 U = ÅÅÅÅÅÅÅÅÅÅ1 so using (7) we can write ± which in the original basis is just and i X 2 1† -1\X † \ y i X † \X† \ y è!!!! i X †H† \ † \L X 0†H† \ †1\L y è!!!! i y j z j z j z j z k X † \X † \ { k X † \X† \ { k X †H† \ † \L X †H† \ † \L { k {

è!!!! i y i y i y j 11z 1 1 11† \ 0 1 j z j z + U ÅÅÅÅÅÅÅÅÅÅ1 k { = ÅÅÅÅÅÅÅÅÅÅ1 - U ÅÅÅÅÅÅÅÅÅÅ1 = ÅÅÅÅÅÅÅÅÅÅ1 in the new basisk as{ we kexpected.{ 2 1 -1 0 2 1 2 1 -1 1 2 -1

è!!!! i yi y è!!!! i y è!!!! i yi y è!!!! i y ` How does† \ this changej the zmatrixj z elementsj z †of\ operators?j Letszj findz out jfor anz operator X , Remembering that ` k {k { ` † k { k {k { k { bi =U ai and hence bi = ai U

† \ † \ X » X » Essential Mathematica for Students of Science © James J. Kelly, 1998 qm12.nb 3

` ` ` ` ` ` † ` ` bk X bl = bk 1 X 1 bl = bk ai ai X a j a j bl = ak U ai ai X a j a j U al = i j i j j ` † ` ` ` ` = ak U X U al or in somewhat symbolic notation where we mean that X ' is just the matrix for X in a new basis or representationY » » ] Y … … ] Y … ‚ † \ Y ° » ⁄ † \ X † \ ⁄ ⁄ X » » \ Y ° » ⁄ † \ X » † \ ` ` † ` ` XX'=U» XU † - \This is known as a similarity transformation in matrix algebra.

` † ` ` † ` ` So to summarize (ket in new basis)=U (ket old basis) AND X'=U X U (8)

Example: ` 01 2) Now S in the ± basis is ÅÅÅÅÑ . Lets write it in the S ; ≤ basis. We again know what the answer has to x 2 10 x 10 be - it has to be ÅÅÅÅÑ . Lets use our methods and see if that is what we get. We use (8) 2 0 -1 ji zy † \ 11j 01z 11 11† \ 1 -1 20 10 ` † ` ` Ñ 1 1 k { Ñ 1 Ñ 1 Ñ U Sx U = ÅÅÅÅ ÅÅÅÅÅÅÅÅÅÅ ÅÅÅÅÅÅÅÅÅÅ = ÅÅÅÅ ÅÅÅÅ = ÅÅÅÅ ÅÅÅÅ = ÅÅÅÅ as we 2 2 i 2 1y -1 10 1 -1 2 2 1 -1 11 2 2 0 -2 2 0 -1 j z expected. k { è!!!! è!!!! i y i y i y i y i y i y i y j z j z j z j z j z j z j z TRACE k { k { k { k { k { k { k {

Now we define a trace of an operator as the sum of the diagonal elements

` ` Tr(X )= ai X ai . This turns out to be INDEPENDENT of representation (9) i

‚ Y » » ] ` ` ` Proof: ai X ai = ai bk bk X bl bl ai = ai bk bk X bl bl ai = i i k l k l i ` ` ` ` ` = bl ai ai bk bk X bl = bl 1 bk bk X bl = dkl bk X bl = bk X bk QED k l i k l k l k ‚ Y » » ] ‚ Y … ‚ † \ Y ° » ⁄ † \ X † \ ‚ ‚ ‚ X » \ Y ° »† \X † \

A‚ few‚ other‚ thingsY † I \willX »ask\ youY °to prove» ] ‚ in the‚ problemY ° » set] Y ° » ] ‚ ‚ Y ° » ] ⁄ Y ° » ]

` ` ` ` Tr(X Y =Tr(Y X ) ` † ` ` ` Tr(U X U = Tr X ) Tr ai a j =dij (10) Tr bi Lai = ai bi L H H† \ X §L H† \ X §L X † \ Example: ` 01 10 3) Now in example 2 we had two representation of S as ÅÅÅÅÑ in the ususal ± and as ÅÅÅÅÑ in the x 2 10 2 0 -1

Sx; ≤ . The trace is the same (i.e. zero) in either basis. i y i y j z † \ j z k { k { †Example:\

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4) now lets try something harder. Let's write Sx; ≤ in the Sy; ≤ basis. So we have

± old basis and Sy± new basis † \ † \ ` old1 new1 old1 new2 + Sy + + Sy - UU† \ = † \ == old2 new1 old2 new2 - Sy + - Sy - + + + i - + + - i - 11 ÅÅÅÅÅÅÅÅÅÅ1 = ÅÅÅÅÅÅÅÅÅÅ1 2 i X - +† + i \X- -† + -\ iy -i X † 2 \Xi -† i \ y j z j z ` † X 1† -i \X † \ X † \X† \ U =kÅÅÅÅÅÅÅÅÅÅ1 { k { è!!!! i X2 †H† 1\ i † \L X †H† \ † \L y è!!!! i y j z j z k X †H† \ † \L X †H† \ † \L1 { 1 k { 1 1 In the original ± basis Sx + U ÅÅÅÅÅÅÅÅÅÅ and Sx - U ÅÅÅÅÅÅÅÅÅÅ è!!!! i y 2 1 2 -1 j z k { 1 1 1 -i 1 1 1 - i 1 1 1 -i 1 1 1 + i In the new basis Sx + U ÅÅÅÅÅÅÅÅÅÅ ÅÅÅÅÅÅÅÅÅÅ = ÅÅÅÅ - U ÅÅÅÅÅÅÅÅÅÅ ÅÅÅÅÅÅÅÅÅÅ = ÅÅÅÅ 2 2 è!!!!1 i iy 1 2 1 + i è!!!! i y2 2 1 i -1 2 1 - i † \ » \ j z » \ j z k { k { What does this mean? It means è!!!! è!!!! i yi y i y è!!!! è!!!! i yi y i y » \ j zj z j z † \ j zj z j z S + = ÅÅÅÅ1 (1-i) S + + ÅÅÅÅ1 (1+i) S - and S - = ÅÅÅÅ1 (1+i) S + + ÅÅÅÅ1 (1-i) S - x 2 y 2 y k {k { x k 2 { y 2 k y {k { k {

Lets plug in for Sy ≤ and see if it makes sense 1 1 1 1 1 1 1 1 » Sx +\= ÅÅÅÅ ÅÅÅÅÅÅÅÅÅÅ (1-i)(» +\ + i - )+ ÅÅÅÅ» ÅÅÅÅÅÅÅÅÅÅ (1+i)(\ + -i -» )= ÅÅÅÅ \ÅÅÅÅÅÅÅÅÅÅ ( + + »i - -i +\ + - + » + - i\- +i + + - )= ÅÅÅÅ ÅÅÅÅÅÅÅÅÅÅ (2 + 2 2 2 2 2 2 2 2 +2 - ) » \ è!!!! è!!!! è!!!! è!!!! 1 = ÅÅÅÅÅÅÅÅÅÅ ( + + - ) So it checks out. You can do the same for Sx - » \ 2 † \ † \ † \ † \ † \ † \ † \ † \ † \ † \ † \ † \ † \ † \ è!!!! ` 01 2) Now S in the† \± † basis\ is ÅÅÅÅÑ . Lets write it in the S ; ≤ basis.» \ x 2 10 y

` † ` ` Ñ 1 1 1 -i 01 11 Ñ 1 1 -i i -i Ñ 1 0 -2 i Ñ 0 -i U Sx U = ÅÅÅÅ ÅÅÅÅÅÅÅÅÅÅ ÅÅÅÅÅÅÅÅÅÅ = ÅÅÅÅ ÅÅÅÅ = ÅÅÅÅ ÅÅÅÅ = ÅÅÅÅ is 2 2 2 1 i i 10y i -i 2 2 1 i 11 2 2 2 i 0 2 i 0 † \ ` j` z † \ the representation of Sx in the kSy ≤ { basis. è!!!! è!!!! i y i y i y i y i y i y i y j z j z j z j z j z j z j z Diagonalization: k { k { k { k { k { k { k { » `] ` Suppose we know the matrix elements a B a where the a are not the eigenkets of B. How do we find the basis in i j ` j ` which a matrix is diagonal? i.e. we want to find B in the b basis. OK - so let's first lets look at an operator B in the ` j basis of its eigenkets b j where B b j = b j b j . Now do the following ` Y ° † \ ` † \ bi B b j = b j bi b j = b j dij So in this basis B is diagonal.† \ If we find the unitary operator where ` ` † ` ` ` ` b = U a then we know† \ that U †BU\ is diagonal.† \ So we have to find U such that b = U a and the b are i i ` i i i Yeigenvectors» » ] of X B.» \ ` ` So first we want to find the eigenvectors and eigenvalues of an operator B where we know the matrix elements a B a † \ † \ ` † \ † \ † \ i j and where the a j are not the eigenkets of B. This means we want to find the b j's such that ` ` B b j = b j b j where the b j 's are eigenkents of B Y ° † \ First lets do some† \ things to this: ` ` ` ` a†i B\ b j †= \b j ai b j ï† \ ai B 1 b j = b j ai b j ï k ai B ak ak b j = b j ai b j

writing this in matrix notation for a 2x2 for example we have two eqn, one for each eigenvalue b j = b1 and b2 Y » » ] X » \ Y … … ] X » \ ⁄ X » … ] X » ] X » \

` ` a1 B a1 a1 B a2 a1 b j a1 b j = b ` ` a b j a b a2 B a1 a2 B a2 2 j 2 j i y thenjY »from» knowing]Y » a »bit of] z lineariX » algebra\y weiX can» solve\y this as j z j z j z j ` ` z kX » \{ kX » \{ det(kY B»-l1»)= 0]Y where» »the ]values{ of l will be the eigenvalues b j. Once we solve for the b j we can find the eigenvectors b j a1 b j which in matrix notation is . Now define x = a b .But from (4) x is just the expression for the unitary a b ij i j ij ` 2 j operator we want since a U a = a b . So all we have to do is to put the eigenvectors b in the a basis, side by† \ ` i j i j ` j ` i iX » \y side and we get U. Note thatj for thisz procedure to work itX is »important\ that B be hermitian. e.g. S+ in the Sz basis is 01 kX » \{ ` ` a1 b j 0 Ñ and the procedure will fail since (B-l1) = 0 just give 0=0. (This is since one of the eigenkets is ). 00 X » † \ X » \ a2 b j † \ † \ 0

i y iX » \y i y j z So to summarize: To diagonalize anj operatorz j z k { 1) Find the eigenvalues b and eigenvectorskX » \b{ in the original basis a k { ` ` i i i (11) 2) form U where ai U a j = ai b j ` ` † ` ` ` 3) then diagonalize as B' =U BU where B' is now diagonal » \ » \ X » † \ X » \ Example: ` 5) now lets start with Sx in the Sz representation and change it to the Sx representation where it should be diagonal. 1 First we have that the ai = ± and the bi = Sx ≤ = ÅÅÅÅÅÅÅÅÅÅ ( + ± - ) 2 ` a1 b1 a1 b2 + Sx; + + Sx; - +1 +1 ` † +1 +1 U= = = ÅÅÅÅÅÅÅÅÅÅ1 and U = ÅÅÅÅÅÅÅÅÅÅ1 a2 b1 a2 b2 - Sx; + - Sx; - 2 è!!!!+1 -1 2 +1 -1 † \ † \ † \ † \ † \ † \ check unitarity i X † \X † \ y i X † \X† \ y è!!!! i y è!!!! i y ` †j` +1 +1 z j +1 +1 +20z j ` z j z U kUX= ÅÅÅÅÅÅÅÅÅÅ†1 \X † \ { ÅÅÅÅÅÅÅÅÅÅk1X † \X »= ÅÅÅÅ1 \ { k= 1 it is{ k { 2 +1 -1 2 +1 -1 2 0 +2

check thatè!!!! è!!!! ji zy ji zy ji zy ` j 1 z+1 +j1 1 z 1 1j z 1 +1 +1 0 1 1 b j = U aki ÅÅÅÅÅÅÅÅÅÅ { k ={ ÅÅÅÅÅÅÅÅÅÅ k U Sx; + { and ÅÅÅÅÅÅÅÅÅÅ = ÅÅÅÅÅÅÅÅÅÅ U Sx; - good 2 +1 -1 0 2 1 2 +1 -1 1 2 -1

Ñ 10 ` † ` ` ` ÅÅÅÅ . That èis!!!! wei expect thaty i they èb!!!!j iarey eigenkets of UèS!!!!xiU with they samei y eigenvaluesè!!!! i y as the Sx eigenvalues. †2 \0 -1† \ j z j z j z † \ j z j z j z † \ k { k { k { k { k { k { ` ` † ` ` 1 +1 +1 Ñ 01 1 +1 +1 1 Ñ +1 +1 +1 -1 1 Ñ 20 Ñ 10 Sx'=U SxU= ÅÅÅÅÅÅÅÅÅÅ ÅÅÅÅ ÅÅÅÅÅÅÅÅÅÅ = ÅÅÅÅ ÅÅÅÅ = ÅÅÅÅ ÅÅÅÅ = ÅÅÅÅ so it i y 2 +1 -1 2 10 2 +1 -1 2 2 +1 -1 +1 +1 2 2 0 -2 2 0 -1 j z † \ works!k { è!!!! i ` y ` i y è!!!!10i y 01i y i y i y i y We should checkj Tr(S ')=Tr(z Sj') orz Tr j =Trz j=0 so it worksz j z j z j z k x { xk { 0k -1 { 10k { k { k { k {

i y i y j z j z Example: k { k { ` 1 -1 Diagonalize X = So we need to find the eigenvalues and eigenkets of this matrix. The eigenkets will -11 serve as the "new". The old basis is just ± 1 -1 a i a y 1 - b -1 a 1 - b -1 =bj ïz =0 ï det =0 -11 b k b { -11- b b -11- b † \ 1 - b 2-1=0 b=0 or 2 are the eigenvalues i y i y i y i y i y i y j z j z j z j z j z j z k { k { k { k { k { k { H L Essential Mathematica for Students of Science © James J. Kelly, 1998 6 qm12.nb

1 for b=0 a-b=0 a=-b and using a2 + b2 = 2 we get that the eigenket is ÅÅÅÅÅÅÅÅÅÅ1 2 1 1 for b=2 -a-b=0 and a=-b and using a2 + b2 = 2 we get that the eigenket is ÅÅÅÅÅÅÅÅÅÅ1 è!!!! i y2 -1 j z k { So old basis is ± the new basis is ÅÅÅÅÅÅÅÅÅÅ1 + + - ) and ÅÅÅÅÅÅÅÅÅÅ1 + - - ) 2 2 è!!!! i y j z ` 11 ` † 11 j z U= ÅÅÅÅÅÅÅÅÅÅ1 and U = ÅÅÅÅÅÅÅÅÅÅ1 k { 2 1 -1 2 1 -1è!!!! è!!!! † \ H† \ † \ H† \ † \ ` † ` ` 11 1 -1 11 11 02 00 00 U X U = ÅÅÅÅ1 = ÅÅÅÅ1 = ÅÅÅÅ1 = è!!!! i 2 1y -1 -11è!!!! 1i -1 y 2 1 -1 0 -2 2 04 02 j z j z k { k { Note that the trace is 2 for the old and new basis and that for the new basis, the eigenvalues are down the diagonal. i yi yi y i yi y i y i y j zj zj z j zj z j z j z k {k {k { k {k { k { k { Unitary Equivalent Observables ˆ Theorem : If we have two basis sets connect by an operator U as bi = ˆ ˆ ˆ ˆ−1 ˆ U ai then we may construct a unitary transform of A as U A U .A ˆ ˆ ˆ− and U A U 1 are said to be unitary equivalent observables † \ meaning that they have the same sets of eigenvalues. † \

` -1 ` ` -1 ` Note first that the definition of U is the inverse of U so that U U=1 ` ` ` -1 ` ` ` ` -1 ` ` ` ` ` -1 Proof of theorem: A al = al al ï A U U al = al al ï U A U U al = al U al ï U A U bl = al bl ` ` ` ` -1 ` We can by definition write B bl = bl bl and comparing this with U A U bl = B bl = al bl we recognize ` ` ` ` -1 that B and U A U † are] simultaneously† \ diagonalizable† ] † with\ the same eigenvalues.† ] Often †in the\ cases of physical† ] interest,† \ they are the same operator. † ] † \ † ] † ] † \

` ` ` ` ` 1 +1 +1 ` -1 ` † 1 +1 +1 In our example bi = Sx ≤ B = Sx ai = ± A = Sz and U= ÅÅÅÅÅÅÅÅÅÅ U = U = ÅÅÅÅÅÅÅÅÅÅ 2 +1 -1 2 +1 -1

` ` ` -1 1 +1 +1 Ñ 10 1 +1 +1 1 Ñ +1 +1 +1 +1 1 Ñ 02 Ñ 01 U Sz U = ÅÅÅÅÅÅÅÅÅÅ ÅÅÅÅ ÅÅÅÅÅÅÅÅÅÅ = ÅÅÅÅ ÅÅÅÅ = ÅÅÅÅ ÅÅÅÅ = ÅÅÅÅ 2 +1 -1 2 0 -1 2 +1 -1 2 2 +è1!!!! i-1 -1y +1 2 2 è20!!!! i 2 10y † \ † \ † \ † \ j z j z ` ` 0 +1 k 1 { 1 k { Now B = S U and has eigenvalues ± ÅÅÅÅÑ with eigenvectors and Lets check and see if these are xè!!!! +i10 y i y è!!!! i 2 y i y i1 -y 1 i y i y j z j z j z j z j z j z j z k ` ` ` -1{ Ñ k 01 { k ` { k { k { k { k { eigenvectors of U Sz U = ÅÅÅÅ This is just Sx again, so its obvious that they are. As stated, the two operators i y 2 10 i y i y ` ` ` ` j-1 z j z j z B and U A Uk are often{ the same operator in cases of interest. k { k { i y j z k `{ ` ` ` ` -1 ` 1 +1 +1 So what this says in this case is that Sz and Sx=U Sz U are related by a unitary transformation U= ÅÅÅÅÅÅÅÅÅÅ and 2 +1 -1 ` ` have the same eigenvalues ÅÅÅÅÅÅÅÅ≤Ñ . Now what is interesting is that we know what sort of transformation changes S to S - its 2 z x ` è!!!! i y a rotation. But is U really a rotation? Yes it is. It tells us how the 2 dimensional space of spin transforms underj rotationz in 3-space. k {

Note of interest: ` ` The transformation that changes Sz to Sx is ` ` ` ` -1 ` SizU Si U is the same as RijS j where Rij is just a rotation. This can be done because the U's are just 2x2 matrices. But actually there are 2 U's that will do it. U, and -U this is just the double covering aspect of SU(2) to SO(3) which then

means that fermions have a funny thing that a rotation in 2p does not bring you back to the same thing but you have to go through 4p.