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 8 < 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 Essential Mathematica for Students of Science © James J. Kelly, 1998 2 qm12.nb 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 Yrepresentation» » ] 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:† \ Essential Mathematica for Students of Science © James J. Kelly, 1998 4 qm12.nb 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.

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