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x y z z′ R ( 1, 2, 3 ) = 11 y′ 11 =x′ 22 22 22 x′′ z′′ =y′′ 33 y′′′ z′′′ = x′′′ 33 33

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R ( 1, 2, 3 ) = These operators DO NOT COMMUTE! about x-axis about y ′ -axis about z ′′ -axis 1 st 2 nd 3 rd Recall: the “generators” of rotations are angular momentum operators and they don’t commute! but as n n Infinitesimal rotations DO commute!!

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1 3 0 - 3 1 0 0 0 1 1 0 - 2 0 1 0 2 0 1 1 0 0 0 1 1 0 - 1 1 R( 1, 2, 3 ) = 1 3 - 2 - 3 1 0 2 0 1 1 0 0 0 1 1 0 - 1 1 = 1 0 - 2 0 1 1 2 - 1 1 = 1 3 0 - 3 1 0 0 0 1 or 1 3 - 2 - 3 1 1 2 - 1 1 =

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R( 1, 2, 3 ) = 1 3 - 2 - 3 1 1 2 - 1 1 R( 1, 2, 3 ) = 1 2 + 3 R( 1, 2, 3 ) = If we imagine building up to full rotations by applying this repeatedly N times ℓim [R( 1, 2, 3 )] N N ℓim N

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R(1, 2, 3 )R(1, 2, 3 ) ℓim N Which we can re-write in the form by slightly re-writing the “vector” components: Check THIS out: We have found a “new” representation of L x, L y, L z !!

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U 3 U = †

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Our alternate approach to motivating where the generator was an honest-to-goodness matrix! gave representing 3-dimensional rotations with the basis: A B C ABCABC satisifes all the same arithmetic as L x, L y, L z which we argue

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Notice can try diagonalizing the z th matrix: Eigenvalues of 1, 0, 1 We should be able to diagionalize 3 by a SIMILARITY TRANSFORMATION ! U 3 U = †

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For J=1 states a matrix representation of the angular momentum operators

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generates 3-dimensional rotations and with the basis: serve as the angular momentum operators! and with the simpler basis: And of course we already had

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U 3 U = † U 1 U = † U 2 U = †

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[ 1 2 2 1 ] = i 3 U † [ 1 2 2 1 ]U =U † i 3 U U † 1 2 U U † 2 1 U = iU † 3 U U † 1 UU † 2 U U † 2 UU † 1 U = iU † 3 U Since U † U = UU † = I JxJx J x J y J y J x iJ z

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000000 000000 000000 000000 000000 000000 000000 000000 000000 000000 000000 000000 abcxyzabcxyz, This 6×6 matrix also satisfies the same algebra: 000000 000000 000000 000000 000000 000000 000000 000000 000000 000000 000000 000000 3-dimensional transformations (like rotations) are not limited to 3-dimensional “representations”

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Besides the infinite number of similarity transformations that could produce other 3×3 matrix representations of this algebra

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The 3-dimensional representation in the orthonormal basis that diagonalizes z is the “DEFINING” representation of vector rotations R ( ) = e iJ· /ħ ^ can take many forms

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m s = ± 1212 spin “up” spin “down” s = ħ = 0.866 ħ 3232 s z = ħ 1212 | n l m > | > = nlm 1212 1212 1010 ( ) “spinor” the most general state is a linear expansion in this 2-dimensional basis set 1 0 0 1 ( ) = + ( ) with 2 + 2 = 1 spin : 1212 p, n, e, , , e, , , u, d, c, s, t, b leptons quarks the fundamental constituents of all matter!

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obviously: How about the operators s x, s y ? eigenvalues of each are also ħ/2 but their matrices are not diagonal in this basis s-s- = 1ħ s+s+ s-s- ħ s+s+ ħ obviously work on the basis we’ve defined

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You already know these as the Pauli matrices obeying the same commutation rule: but 2-dimensionally! What if we used THESE as generators?

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Should still describe “rotations”. Its 3-components will still require 3 continuously variable independent parameters: x, y, z But this is not the defining representation and can not act on 3-dimensional space vectors. These are operators that obviously act on the SPINORS (the SPIN space, not the 3-dimensional wave functions.

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Spinors are 2-component objects intermediate between scalars (1-component) and vectors (3-component) When we rotate the “coordinate system” scalars are unchanged. Vector components are mixed by the prescriptions we’ve outlined. What happens to SPINORS? actually can only act on the spinor part

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The rotations on 3-dim vector space involved ORTHOGONAL operators These carry complex elements, so the R would not be orthogonal! R t = R -1 i.e. R t R = RR t = 1 As we will see later this is a UNITARY MATRIX of determinate = 1 Let’s stay with the simplified case of rotation = z (about the z-axis) ^

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Notice: z z = and obviously: z z z = z ( ) ( ) + i z

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= cos + isin an operator analog to: e i = cos + isin 1 0 0 1 ( ) 1 0 0 -1 ( ) Let’s look at a rotation of 2 (360 o ) = 1 + 0 This means A 360 o rotation does not bring a spinor “full circle”. Its phase is changed by the rotation.

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Limitations of Schrödinger’s Equation 1-particle equation 2-particle equation: mutual interaction But in many high energy reactions the number of particles is not conserved! n p+e + + e n+p n+p+3 e + p e + p + 6 + 3

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Let’s expand the DEL operator from 3- to 4-dimensions i.e. lowered since operates on x then and as we’ve argued before, the starting point for a relativistic QM equation: The Klein-Gordon Equation or if you prefer:

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But this equation has a drawback: Look at Schrödinger’s Equation for a free particle * ( ) ( ) probability density probability current density The Continuity Equation

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starting from the Klein-Gordon Equation * ( ) ( ) * = 0 not positive definite The Klein-Gordon Equation is 2 nd order in t! Need to know initial (t=0) state as well as (0) much more complicated time evolution

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The Klein-Gordon Equation or if you prefer: We noted difficulties trying to work with: stemming from the fact that its 2 nd order in t

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In 1-dim, might try factoring p 2 m 2 c 2 = 0 (p + mc)(p mc) = 0 two equations, 1 st order in t but can’t add a vector and scalar! If either p +mc = 0 or p mc = 0, the Klein Gordon Equation is satisfied, and the relativistic energy-momentum relation automatically holds!

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are all equal!

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p 1, p 2, p 3 operators do commute! (It’s angular momentum operators which don’t!) For this factoring exercise to work, need ( 0 ) 2 = 1 ( 1 ) 2 = 1 ( 0 1 + 1 0 ) = 0

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For this factoring exercise to work, need ( 0 ) 2 = 1 ( 1 ) 2 = 1 ( 0 1 + 1 0 ) = 0 but if you naively choose 0 = 1 and 1 = i then ( 0 1 + 1 0 ) = i More generally these quantities must satisfy ( 0 ) 2 = I ( i ) 2 = I { , } = ( + ) = 2g (anti-commute!) they are all UNITARY: † = † = I but not HERMITIAN: = †

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Some properties: ( 0 ) 2 = I ( i ) 2 = for i=1,2,3 { , } = ( + ) = 2g (anti-commute!) they are all UNITARY: † = † = I but not HERMITIAN: = † since: g 2g for = 0 2 ( 1 ) for = 0 2 ( 0 ) which means: 1 ? ?

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0 = I 0 0 -I i = 0 i - i 0 0 = 1 0 0 0 0 1 0 0 0 0 -1 0 0 0 0 -1 1 = 0 0 0 1 0 0 1 0 0 -1 0 0 -1 0 0 0 2 = 0 0 0 -i 0 0 +i 0 0 +i 0 0 -i 0 0 0 3 = 0 0 1 0 0 0 0 -1 -1 0 0 0 0 1 0 0 Dirac discovered this could only be resolved with matrices of at least order 4×4

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With now a 4-element column matrix, for example with the basis: 12341234 10001000 00100010 01000100 00010001 = = 11 +2+2 +3+3 +4+4 and Dirac’s equation is now a matrix equation a set of 4 simultaneous equations Although the columns above carry four components they are NOT 4-vectors! They are “Dirac spinors” or “bi-spinors.” does not act on them directly.

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0 = I 0 0 -I i = 0 i - i 0 0 = 1 0 0 0 0 1 0 0 0 0 -1 0 0 0 0 -1 1 = 0 0 0 1 0 0 1 0 0 -1 0 0 -1 0 0 0 2 = 0 0 0 -i 0 0 +i 0 0 +i 0 0 -i 0 0 0 3 = 0 0 1 0 0 0 0 -1 -1 0 0 0 0 1 0 0 The block diagonal form suggests it may sometimes be simpler to work with the “reduced” notation of = AA BB where AA = 11 22 BB = 33 44

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