Understanding Special Theory of Relativity in Quantum Mechanics
Delve into the concepts of the Dirac equation, energy-momentum relationships, and the application of special relativity in quantum mechanics. Explore Lorentz transformations, 4-vectors, and the velocity Lorentz transformation, enhancing your understanding of this fascinating field.
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PHY 741 Quantum Mechanics 12-12:50 AM MWF Olin 103 Plan for Lecture 32: Chap. 20 in Shankar: The Dirac equation 1. Some simple concepts of special theory of relativity 2. Energy and momentum relationships 3. Dirac equation for a free particle 11/17/2017 PHY 741 Fall 2017 -- Lecture 32 1
11/17/2017 PHY 741 Fall 2017 -- Lecture 32 2
Notions of special relativity The basic laws of physics are the same in all frames of reference (at rest or moving at constant velocity). The speed of light in vacuum c is the same in all frames of reference. y y v x x y x y x 11/17/2017 PHY 741 Fall 2017 -- Lecture 32 3
Convenient v notation : Lorentz transformations c 1 2 1 Stationary frame Moving ct + frame y y ( x ) = + ' ' ct x ( ' ) = x ' ' ct v = y y x = ' z z x y x y x 11/17/2017 PHY 741 Fall 2017 -- Lecture 32 4
1 v c Lorentz transformations -- continued 2 1 = v : v x For the moving frame with 0 0 0 0 0 0 0 0 = = 1 - L L 0 0 1 0 0 0 1 0 0 0 0 1 0 0 0 1 ' ' ct ct ct ct ' ' x x x x = = 1 - L L ' ' y y y y ' ' z z z z Notice : = 2 2 2 2 2 2 2 2 2 2 ' ' ' ' c t x y z c t x y z 11/17/2017 PHY 741 Fall 2017 -- Lecture 32 5
Examples of other 4-vectors applicable to the Lorentz transformation: = v : v x 1 For the moving frame with v c 2 1 0 0 0 0 0 0 0 0 = = 1 - L L 0 0 1 0 0 0 1 0 0 0 0 1 0 0 0 1 / / ' k / ' k / c c c c ' ' k k x x x x = = = 1 - L L k r k r Note : ' ' t ' ' t ' ' k k k k y y y y ' ' k k k k z z z z ' ' E E E E ' ' p c p c p c p c x x x x = = = 1 2 2 2 2 2 2 - L L Note : ' ' E p c E p c ' ' p c p c p c p c y y y y ' ' p c p c p c p c z z z z 11/17/2017 PHY 741 Fall 2017 -- Lecture 32 6
Lorentz transformation of the velocity Stationary frame Moving ct + frame ( x ) = + ' ' ct x ( ' ) = x ' ' ct = y y = ' z z For an infinitesimal increment: frame Stationary Moving frame ( ) = + ' ' cdt cdt dx ( ) = + dx dx' cdt' = dy dy' = dz dz' 11/17/2017 PHY 741 Fall 2017 -- Lecture 32 7
Lorentz transformation of the velocity -- continued frame Stationary Moving frame ( ) = + ' ' cdt cdt dx ( ) = + dx dx' cdt' = dy dy' = dz dx dz' dy dz Define : u u u x y z dt dt dt ' ' ' dx dy dz ( ( dt ' ' ' u u u x y z ' ' ' dt dt v / dt ) ) + ' + ' u dx dx' cdt' / ' = = = x u x + ' + 2 1 dt dx c vu c x ' u ' dy dy + ' y = = = u ( 1 ) ( ) y + 2 / ' ' / dt dt dx c vu c x 11/17/2017 PHY 741 Fall 2017 -- Lecture 32 8
+ ' u + v = u Example of velocity variation with x x x 1 / u c ux/c ' u + y u = u ( ) x y 1 / c uy/c 1 = 2 1 11/17/2017 PHY 741 Fall 2017 -- Lecture 32 9
Velocity transformations continued: ' Consider: 1 ' / x vu + ' u vu + + ' u v u vu y = = = . x z u u u ( ) ( ) x y z 2 + 2 2 c 1 ' / x 1 ' / x c c v v + 2 1 ' / x 1 vu c ( ) = = + 2 Note that 1 ' / x vu c ' u v u ( ) ( ) ( ) 2 2 2 1 / 1 '/ 1 / u c u c v c ( ) = = = = + ' c c v u u v + ' ' u v u u ' y x ( u ) ( ) = + v u c ' ' u u ' c u u u ' ' ' ' u x v u x u v u u x u u ' ' u y u u z u z c ' u u u u u ' ' ' ' u x u x = L Velocity 4-vector: u ' u y u y PHY 741 Fall 2017 -- Lecture 32 ' u z u z 11/17/2017 10
Some details: 2 2 2 2 ' u c v v c u c u c ( ) = + = + 2 1 ' / x 1 1 1 1 x vu c ' u v u 2 2 2 2 ' u vu + + ' ' u v u vu y = = = where . x z u u u ( ) ( ) x y z + 2 + 1 ' / x 2 2 vu c 1 ' / x 1 ' / x c c v v 2 u c 2 2 y 2 y ' u c u c 2 x 2 z 2 z 2 ' ' u c u c v u u c v c u c v c + + + = + + + 1 1 x x 2 2 2 2 2 2 2 c 2 2 2 2 2 2 ' u c v u c v u c v c v c + = + + 1 1 1 1 x x 2 2 2 2 2 2 2 2 2 2 ' u c v u c u c v c + = 1 1 1 1 x 2 2 2 2 11/17/2017 PHY 741 Fall 2017 -- Lecture 32 11
c u u Significance of 4-velocity vector: u x u u y u u z Introduce the rest mass m of particle characterized by velocity u: 2 c E mc u u u p c mu c u x x = = u x mc u p c mu c u y y u y u p c mu c u z z u z Properties of energy-moment 4-vector: ' ' E E E E ' ' p c p c p c p c x x x x = = = 1 2 2 2 2 2 2 - L L Note : ' ' E p c E p c ' ' p c p c p c p c y y y y ' ' p c p c p c p c z z z z 11/17/2017 PHY 741 Fall 2017 -- Lecture 32 12
Properties of Energy-momentum 4-vector -- continued = c mu y u 2 E mc u p c mu c x u x p c y p c mu c z u z ( 1 ) 2 2 2 2 u ( ) 2 u mc u 2 0, y = = = 2 2 2 2 2 2 2 Note : 1 ' ' x z E p c mc E p c 2 c c c E u = 2 p Notion of "rest energy": For mc = + 2 2 2 2 4 2 Define kinetic energy: E E mc p c m c mc K 2 p p = + 2 Non-relativistic limit: If 1, 1 1 E mc K mc mc 2 p 2 m 11/17/2017 PHY 741 Fall 2017 -- Lecture 32 13
Summary of relativistic energy relationships = c mu c p z u z 2 E mc u p c mu c x u x p c mu c y u y = + = 2 2 2 4 2 E p c m c mc u 2 2 + = + = 2 2 2 4 2 2 Check : 1 p c m c mc mc u u u = 2 Example electron an for : c E 5 . 0 MeV m = for 200 GeV E = = 5 4 10 u 2 mc 1 1 12 = 1 1 1 3 10 u 2 2 2 u u 11/17/2017 PHY 741 Fall 2017 -- Lecture 32 14
How do these relationships effect quantum mechanics? Focusing on treatment of Fermi particles Relativistic mechanics Non-relativistic mechanics 2 p ( ) 2 E = = 2 2 2 c 2 p E mc 2 m (with some licens e) = ) ( = ) 2 ( )( p + p 2 E c E c mc i H t p + p i c i c t t ( ) 2 = 2 mc 11/17/2017 PHY 741 Fall 2017 -- Lecture 32 15
Relativistic relationships continued Ref: J. J. Sakurai, Advanced Quantum Mechanics ( ) 2 p + p = 2 i c i c mc t t + p 2 R L Let i c mc t Factored equations: + = 2 L R p i c mc t = 2 R L p i c mc t 11/17/2017 PHY 741 Fall 2017 -- Lecture 32 16
Relativistic relationships continued Ref: J. J. Sakurai, Advanced Quantum Mechanics Factored equations: + = = 2 L R p i c mc t 2 R L p i c mc t = = + U R L Dirac's rearrangement: L R L U U p i c t = 2 m c p c i L L t 11/17/2017 PHY 741 Fall 2017 -- Lecture 32 17
Relativistic relationships continued U U p i c t = 2 mc p c i L L t Further rearrangements: 2 U U p mc p c = i 2 L L t c mc = i H t 11/17/2017 PHY 741 Fall 2017 -- Lecture 32 18
Four component wavefunction of free Fermi particle 2 U mc i = = = + = k U p c = 2 L L t p c mc U U k ( ) ( ) k k r / i iEt Assume e L L k c U L ( ) k k ( ) 2 E k mc c mc L U ( ) k ( ) k 2 E + 2 2 2 c 2 2 4 c E m = + 2 2 c 2 2 4 k E m c 11/17/2017 PHY 741 Fall 2017 -- Lecture 32 19
0 1 1 0 0 i 1 0 0 i = = = Pauli matrices: x y z 0 1 k + k ik z x y = k k ik k k x y z c c c c k = = U L L U ( ) k ( ) k ( ) k ( ) k + 2 2 E m E m = + 2 2 c 2 2 4 c k c + k Positive energy solutions: 1 ( ) 0 = E m z z = = U L N N k ( ) k z 2 E mc k m + + c 0 1 2 E c = U L N N ( ) k ( ) k z 11/17/2017 PHY 741 Fall 2017 -- Lecture 32 20
0 1 1 0 0 i 1 0 0 i = = = Pauli matrices: x y z 0 1 k + k k ik z x y = k k ik k x y z c c k = = U L L U ( ) k ( ) k ( ) k ( ) k + 2 2 E mc E mc = + 2 2 c 2 2 4 c k Negative energy solutions: E m k c 1 0 z z = = U L N N ( ) k ( ) k z 2 E mc k m + c 0 1 2 E c = = U L N N ( ) k ( ) k z 11/17/2017 PHY 741 Fall 2017 -- Lecture 32 21
What does this all mean? Positive energy solutions: = = = + 2 2 c 2 2 4 k E m c 1 0 k c + z = U L N N ( ) k ( ) k z z 2 E mc k m + + c 0 1 = U L N N ( ) k ( ) k 2 E c z 2 2 k + 2 2 k For c m c E mc 2 m 1 0 0 0 = U L N N ( ) k ( ) k 0 1 0 0 = U L N N ( ) k ( ) k 11/17/2017 PHY 741 Fall 2017 -- Lecture 32 22
