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Course-2, PHY502, Block-VI Relativistic

Prof O. P. S. Negi Vice Chancellor Uttarakhand Open University Haldwani (Nainital) Uttarakhand Quantum Mechanical Operators Physical Quantity  Operators symbol  actual operation Motivation?   Momentum p  pˆ  x x i x  Total Energy E  Eˆ  i t Coordinate x  xˆ  x

Potential Energy U (x) Uˆ (x)  U (x) Schrödinger Equation • Schrödinger, building on the formulation of Heisenberg, realized the state of a could be described as a complex-valued . • Any wave function could be expressed as linear combinations of eigen functions, whose evolution through was given by the Schrodinger’s Equation • Schrödinger developed the wave equation which can be solved to find the wave-function by translating the equation for energy of classical physics into the language of waves 2 p 2 2 x  U(x)  E  U(x) x  i  x 2m 2m 2 x t • For fixed energy, we obtain the time-independent Schrödinger equation, which describes stationary states 2 2 x  U(x) x  E x 2m 2x • the energy of such states does not change with time – ψn(x) is an eigen-function or eigen-state – U is a potential function representing the particle interaction with the environment Special Relativity: Postulates

The two fundamental postulates of special relativity are : 1. The principle of relativity: The laws of physics take the same form in all inertial reference frames. 2. The constancy of the : The speed of light in is same in all inertial reference frames.

• Lorentz Transformations 1   v2 1 c2 Special Theory of Relativity: Four vector • We live in a four-dimensional world of space-time continuum. • Einstein introduced the concept of four vectors such that the scalar product of any two four-vectors is invariant under Lorentz transformations. • It is similar to the concept that the scalar product of any two three-vectors in the three dimensional space is invariant under rotation of . • We list below some of the four-vectors. • Space-time four vector (four-space) x  (x, y, zx ,ict)( 1,2,3,4) • Four differential          ( , , ,i ) (Space-time gradient) x x y z ct • Energy-momentum four vector iE p  ( p , p , p , ) (Four-momentum)  x y z c • Scalar product of four vectors (length) listed above and given below remain invariant under Lorentz Transformations. ds2  dx dx  dx2  dy 2  dz 2  c2dt 2 • Line element   2 2 2 1 2 2        • D’ Alembertian operator   x2 y 2 z 2 c2 t 2 E 2 • Conservation of energy p  p  p2  p2  p2   m c2 momentum   x y z c2 0 Relativistic Limits

• The Schrödinger equation breaks down at relativistic limits. • This is because it is not Lorentz covariant. • Time and space do not enter the equation symmetrically.

Question-What is the remedy? Friday, November 5, 2010 Drawbacks of Schrödinger Equation

2 2 x   U(x) x  i  x 2m 2 x t

• It does not confirm with the principle of relativity • It is not invariant under Lorentz transformations • It is one particle non relativistic equation • Space and time are not on equal footings • Homogeneity of space and time is lost • and magnetic moments are silent features • Exchange of is difficult. • Question-What is the remedy? Natural Units Role of c and Reconciliation of ћ are c  1 Relativity with respectively Quantum Known as the c = 1 → [ L ] = [ T ] mechanics scales used  = 1 → [ E ] = [ T ]1 for Relativity Relativistic and Choosing [ E ] = MeV, we have Quantum quantum • [ M ] = [ P ] = [ E ] = MeV Mechanics mechanics • [ L ] = [ T ] = [ E ] 1 = (MeV) 1 c = 2.997  108 m/s T( s ) = 6.582  1022 T( MeV 1 )  = 6.582  1022 MeV  s L( m ) = 1.937  1013 L( MeV 1 ) 1 MeV = 1.602  1013 J M( kg ) = 1.783  1030 M( MeV ) Domains of Physics

Speed c Relativistic Quantum Relativity Relativistic Mechanics Cosmology c/10

Quantum Classical Newtonian Mechanics Mechanics Cosmology

Size Nucleus Galaxy (10-14 m) (10-10 m) (1020 m)

Klein-Gordon Equation: Relativistic Generalization of S.E. • Using Einstein’s relation for a free particle:  E 2  p2c2  m2c4; (E  Eˆ  i ; p  pˆ  iˆ ) 0 t • Einstein’s relation is rewritten in a covariant form: 2 p p  (m0c) ˆ ˆ  2  2  • one obtains: E(E (r,t)  c pˆ  pˆ (r,t)  m0 (r,t)     2 ˆ ˆ 2  (i )(i )  c (i)(i)  m0 • K. G. Equation: t t 2  m2c2   (2  1 ) (x,t)  0  (x,t) c2 t 2 2 • Now let’s introduce covariant derivative:          p  i  (i,i )    ( , , , )  x x y z ict x x4 • In this notation, Gordon-Klein equation becomes:

2     (x,t)  0 m c Here (  0 ) is inverse Compton wavelength) 

Problems of indefinite Probability with Klein-Gordon Equation: 1. Unlike the Schrödinger equation, it is the second order in time  and therefore the solutions are specified on both and everywhere t • Probability and current source densities are  i    *  (r,t)   *   2m c2  t t  2. Probability is not positive definite 0      and can not be interpreted as J (r,t)   *  ( *)  2m i position probability 0

• But satisfy Continuity equation  (r,t)    guarantees that “No probability is lost”    J(r,t)  0 t Problems of Negative energies with with Klein Gordon Equation Free particle K.G. Equation 2  2  (2c22  m2c4 ) t 2 0    i ( pr Et ) Plane wave solution  (r,t) Ae    E ; p  k Energy eigen values are 2 4 2 2 E   (m0 c  p c ) • We have additional ‘’ solutions and energy spectrum is not bound from below. • Then arbitrary large energy can be extracted from the system if external perturbation leads to transition between positive and negative energy states. • K.G. Equation may be interpreted as equation as the relativistic wave equation for spin less particle in the frame work of may particle theories. Non Relativistic limit of K.G.Equation • To study the nonrelativistic limit, we use the ansatz   ( i m c2t)  (r,t) (r,t) e  0 2 2 E  E'm0c where E' m0c 2 i m c2 2 i m c2   i m c ( 0 t) i m c ( 0 t) • In nonrelativistic limit  (  0 )e    0 e  t t   We get 2 2 i m c2     i m c ( 0 t)  (  0 )e  t 2 t t  2 2 4 i m c2   ( 0 t) • Which on inserting to i m0c  m0 c    2  2 e  t  K.G. Equation leads to the    2 Free Schrödinger equation for spin less particle.i   2 • So, the wave equation does not depend t 2m upon whether the particle is relativistic or non-relativistic, the Klein Gordon equation describes spin- zero particles. This is also known as the Schrodinger form of the free Klein-Gordon equation, Issues with Klein-Gordon

• Klein-Gordon Equation was obtained in 1927 by Oskar Klein and Walter Gordon. • But certain problems arise which stand in its way of being a complete description of the dynamics of a relativistic .

• Klein-Gordon Eqn. is Lorentz covariant . • It fails some of the requirements of the QM postulates. • It is second order differential equation with respect to the time unlike Schrödinger's equation of . Being second order in time, determining a particular solution required information about both ψ as well as ∂ψ/∂t. But QM says wave- function must be a complete description.  Also, Klein-Gordon admits solutions where norm and hence, probability is not conserved. So, Klein-Gordon is necessary but not sufficient. Klein-Gordon Equation: Problems

• It was actually found by Schrödinger before his non-relativistic equation. • K.G. equation is the relativistic generalization of Schrödinger’s equation and thus it may be written in covariant way. • Unlike Schrödinger’s equation it is second order differential equation in time. • However, didn’t work for the , and hence was abandoned – This is because it’s an equation for spin-0, not spin ½ particles • Also incompatible with statistical interpretation of |Y(r)|2 as the probability of finding particle at point r • However, the Klein-Gordon equation does not lead to a positive definite probability density and admits positive and negative energy solutions – these features led to it being abandoned as a viable candidate for a relativistic quantum mechanical theory. • Consequently, the arbitrary large energies can not be extracted from the system if external perturbation leads to transition between positive and negative energies. . So, Klein-Gordon is necessary but not sufficient and hence can not be accepted as the equation of of a particle with spin angular momenta. • Dirac’s strategy was to find first order equation still compatible with the relativistic energy/momentum relation. Dirac’s Insight •Dirac placed emphasis on two constraints: 1. Relativistic equation must be first order in time derivative. 2. Elements of wave- function must obey Klein-Gordon equation.

 We need an equation that was both first-order in time as well as space.  realized that this might be possible if we interpret the square root in the Einstein formula differently. Significance: • Among other virtues, Dirac's equation accounted for spin, discovered experimentally in 1925 but without a theoretical basis at the time. • Dirac's theory also predicted an for every particle, which has since been found to be correct

Paul Dirac (1902-84) • First to try to combine quantum mechanics with special relativity • Obtained the relativistic version of Schrödinger’s equation in 1928   i  E  c p  p  p  m c2  H t x x y y z z 0      i  ic(   )  m c2 t x x y y z z 0 where

3    2     2 H  c j p j m0c   ic( x  y  z )  m0c  j1  x y z • Known as the : The - yet unknown - coefficients α, and β cannot be simple numbers. Dirac equation……….

• ThenΨ cannot be a simple scalar, but has to be a column vector   (r,t)   1    2 (r,t)    (r,t)     3            N (r,t) • So that the Dirac equation reduces to be N N N       2 i  ic( x  y  z )   m0c   (H)  t  1 x y z  1  1 • Thus the Schrodinger-like equation represents a system of N coupled first-order differential equations of the componentsΨ for i = 1,2,... ,N How to determine α and β in Dirac equation………. • To continue, we demand the following natural properties: 1. The correct energy-momentum relation for a relativistic free particle 2 2 2 2 2 2 4 E  c ( px  py  pz )  m0 c  (r,t)    2. the continuity equation for the density    J(r,t)  0 and t 3. the Lorentz covariance (i. e. Lorentz form-invariance) for various form of Dirac equation 2  2   2c22  m2c4 • To fulfil the t 2  0  requirement that or E 2  p2c2  m2c4 every single 0 which on iteration gives rise to component of the     2 3     2 spinor Ψ has to  2   2c2 j k k j 2  satisfy the Klein- t j,k 1 2 x jxk Gordon equation 3     im c3 (    )   2m2c4 i.e. 0  j j 0 j1 x j Dirac’s Insight   Comparing coefficients ,  2 2  j  1  ( j 1,2,3);  1           2 1 we have:     j k k j jk  jk  k j  0 j  k • These anti-commutation      j    j  0 relations define an or in general algebra for the matrices           2 1 • Hermiticity of the Dirac Hamiltonian leads        j   j and   

 2  2 Since,  j  1  ( j  1 , 2 ,3 );   1 ; the eigenvalues can only have the values ±1. • This has no solutions in scalars, but it does have if we allow α and β to be square matrices. What should be the minimum rank for α and β in Dirac equation ?

• The anti-commutation relations       j   j  and    j  j

• leads the (the sum of the diagonal elements of the matrix) of each α and β has to be zero i.e.  2       tr  j  tr   j  tr  j   tr  j  tr  j  0

• Because the eigenvalues of α and β are equal to ±1, each matrix α and β has to possess as many positive as negative eigen values, and therefore has to be of even dimension. The smallest even dimension, N= 2, cannot be right, because only three anti-commuting matrices exist, namely the three . Therefore, the smallest dimension for which the requirements must be fulfilled is N =4. Dirac-Pauli Representation of α and β • One of possible explicit representation of the Dirac matrices, is    02  j    1 0      and    2 2  where the Pauli matrices j  0     j 2  02  12  0 1 0  i 1 0   1 0 0 0 are 1    ; 2   ; 3    with 12    and 02    1 0  i 0  0 1 0 1 0 0 which may be written in its explicit form in terms of 4 x 4 matrices as 0 0 0 1 0 0 0  i      0 0 1 0  0 0 i 0         1 0 1 0 0 2 0  i 0 0      1 0 0 0 i 0 0 0  0 0 1 0  1 0 0 0       0 0 0 1  0 1 0 0         3 1 0 0 0  0 0 1 0      0 1 0 0  0 0 0 1 • This representation is also referred as the Dirac Pauli representation • In contrast to KG Eqn. DE is of first order in all coordinates . • Space and time are on the same footings unlike the Schroedinger wave eqn. Unitary transformations for Dirac Equation • Dirac Pauli representation of α and β has been chosen to satisfy the properties of these matrices • The representation can be changed to another representation by means of unitary transformations to yield the properties of α and β    1   1 k ' Sk S ; ' SS Which leads to

    1     1    1 S j k S  S k j S  2S jk 1S    1    1    1    1   1   1  S j S S k S  S k S S j S  2S jk S S1S       j ' k ' k ' j ' 2 jk and    1     1  S j S  S j S  0    1   1   1    1   1  S j S SS  SS S j S  2S0S       j ' ' j ' 0

Dirac equation: Probabilistic interpretation    2 H  c.p  m0c 3   2 i  ick  m0c t k1 xk

     † (1 , 2 , 3 , 4 )

† 3 †   † 2 † † i  ic k  m0c   t k1 xk † † k k ,   

† 3 †   † 2 † †  i   ic k  m0c    t k1 xk 3 †  †  2 † i  ic k  m0c   t k1 xk Dirac equation: Probabilistic interpretation • subtraction with the use of hermiticity provides † 3 3 † †    †    i(   )  ic  k    k  t t  k 1 xk k 1 xk  3  †  †  i (  )  ic (  k ) t k 1 xk 3  †  †  (  )   (c  k )  0 t k 1 xk     div J  0 t 4 † * • Positive definite probability /charge source density is        1 J  c †  • k k Current source density is   or J  c †   • So the conservation law is described as d 3x  †   div J d 3x  0 t   •Thus we can accept the interpretation of ρ as a probability density [in contrast to the density for the Klein-Gordon equation which was not positive definite]. •Accordingly, we call J the density. Probability Conservation

• That the Dirac equation is first-order in time and space is straightforward. • The equation permits, when the matrices αk and β are Hermitian, a continuity equation. • The density is Y†Y, which consists of a positive definite entry that can be interpreted as probability density. • Probability can hence remain conserved. • Dirac equation is thus a happy marriage of QM and STR. Arpan Saha, Sophomore, IITB, Engineering Physics with Nanoscience The Dirac Equation : Covariant formulation • Before deriving the consequences of Dirac eqn. we first cast into more symmetric form. Let us introduce the new operators   k  ik (k 1,2,3);  4  ; x  (x,ict) • On multiplying from left by β to the Dirac equation and solving we  3  get   2   i  ic k  m0c    t k 1 xk  3   2 2  i  c(ik )   m0c  t k 1 xk 3 2   m0c    (ik )   ict k 1 xk c 3   m0c   4   k  (  ) x4 k 1 xk  4       0 1 x • Then we get the completely symmetric covariant form of Dirac eqn.   m c   0 (  )  0 (  )     (x,t)  0    x  x   The Dirac Equation : Covariant formulation • Dirac equation in covariant form can also be written in slash notations as ( )  0 (   )

• Dirac equation may also be written in covariant form as equation (  p im0c)  0 or (p im0c)  0 • Derivation   2 (c  p  m0c )  E 3 2 2  (ic k pk  i m0c )  iE k 1 3 2  (c k pk  im0c )  c 4 p4 k 1 3  ( k pk   4 p4  im0c)  0 k 1 4  (  p  im0c)  0 1

 (  p  im0c)  0 Dirac Algebra • Dirac matrices satisfy the property       2 (, 1,2,3,4)       • So we may define the entity 5 1 2 3 4

• It also ant commutes with other gammas   5  5   25 • Next we may consider the following sixteen identities   1; (1      ;)

1,  2  3  4 ;

i 2 3 , i 31, i1 2 , i 1 4 , i 2 4 , i 3 4 ;

i 1 2 3 , i1 2 4 , i 31 4 , i 2 3 4 ;

1 2 3 4 • Denoting the general element of this array by Г (A=1,2,3…….16), it 2  A can easily be verified that A  1    a a  1,  i A B C       (if A  1) B A B A  Tr A  0; Tr 1  d Dirac Algebra

• The sixteen quantities ГA (A=1,2,3…….16), are linearly independent

i.e. 16 a   0 unless a  a  ......  a  0 •  A A 1 2 16 ГA form the Dirac algebra A1 • This algebra has a unit element but is not commutative • If ψ denotes a n-dimensional “vector” in the representation space, then any operator X acting on ψ can be expressed as a of the sixteen Г i.e. A  16    1 X   xAA; xA  d Tr(XA ) a1

• If , for an algebra det[ԌAB ] ≠0 where ԌAB ≡ Tr ГA ГB , then the algebra is called semi simple algebra.

• Here Tr ГA ГB = δAB , Tr I= d δAB 16 • Hence det[ԌAB ] =d ≠0 , so that the Dirac Algebra is semi simple. Dirac Algebra • Frobenius theorem states that the number ‘r’ of possible irreducible representation of a (finite) semi simple algebra (which possesses an unit element) is equal to the number of elements in the algebra that commute with all other elements of the algebra 2 2 2 N  d1  d2 ......  dr

where N is the total number of elements of the algebra and dn is the dimensionality of nth irreducible representation. • For Dirac algebra the only commutating element is unit matrix. • So, up to the equivalence ,there exists only one irreducible 2 representation. Further N=16 so that 16=(d1 ) gives d1 =4. • Hence up to the equivalence , the Dirac algebra has only one irreducible representation and this is four dimensional. Derivation of Spin from Dirac equation • Dirac equation can be shown to describe a particle with spin ½. • For this we start from the commutation relation between the Dirac Hamiltonian and the orbital momentum of the particle

    2   H, Lz  (c  p  m0c ),(r  p)z       H, Lz  ic( x py  y px )  ic(  p)z  0 Similarly     H, Lx  ic( y pz  z py )  ic(  p)x  0     H, Ly  ic( z px  x pz )  ic(  p)z  0 or in general     H, L ic(  p)  0 • The commutators are non vanishing and neither total nor its x-,y- and z- components are constants of motion in relativistic quantum mechanics • Hence in contrast to what we expect from a free particle, the orbital angular momentum L is not a constant of motion Derivation of Spin from Dirac equation….. • Law of conservation of angular momentum fails and to save the law of conservation of angular momentum, we must amend the orbital angular momentum by an additional intrinsic angular momentum. • So, we must assume that the particle carries an intrinsic angular momentum S besides its orbital angular momentum such that total angular momentum J=L+S is conserved so that [H,J]=0. • Here S must be determined in such a way that    1. It satisfies the commutation rules S j ,Sk  i jkl Sl 2. It commutes with L (so that the familiar rules of angular momentum should be applicable to obtain J from L and S. 3. The total angular momentum J=L+S commutes with the Dirac Hamiltonian H that is it is constant of motion. • It can easily be verified that we may define S from four dimensional representation of Pauli matrices    j 0   1   j     and S j   j  0  j  2 Derivation of Spin from Dirac equation…….. • Contrary to L, the intrinsic angular momentum S does not depend on the state of motion (i.e. on space time coordinates) • The third component of S i.e. S is diagonal 1 0 0 0  z    1 0 1 0 0 S    z 2 0 0 1 0    0 0 0 1 • with the eigen values sz =±½ ћ each being two fold degenerate. • Therefore the first and third rows and columns correspond to spin up (sz =+1/2 ћ) while the second and fourth columns leads to spin down (sz =-1/2 ћ) and 1 0 0 0  0 1 0 0 S 2  S 2  S 2  S 2  3 2    s(s 1)2 x y z 4 0 0 1 0   0 0 0 1 • It shows that total intrinsic angular momentum S2 has the eigen values s(s+1) ћ² where s=1/2 Derivation of Spin from Dirac equation……….. • Thus, we are led to consider as S the spin operator   1    j 0     S   with      i ( )  ( , , ) j j j    2 1 2 3 2  0  j  • Where 1  i 2 3; 2  i 31; 3  i 1 2        2 y zi  2 z x j  2 x yk   1 1    S  2    4 i( )         S  S  iS or S j , Sk  i jkl Sl • Hence the Dirac equation admits spin and thus describes the particles of spin ½. • The spin has been automatically incorporated into the wave equation because of the unavoidable increase in the number of components of Dirac spinor ψ

    2   Consequently, total angular H, Lz  (c  p  m0c ),(r  p)z      momentum commutes with  H, Sz  ic( x py  y px )  ic(  p)z      Dirac Hamiltonian and is thus H, Sx  ic( y pz  z py )  ic(  p)x  0 verifies the law of     H, S y  ic( z px  x pz )  ic(  p)z  0 conservation of angular      H, S ic(  p)  0 momentum            H, J  H, L H, S ic(  p)  ic(  p)  0 Free Motion of a Dirac Particle • Let us examine the solution of the free Dirac equation (that is, the Dirac equation without potentials)

• Its stationary states are found with the ansatz   ( i Et )  (r,t)  (r) e    2 • Which transforms D.E. into [ic(.)  m0c ] (r)  E (r) • Again the quantity E describes the time evolution of the ψ (r). • For many applications it is useful to split up the four-component

spinor into two two-component  ф and χ i.e.  (r)   1      2 (r) (r)  (r)        (r)  (r)  3        4 (r)     1(r)    3 (r) (r)    ; (r)      2 (r)  4 (r) Free particle solutions of Dirac equation………. • Note that, since Ĥ is only a function of P̂ then [P̂, Ĥ ]=0 so that the

eigenvalues pj of P can be used to characterize the states. • In particular, we look for free-particle (plane-wave) solutions of the     i ( p.r ) form  (r)  u( p)e    ( p) • u( p)     Where u(p) is a four-component vector which satisfies  ( p)   2     • (c.p  m0c )u( p)  Eu(p) and • Therefore, writing the equation in matrix form, we find      m c2 c(  p)   E 0    E  m c2 c(  p)    0         0    0    2         2   c(  p)  m0c     0 E    c(  p) E  m0c      2   c(  p) (E  m0c )  c(  p)  0    2  • which yields two equations (E  m0c )     2 c(  p)  c(  p)  (E  m0c )  0    2  (E  m0c ) which gives either 2 2 2   2 (E  m0c )(E  m0c )  c (  p)  2 2 2   2 or (E  m0c )(E  m0c )  c (  p)  Free particle solutions of Dirac equation………. • Thus a free solution of Dirac equation can be written with the   spinor amplitude expressed in either  of the two alternative c forms p           u( p)  N c  p  or u( p)  N 2    E  m0c   E  m c2     • Where N is normalization constant. 0    2   2 2          2 • However c (  p)  c (  p)(  p)  p p i (p p)  p          • Here we have used the identity (  A)(  B)  A B i (A B) 2 2 2 2 4 2 2 2 2 4 • Hence, we get E   (c p  m0 c ) or E   (c p  m0 c ) • It means the relativistic energy eigen values in terms of relativistic energy relationship are obtained either for φ or χ or for both.

2 2 2 4 • Thus we get the energy eigen values E   (c p  m0 c )   • We see that the eigenvalues can be positive or negative and E is the energy of a particle described by the spinors Free particle solutions of Dirac equation……….

• Here, we also obtain a negative energy as a possible value like the KG equation. • These negative energy eigen values are not accepted in classical and non relativistic mechanics since we do not observe a free particle with negative energy and that’s why we have rejected them. • Thus, the Dirac equation admits positive as well as negative energy solutions and both are compatible with Dirac spinors. • So it is customary to write the positive energy solutions (E=+ε)using    first form for φ i.e.    ()  c  p  u ( p)  N   (E   )  2     m0c  • And the negative energy solutions (E=-ε)using second form for χ     c  p  i.e.     u() ( p)  N 2 (E   )    m0c       Free particle solutions of Dirac equation………. • This convention has the advantages 2 – Since cp< ε the quantity c σ.p/(ε +m0 c ) is always smaller than one and tends to zero in non relativistic limit. – Therefore the lower components of the spinor amplitude u(+) are negligible in non relativistic limit. – The same happens to the upper components of u(-) which is negligible in NR

limit 1 0     or     – In the special case in which φ and χ are each equal to 0 1        c(  p)           • We have u() ( p)  N c(  p) ; and u() ( p)  N 2          m0c   2        m0c      1   0       0   1  • Which expands to cp ( p  ip ) (1)   z  (2)   x y  u ( p)   2 (E  ,   ); u ( p)   2 (E  ,   )   m0c   m0c  ( p  ip )     x y    cp z   2   2     m0c     m0c 

  cp z   ( px  ip y )   2   2     m0c     m0c 

(3)   ( px  ip y )  (4)   cp z  u ( p)   2 (E  ,   ); u ( p)   2 (E  ,   )    m0c     m0c   1   0       0   1  Free particle solutions of Dirac equation……….   • Here we have used   p   x px  y py  z pz  0 p   0 ip   p 0   p p  ip    x    y    z    z x y    ip 0     p  ip  p   px 0   y   0  pz   x y z 

   ()  u() exp(i pr ) ( 1,2,3,4) • Obviously in each case 

• Although the expression of φ and χ as either χ + or χ – imply that all the four spinors correspond to spin up or spin down relative to the

z- axis. These spinors are not eigen function of Σ z which does not commute with Dirac Hamiltonian unless its unit vector is parallel to the momentum. • So the direction of propagation must be considered as Z-axis which

is placed parallel to momentum axis where p x = p y =0 in all of these four spinors. So we must consider     p  z p ; z p   p Free particle solutions of Dirac equation…… • As such we get the spinor amplitudes

 1   0            1      0        u(1) ( p)  N cp  N cp (,); u(2) ( p)  N cp  N (,)         0    m c2      m c2     m c2     cp   0  0  0     0   2       m0c    cp   z   0    m c2  cp    cp   0   cp    (3)   2   (4)   2   2 u ( p)  N   m c  N 0 (,); u ( p)  N   m c  N   m0c (,)  0     0     1     0             1   0    • In order to obtain the normalization constant N, we use () ( )  u† u  ( 2 )  m0c • Which yields the constant of normalization as  ( ) 2 2 m0c (  m0c ) N   2 2 2  c p  2m0c 1 2 2   (  m0c )  Free particle solutions of Dirac equation…… • However, the complete spinor ψ(r) is normalized as  † d 1 (        )d 1   1 1 2 2 3 3 4 4 • So, the complete free particle Dirac Spinor is written as

 m c2 i    () (r,t)  0 u() exp .[ ( pr  Et)] (E  ) V  • The most general free particle solution of Dirac ‘s equation is

4 2  m0c  ()  i    (r,t)   c ( p)u ( p)exp .[ ( p r  Et) p 1 V 

2 2 m0c  ()  i    (2)  i     {c ( p)u ( p)exp .[ ( p r t)  c2 ( p)u ( p)exp .[ ( p r  t)} p 1 V  

• It shows the obvious need of both positive and negative energy solutions since it is necessary to have four basic independent spinors to express an arbitrary spinor in terms of them. Dirac equation :-Particle Velocity…

• Recalling the equation of motion for velocity  dx x 1   v    x, H  x dt t i • If x̂ is explicitly independent of time, the equation of motion for    3  velocity reduces to  dx 1  1    2  vx   x, H  x,(c  j p j  m0c )  c x dt i i  • Or in general v̂=cα̂ or α̂ = v̂/c which shows thatj 1α̂ plays the role of the velocity of particle. • Since the eigen values of α̂ are either +1 or -1, it is to be concluded that the possible values of the velocity of a particle are ± c, in contradiction to the experimental fact. • So the fact is that the velocity we measure is the expectation value of v̂=cα̂ under the physical condition of a particle i.e. well defined energy and momentum. For example if the Z-axis is parallel to direction of propagation, the observed velocity is   v  c  c  †  d  c (        )d 3  3  1 3 2 4 3 1 4 2 • The phase and group velocity then accommodate the positive and  c2 p negative energies v   and v   ( for E  ) ph p g  Dirac’s interpretation of the negative energy solution • In fact, it is impossible to ignore the negative energy solutions . • The Dirac equation gives a non-zero probability of transition between a positive energy state and a negative energy state, so there must be some reason to explain why we do not observe such transitions. • Moreover, we require a complete set of wave functions in order to be able to represent an arbitrary wave function as a (Fourier) expansion. • If we exclude the negative energy states, we no longer have a complete set of wave functions. • So, if negative energy states do exist, one had to explain why did not fall down to lower energy levels. • To handle this problem, Dirac appealed to the fact that electrons are and no two electrons could take the same quantum state (the Pauli Exclusion Principle). • Interpretation of Negative Energy Solutions • Therefore, Dirac argued, there would be no problem if one could imagine that all the negative energy states were already filled with electrons, forming a so-called of electrons with negative energies. • Therefore, the exclusion principle forbids transitions from positive energy states to the negative energy states. • Consequently, electrons with positive energy would not be able to fall down to the lower level energy levels. • The existence of the “Dirac sea" of an infinite number of filled states may be philosophically unappealing, but it is at least mathematically consistent. • Since these negative energy levels were inaccessible to the few “real” electrons, the existence of such sea would not be observable. • However, Dirac himself soon noticed that there would be at least one observable manifestation of such sea. Interpretation of Negative Energy Solutions…….. • Let us assume that real electrons are described only by positive energy states , these are the states with E = +E. • All states of negative energy are occupied by electrons, one electron in each state of negative energy. • This is illustrated in Fig. where the negative 2 energy states with E < m 0 c are occupied by electrons and form the "Dirac sea". • Negative energy electrons represent the vacuum state and are unobservable; whereas real,( i.e. observable) electrons in general exist only in states of positive energy . • If to the vacuum a positive energy electron is added, the electron can suffer transitions only to the states 2 of energy between + m 0 c to + ∞, but transition into the negative energy states are forbidden due to Pauli Exclusion Principle. Dirac’s electron holes theory

2 • However, it is possible to communicate an energy E ≥ 2 m0 c to the ‘vacuum’ in the form of a (γ ray) of energy. • So, the energy gap separating the negative and positive energy 2 solutions is exactly 2 m0 c leads to the possibility to ‘raise’ one negative energy electron from the vacuum to the positive energy state. • As a result a “HOLE” is produced in the vacuum (i.e. the sea of negative energy electrons) and a positive energy electron is created which becomes observable. • Such a ‘hole’ behaves, for all purposes, exactly as a particle with same mass as that of the electron but having an opposite sign of electric charge. • Hence it may be identified with “positron” which has positive energy , because the “hole” which represents it means the missing of a certain amount of negative energy . Dirac’s electron holes theory • Thus the overall effect of the process will be the transformation of (radiative ) energy into the creation of electron-hole pair. • If -p, -ε, and –e are respectively the momentum, energy and charge of the electron in its original negative energy state, the hole in the vacuum acts as a particle of momentum +p, energy + ε, and charge +e, which is called a ‘positron’. • The process is described as e (E0)   e (E0) or   e (E0)  e (E0) which is called ‘Pair Production ’since two observable particles are simultaneously created. • The reverse process would be a transition of a positive energy electron into a vacant negative energy state or hole which can further be expressed as e (E0)  e (E0)  2 or e (E0)  e (E0)  2 • Since as a result, both the electron and hole become unobservable the process is called ‘electron-positron annihilation’ so that two photon are produced to conserve both energy and momentum. • Positrons are observed by C. D. Anderson in 1933 in cosmic rays. Concluding Remarks

• While the formulation in terms of ‘holes’ might seem to indicate some asymmetry, particles and their are absolutely symmetrical. • An operation called charge conjugation takes one from a particle’s wave function to that of its antiparticle in the same potential. • In fact, as Feynman showed we might even regard antiparticles as particles traveling backwards through time.

Arpan Saha, Sophomore, IITB, Engineering Friday, November 5, 2010 Physics with Nanoscience , The equation of continuity • To derive an equation of continuity, we first introduce the so called “Adjoint Dirac Spinor and adjoint Dirac equation”.

†         • Adjoint spinor is denoted as    4  (1 , 2 , 3 , 4 )  (1 , 2 , 3 , 4 ) (  )   0 • So the adjoint of DE (  )  0 is      • Derivation  (   )  0

(    )  0  ( j j   44 )  0 Taking its hermitian conjugate † † † † † ( j ) ( j )  (4 ) ( 4 )   0  †  † †  ( j ) j  (4 ) 4   0 † †  ( j ) 4 j  (4 ) 4 4  4  0 (   4  4  )

 ( j ) 4 j 4  (4 ) 4   0

 ( j ) j  (4 ) 4   0

 ( j ) j  (4 ) 4   0  (  )     0 The equation of continuity in covariant notations • Multiplying D. Eqn. from the left by ψ̅ and its adjoint equation from right by ψ and then adding, we get  (  )  0 • Introducing the four current density

j  ic  • We get the covariant form of equation of continuity as

 j  0 where † ( jk  ic k ; j4  ic 4  ic; &    ) Dirac particle in electromagnetic field  Here we introduce the case of an electromagnetic field whose effects are defined by the four potential {Aμ} ≡(Ā, iФ) where the vector potential Ā (r,t) and the scalar potential Ф(r,t) modify the canonical momentum and energy with electric charge e as   e   p  p  A  ; E  E  e   c   e   So the Dirac equation (E  e)  [c ( p  A)  m c2 ] c 0 changes to   e    E  [c ( p  A)  m c2  e]  H em c 0 D   em   e 2  H D  [c ( p  A)  m0c  e]  Thus the Dirac equation for a particle interactingc with EM field is 3    2  i  [k ( pk  eAk )  m0c  e] t k1  Which can also be e (    im0c)  0 or (  im0c)  0 where    ( p  A ) written as c OR ie ( D )  0 or (D )  0 where D  (  A )     c  Dirac particle in electromagnetic field…… • So, the Dirac equation in an electromagnetic field is described as  ie    (   A )    0    D   0  c   e  or   ( p  A )  im0c  0      im0c  0 ie  c  D  (  A ) • Where    c  is known as the covariant derivative. Here, one can also obtain the DE in EM field on replacing the partial derivative by covariant derivative. • So, the adjoint Dirac equation becomes ie  (  A )   0   D   0  c     e  or ( p  A )  im c  0     im c 0  c   0   0 • Which are the equations for anti electron (positron)and may also be obtained on replacing the spinor by adjoint spinor and charge –e by +e showing that if ψ is used for particle its adjoint is associated with anti particle. Gauge invariance of Dirac equation in electromagnetic field • Dirac equation in electromagnetic field  ie    (   A )    0    D   0  c   e  or   ( p  A )  im0c   0      im0c  0  c  is invariant under the following local gauge transformations  ie 1 2 A  A  &   e c with (2  )  0   x c2 t 2 • It shows that   ie    (   A )     0    D     0  c   e  or   ( p  A )  im0c    0      im0c   0  c  • It supports the invariance under local U(1) gauge symmetry of electromagnetic interactions. • The symmetry group U(1) is the group of all 1X1 unitary matrices. Non Relativistic limit of Dirac equation…… 2 • Consider a free electron at rest p=0 and E=m0c where m0 is the rest mass of the Dirac free particle and the Dirac equation reduces to  i  m c2 t 0 • Then the Four of the eigen spinors satisfying Dirac equation are 1 0   m c2   m c2 0 i 0 t 1 i 0 t  1(t)  e  (;);  2 (t)  e  (;) 0 0         0 0 0 0   m c2   m c2 0 i 0 t 0 i 0 t  3 (t)  e  (;);  4 (t)  e  (;) 1 0         0 1 • Which are known for the free particle spinor

Arpan Saha,amplitudes Sophomore, IITB, Engineering of a Dirac particle at rest . Friday, November 5, 2010 Physics with Nanoscience , Non Relativistic limit of Dirac equation…… We may thus express the four component Dirac spinor as   (r,t)   1        2 (r,t) (r,t)   1(r,t)    3 (r,t)  (r,t)       with (r,t)    & (r,t)      (r,t)  (r,t)  (r,t)  (r,t)  3     2   4      4 (r,t)        c.p(r,t)  c.p(r,t)   (r,t)  2 ;(r,t)  2 and H  E  i E  m0c E  m0c t • In non relativistic limit, we have for positive energy solutions   E  E    m c2 where m c2 c.p p 0 0   2   • Then 2 m 0 c shows that if the particle velocity is small compared to the velocity of light then two of the four component wave function are small compared to other two. • For E=+ε, φ represents the large component and χ denotes the small component while for E=-ε, φ denotes small component and χ is used for large component. In case the state of the particle does not have well defined value of momentum, we have the relation ic(.)   2 2m0c Non Relativistic limit of Dirac equation…… • For Dirac particle in electromagnetic field, we have   e  p  p  A; E  E  e c   e  (E  e  m c2 )  c{.( p  A)} 0 c   e  (E  e  m c2 )  c{.( p  A)} 0 c • For non relativistic limit in weak field approximation ,we have 2 2 E    m0c  (  e) m0c   e     c{.( p  A)}  (e) and c   e    e  c{.( p  A)} c{.( p  A)} c c   2   2    2m0c  e 2m0c • Which reveals for positive energy that χ represents the small component of the Dirac spinor ψ compared to large component φ .   e  • So, inserting c  .( p  A ) we get   c  2  2m0c    e 2  {.( p  A)}   2  c  (. )      e    e  2m0   2m0    Non Relativistic limit of Dirac equation……        • Here we have (. )2  (. )2  i.(  ) where    e   e   ie    ie     ( p  A)( p  A)  ( A)   c c c c      1 A    ( A) and      c t     e    (. )2  (. )2  [.( A)] c   e   ( p  A)2  c e        (.)  e  2m0 2m0c    • Which is recognized as where the second term on right hand side is potential energy associated with a magnetic dipole momentum  e   e    with its eigen values     B (Bohr magnetron) 2m0c 2m0c  e     • Furthermore, since the charge of electron is negative 2 m c is antiparallel to the spin. Pauli equation not only establishes a meaningful0 non relativistic analogue of Dirac equation but also shows that the Dirac particles (at least with positive energy ones) are electron. Non Relativistic limit of Dirac equation……   e   ( p  A)2 • Pauli equation  c e        (.)  e  2m0 2m0c    • Can also be obtained for first order approximation. For pure weak uniform magnetic field (i.e. in the absence of electric field), we have  1      A    r;    A 2  e    e   e2    ( p  A)2  p.p  2 p.A  A.A c c c2  e  e  1   1    ( p  A)2  p2  2 p.(   r) ( A2  (   r)2  0) c c 2 2  e  e    e    ( p  A)2  p2  .(r  p)  p2  L. c c c   e   ( p  A)2  c e           (.) (  0    0)  2m0 2m0c     p2 e   e         L. (2S.) 2m0 2m0c 2m0c   p2 e          {(L  2S).} 2m0 2m0c  Non Relativistic limit of Dirac equation…… • The coefficient of the interaction of the spin with magnetic field gives the magnetic moment of the electron corresponding to a ‘g’ factor with value ‘g=2’.  e  • The magnetic moment    is then be called as spin magnetic 2m0c moment as it occurs only for the particle with spin. • Thus the Hamiltonian of Dirac equation in non relativistic limit contains a term which takes the intrinsic properties of the electron into account showing that the Dirac particle with positive energy is an electron. • The consequence of the theory is in the excellent agreement with experiments for electron. • As such the Dirac equation is a starting point to construct a relativistic theory of a particle like electron. • The ratio of the to the spin angular momentum is equal to e/2m0 c which is twice the value for orbital angular momentum. Spin and magnetic moment from Dirac equation

• We may establish the second order differential equation by iterating the Dirac equation as  ie   ie    (   A )    (  A )    0  0     i   c   c   z y x     z 0  x  i y  1 F   D2    (D D  D D )  0      0  i          y x z  2  i  i  i 0   x y z  2 1 ie  D    ( F)  0  Fj4  i j ; Fjk   jkll (j,k,l 1,2,3) 2 c    1 A    ie     ;    A (D , D   F; F    A   A ) c t c ie   e   • So, we get  2 2  D   (.)  (.)  0  c c  1 ie ie   e      ( F )   (.)  (.) 2   c  c c • Restricting ourselves for pure magnetic field (i.e. E=0↔Φ=0), we get 3   2 2 2  e   D j  4    (.)  0  j1  c

   i ;  j 4   j4  i j ;  j k   jk  i jkll Spin and magnetic moment from Dirac equation ….

• Multiplying this equation by -ħ²/2m0 and substituting iħ∂ψ∕∂t=E’ ψ, we obtain 2     ie 2 1 2 1 2 e   ( j  Aj )  2 E'  m0c  (.)  0 2m c 2m c 2 2m c • For non relativistic limit  0 0 0  2 2 2 2 4 2 E' 'm0c ; ' m0c  E'  m0 c  2'm0c • We then get   2 e      e     (.)  ' (  p  A); p j  i ) 2m0 2m0c  c x j • This equation can be easily compared with the corresponding Schrödinger energy equation and shows that the particle in the magnetic field has a magnetic energy e      e  Vm   .  .; (  ) 2m0c 2m0c   2  V   ' • Consequently , we have  m  which implies that the particle has a 2m0   e  magnetic dipole moment    . Since the eigen values of Σ are ±1, this 2m c means that this intrinsic (i.e. independent0 of the motion) magnetic dipole moment is capable of taking on the values ±1 Bohr magnetron. • Since the charge ‘e’ of the electron is negative , the direction of its magnetic moment is antiparallel to the spin orientation. All these findings are in agreement with the experimental of spin ½ particles. Solvay Conference, 1927

Heisenberg Schrodinger Pauli

Debye

Bragg Compton

Dirac Born Bohr de Broglie Curie Wilson Planck Einstein Lorentz Concluding Remarks “Anyone who is not shocked by has not understood it.” ―