Relativistic Quantum dipankar/ ¢  Relativistic Quantum Mechanics Dipankar Chakrabarti

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Text of Relativistic Quantum dipankar/ ¢  Relativistic Quantum Mechanics Dipankar Chakrabarti

  • Relativistic Quantum Mechanics

    Dipankar Chakrabarti

    Department of Physics, Indian Institute of Technology Kanpur, Kanpur 208016, India

    (Dated: August 6, 2020)

    1

  • I. INTRODUCTION

    Till now we have dealt with non-relativistic quantum mechanics. A free particle satisfying

    Schrodinger equation has the non-relatistic energy E = ~p22m . Non-relativistc QM is applicable

    for particles with velocity much smaller than the velocity of light(v

  • where gµν is the metric of the space-time. In Minkowski space

    gµν =

    

    1 0 0 0

    0 −1 0 0

    0 0 −1 0

    0 0 0 −1

     . (2)

    So,

    ds2 = (cdt)2 − ((dx1)2 + (dx2)2 + (dx3)2) = (cdt)2 − ( ~dr)2 (3)

    Under Lorentz transformation xµ transforms as x′µ = Λµνxν where Λµ ν is a 4 × 4 matrix

    representing the Lorentz transformation operator. For example, the operator for boost along

    x1 axis

    Λµν =

    

    γ −γβ 0 0

    −γβ γ 0 0

    0 0 1 0

    0 0 0 1

     (4)

    where β = v/c and γ = 1/ √

    1− (v/c)2. So, the transformed coordinates under the boost along

    x1:

    ct′ = γ(ct− v c x1), x′1 = γ(x1 − v

    c ct), x′2 = x2, x′3 = x3. (5)

    Check that

    ds′2 = (cdt′)2 − ((dx′1)2 + (dx′2)2 + (dx′3)2) = γ2(cdt− βdx′1)2 − γ2(dx1 − βcdt)2 − (dx2)2 − (dx3)2

    = ds2 (6)

    i.e., ds2 is Lorentz invariant. ds2 can be both positive or negative unlike spatial distance ( ~dr)2

    which is always positive. If

    ds2 > 0 i.e., (cdt)2 > ( ~dr)2, the interval is called "time-like"

    ds2 < 0 i.e., (cdt)2 < ( ~dr)2, the interval is called "space-like"

    ds2 = 0 i.e., (cdt)2 = ( ~dr)2, the interval is called "light-like".

    3

  • covariant & contravariant vectors: Any quantity which transforms like xµ under Lorentz

    transformation is called a contravariant vector while anything which transforms like ∂ ∂xµ

    is called

    covariant vector. General convention for contravariant vector is aµ (i.e.,µ is in the superscript)

    and for covariant vector aµ (i.e, µ is in the subscript) i.e, ∂∂xµ = ∂µ. The inner product of a

    covariant vector and a contravariant vector is a Lorentz invariant(i.e., scalar). The contra and

    covariant vectors are related by

    xµ = ∑ ν

    gµνx ν . (7)

    Using the convention of summation over repeated indices we can write the above eqn as xµ =

    gµνx ν where ν in gµν is repeated again in xν and hence is summed over. Similarly, xµ = gµνxν .

    In Minkowski space, gµν = gµν . So, we have

    x0 = g0νxν = g00x0 + g01x1 + g02x2 + g03x3 = g00x0 = x0 (8)

    x1 = g1νxν = g10x0 + g11x1 + g12x2 + g13x3 = g11x1 = −x1. (9)

    Similarly x2 = −x2 and x3 = −x3.

    Inner product or scalar product of two 4-vectors is defined as

    A ·B = AµBµ = (A0B0 + A1B1 + A2B2 + A3B3) = (A0B0 − A1B1 − A2B2 − A3B3) (10)

    = A0B0 − ~A · ~B = gµνAµBν = gµνAµBν . (11)

    Differential operators:

    ∂µ = ∂

    ∂xµ = (1

    c

    ∂t , ∂

    ∂x1 , ∂

    ∂x2 , ∂

    ∂x3 ) (12)

    = (∂0, ∂1, ∂2, ∂3) = ( 1 c

    ∂t , ~∇) (13)

    ∂µ = gµν∂ν = ( 1 c

    ∂t ,−~∇) (14)

    The Lorentz invariant second order differential operator or the d’Alembertian operator is

    � = ∂µ∂µ = ( 1 c2 ∂2

    ∂t2 ,−( ∂

    2

    ∂x2 + ∂

    2

    ∂y2 + ∂

    2

    ∂z2 )) = ( 1

    c2 ∂2

    ∂t2 ,−∇2). (15)

    We know the relativistic mass mr = γm and energy E = mrc2 = γmc2. The energy-momentum

    4-vector is pµ = (E/c, ~p) where ~p = γm~v. So,

    p2 = gµνpµpν = pµpµ = ( E

    c )2 − (~p)2 = (γmc

    2)2 c2

    − (γm~v)2 = m2c2 (16)

    4

  • (in the natural unit ~ = c = 1, p2 = m2). So, the relativistic energy momentum relation is

    given by E2 = (~p)2c2 +m2c4. Another useful quantity is

    p · x = pµxµ = Et− ~p · ~x. (17)

    For non-relativistic particle (v

  • scalar. Thus KG equation describes the relativistic dynamics of a scalar particle. The plane

    wave solution of the KG eqn is

    φ(x) = Ne−i(Et−~p·~x) (25)

    where N is the normalization constant and energy E = ± √ ~p2c2 +m2c4 i.e., energy can be both

    positive and negative.

    Continuity Equation:

    Pre-multiply Eq.(23) by φ∗(x) to get

    φ∗(x) ( 1 c2 ∂2

    ∂t2 − ~∇2

    ) φ(x) = −m

    2c2

    ~2 φ∗(x)φ(x) (26)

    Now take the complex conjugate of Eq.(23) and post-multiply with φ(x), which gives

    ( 1 c2 ∂2

    ∂t2 φ∗)φ− (~∇2φ∗)φ = −m

    2c2

    ~2 φ∗(x)φ(x) (27)

    Eq(26)-Eq(27) gives:

    φ∗ 1 c2 ∂2φ

    ∂t2 − 1 c2 ∂2φ∗

    ∂t2 φ− (φ∗∇2φ− φ∇2φ∗) = 0 (28)

    ⇒ 1 c

    ∂t

    [ i~

    2mc

    ( φ∗ ∂φ

    ∂t − ∂φ

    ∂t φ )]

    + ~∇ · [ ~ 2im

    ( φ∗~∇φ− (~∇φ∗)φ

    )] = 0 (29)

    ⇒ 1 c

    ∂t ρ+ ~∇ ·~j = 0 (30)

    ⇒ ∂µjµ = 0 (31)

    This is the continuity equation for the KG eqn, where

    j0 = ρ = i~2mc

    ( φ∗ ∂φ

    ∂t − ∂φ

    ∂t φ )

    (32)

    ~j = ~2im

    ( φ∗~∇φ− (~∇φ∗)φ

    ) . (33)

    Recall the continuity eqn for Schrodinger equation, ρ is the probability density and ~j is the

    probability current. Continuity equation has the interpretation of conservation of probability.

    It tells that if the probability of finding a particle in some region decreases, the probability of

    finding it out side that region increases, i.e., there is a flow of probability current so that the total

    probability remains conserved. Since the KG eqn also satisfies the same continuity eqn, it is

    natural to interpret ρ as the probability density and ~j as the probability current. [Note: Density

    6

  • transforms like the 0th component of a 4−vector (jµ) under Lorentz transformation. Since φ

    is a Lorentz invariant quantity, φ2 does not transform like a density, but ρ defined in Eq.(32)

    does.] The probability density corresponding to the plane wave solution reads ρ = 2|N |2E.

    There are two major problems with the KG equation.

    (1) The eqn has both positive and negative energy solutions. The negative energy solution

    poses a problem! For large |~p| we can have large negative energy, i.e., the system become

    unbounded from below. So, we can extract any arbitrary large amount of energy from the

    system by pushing it into more and more negative energy states. One may say, we truncate

    the physical space to be the positive energy states only i.e, only E = + √ ~p2c2 +m2c4 are

    physical. But then (a) the eigenstates don’t form a complete basis states, (both +ve and -ve

    energy states are Fourier modes of φ); if we don’t have completeness relation, we cannot have

    superposition principle too ie., we cannot expand a state χ in the basis of φ ( i.e., χ = ∑i ciφi is no longer valid) and (b) a perturbation may cause the system to jump to a negative energy

    states. Since -ve energy states are valid solutions of the KG equation, we can not stop that.

    So, just interpreting negative energy states as unphysical does not work.

    (2) The second problem is associated with the probability density. As we have seen ρ =

    2|N |2E, i.e, ρ is negative if E is negative. But to interpret ρ as the probability density, it must

    be positive definite.

    [Though in QM, KG equation looks awkward at this moment, but in QFT this is a valid

    equation for scalar (spin=0)particles. Feynman and Stückelberg interpreted the positive energy

    states as particles propagating forward in time and negative energy states are propagating back-

    ward in time and thus represent antiparticles propagating forward in time. But we’ll not discuss

    those developments here.]

    7

  • B. Dirac Equation:

    The probability density in KG eqn depends on energy and becomes negative for negative

    energy. The energy in the expression of ρ appears due to the time derivative in Eq.(32). Dirac

    realised that this is due to the fact that KG eqn involves second order time derivative. Notice

    that Schrodinger equation invoves first order time derivative, and ρ does not involve any time

    derivative.. So, if we want to write a relativistic wave equation with positive definite probability

    density, the equation should be first order in time derivative. To be consistent with the Lorentz

    transformations in special theory of relativity, the wave equation with first order time derivative

    must also be first order in space derivatives. So, Dirac wrote the Hamiltonian as

    H = α1p1c+ α2p2c+ α3p3c+ βmc2. (34)

    Writing the momentum in differential operator form in the position space, we must have the

    wave equation

    i~ ∂ψ(x) ∂t

    = ( − i~c(α1

    ∂x1 + α2

    ∂x2 + α3