Time reversal operator Masatsugu Sei Suzuki, Department of State University of New York at Binghamton (Date: December 08, 2016)

If a movie film is run backwards, all movement reversed can be seen: a falling stone will be replaced by a stone projected upwards. This is true because Newton’s equation of in a uniform gravitational field is unchanged when the time t is replaced by –t. However, in the Schrödinger picture of quantum , the simple time inversion operator t → −t also changes the sign of the energy, because the energy operator is given by ih ∂ /( ∂t) . Thus in , the time inversion operation has a much more drastic effect than simply reversing the motion. Since the idea of reversing the motion is perfectly sensible in classical mechanics, i.e., one which reverses directions of motion but does not change the sign of the energy. Here we construct such an operation, called time-reversal and denoted by Θˆ . It will turn out that Θˆ is a very unusual operator, being non-linear.

Θˆ = UKˆˆ (Uˆ and Kˆ decomposition) where Θˆ is an anti-unitary operator, Uˆ is the unitary operator, and Kˆ is the complex conjugate operator.

______H.A. Kramers Hendrik Anthony "Hans" Kramers (2 February 1894 – 24 April 1952) was a Dutch who worked with to understand how electromagnetic waves interact with matter.

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https://en.wikipedia.org/wiki/Hans_Kramers

______

1. Classical mechanics The time-reversal invariance is the only one that can be discussed within the framework of classical theory. A symmetric principle or a conservation law is a result of an invariance property of the theory under a certain transformation group. Some correspondences between invariance properties of a system and the conservation laws resulting from them are:

Transformation group Conservation law translation in time energy translation in space momentum rotation angular momentum

The time reversal transformation is the only discontinuous transformation appearing in classical theory. (K. Nishijima, Fundamental particles W.A. Benjamin, 1963, p.20).

For example, we consider the motion of a charged particle of charge q and mass m in an electric field E t)( ,

d 2 m r t)( = qE t)( . (1) dt 2

If r t)( is a solution of this equations, then so is r(−t) as follows easily from the fact that the equations are second order in time, so that the two changes of sign coming from t → −t cancel.

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In other words, if we take a movie of a motion which is physically allowed according to Eq. (1) and run it backwards, the reversed motion is also physically allowed. However, this property does not hold for magnetic forces, in which case the equations of motion include first order time derivatives:

d 2 q d m r t)( = r t)( × B t)( , (2) dt 2 c dt

In these equations, the left-hand side is invariant under t → −t , while the right-hand side changes sign. For example, in a constant magnetic field, the sense of the circular motion of a charged particle (clockwise or counterclockwise) is determined by the charge of the particle, not the initial conditions, and the time-reversed motion r(−t) has the wrong sense. We note, however, that if we make the replacement B → −B as well as t → −t , then time-reversal invariance is restored to Eq.(.2). In other words, the time reversed motion is physically allowable in the reversed magnetic field. The time reversal is an odd kind of symmetry. It suggests that a motion picture of a physical event could be run without the viewer being able to tell something is wrong. Application of the time reversal operation classically requires reversing all velocities and letting time proceed in the reverse direction so that a system runs back through its past history. That this is a classically allowed symmetry follows from the fact that F = ma involves only a second derivative with respect to time. A similar situation holds in quantum mechanics provided that H is real, as it is expected to be, since it represents energy (Tinkham, Group theory and quantum mechanics). We consider a set S of dynamical variables - a point in phase space- which specifies the state of a dynamical system. For the case of a single particle, the point S is described by the particle co-ordinate and the momentum.

S = S x p),( .

Suppose the system to be initially in the state Si (xi , pi ) , and after a time T in the state S f (x f , p f ) ,

T Si Sf

The system is said to be reversible if there exists a transformation R, on the collection of states S, with the property that to every state S there corresponds one and only one state S R , such that

T SR SR f i

Thus R is seen to be a mapping of the phase space on itself, as illustrated in Fig. If R does not exist, the system is said to be irreversible.

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p

Si

Sf

x

R Sf

R Si

Fig. The mapping S → SR in the phase space ( x, p) for a reversible system.

For example, we consider the circular motion of a particle in the presence of a certain force field. The particle moves from the point A to the point B after time T. Let the particle reverse its motion at the point B. Then the particle traverses backward along the same trajectory from the point B to the point A after time T.

R S S A i A i T T R B Sf B Sf

Fig. Classical trajectory of a particle undergoing the circular motion.

Symbolically, we describe this fact by saying

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T Si rA,pi Sf rB,pf and

T R R Sf rB, pf Si rA, pi

The operation R, called the time-reversal operation, is defined by

r → r R = r , and p → p R = − p

We define the time-reversed state as one which the position is the same but the momentum is reversed.

(a)

p R −= p (R: time reversal).

H R = H , and

r R = r .

Hamiltonian H and r are invariant under the time reversal.

In addition to the rule (a), we also a set of rules of the R-operation.

(b) Physical quantities that are not dynamical variables are not changed by the R operation, e.g., mass, charge, etc.

(c) If F is a function of the dynamical variables A, ..., then

F A,( B,...)R = F(AR , BR ,...).

Here we also have the following theorems (the proof is given in K. Nishijima, Fundamental particles W.A. Benjamin, 1963, p.20).

(( Theorem-1)) If Q is an arbitrary dynamical variable, then

{Q, H}R +{Q R , H} = 0

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∂u ∂v ∂u ∂v vu },{ = ∑( − ) i ∂qi ∂pi ∂pi ∂qi

(( Proof ))

dQ ∂Q & ∂Q & = ∑( qi + pi ) dt i ∂qi ∂pi ∂Q ∂H ∂Q ∂H = ∑( − ) i ∂qi ∂pi ∂pi ∂qi = {Q, H}

After an infinitesimal time dt , Q changes into Q’,

Q'= Q +{Q, H}dt

Time-reversal invariance requires Q'R to be changed into QR after time dt .

QR = Q'R + Q R ,'{ H}dt = QR + {Q, H}R dt + Q R ,'{ H}dt

= QR + {Q, H}R dt +{QR +{Q, H}R dt, H}dt = QR + {Q, H}R dt +{QR , H}dt + O(dt)2

Then we have

{Q, H}R +{QR , H} = 0

(( Theorem-2))

{F,G}R −= {F R ,G R} = 0

(( Theorem-3))

R  dQ  dQ R   −=  dt  dt

(( Theorem-4))

LR = L

6 where L is the Lagrangian of the system. From these three rules and four theorems, we can develop the theory of time reversal in classical physics.

2. Approach from the Electromagnetic theory (i) Hamiltonian in the presence of E and B The Hamiltonian of a charged particle with mass m and charge q is given by

1 q H = [ p − A(r)]2 + qφ r)( 2m c where A and φ are the usual potentials of the electromagnetic field. Since p R −= p , H R = H ,and r R = r , we have

A r)( R −= A r)( , and

φ r)( R = φ r)( .

Noting that

B = ∇× A,

1 ∂A E −= − ∇φ . c ∂t then we have

E R = E ,

B R = −B .

(ii) Maxwell's equation Maxwell’s equation is given by

 ⋅∇ E = 4πρ  1 ∂B ∇ × E −=  c ∂t , ⋅∇ B = 0  4π 1 ∂E ∇ × B = j +  c c ∂t

From this, we get the expressions

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ρ R = ρ

and

j R −= j

(iii) Lorentz force on a particle of charge q The Lorentz force is given by

1 F = q(E + v × B) c

Then we get

1 F R = q(E R + v R × B R ) c 1 = q(E + v × B) c = F which means that F is invariant under the time reversal, where we use

R  dr  dr R dr v R =   −= −= −= v  dt  dt dt

(iv) The poynting vector S

c S = E × B 4π

So we get

c c S R = E R × B R −= E × B −= S 4π 4π under R. Under the time-reversal R an electromagnetic wave reverses its direction of propagation leaving the polarization vector ( E R = E ) unchanged.

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3. Summary In summary, there are two kinds of operators.

(i) Even: Classical variables which do not change upon time reversal include:

r, the position of a particle in three-space a, the acceleration of the particle F, the force on the particle H, the energy of the particle φ, the electric potential (voltage) E, the electric field D, the electric displacement ρ, the density of electric charge P, the electric polarization All masses, charges, coupling constants, and other physical constants, except those associated with the weak force

(ii) Odd: Classical variables which are negated by time reversal include:

t, the time when an event occurs v, the velocity of a particle p, the linear momentum of a particle L, the angular momentum of a particle (both orbital and ) A, the electromagnetic vector potential B, the magnetic induction H, the magnetic field J, the density of electric current M, the magnetization S, Poynting vector

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4. Irreversible case If an equation is not invariant, then the system is irreversible.

A simple example is given by Ohm’s law;

j = σE : irreversible on time reversal. because j R −= j and E R = E , where σ is the conductivity. In this case Joule heat is produced so that the system is not reversible.

5. Time-reversal in quantum mechanics A system is said to exhibit symmetry under the time reversal if, at least in principle, its time development may be reversed and all physical processes run backwards, with initial and final states interchanged. Symmetry between the two directions of motion in time implies that to every state ψ there corresponds a time-reversed state Θψ and that the transformation Θ preserves the value of all probabilities, thus leaving invariant the absolute value of any inner product between states. We start with the Schrödinger equation given by

∂ ih ψ r t),( = Hψ r t),( . ∂t

Suppose that ψ r t),( is a solution. We can easily verify that ψ r,( −t) is not a solution because of the first-order time derivative. However,

∂ − ih ψ * r t),( = H ψ ** r t),( = Hψ * r t),( ∂t

Here we use the reality of H. When t → −t

∂ ih ψ * r t),( = Hψ * r,( −t) . ∂t

This means that ψ * r,( −t ) is a solution of the Schrödinger equation. The time reversal state is defined by

ψ R r t),( =ψ * r,( −t)

ψ R r t),( = r Θˆ ψ (−t) = Θψ r,( −t) = r ψ (−t) * where Θˆ is the time reversal operator and Θ is the same operator in the r representation.

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If we consider a ,

i − Et ψ r t),( = e h ψ r )0,( and

i − Et * * ψ r,( −t) = e h ψ r )0,(

Then we have

i − Et R * * ψ r t),( =ψ r,( −t) = e h ψ r )0,( or

i i − Et − Et * Θe h ψ r )0,( = e h ψ r )0,( .

When t = 0, we have

ψ R r )0,( = Θψ r )0,( =ψ * r )0,(

((Example-1)) Plane wave

As a simple example, we consider the plane wave propagating with the momentum hk .

ψ r t),( = ei( rk −⋅ ωt) = e− ωti ψ r )0,( , with ψ r )0,( = ei ⋅rk . Then we get

ψ R r t),( =ψ * r,( −t) = [ei( rk +⋅ ωt ]*) = e i( rk +⋅− ωt) = e− ωti ψ * r )0,( , implying that the ψ * r,( −t ) is the plane wave propagating with the momentum − hk .

(( Example-2)) Θˆ iΘˆ −1 = −i1ˆ Using the formula

ψ R r )0,( = Θψ r )0,( =ψ * r )0,( , with the change of ψ r )0,( → iψ r )0,( , we get

Θ[iψ r )]0,( = [iψ r )]0,( * −= iψ * r )0,( −= iΘψ r )0,( ,

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For arbitrary wavefunction ψ r )0,( , we have

Θi = −iΘ or

ΘiΘ−1 −= i .

In the operator in the quantum mechanics, this can be rewritten as

Θˆ iΘˆ −1 = −i1ˆ

6. Quantum mechanical expression: analogy from classical mechanics Time reversal is really just reversal of motion. Under time reversal

xˆ '= xˆ , pˆ '= − pˆ and we also switch initial and final states. Imagine taking a movie of some process and then playing the movie backwards. Does the motion in the backward played movie obey the laws of physics? If yes, then the process is invariant under time reversal. If not, then it is non-invariant under time reversal.

In particular, setting t = 0,

ψ R )0( = Θˆ ψ )0( , or ψ R = Θˆ ψ we see that Θˆ maps the initial conditions of the original motion into the initial conditions of the time-reversed motion.

(a)

We start with the definition

ψ R (t = )0 = Θˆ ψ (t = 0

The analogy from the classical mechanics shows that

r R = r

In quantum mechanics, this can be rewritten as

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ψ R (t = )0 rˆψ R (t = )0 = ψ (t = )0 rˆψ (t = )0 or

ψ (t = )0 rˆψ R (t = )0 = ψ (t = )0 Θˆ +rˆΘˆ ψ (t = )0 = ψ (t = )0 rˆψ (t = )0 where

Θˆ + = Θˆ −1

Then we have

Θˆ +rˆΘˆ = rˆ, Θˆ rˆΘˆ −1 = rˆ , Θˆ rˆ = rˆΘˆ

We also get the similar expression for pˆ ,

Θˆ pˆΘˆ −1 −= pˆ Θˆ pˆ −= pˆΘˆ from the analogy of classical mechanics, pR −= p

(b) Since rˆΘˆ r' = Θˆ ˆ rr ' = Θˆ rr '' = r'Θˆ r'

pˆΘˆ p' = Θ− ˆ pˆ p' = Θ− ˆ p p'' = − p'Θˆ p'

Θˆ r' is the eigenket of rˆ with the eigenvalue r', leading to the expression

Θˆ r' = r' ,

Θˆ p' is the eigenket of pˆ with the eigenvalue p', leading to the expression

Θˆ p' = − p' .

7. Wigner’s time reversal transformation We start with Schrödinger equation given by

∂ ih ψ t)( = Hˆ ψ t)( (1) ∂t

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1 Hˆ = pˆ 2 + V (rˆ) 2m where

[H ,Θˆ ] = 0 , or ΘHΘˆˆˆ −1 = Hˆ since

Θˆ pˆΘˆ −1 −= pˆ , Θˆ rˆΘˆ −1 = rˆ

The application of the operator Θˆ to both sides of the Schrödinger equation gives

∂ Θˆ ih ψ t)( = ΘHˆˆ ψ t)( ∂t or

∂ hΘiΘˆˆ −1Θˆ ψ t)( = HΘˆˆ ψ t)( ∂t or

∂ − ih Θˆ ψ t)( = HΘˆˆ ψ t)( (2) ∂t

If we change t into –t in Eq.(2),

∂ ih Θψ (−t) = HˆΘψ (−t) (3) ∂t which satisfies the original Schrödinger equation. On the time-reversal t → t'= −t , the state vector changes into a new state defined by

ψ R t)( = Θˆ ψ (−t) or

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r ψ R t)( = r Θˆ ψ (−t) = Θ r ψ (−t) = r ψ (−t) * or

* r ψ R t)( = r ψ (−t)

Here we note that

ψ (−t) = ∫ d rr '' r' ψ (−t)

Θˆ ψ (−t) = ∫ dr r'' ψ (−t) *Θˆ r'

= ∫ d rr '' r'ψ (−t) *

r Θˆ ψ (−t) = dr r'' ψ (−t) * Θˆ r' ∫ = r ψ (−t) *

8. The at t = 0 At t = 0, we have the relation

r Θˆ ψ = r ψ *

Here we note that

ψ = ∫ d rr '' r' ψ

* * Θˆ ψ = ∫dr r'' ψ Θˆ r' =∫d rr '' r'ψ

* * r Θˆ ψ = ∫ dr' rr ' r' ψ = r ψ since

Θˆ (c r )' = c*Θˆ r' = c* r' and

Θˆ r' = r' .

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where Θˆ is the anti-unitary operator,

Θˆ =UKˆˆ , Θˆ (c r )' = UKˆˆ (c r )' = c*UKˆˆ r )'( = c*Θˆ r'

ˆ KcK ˆ −1 = c*1ˆ , where c is a complex number. Kˆ (the complex conjugate operator) is an operator which takes any complex number into its conjugate complex.

Kˆ −1 = Kˆ .

Θˆ = UKˆˆ : Uˆ and Kˆ decomposition.

* (( Note )) p Θˆ ψ = − p ψ

We also note that

ψ = ∫ dp p'' p' ψ

Θˆ ψ = ∫ dp' p' ψ * Θˆ p' = ∫ dp' − p' p' ψ * = ∫ dp p'' − p' ψ *

* * p Θˆ ψ = ∫ dp' p p' − p' ψ = − p ψ since

Θˆ (c p )' = c*Θˆ p' = c* − p' and

Θˆ p' = − p' .

9. Property of the state vector under the time

i ψ r t),( = r exp(− ˆtH )ψ )0( h

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i * i ψ R tr ),( =ψ * r,( −t) = r exp( ˆtH )ψ )0( = ψ )0( exp(− ˆtH ) r h h

We show that

~ φ )0( ψ~ )0( = ψ )0( φ )0(

~ φ )0( = φ R , ψˆ )0( = ψ R

(( Proof ))

ψ~ )0( = Θψ )0( = Θ∑ n n ψ )0( =∑Uˆ n n ψ )0( * =∑Uˆ n ψ )0( n n n n

~ * φ )0( = Θ φ )0( = Θ∑ m m φ )0( =∑Uˆ m m φ )0( =∑Uˆ m φ )0( m m m n

~ * φ )0( = ∑ φ m m Uˆ + n

~ * φ )0( ψ~ )0( = ∑ φ )0( m m Uˆ +Uˆ n ψ n )0( ,nm * = ∑ φ )0( m δ ,nm ψ )0( n ,nm = ∑ φ )0( n * ψ )0( n n = ∑ ψ )0( n n φ )0( n = ψ )0( φ )0( where

Θˆ =UKˆˆ , Θˆ −1 = KUˆˆ + .

10. Note (Sakurai and Napolitano) We mentioned earlier that it is best to avoid an anti-unitary operator acting on bras from the right. Nevertheless, we may use

β Θˆ α

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which is to be understood always as

( β )(Θˆ α ) and never as

( β Θˆ )(α )

11. Time reversed state vector ψ R t)( i We apply the time reversal operator to the time-evolution operator Uˆ = exp(− ˆtH ) . T h

i ψ t)( = Uˆ t)( ψ )0( = exp(− ˆtH )ψ )0( T h

i where Uˆ t)( = exp(− ˆtH ) is the time evolution operator. The time reversed state vector is given T h by

ψ R t)( = Θˆ ψ (−t) = ΘUˆˆ (−t)ψ )0( i = Θˆ exp( ˆtH )Θˆ −1Θˆ ψ )0( h i = exp(Θˆ tH Θˆˆ −1)Θˆ ψ )0( h t = exp[ Θˆ Hi Θˆˆ −1]Θˆ ψ )0( h t = exp[ ΘiΘˆˆ −1ΘHΘˆˆˆ −1]Θˆ ψ )0( h i = exp(− ˆtH )Θˆ ψ )0( h = Uˆ t)( Θˆ ψ )0( where

Θˆ iΘˆ −1 = −i1ˆ , ΘHΘˆˆˆ −1 = Hˆ

Then we have

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i ψ R t)( = Θˆ ψ (−t) = exp( − ˆtH )Θˆ ψ )0( = Uˆ t)( Θˆ ψ )0( h

We note that

Θˆ ψ (−t) = Θˆ ∫ d rr '' r' ψ (−t) * = ∫ dr'Θ r' r' ψ (−t) = ∫ d rr '' r' ψ (−t) * and r Θˆ ψ (−t) = r ψ (−t) *

12. Property of the time-reversal operator Θˆ

We consider a state represented by ψ )0( at t = 0. First, we apply Θˆ to the state ψ )0( , and then let the system evolve under the influence of the Hamiltonian Hˆ .

i (1ˆ − Hˆδt)Θˆ ψ )0( h

(see Fig.(a)). This state is the same as

i Θˆ ψ (−δt) = Θˆ 1[ − Hˆ (−δt)] ψ )0( , h if the motion obeys the symmetry under time reversal (see Fig.(b)). In other words, we have

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i i Θˆ (1ˆ + Hˆδt)ψ )0( = 1( − Hˆδt)Θˆ ψ )0( , h h or

Θˆ Hi ˆ ψ )0( −= Hi ˆΘˆ ψ )0( .

Since ψ )0( is arbitrary state, we have

Θˆ Hi ˆ = − Hi Θˆˆ .

Using the property

Θˆ iΘˆ −1 = −i , we get

Θˆ Hi ˆ = ΘiΘˆˆ −1ΘHˆˆ

−= iΘˆ H

Then we have

HΘˆˆ = ΘHˆˆ , or [Hˆ ,Θˆ ] = 0

(( Example ))

Suppose that n is the eigenket of Hˆ with the eigenvalue En. Then we have

ˆˆ ˆˆ ˆ ˆ ΘH n = HΘ n = ΘEn n = EnΘ n

Thus Θˆ n is also the eigenket of Hˆ with the eigenvalue En (degenerate state)

13. Orbital angular momentum operator We now consider the time-reversal for the spherical harmonics given by

m n l,m = Yl θ φ),( ,

Using the formula

r Θˆ ψ = r ψ * ,

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we get the relation

* m * m −m n Θ l, m = n l, m = Yl θ φ),( = (− )1 Yl θ φ),( , where,

(− )1 l 2( l + )1 (l + m)! 1 d −ml Y m θ φ),( = eimφ (sinθ )2l , l 2l l! 4π (l − m)! sin m θ d(cosθ ) −ml

− m for m≥0 and Yl (θ,φ) is defined by

− m m m * Yl (θ,φ) = (−1) [Yl (θ,φ)] .

Thus we have

n Θ l,m = (− )1 m n l,−m

This means that

ˆ h Lz l,m = m l,m , (1)

Θˆ l,m = (− )1 m l,−m , and

ˆˆ h m LzΘ l,m −= m (− )1 l,−m , (2)

We also have

Θˆ 2 l,m = (− )1 m Θˆ l,−m = (− m ()1 − )1 −m l,m = l,m or

Θˆ 2 = 1ˆ.

From Eqs.(1) and (2), we have

ˆˆ h ˆ h m ˆˆ ΘLz l,m = m Θ l,m = m (− )1 l,−m = −LzΘ l,m

ˆˆˆ −1 ˆ ˆˆ ΘLzΘ Θ l,m = −LzΘ l,m

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Thus we have

ˆˆˆ −1 ˆ ΘLzΘ −= Lz

More generally

Θˆ LˆΘˆ −1 −= Lˆ

14. Angular momentum Jˆ and rotation oiperator Rˆ Lˆ is odd under time reversal. It is natural to consider that for spin angular momentum,

Θˆ SˆΘˆ −1 = −Sˆ

The spin operator is odd under the time reversal.

The operator Θˆ anti-commutes with the component of the total angular momentum Jˆ as

ΘJΘˆˆˆ −1 = −Jˆ

i Consequently it commutes with the rotation operator Rˆ = exp[− ϕ(Jˆ ⋅ n)] h

ΘRΘˆˆˆ −1 = Rˆ , or RΘˆˆ = ΘRˆˆ since

i ΘRΘˆˆˆ −1 = Θˆ exp[− ϕ(Jˆ ⋅n)]Θˆ −1 h 1 = exp[− ϕΘˆ i(Jˆ ⋅n)Θˆ −1] h 1 = exp[− ϕ(ΘiΘˆˆ −1)Θˆ (Jˆ ⋅n)Θˆ −1] h 1 = exp[− ϕ(−i)(−Jˆ ⋅n)Θˆ −1] h i = exp[− ϕ(Jˆ ⋅n)] h

15. Position and momentum operators Similar relation is valid for pˆ ,

Θˆ pˆΘˆ −1 −= pˆ

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pˆ is an odd under time reversal. This implies that

pˆΘˆ p = Θ− ˆ pˆΘˆ −1Θˆ p Θ−= ˆ pˆ p −= pΘˆ p

Thus the state Θˆ p is the momentum eigenket of pˆ with the eigenvalue (-p).

Θˆ p = − p

Similarly, we have the relation for rˆ ,

Θˆ rˆΘˆ −1 = rˆ

This implies that

rˆΘˆ r = Θˆ rˆΘˆ −1Θˆ r = Θˆ ˆ rr = rΘˆ r

Thus the state Θˆ r is the position eigenket of rˆ with the eigenvalue ( r ).

Θˆ r = r

16. Commutation relation [xˆ, pˆ ]= ih1ˆ We note that the relation Θˆ iΘˆ −1 = −i can be derived directly from the commutation relation [P. Stehle, Quantum Mechanics (Holden-Day, Inc., 1966)] as follows.

Θˆ ihΘˆ −1 = Θˆ [xˆ, pˆ]Θˆ −1 ˆ ˆ ˆ −1 ˆ ˆ ˆ −1 ˆ ˆ ˆ −1 ˆ ˆ ˆ −1 = ΘxΘ ΘpΘ − ΘpΘ ΘxΘ −= pxˆˆ + xpˆˆ −= ih1ˆ or

ΘiΘˆˆ −1 −= i1ˆ where

[xˆ, pˆ ]= ih1ˆ , and

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Θˆ xˆΘˆ −1 = xˆ , Θˆ pˆΘˆ −1 −= pˆ .

ˆ ˆ h ˆ 17. Commutation relation [J x , J y ] = Ji z

ˆ ˆ ˆ ˆ −1 ˆˆˆ −1 ˆˆˆ −1 ˆˆˆ −1 ˆˆˆ −1 Θ[J x , J y ]Θ = ΘJ xΘ ΘJ yΘ − ΘJ yΘ ΘJ xΘ = (−Jˆ )(−Jˆ ) − (−Jˆ )(−Jˆ ) x y y x ˆ ˆ = [J x , J y ] h ˆ = Ji z

ˆ h ˆˆ −1 ˆ h ˆ −1 ˆˆˆ −1 h ˆ h ˆ Θ Ji zΘ = Θi Θ ΘJ zΘ −= i (−J z ) = Ji z .

So we should expect that Jˆ transforms just as i does, which changes sign under the time reversal

18. Even and odd under the time reversal

α~ Θˆ AˆΘˆ −1 α~ = α Aˆ α ( Aˆ is Hermite operator)

When

Θˆ AˆΘˆ −1 ±= Aˆ [plus (+); even and minus (-): odd)] we get

α~ Aˆ α~ ±= α Aˆ α which means that

Aˆ ' ±= Aˆ

The expectation value taken with respect to the time-reversed state.

(( Note ))

Θˆ AˆΘˆ −1 = Aˆ : the observable Aˆ is even under the time reversal symmetry.

Θˆ AˆΘˆ −1 = −Aˆ : the observable Aˆ is odd under the time reversal symmetry.

24

(1) Θˆ iΘˆ −1 = −i1ˆ (i is a pure imaginary, 1ˆ is the identity operator).

(2) Θˆ pˆΘˆ −1 −= pˆ : Θˆ p = − p .

(3) Θˆ pˆ 2Θˆ −1 = pˆ 2 .

(4) Θˆ rˆΘˆ −1 = rˆ : Θˆ r = r ).

(5) Θˆ V (rˆ)Θˆ −1 = V (rˆ) , (V (rˆ) is a potential).

(6) ΘSΘˆˆˆ −1 −= Sˆ ( Sˆ is the spin angular momentum).

pˆ 2 (7) Hˆ , when Hˆ = +V (xˆ) and V (xˆ) is a potential energy. 2m

Θˆ HˆΘˆ −1 = Hˆ

The relation is independent of the form of V (xˆ) .

ˆ (8) Tx a )( ; translation operator

ˆˆ ˆ −1 ˆ ΘTx a)( Θ = Tx a)( since

i ΘTˆˆ a)( Θˆ −1 = Θˆ exp(− ˆap )Θˆ −1 x h 1 = exp[− aΘˆ piˆΘˆ −1] h 1 = exp[− a(ΘiΘˆˆ −1)(Θˆ pˆΘˆ −1)] h i = exp[− ˆap ] h ˆ = Tx a)(

From the above, we have

ˆˆ ˆ ˆ ΘTx a)( = Tx a)( Θ .

25

ΘTˆˆ a)( x = Θˆ x + a = x + a ,

Tˆ a)( Θˆ x = Tˆ a)( x = x + a

(9) Rˆ : rotation operator

Θˆ RˆΘˆ −1 = Rˆ or ΘRˆˆ = RΘˆˆ .

RˆΘˆ r = Rˆ r = ℜr , Θˆ Rˆ r = Θˆ ℜr = ℜr

(10) Θˆ πˆΘˆ −1 = πˆ , or Θˆ πˆ = πˆΘˆ . since

πˆΘˆ r = πˆ r = − r , Θˆ πˆ r = Θˆ − r = − r

19. Anti-unitary operator and anti-linear operator

ψ → ψ~ = Θˆ ψ

The time reversal operator acts only to the right because it entails taking the complex conjugate. If one define the time-reversal operator in terms of bras and kets,

~ * β α~ = β α = α β (fundamental property) where

α~ = Θˆ α , and

~ β = Θˆ β , are the time-reversal states. The operator Θˆ is said to be anti-unitary if

~ * β α~ = β α = α β and

ˆ * ˆ * ˆ Θ(C1 α + C2 β ) = C 1Θ α + C2 Θ β )

26

(anti-linear operator)

20. Property of Θˆ (I)

~ α~ = Θˆ α , β = Θˆ β

Suppose that β = Bˆ µ . Then we have

~ β = Θˆ β = ΘBˆˆ µ = ΘBΘˆˆˆ −1Θˆ µ = ΘBΘˆˆˆ −1 µ~

The scalar product

~ β α = α~ β = α~ ΘBΘˆˆˆ −1 µ~

21. Property of Θˆ (II) The time-reversal operator is expressed by

Θˆ =UKˆˆ (Uˆ and Kˆ decomposition) where Uˆ is the unitary operator and Kˆ is the complex-conjugate operator that forms the complex conjugate of any coefficient that multiplies a ket.

α~ = Θˆ α = UKˆˆ α = ∑UKˆˆ α' α' α α ' = ∑Uˆ α' α' α * α '

~ β = Θˆ β

= UKˆˆ β = ∑UKˆˆ α" α" β α" = ∑Uˆ α" α" β * α '

27

~ * β α~ = ∑ α" β α" Uˆ +Uˆ α' α' α αα ",' * = ∑ α" β δ αα ",' α' α αα ",' = ∑ α α' α' β α ' = α β = β α *

22. The density operator and the expectation One can then show that expectation operators must satisfy the identity

~ * β Θˆ AˆΘˆ −1 α~ = α Aˆ + β = β Aˆ α

(( Proof ))

γ = Aˆ α , γ~ = Θˆ γ = Θˆ Aˆ α or

γ = α Aˆ +

α Aˆ + β = γ β ~ * = γ~ β ~ = β γ~ ~ = β Θˆ Aˆ α ~ = β Θˆ AˆΘˆ −1Θˆ α ~ = β Θˆ AˆΘˆ −1 α~ or

~ α Aˆ + β = β Θˆ AˆΘˆ −1 α~ (1)

For the Hermitian operator Aˆ , we get

~ α Aˆ β = β Θˆ AˆΘˆ −1 α~ ( Aˆ : Hermitian operator)

28

When we take the comple conjugate of Eq.(1), we get

* ~ * α Aˆ + β = β Θˆ AˆΘˆ −1 α~ or

β Aˆ α = αˆ (Θˆ AˆΘˆ −1)+ β (2)

((Average of the operator Aˆ over the density operator ρˆ )) If we look at the average of a Hermitian operator Aˆ over the density operator ρˆ

A = Tr[ρˆAˆ] = ∑ n ρˆ ˆ nA = ∑ nˆ (Θˆ ρˆAˆΘˆ −1)+ n~ n n~

Since the ket n~ is also complete,

A = Tr(Θˆ ρˆAˆΘˆ −1)+ = Tr(Θˆ ρˆΘˆ −1Θˆ AˆΘˆ −1)+

If the density operator is invariant under the time reversal,

Θˆ ρˆΘˆ −1 = ρˆ then we have

A = Tr(ρˆΘˆ AˆΘˆ −1)+ = Tr[(Θˆ AˆΘˆ −1)+ ρˆ + ]

Since the density operator is Hermitian operator ( ρˆ + = ρˆ ),

A = Tr[ρˆ(Θˆ AˆΘˆ −1 + ]) = Θˆ AˆΘˆ −1)+ .

23. Property of Θˆ (IV) Most operators of interest are either even or odd under the time reversal.

Θˆ AˆΘˆ −1 ±= Aˆ . where Aˆ is the Hermitian operator. Suppose that Θˆ AˆΘˆ −1 ±= Aˆ

29

~ α Aˆ β = β Θˆ AˆΘˆ −1 α~ ~ = β ± Aˆ α~ ~ = ± β Aˆ α~ ~ * = ± α~ Aˆ β

If α = β

α Aˆ α ±= α~ Aˆ α~

This is consistent with the result expected from the classical mechanics.

21. Property of Θˆ (V) We can now see the invariance of the fundamental commutation relations under the time reversal

[xˆ, pˆ] = ih1ˆ

Using

~ α Aˆ β = β Θˆ AˆΘˆ −1 α~ we have

~ α [xˆ, pˆ] β = β Θˆ [xˆ, pˆ]Θˆ −1 α~ ~ = β Θˆ xˆΘˆ −1Θˆ pˆΘˆ −1 − Θˆ pˆΘˆ −1Θˆ xˆΘˆ −1 α~ ~ −= β [xˆ, pˆ]α~ or

~ ~ α ih β −= β ih α~ = β Θˆ ihΘˆ −1 α~ or

Θˆ ihΘˆ −1 = −ih1ˆ

Θˆ [xˆ, pˆ]Θˆ −1 −= [xˆ, pˆ ]

30

The operator ih1ˆ flips sign under the time-reversal operator.

How about the commutation relation of the angular momentum?

ˆ ˆ h ˆ [Lx , Ly ] = Li z ?

ˆ ˆ ˆ + ~ ˆ ˆ ˆ −1 ~ [Lx , Ly ] is not Hermitian. using the formula α A β = β ΘAΘ α

ˆ ˆ + ~ ˆ ˆ ˆ ˆ −1 ~ α [Lx , Ly ] β = β Θ[Lx , Ly ]Θ α ~ ˆˆˆ −1 ˆˆˆ −1 ˆˆˆ −1 ˆˆˆ −1 ~ = β ΘLxΘ ΘLyΘ − ΘLyΘ ΘLxΘ α ~ ˆ ˆ ~ = β [Lx , Ly ] α

h ˆ ~ h ˆ ~ α − Li z β = β Li z α

~ ˆ h ˆ ˆ −1 ~ = β Θ( Li z )Θ α or

ˆ h ˆ ˆ −1 h ˆ Θ( Li z )Θ = Li z

ˆ ˆ ˆ ˆ −1 ˆ ˆ Θ[Lx , Ly ]Θ = [Lx , Ly ] since

ˆ ˆ h ˆ [Lx , Ly ] = Li z

Finally we now consider the raising and lowering operator for the simple harmonics. These operators are even under the time reversal.

Θˆ aˆΘˆ −1 = aˆ , and Θˆ aˆ+Θˆ −1 = aˆ+ where

β  piˆ  aˆ =  xˆ +  2  mω0 

+ β  piˆ  aˆ =  xˆ −  2  mω0 

31

The Hamiltonian are usually invariant under the time reversal.

ΘHΘˆˆˆ −1 = Hˆ .

Here we list a few terms which might appear in a Hamiltonian and discuss whether they violate the time reversal or parity.

pˆ 2 1. is invariant under both 2m

2. pˆ ⋅ rˆ is invariant under parity but not time reversal.

3. Lˆ ⋅ pˆ is invariant under time reversal but not parity.

4. S ⋅ B and pˆ ⋅ Aˆ is invariant under both

5. Quenching of the orbital angular momentum

22. Property of Kˆ (I): Kˆ 2 = 1 , Kˆ + Kˆ = 1, Kˆ + = Kˆ = Kˆ −1 We start from the following equation

Kˆ ψ = Kˆ ∑ α' α' ψ = ∑ α' α' ψ * α ' α ' where

∑ α ' α ' = 1ˆ (closure relation) α '

Then we get

Kˆ 2 ψ = KK ˆˆ ψ = Kˆ ∑ α' α' ψ * = ∑ α' α' ψ = ψ α ' α ' leading to the relation

Kˆ 2 = 1

We also have the relation

32

ψ Kˆ + Kˆ ψ = (∑ α' ψ α )(' ∑ α" α" ψ * ) α ' α" = ∑ ψ α" α' ψ α' α" αα "', . = ∑ ψ α' α' ψ α ' = ψ ψ leading to the relation

Kˆ + Kˆ = 1ˆ , since

Kˆ ψ = Kˆ (∑ α" α"ψ ) = ∑ α" α"ψ * α" α "

ψ Kˆ + = ∑ α" α" ψ α "

In conclusion, we have

Kˆ 2 = 1 , Kˆ + Kˆ = 1, Kˆ + = Kˆ −1 = Kˆ .

23. Property of Kˆ (II): ˆ KcK ˆ = c* Here we use the formula

Kˆ 2 = 1ˆ .

For arbitrary state vector ψ , we have

Kˆ 2 (c* ψ ) = c* ψ or

Kˆ c*2 ψ = ˆ ˆcKK * ψ

= KˆcK ψ

33

Then we get the relation

ˆ KcK ˆ = c* , or ˆ KcK ˆ −1 = c*

24. Property of Kˆ (III): KˆAˆKˆ = Aˆ * Suppose that Aˆ is an arbitrary operator. Then we get

ˆ ˆKAK ˆ ψ = ˆ ˆKAK ˆ ∑ n n ψ n = Kˆ ∑ ˆ nA n ψ * n

= Kˆ ∑ m m ˆ nA n ψ * ,mn = ∑ m m ˆ nA * n ψ ,mn where n m = δ ,mn . We note that

T Aˆ + = (Aˆ * ) or

T Aˆ * = (Aˆ + ) ,

So we have

T n Aˆ * m = n (Aˆ + ) m = m Aˆ + n

Then we get

ˆ ˆKAK ˆ ψ = ∑ m n Aˆ + m n ψ ,mn = ∑ m m Aˆ * n n ψ ,mn = Aˆ * ψ or

KˆAˆKˆ = Aˆ *

34

where we use the closure relation.

25. Note on the property of Kˆ

Kˆ ψ = Kˆ ∑ n n ψ = ∑ Kˆ n n ψ * n n

We choose a basis that Kˆ n = n . Then we get

Kˆ ψ = ∑ n n ψ * . n

Note that this definition is basis-dependent. But our definition of the time reversal operator is also basis-dependent, so the dependency actually cancels and there is no problem. In general, we can always write an anti-unitary operator in to the product of a unitary operator with Kˆ ;

26 Schrödinger equation for the time-reversed state (using operator) We start with the Schrödinger equation

∂ ih ψ = Hˆ ψ ∂t

When t → t'= −t , ψ → ψ R

∂ ∂ ih ψ t)( = Hˆ ψ t)( , ih ψ R = Hˆ ψ R ∂t ∂t'

∂ Θˆ ihΘˆ −1Θˆ Θˆ −1Θˆ ψ t)( = ΘHΘˆˆˆ −1Θˆ ψ t)( ∂t

We use

Θˆ = UKˆˆ

∂ Uˆ ( ˆiK hKˆ −1 )Uˆ −1UKˆˆ Kˆ −1Uˆ −1UKˆˆ ψ t)( = ΘHΘˆˆˆ −1Θˆ ψ t)( ∂t

Using the relations

35

ˆiK hKˆ −1 −= ih , we get

∂ − ihUUˆˆ −1UKˆˆ Kˆ −1Uˆ −1Θˆ ψ t)( = ΘHΘˆˆˆ −1Θˆ ψ t)( ∂t or

∂ − ihUKˆˆ Kˆ −1Uˆ −1Θˆ ψ t)( = ΘHΘˆˆˆ −1Θˆ ψ t)( ∂t or

∂ − ih Θˆ ψ t)( = ΘHΘˆˆˆ −1Θˆ ψ t)( ∂t

∂ ih Θˆ ψ (−t )' = ΘHΘˆˆˆ −1Θˆ ψ (−t )' ∂t'

Using the relation

ΘHΘˆˆˆ −1 = Hˆ we have

∂ ih Θˆ ψ (−t )' = HΘˆˆ ψ (−t )' ∂t'

Switching t'→ t

∂ ih Θˆ ψ (−t) = HΘˆˆ ψ (−t) ∂t'

This means that the time-reversed state is defined by

ψ R t)( = Θˆ ψ (−t) .

36

When t = 0,

ψ R = Θˆ ψ )0(

27. Spin-less particle (I) We state an important theorem on the reality of the energy eigenfunction of a spinless particle.

(( Theorem ))

Suppose that the Hamiltonian is invariant under time reversal and the energy eigenkets φn is nondegenerate. Then the corresponding energy eigenfunction is real (or more generally, a real function times a phase factor independent of r).

(( Proof )) The Hamiltonian Hˆ is invariant under the time reversal.

ΘHˆˆ = HΘˆˆ

ˆˆ ˆˆ HΘ φn = ΘH φn = EnΘ φn

~ ˆ Thus φn = Θ φn and φn have the same energy. The non-degeneracy assumption prompts us to ˆ conclude that Θ φn and φn must have the same state.

~ ˆ φn = Θ φn = φn

We have

~ ~ ~ * * r = Θ r = r , r φn = r φn = φn r)(

We note that

~ ~ r φn = r φn = φn r)(

Then we have

* φn r)( = φn r)(

The wave function φn r )( is real.

28. Spin-less particle (II): Quenching of orbital angular momentum

37

Theorem: quenching of orbital angular momentum

ˆ When H is invariant under time reversal and φn is nondegenerate state, the orbital angular momentum is quenched;

ˆ φn L φn = 0

(( Proof-1))

ˆ ~ ˆ ˆ ˆ −1 ~ ~ ˆ ~ ˆ φn L φn = φn ΘLΘ φn −= φn L φn −= φn L φn since

~ ˆ φn = Θ φn = φn

Then we have

ˆ φn L φn = 0

(( Proof-2))

From the definition

ˆ * ˆ+ ˆ φn L φn = φn L φn = φn L φn

* When r φn = φn r)( = φn r)( is real,

h h φ Lˆ φ * = [ φ r (r × ∇) r φ ]* dr = [ φ r * (r × ∇) r φ * ]* dr n n ∫ n i n ∫ n i n or

h φ Lˆ φ * = φ r [− (r × ∇)] r φ dr −= φ Lˆ φ n n ∫ n i n n n

Therefore we have

ˆ φn L φn = 0.

The expectation value of L for any non-degenerate state is zero. If the crystal field has sufficiently low symmetry to remove all the orbital degeneracy, then, to lowest order, the orbital angular

38 momentum is zero and we say that the crystal field has completely quenched it. For this reason, the static susceptibility of iron-group is found experimentally to arise predominantly from the spin.

((Note )) Curie law The Curie law indicates that the ground state in the absence of the magnetic field is degenerate.

29. Real-space wave function at t = 0 of the time-reversal state Real-space wave function of the time reverse state is obtained as follows.

r~ = Θ r = r ~ ˆ φn = Θ φn ~ ~ * * r φn = r φn = φn r)( from the definition. Since rˆ = r , we have

~ ~ ~ * r φn = r φn = φn r)(

In summary, the real-space wave function of the time reversed state:

~ * r φn = φn r)(

(( Example )) m Yl θ φ),( is complex because the states l,±m are degenerate. Wavefunction of a plane wave ip ⋅ r ip ⋅ r exp( ) for the state p is complex. It is degenerate with exp(− ) for the state − p . h h

30. Momentum-space wave function at t = 0 of the time-reversal state It is apparent that the momentum-space wave function of the time-reversed state is not just the complex conjugate of the original momentum-space wave function.

~ k = Θ k = − k , ~ ˆ φn = Θ φn

~ ~ * * k φn = k φn = φn k)(

Since

39

~ k = − k we have

~ * − k φn = φn k)(

When k → −k , we have the momentum-space wave function of the time reversed state as

~ * * k φn = φn (−k) = − k φn

31. Degeneracy of Bloch states: E k)( = E(−k) We would expect the relation E k)( = E(−k) for the energy dispersion relation, to hold only when the direct (and hence the reciprocal) lattice admits of a center of inversion. A rather surprising result is that

E k)( = E(−k) always, whether or not the crystal is centro-symmetric. We start with a Bloch eigenfunction (one-dimensional case), which is defined as

ikx ψ k x)( = e uk x)( and

Hψ k x)( = E ψ kk x)(

where uk x )( satisfies the periodic boundary condition.

uk (x + a) = uk x)(

ψ k x)( is the energy eigenfunction of the Hamiltonian H with the energy.

ikx Tx a)( ψ k x)( = Tx (a)[e uk (x)] = e ( +axik )u (x + a) k ika ikx = e e uk x)( ika = e ψ k x)(

40 where Tx a )( is the translation operator. We now consider the time-reversed state at t = 0, which is defined as

R * −ikx * ψ x)( =ψ k x)( = e uk x)(

We note that

* * * * H ψ k x)( = Ek ψ k x)(

Both the Hamiltonian H and the energy eigenvlalue Ek are real. Thus we have

* * Hψ k x)( = E ψ kk x)(

* So it follows that ψ k x )( and ψ k x )( are wave functions with degenerate energy eigenvalue. Since

* −ikx * Tx a)( ψ k x)( = Tx (a)[e uk (x)] = e− ( +axik )u *(x + a) k −ika −ikx * = e e uk x)( −ika * = e ψ k x)(

* The wave function ψ k x )( belongs to an eigenvalue with –k.

* ψ k x)( =ψ −k x)( .

* while ψ k x )( belongs to an eigenvalue with k. Thus ψ k x )( and ψ k x )( are degenerate energy eigenstate with different k. The change of k → −k in the relation given byHψ k x)( = E ψ kk x )( leads to

* * Hψ −k x)( = E−kψ −k x)( , or Hψ k x)( = E− ψ kk x )(

Then we have

E−k = Ek .

41

So we can say that the property of E−k = Ek can be derived from the time reversal property.

((The relation, E +Gk = Ek )) This relation can be derived directly from the Bloch theorem.

From the Bloch theorem

ikx ψ k x)( = e uk x)( , we get the relation

ika ψ k (x + a) = e ψ k x)( . since

( + axik ) ika ikx 1ka ψ k (x + a) = e uk (x + a) = e e uk x)( = e ψ k x)(

We know that the reciprocal lattice G is defined by

2π G = n , ( n: integer). a

When k is replaced by k + G,

( + )aGki ika ψ +Gk (x + a) = e ψ +Gk x)( = e ψ +Gk x)( ,

iGa 2πni since e = e = 1. This implies that ψ +Gk x )( is the same as ψ k x )( .

ψ +Gk x)( =ψ k x)( .

or the energy eigenvalue of ψ +Gk x )( is the same as that of ψ k x )( ,

E +Gk = Ek .

32. Wave function at t = 0 for the free particle; Symmetry of the energy dispersion with

Ek = E−k

We assume that

ΘHΘˆˆˆ −1 = Hˆ

42

The energy eigenvalue problem

ˆ H k = Ek k

ˆˆ ˆ ΘH k = ΘEk k

Then we have

ˆˆˆ −1 ˆ ˆ ΘHΘ Θ k = ΘEk k or

ˆˆ ˆ HΘ k = Ek Θ k since ΘHΘˆˆˆ −1 = Hˆ . Here we note that

Θˆ k = − k

Then we have

ˆ H − k = Ek − k

When k is changed into –k, we get

ˆ H k = E−k k leading to the property

E−k = Ek .

33. Time reversal operator for spin 1/2 system (I) We show that the time reversal operator for the spin 1/2 system is given by

ˆ ˆ Θ −= iσˆ y K

(( Proof ))

43

ˆ ˆ −1 ˆˆ ˆ −1 ˆ + ˆ * ˆ + ˆ ˆ + Θσˆ xΘ = UKσˆ x K U = Uσˆ x U = Uσˆ xU = −σˆ x

ˆ ˆ −1 ˆˆ ˆ −1 ˆ + ˆ * ˆ + ˆ ˆ + Θσˆ yΘ = UKσˆ y K U = Uσˆ y U = −Uσˆ yU = −σˆ y

ˆ ˆ −1 ˆˆ ˆ −1 ˆ + ˆ * ˆ + ˆ ˆ + Θσˆ zΘ = UKσˆ z K U = Uσˆ z U = Uσˆ zU = σˆ z or

ˆ ˆ + Uσˆ xU = −σˆ x

ˆ ˆ + Uσˆ yU = σˆ y

ˆ ˆ + Uσˆ zU = σˆ z

We assume that the unitary operator is expressed by

ˆ ˆ U = aσ x + βσ y + γσ z +δ1 where a, b, g, and d are complex numbers. Using the Mathematica, we find that

ˆ U = βσ y = −iσ y

ˆ ˆ Θ = −iσ y K

Note that

ˆ 2 ˆ ˆ ˆ ˆ * * ˆ Θ = (−iσˆ y K)(−iσˆ y K) = iσˆ y ( iK σˆ y K) = iσˆ y (−iσˆ y ) = σ σˆˆ yy = −1

34. The time reversal operator for spin 1/2 system (II) We show that Θˆ for the spin 1/2 system is given by

i Θˆ =η exp(− Sˆ π )Kˆ −= i ση ˆ Kˆ h y y

We start with the relations

44

ˆ ˆ ˆ −1 ˆ ΘSzΘ −= Sz

ˆ ˆ ˆ ˆ ΘS z −= S zΘ

h Sˆ Θˆ + z Θ−= ˆ Sˆ + z −= Θ + z z z 2

h The time reverse state Θˆ + z is the eigenket of Sˆ with an eigenvalue − . Then we have z 2

Θˆ + z =η − z where η is a phase factor (a complex number of modulus unity). Similarly, we have

Θˆ − z =η + z .

i i In general, for the rotation operator Rˆ = exp(− Sˆ φ)exp(− Sˆ θ ) , h z h y

i i + n = Rˆ + z = exp(− Sˆ φ)exp(− Sˆ θ ) + z h z h y

i i Θˆ + n = Θˆ exp(− Sˆ φ)Θˆ −1Θˆ exp(− Sˆ θ )Θˆ −1Θˆ + z h z h y

We note that

i i Θˆ exp(− Sˆ φ)Θˆ −1 = exp(− Sˆ φ) h z h z

i i Θˆ exp(− Sˆ θ )Θˆ −1 = exp(− Sˆ θ ) h y h y

ΘSΘˆˆˆ −1 = −Sˆ and

Θˆ iΘˆ −1 = −i1ˆ .

Thus we have

ΘRΘˆˆˆ −1 = Rˆ , or ΘRˆˆ = RΘˆˆ

45

and

Θˆ + n = Θˆ Rˆ + z = RΘˆˆ + z

= ηRˆ − z = η − n

Here we note that

i i − n = exp(− Sˆ φ)exp(− Sˆ θ ) − z h z h y

i − z = exp(− Sˆ π ) + z h y where

1 Rˆ θ )( = exp(− Sˆ θ ) y h y θ = exp(−iσˆ ) y 2 θ θ = cos 1ˆ − iσˆ sin 2 y 2  θ θ  cos − sin  =  2 2  θ θ  sin cos   2 2 

Then

Θˆ + n = RˆΘˆ + z =η − n i i =η exp(− Sˆ φ)exp(− Sˆ θ ) − z h z h y i i =η exp(− Sˆ φ)exp[− Sˆ (θ +π )] + z h z h y or

i i i i exp(− Sˆ φ)exp[− Sˆ (θ )]Θˆ + z = η exp(− Sˆ φ)exp[− Sˆ (θ + π )] + z h z h y h z h y

46

i Θˆ + z = η exp(− Sˆ π ) + z h y

Therefore we have

i Θˆ =η exp(− Sˆ π )Kˆ −= i ση ˆ Kˆ h y y with

i i h exp(− Sˆ π ) = exp(− σˆ π ) h y h 2 y i = exp(− σˆ π ) 2 y

−= iσˆ y and

 θ θ  θ cos − sin  θ θ exp(−iσˆ ) =  2 2  = 1ˆcos − isin σˆ y 2 θ θ 2 2 y  sin cos   2 2 

ˆ K is an operator which takes the complex conjugate and − iσˆ y is a unitary operator. We now calculate

ψ = C+ + z + C− − z

ˆ ˆ Θψ = −i ση ˆ y K(C+ + z + C− − z ) = −i ση ˆ (C * + z + C * − z ) y + − * * = −iη(C+ σˆ y + z + C− σˆ y − z ) * * = η(C+ − z − C− + z ) since

σˆ y + z = i − z , and σˆ y − z −= i + z

Let us apply Θˆ to ψ , again

47

ˆ 2 ˆ * * Θ ψ = Θ η([ C+ − z − C− + z )] ˆ * * = −i ση ˆ y K η([ C+ − z − C− + z )] * = −i ση ˆ y[η (C+ − z − C− + z )]

= −i[(C+σˆ y − z − C−σˆ y + z )]

= −i[(C+ (−i) + z − −iC + z )]

= −(C+ + z + C− − z ) = −ψ or

Θˆ 2 = −1ˆ

This is an extraordinary result.

(( Note ))

Θˆ −1 Θ−= ˆ

We show that

ˆ ˆ −1 Θσˆ xΘ −= σˆ x , ˆ ˆ −1 Θσˆ yΘ −= σˆ y ˆ ˆ −1 Θσˆ zΘ −= σˆ z

(a)

Θˆ σˆ Θˆ −1 ψ Θ−= ˆ σˆ Θˆ (C + z + C − z ) x x + − ˆ ˆ −= (−i ση ˆ y K)σˆ x (−i ση ˆ y K)(C+ + z + C− − z ) or

48

ˆ ˆ −1 * * Θσˆ xΘ ψ = −(−i ση ˆ y K)σˆ x (−i ση ˆ y )(C+ + z + C− − z ) * * = i ση ˆ y Kσˆ x (−iη)(iC+ − z − iC− + z ) * * = i ση ˆ y Kσˆ x (η)(C+ − z − C− + z ) * * = i ση ˆ y Kη(C+ σˆ x − z − C− σˆ x + z ) * * = i ση ˆ y Kη(C+ + z − C− − z ) * = i ση ˆ yη (C+ + z − C− − z ) 2 = iη (C+σˆ y + z − C−σˆ y − z )

= i(iC+ − z + iC− − z )

= −(C+ − z + C− + z )

= −σˆ x (C+ + z + C− − z )

= −σˆ x ψ

or

ˆ ˆ −1 Θσˆ xΘ −= σˆ x

(b)

Θˆ σˆ Θˆ −1 ψ = Θ− ˆ σˆ Θˆ (C + z + C − z ) y y + − = −(−i ση ˆ y K)σˆ y (−i ση ˆ y K)(C+ + z + C− − z ) or

ˆ ˆ −1 * * Θσˆ yΘ ψ = −(−i ση ˆ y K)σˆ y (−i ση ˆ y )(C+ + z + C− − z ) 2 * * = i ση ˆ y K(−iη)σˆ y (C+ + z + C− − z ) * * = i ση ˆ y K(−iη)(C+ + z + C− − z ) * = i ση ˆ yiη (C+ + z + C− − z ) 2 = −η σˆ y (C+ + z + C− − z )

= −σˆ y (C+ + z + C− − z )

= −σˆ y ψ or

49

ˆ ˆ −1 Θσˆ yΘ −= σˆ y

(c)

Θˆ σˆ Θˆ −1 ψ = Θ− ˆ σˆ Θˆ (C + z + C − z ) z z + − = −(−i ση ˆ y K)σˆ z (−i ση ˆ y K)(C+ + z + C− − z ) or

ˆ ˆ −1 * * Θσˆ z Θ ψ = −(−i ση ˆ y K )σˆ z (−i ση ˆ y )(C+ + z + C− − z ) * * = i ση ˆ y Kσˆ z (−iη)(iC + − z − iC − + z ) * * = i ση ˆ y Kσˆ zη(C+ − z − C− + z ) * * = i ση ˆ y Kη(C+ σˆ z − z − C− σˆ z + z ) * * = i ση ˆ y Kη(−C+ − z − C− + z ) * = i ση ˆ yη (−C+ − z − C− + z ) 2 = −iη (C+σˆ y − z + C−σˆ y + z )

= −i(−iC + + z + iC − − z )

= −C+ + z + C− − z )

= −σˆ z (C+ + z + C− − z )

= −σˆ z ψ or

ˆ ˆ −1 Θσˆ zΘ −= σˆ z

Finally we show that ψ and ψ~ = Θˆ ψ are orthogonal.

C   +  ψ = C+ + + C− − =   C− 

−ηC *  ψ~ = Θˆ ψ =η(C * − − C * + ) =  −  + −  *   ηC+ 

C  ~ * *  +  ψ ψ = ()−η C− η C+   = 0 C− 

50

Suppose that [Hˆ ,Θˆ ] = 0ˆ . In this case, ψ and ψ~ = Θˆ ψ are the eigenkets of Hˆ with the same energy eigenvalue. Since ψ and ψ~ = Θˆ ψ are orthogonal to each other, these two states are degenerate. (Kramers’ theorem).

35. Time reversal state: general case More generally

Θˆ 2 j = half − int eger = − j = half − int eger

Θˆ 2 j = int eger = j = int eger or

Θˆ 2 j,m = (− )1 2 j j,m

(( Proof ))

We first note that

iπ Θˆ =η exp(− Jˆ )Kˆ (generalization) h y

We now consider a state

α = ∑ j,m j,m α m

51

ΘΘˆˆ α = Θˆ (Θˆ ∑ j,m j, m α ) m ˆ iπ ˆ ˆ = Θ(η∑exp(− J y )K j, m j, m α ) m h ˆ iπ ˆ * = Θ(η∑exp(− J y j,) m j, m α ) m h

* ˆ iπ ˆ * =η Θ(∑exp(− J y j,) m j,m α ) m h

* iπ ˆ ˆ * =η (∑exp(− J y )Θ j, m j,m α ) m h

* iπ ˆ iπ ˆ ˆ * =η ∑exp(− J y )η exp(− J y )K j, m j, m α m h h * iπ ˆ iπ ˆ =η ∑exp(− J y )η exp(− J y j,) m j, m α m h h 2 2iπ ˆ = η ∑exp(− J y j,) m j, m α m h 2iπ ˆ = ∑exp(− J y ) j,m j,m α m h = ∑(− )1 2 j j, m j, m α m = (− )1 2 j α where

2iπ exp(− Jˆ j,) m = Rˆ 2( π j,) m = (− )1 2 j j,m (in general) h y y

Then we have

Θˆ 2 α = (− )1 2 j α or

Θˆ 2 = (− )1 2 j1ˆ

(( Note )) Obviously, a double reversal of time, corresponding to the application of Θˆ 2 to all states, has no physical consequence.

Θˆ 2 ψ = C ψ ,

52

for all ψ ( C = )1 , where C is a phase factor. When the replacement of ψ → Θˆ ψ is made in the above equation, we get

C(Θˆ ψ ) = Θˆ 2 (Θˆ ψ ) = Θˆ 3 ψ = ΘΘˆˆ 2 ψ = Θˆ C ψ = C*Θˆ ψ

Then we have

C = C*

So C is real.

C = 1 or -1, depending on the nature of the system.

36. Kramers’ Degeneracy (I)

For Θˆ 2 = −1ˆ , ψ and its time reversed state Θˆ ψ ) are orthogonal to each other.

(( Proof )) We use the formula

~ * β α~ = α β = β α where

α = Θˆ ψ , α~ = Θˆ α = Θˆ (Θˆ ψ ) = Θˆ 2 ψ

~ β = ψ , β = Θˆ ψ = α

~ β = α

Since Θˆ 2 ψ = C ψ with C = ±1

α~ = Θˆ 2 ψ = C ψ

Then we have

~ β α~ = α C ψ = C α ψ and from definition

53

~ β α~ = α β = α ψ

~ since β = α . In the case C = -1,

α ψ = 0 showing that for such systems, time-reversed states ( ψ and α = Θˆ ψ ) are orthogonal.

((Kramers’ theorem)) As a corollary, if C = -1 and the Hamiltonian is invariant under time reversal, the energy eigenstates may be classified in degenerate time reversed pairs (Kramers doublet or Kramers degeneracy)

(( Proof )) Since Hˆ is invariant under time reversal,

[Hˆ ,Θˆ ] = 0ˆ .

~ ˆ Let φn and φn = Θ φn be the energy eigenket and its time-reversed states, respectively.

ˆˆ ˆˆ ˆ ˆ HΘ φn = ΘH φn = ΘEn φn = EnΘ φn

ˆ ˆ 2 ˆ ˆ Θ φn and φn belong to the same energy eigenvalue. When Θ −= 1 (half-integer), Θ φn and ˆ φn are orthogonal. This means that Θ φn and φn (having the same energy) must correspond to distinct states (degenerate states).

When the magnetic field B is applied, Hˆ may then contain terms like

54

Sˆ ⋅ B

Pˆ ⋅ Aˆ + Aˆ ⋅ Pˆ

Sˆ, Pˆ are added under time reversal, we have [Hˆ ,Θˆ ] ≠ 0ˆ

(( Note )) Suppose that there are N . N is an even or an odd integer number. We use η = 1.

Φ = u + z + z + z + z .... + z + z 1 1 2 3 4 N −1 N + u + z + z + z + z ..... + z − z 2 1 2 3 4 N −1 N + u + z + z + z + z .... z − z + 3 1 2 3 4 N −1 N ...... + u + z − z − z − z .... − z − z N −1 1 2 3 4 N −1 N + u − z − z − z − z .... − z − z N 1 2 3 4 N −1 N

ˆ − iσˆ y K + z = − z , and

ˆ − iσˆ y K − z +−= z

~ * * N Φ Φ = u1 u2 N 1[ + (− ])1

~ When N is odd, Φ Φ = .0

37. Kramers degeneracy (II) We use this formula

~ φ ψ = ψ~ φ

Suppose that

~ φ = Θˆ ψ = ψ~ , φ = Θˆ 2 ψ

Then we get

55

ψ~ ψ = ψ~ Θˆ 2 ψ

For Θˆ 2 = −1ˆ , we get

ψ~ ψ −= ψ~ ψ = 0

Thus, for odd system, the time reversal converts a state ψ into an independent state. This is the basis for the Kramers’ degeneracy. All energy levels of s system containing an odd number of electrons must be at least doubly degenerate regardless of how low the symmetry is, provided that there are no magnetic fields present to remove the time-reversal symmetry. This theorem follows from the fact that, if

Θˆ 2 ψ −= ψ then Θˆ ψ is orthogonal to ψ .

38. Physical explanation of the Kramers theorem (III) Under the time reversal procedure, the spin state changes from ± z to m z .

Θˆ + z =η − z , Θˆ − z =η + z

The spin Hamiltonian is invariant under the time reversal.

HΘˆˆ = ΘHˆˆ , or ΘHΘˆˆˆ −1 = Hˆ

Then the energy of the reversed spin state is the same as that of the original spin state before the time reversal. Since

h h Sˆ + z = + z , Sˆ − z = − z . z 2 z 2

Since + z − z = 0 , + z and Θˆ + z =η − z are different states with the same energy eigenvalue. In other words, these two states are degenerate states with the same energy eigenvalue, forming the Kramers doublet. We note that the magnetic moment µˆ is given by

56

2µ µˆ −= B Sˆ . z h z

So the magnetic moments are different for the two states + z and Θˆ + z .

2µ µˆ ± z = − B Sˆ ± z = mµ ± z z h z B

39. Example of the Kramers theorem for spin systems In the absence of an external magnetic field, there exists at least one doublet state (with double-degeneracy) among the energy levels of spin of odd number of electrons in the crystal field. This doublet is called the Kramers doublet. This theorem in general holds for any spin systems. Here we explain this theorem for the spin Hamiltonina given by

ˆ ˆ 2 H = SD z with spin S = 1, 1/2, 1, 3/2, and 2. When an external magnetic field is applied, the Kramers doublet is split. We consider the spin-orbit coupling

ˆ ˆ ˆ H so = λso L⋅ S

Because both angular momenta are odd, the Hamiltonian is even and symmetric under time- reversal. When there is no spin-orbit coupling, we know that the atom has 2 s + 1 = 2 degeneracy, but with spin-orbit coupling there is still 2-fold degeneracy because of the Kramer's degeneracy.

(a) Spin 1/2 There is one doublet (Kramers doublet) with S = ,2/1 m = ± 2/1 (having an Dh 2 ). 4

57

E S 1 2

1 2 m D 2 4

0

(b) Spin S = 1. The energy level (energy 0) for the state S = ,1 m = )0 is singlet (the ground state). The energy level (energy Dh2 ) for the state S = ,1 m = ±1 is doublet.

E

S 1

m 1 D 2

m 0 0

58

Suppose that the spin Hamiltonian is given by

ˆ ˆ 2 ˆ 2 ˆ 2 H = SD z + E(Sx − S y ) with S = 1.

We note that [Hˆ ,Θˆ ] = 0

The Hamiltonian is given by

 D 0 E  2   Hˆ = h  0 0 0     E 0 D

Hˆ 1,1 = h2[A 1,1 + B − ]1,1 (1)

Hˆ 0,1 = 0 ( 0,1 is the eigenstate of Hˆ with the eigenvalue 0)

Hˆ −1,1 = h2[B 1,1 + A − ]1,1 (2)

In the subspace of { 1,1 and −1,1 }, the Hamiltonian can be written as

 D E  ˆ h2 H sub =    E D 1 0 0 1 h2   h2   = D   + E   0 1 1 0 h2 ˆ = (D1+ Eσˆ x )

ˆ h2 ˆ h2 H sub ± x = (D1+ Eσˆ x ) ± x = (D ± E) ± x with

1 1 h2 + x =   (eigenvalue: (D + E) ) 2 1

 1  1 h2 − x =   (eigenvalue: (D − E) ) 2 −1

In summary we have three states

59

φ1 = 0,1 (energy eigenvalue, 0)

1 φ = 1,1[ + − ]1,1 (energy eigenvalue: (D + E)h2 ) 2 2

1 φ = 1,1[ − − ]1,1 (energy eigenvalue: (D − E)h2 ) 3 2

The time reversal states:

ˆ ˆ Θ φ1 = Θ 0,1 = 0,1

1 1 Θˆ φ = Θˆ 1,1[ + − ]1,1 = [(− −1,1)1 + (− )1 −1 ]1,1 = − φ 2 2 2 2

1 1 Θˆ φ = Θˆ 1,1[ − − ]1,1 = [(− −1,1)1 − (− )1 −1 ]1,1 = φ 3 2 2 3

(c) Spin S = 3/2 3 3 The energy level (energy Dh2 4/ ) for the state S = ,m ±= ) is Kramers doublet (the 2 2 9Dh2 3 1 ground state). The energy level (energy ) for the state S = ,m ±= ) is Kramers 4 2 2 doublet.

60

E 3 S 2

3 9 D 2 m 2 4

2 1 D m 4 2

0

(d) Spin 2

There are two energy levels. The energy level (energy 4Dh2 ) for the state S = ,2 m ±= )2 is doublet, and the energy level (energy Dh2 ) for the state S = ,2 m = ± )1 is also doublet. The energy level (energy 0 ) for the state S = ,2 m = )0 is singlet.

61

E S 2

4D 2 m 2

2 D m 1

m 0 0

40. Θˆ 2 = ±1ˆ

Suppose that

Θˆ 2 = η1ˆ

Using the definition of Θˆ (= UKˆˆ ) , we have

62

Θˆ 2 =UKUKˆˆˆˆ =UUˆˆ *

We note that

Uˆ −1 = Uˆ + = (Uˆ T )* = (Uˆ * )T or

T −1 Uˆ * = Uˆ −1 = Uˆ T

Then we get

−1 Θˆ 2 = UUˆˆ * = UUˆˆ T =η1ˆ

From this we have

Uˆ =ηUˆ T , or Uˆ T = Uˆη .

Then we get

Uˆ =ηUˆ T = ηUˆη = η 2Uˆ leading to

η 2 = 1, or η = ±1

We may also write

Θˆ 2 = ±1ˆ

The sign of Θˆ 2 is determined by the properties of Uˆ , which in turn are governed by the nature of the complete set of kinematic variables required by the definition of a given physical system. Therefore these are two classes of quantum mechanical systems, even systems and odd systems.

ˆ 41. U =η(−iσˆ y ) for spin 1/2 system

63

We consider the expression of the unitary operator Uˆ for the spin 1/2 system. We start with the relation

Θˆ σˆΘˆ −1 = −σˆ with

Θˆ σˆΘˆ −1 = UˆKσˆ(UˆK)−1 = UˆKσˆK −1Uˆ −1

= UˆKσˆK −1Uˆ + = Uˆσˆ *Uˆ +

Then we have

ˆ ˆ −1 ˆ * ˆ + ˆ ˆ + Θσˆ xΘ = Uσˆ x U = Uσˆ xU = −σˆ x

ˆ ˆ −1 ˆ * ˆ + ˆ ˆ + Θσˆ yΘ = Uσˆ y U −= Uσˆ yU −= σˆ y

ˆ ˆ −1 ˆ * ˆ + ˆ ˆ + Θσˆ zΘ = Uσˆ z U = Uσˆ zU −= σˆ z where

0 1 0 − i 0 1 ˆ ˆ * ˆ ˆ * ˆ ˆ * σ x = σ x =   , σ y −= σ y =   , σ z = σ z =   1 0  i 0  1 0

ˆ Hence we find that the unitary operator U anti-commutes with σˆ x ;

ˆ ˆ ˆ {U,σˆ x} = Uσˆ x + σˆ xU = 0

It also anti-commutes with σˆ z ;

ˆ ˆ ˆ {U ,σˆ z } = Uσˆ z + σˆ zU = 0 .

ˆ In contrast, U commutes with σˆ y ;

64

ˆ ˆ ˆ [U,σˆ y ] = Uσˆ y −σˆ yU = 0 .

ˆ These commutation relations suggest that U is a function of σˆ y .We assume that

ˆ U =η(−iσˆ y ) , where η is a phase factor with η = 1. Since

ˆ + * + * U =η (iσˆ y ) =η iσˆ y so we have

ˆˆ + * 2 2 ˆ UU = η(−iσˆ y )η iσˆ y = η σˆ y = 1

(( Note )) The commutation relations

σ σˆˆ yx −= σ σˆˆ xy = iσˆ z ,

σ σˆˆ zy −= σ σˆˆ yz = iσˆ x ,

σ σˆˆ xz −= σ σˆˆ zx = iσˆ y

______42. Spin 1/2 system

ˆ ˆˆ ˆ Θ = UK =η(−σˆ y )K where Uˆ is the unitary operator, and Kˆ 2 = 1ˆ .

0 −1 ˆ ˆ U =η(−iσ y ) =η  1 0 

0 1 0 −1 1 0 ˆ + ˆ 2      ˆ U U = η    =   = 1 −1 01 0  0 1

0 −1 0 −1 1 0 ˆ 2 ˆˆˆˆ ˆˆ * 2      ˆ Θ = UKUK = UU = η    −=   −= 1 1 0 1 0  0 1

65

We assume that

0 − i ˆ iφ ˆ iφ ˆ 2 U = e σ y = e   , K = 1  i 0 

Θˆ = UKˆˆ

0 − i 0 i ˆ 2 ˆˆˆˆ ˆˆ * iφ   −iφ   ˆ Θ = UKUK = UU = e  e   = −1  i 0  − i 0

((Note))

ˆ ˆ Θ = (−i)σˆ y K

ˆ −1 ˆ −1 ˆ −1 −1 −1 ˆ + ˆ Θ = [(−i)σˆ y K] = K σˆ y (−i) = Kσˆ y i = Kσˆ yi

ˆˆ −1 ˆˆ ˆ 2 2 ˆ ΘΘ = (−i)σˆ y KK σˆ yi = (−i)σˆ y K σˆ yi = (−i)σˆ y i = 1

ˆ −1 ˆ ˆ ˆ ˆ 2 ˆ ˆ 2 ˆ Θ Θ = Kσˆ yi(−i)σˆ y K = Kσˆ y K = K = 1

ˆ 2 ˆ ˆ * * * 2 ˆ Θ = (−i)σˆ y K(−i)σˆ y K = (−i)σˆ y (−i) σˆ y = σˆ σˆ yy = −σˆ y = −1 since

* σˆ y −= σˆ y

______43. N spin (1/2) particles system ( N ≥ 2 ) For simplicity we choose η = 1. The time-reversal operator for the N spin (1/2) particles system can be expressed by

ˆ N ˆ Θ = (−i) σˆ1y ⊗σˆ2 y ⊗σˆ3y ⊗⋅⋅⋅⋅⊗ σˆ Ny K using the Kronecker product.

66

(a) For the 2-electrons system

0 0 0 1   0 0 −1 0 Θˆ = (−i)σˆ ⊗ (−i)σˆ Kˆ −= K 1y 2 y 0 −1 0 0     1 0 0 0

0 0 0 1 0 0 0 1     0 0 −1 0 0 0 −1 0 Θˆ 2 = K K 0 −1 0 0 0 −1 0 0         1 0 0 0 1 0 0 0 = 1ˆ

(Identity matrix with 4 x 4)

(b) For the 3-electrons system

0 0 0 0 0 0 0 −1   0 0 0 0 0 0 1 0  0 0 0 0 0 1 0 0    ˆ ˆ 0 0 0 0 −1 0 0 0  Θ = (−i)σˆ1y ⊗ (−i)σˆ2 y ⊗ (−i)σˆ3y K =  K 0 0 0 1 0 0 0 0  0 0 −1 0 0 0 0 0    0 −1 0 0 0 0 0 0    1 0 0 0 0 0 0 0 

ˆ 2 ˆ ˆ Θ = (−i)σˆ1y ⊗ (−i)σˆ 2 y ⊗ (−i)σˆ3 y K(−i)σˆ1y ⊗ (−i)σˆ2 y ⊗ (−i)σˆ3 y K 0 0 0 0 0 0 0 −1 0 0 0 0 0 0 0 −1     0 0 0 0 0 0 1 0  0 0 0 0 0 0 1 0  0 0 0 0 0 1 0 0  0 0 0 0 0 1 0 0      0 0 0 0 −1 0 0 0  0 0 0 0 −1 0 0 0  =  K K 0 0 0 1 0 0 0 0  0 0 0 1 0 0 0 0  0 0 −1 0 0 0 0 0  0 0 −1 0 0 0 0 0      0 −1 0 0 0 0 0 0  0 −1 0 0 0 0 0 0      1 0 0 0 0 0 0 0  1 0 0 0 0 0 0 0  = −1ˆ

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(Identity matrix with 8 x 8)

(( Note ))

ˆ ˆ ˆ ˆ (A⊗ B)(ψ1 ⊗ ψ 2 ) = Aψ1 ⊗ Bψ 2 .

44. Property of Θˆ 2 Two successive applications of the time reversal operator, i.e., two reversals of motion, leave the physical situation unchanged.

Θˆ 2 ψ = cψ (1) with c =1 (c is a complex number). The change in ψ → Θψ of Eq.(1) leads to

Θˆ 2 (Θˆ ψ ) = Θˆ (Θˆ 2 ψ ) = Θˆ (cψ (2) = c*Θˆ ψ

From Eqs.(1) and (2). we get

Θˆ 2 (ψ + Θˆ ψ ) = Θˆ 2 ψ + Θˆ 3 ψ ) (3) = cψ + c*Θψ )

Equation (1) must hold for any state vector,

Θˆ 2 (ψ + Θψ ) = c (' ψ + Θψ ) for some c’, Then we have

Θˆ 2 ψ + Θˆ 3 ψ ) = c (' ψ + Θψ ) = c'ψ + c'Θψ (4)

Comparing Eq.(3) with Eq.(4), we get

c = c' , c'= c* ,

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c* = c which means that c is real and

c = ±1.

Θˆ 2 ψ = (± )1 ψ

45. Θˆ 2 = ±1 We start with the definition of the time reversal operator

Θˆ = UKˆˆ

Then we have

Θˆ 2 = UKUKˆˆˆˆ = UUˆˆ * = c

This can be rewritten as

Uˆ * = Uc ˆ + = c(Uˆ * )T = ( Ucc ˆ + )T = c2Uˆ * where we use the relation Uˆ * = Uc ˆ + twice. So we get

c2 = 1

Note that c is real. This can be shown as follows.

(UUˆˆ )** = c* , or Uˆ *Uˆ = c*

Since Uˆ * = Uc ˆ + , we get

Uˆ *Uˆ = Uc ˆ +Uˆ = c , or c = c* (c is real)

So that we have

c = ±1

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46. The matrix expression of the unitary operator Uˆ

ΘJΘˆˆˆ −1 = −Jˆ

ˆˆˆ −1 ˆ ˆˆˆ −1 ˆ ˆˆˆ −1 ˆ ΘJ xΘ −= J x , ΘJ yΘ −= J y , ΘJ zΘ −= J z

ˆ ˆ ˆ ˆ −1 ˆ ˆ ˆˆˆ −1 ˆ Θ(J x + Ji y )Θ −= (J x − Ji y ) , ΘJ +Θ −= J −

ˆ ˆ ˆ ˆ −1 ˆ ˆ ˆˆˆ −1 ˆ Θ(J x − Ji y )Θ −= (J x + Ji y ) , ΘJ −Θ −= J +

Note that

ΘJ Θˆˆˆ −1 = U JK ˆˆˆ Kˆ −1Uˆ −1 = JU ˆˆ *Uˆ −1 = JU Uˆˆˆ −1 −= Jˆ ± ± ± ± m

ˆˆˆ −1 ˆˆˆˆ −1 ˆ −1 ˆˆ * ˆ −1 ˆˆˆ −1 ˆ ΘJ zΘ = U JK z K U = JU z U = JU zU −= J z

ˆ where the matrix elements of J ± and are real;

ˆ * ˆ ˆ * ˆ J ± = J ± , J z = J z

So we have

JU ˆˆ + J Uˆˆ = 0 , JU ˆˆ + J Uˆˆ = 0 ± m z z

From these relations, we get the commutation relation

UJˆˆ 2 = Jˆ 2Uˆ with

1 Jˆ 2 = Jˆ 2 + (J Jˆˆ + J Jˆˆ ) z 2 −+ +−

Note that

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1 UJˆˆ 2Uˆ −1 = Uˆ[Jˆ 2 + (J Jˆˆ + J Jˆˆ )]Uˆ −1 z 2 −+ +− 1 = JU Uˆˆˆ −1 JU Uˆˆˆ −1 + ( JU Uˆˆˆ −1 JU Uˆˆˆ −1 + JU Uˆˆˆ −1 JU Uˆˆˆ −1) z z 2 + − − + 1 = Jˆ 2 + (J Jˆˆ + J Jˆˆ ) z 2 −+ +− = Jˆ 2

Following the procedure introduced by Sachs , we determine the matrix element of Uˆ .

ˆˆ ˆˆ (a) JU z + J zU = 0

ˆˆ ˆˆ h ˆ j,m JU z + J zU j m', = (m + m )' j,m U j m', = 0 leading to

j,m Uˆ j m', ≠ 0 only if m + m'= 0,

ˆˆ ˆˆ (b) JU + + J −U = 0

ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ j,m JU + + J −U j m', = ∑[ j,m U j,m" j,m" J+ j m', + j,m J− j,m" j,m"U j m ]', m" ˆ ˆ ˆ ˆ = j,m U j m +1', j m +1', J+ j m', + j,m J− j,m +1 j,m +1U j m', = 0

For m +1' = −m , we have

ˆ ˆ ˆ ˆ j,m U j,−m j,−m J+ j,−m −1 + j,m J− j,m +1 j,m +1U j,−m −1 = 0 or

j,m +1Uˆ j,−m −1 j,−m Jˆ j,−m −1 −= + −= 1 (1) ˆ ˆ j,m U j,−m j,m J − j,m +1 since

ˆ J+ j,m = ( j − m)( j + m + )1 j,m +1

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ˆ J− j,m = ( j + m)( j − m + )1 j,m −1

ˆ J+ j,−m −1 = ( j + m + )(1 j − m) j,−m

ˆ J− j,m +1 = ( j + m + )(1 j − m) j,m leading to

j,−m Jˆ j,−m −1 + = 1. ˆ j,m J − j,m +1

From Eq.(1), we get

j m', Uˆ j,−m' = (− )1 −mm ' = i (2 −mm )' j,m Uˆ j,−m

Since m − m' is an integer, we have the relation; (− )1 m'−m = (− )1 −mm ' . The change of m'→ −m' and m → −m

j,−m'Uˆ j m', = (− )1 m'−m = i m −m)'(2 j,−m Uˆ j,m

j,−m Uˆ j,m = i2m

Therefore, we have

Θˆ j,m = UˆKˆ j,m = ∑ j m', j m', Uˆ j,m m' = j,−m j,−m Uˆ j,m = i2m j,−m or

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Θˆ j,m = i2m j,−m . where m is either an integer or a half-integer. We note that

Θˆ 2 j,m = Θˆ i2m j,−m = i−2mΘˆ j,−m

Thus we have

Θˆ 2 j,m = i−4mΘˆ j,m

47. The expression of U for j = 1 Here we show that for j =1, Θˆ is given by

Θˆ = UKˆˆ

0 0 1   where Uˆ = 0 −1 0   1 0 0

(( Proof ))

ˆˆ ˆˆ ˆˆ ˆˆ ˆˆ ˆˆ JU x + J xU = 0 , JU y − J yU = 0 , JU z + J zU = 0 where

0 1 0 0 − i 0  1 0 0  h   h     ˆ ˆ ˆ h J x = 1 0 1 , J y =  i 0 − i , J x = 0 0 0  2   2     0 1 0 0 i 0  0 0 −1

We assume that the unitary operator is given by

U U U   11 12 13  ˆ U = U 21 U 22 U 23    U31 U32 U33 

We note that

73

2U U 0   11 12  ˆˆ ˆˆ h JU z + J zU =  U 21 0 −U23  = 0    0 −U32 − 2U33 

leading to U11 = U12 = U 21 = U 23 = U32 = U33 = 0

 U +U   0 13 22 0   2  U +U U +U JU ˆˆ + J Uˆˆ = h 22 31 0 13 22  = 0 x x  2 2    U 22 + U31  0 0   2 

ˆ + ˆ ˆ leading to U13 = U31 −= U 22 .From the condition of U U = 1, we have

 0 0 −1   Uˆ =  0 1 0    −1 0 0 

ˆ ˆˆ ˆˆ This form of U satisfies the relation JU y − J yU = 0 . We note that

 0 0 −11 0      Θˆ 1,1 = UKˆˆ 1,1 =  0 1 0 0 −= 0 −= −1,1      −1 0 0 0 1

 0 0 −10 0      Θˆ 0,1 = UKˆˆ 0,1 =  0 1 0 1 = 1 = 0,1      −1 0 0 0 0

 0 0 −10 1      Θˆ −1,1 = UKˆˆ −1,1 =  0 1 0 0 −= 0 −= 1,1      −1 0 0 1 0

Θˆ 2 = UKUKˆˆˆˆ = UUˆˆ * = 1ˆ

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48. Evaluation of Θˆ and Θˆ 2 for electrons with spin 1/2 (a) One particle with spin 1/2

j = 2/1 with m = ± 2/1 1 m = 2/1 i−4m = i−2 = −= 1 i2 m = − 2/1 i−4m = i2 −= 1

Θˆ 2 j,m −= j,−m

(b) Two particles with spin 1/2

D 2/1 × D 2/1 = D1 + D0

j = 1 1 m =1 i−4m = i−4 = = 1 i 4 m = 0 i−4m −= 1 m = −1 i−4m = i4 = 1

j = 0 m = 0 i−4m = 1

Θˆ 2 j,m = j,−m

(c) Three particles with spin 1/2

D 2/1 × D 2/1 × D 2/1 = (D1 + D0 ) × D 2/1 = D 2/3 + 2D 2/1

j = 2/3 1 m = 2/3 i−4m = i−6 = −= 1 i6 m = 2/1 i−4m = i−2 −= 1 m = − 2/1 i−4m = i2 −= 1 m = − 2/3 i−4m = i6 −= 1

j = 2/1

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1 m = 2/1 i−4m = i−2 = −= 1 i2 m = − 2/1 i−4m = i2 −= 1

Θˆ 2 j,m −= j,−m

49. Rotation operator as a unitary operator i The unitary operator is given by Uˆ = exp(− Sˆ π ) which is the rotation operator for spin S h y (=1/2, 1, 3/2, …).

(a) Spin 1/2

i 0 −1 ˆ ˆ ˆ U = exp(− S yπ ) −= iσ y =   , h 1 0 

1 0 ˆ 2 ˆˆˆˆ ˆˆ * ˆ 2   Θ = UKUK = UU = U −=   0 1 where we use the formula

i i θ θ exp(− Sˆ θ ) = exp(− σˆ θ ) = cos − isin σˆ h y 2 y 2 2 y

(b) Spin 1

0 0 1 i   Uˆ = exp(− Sˆ π ) = 0 −1 0 , h y     1 0 0

1 0 0 2 * 2   Θˆ = UKUKˆˆˆˆ = UUˆˆ = Uˆ = 0 1 0   0 0 1

(c) Spin 3/2

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0 0 0 −1   i 0 0 1 0  Uˆ = exp(− Sˆ π ) = , h y 0 −1 0 0      1 0 0 0 

1 0 0 0   0 1 0 0 Θˆ 2 = UKUKˆˆˆˆ = UUˆˆ * = Uˆ 2 −= 0 0 1 0     0 0 0 1 or

Θˆ 2 = (− )1 2 j1ˆ = (− )1 31ˆ −= 1ˆ

(d) Spin 2 ( j = )2

0 0 0 0 1   0 0 0 −1 0 i Uˆ = exp(− Sˆ π ) = 0 0 1 0 0 , h y   0 −1 0 0 0   1 0 0 0 0

1 0 0 0 0   0 1 0 0 0 Θˆ 2 = UKUKˆˆˆˆ = UUˆˆ * = Uˆ 2 = 0 0 1 0 0   0 0 0 1 0   0 0 0 0 1 or

Θˆ 2 = (− )1 2 j1ˆ = (− )1 41ˆ = 1ˆ

(e) Spin 5/2 ( j = )2/5

77

0 0 0 0 0 −1   0 0 0 0 1 0  i 0 0 0 −1 0 0  Uˆ = exp(− Sˆ π ) =   , h y 0 0 1 0 0 0    0 −1 0 0 0 0    1 0 0 0 0 0 

1 0 0 0 0 0   0 1 0 0 0 0 0 0 1 0 0 0 Θˆ 2 = UKUKˆˆˆˆ = UUˆˆ * = Uˆ 2 −=   0 0 0 1 0 0   0 0 0 0 1 0   0 0 0 0 0 1 or

Θˆ 2 = (− )1 2 j1ˆ = (− )1 51ˆ −= 1ˆ

Many particle systems

We consider the time reversal operator Θˆ for N spins (spin 1/2). ˆ 1 2 N−1 N ˆ Θ = (σˆ σˆ yy .....σˆ y σˆ y )K

We note that

ˆ 2 1 2 N −1 N ˆ 1 2 N −1 N ˆ Θ = (σ σˆˆ yy .....σˆ y σˆ y )K(σˆ σˆ yy .....σˆ y σˆ y )K

1 2 N −1 N ˆ 1 ˆ −1 ˆ 2 ˆ −1 ˆ N −1 ˆ −1 ˆ N ˆ −1 ˆ 2 = (σ σˆˆ yy .....σˆ y σˆ y )Kσˆ y K Kσˆ y K ....Kσˆ y K Kσˆ y K )K

Using the relation,

ˆ 1 ˆ −1 *1 1 ˆ 2 ˆ Kσˆ y K = (σˆ y ) = −σˆ y , K = 1 we have

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ˆ 2 1 2 N −1 N ˆ 1 2 N −1 N ˆ Θ = (σˆ σˆ yy .....σˆ y σˆ y )K(σˆ σˆ yy .....σˆ y σˆ y )K N 21 22 N 2 = (− ()1 σˆ y () σˆ y ) ....(σˆ y ) = (− )1 N

For N = even Θˆ 2 = 1

For N = odd Θˆ 2 = − 1

(a) N = 2

D 2/1 × D 2/1 = D1 + D0

leading to the total spin with S = 1 and 0 (integer)

(b) N=3

D 2/1 × D 2/1 × D 2/1 = D 2/1 × (D1 + D0 ) = D 2/3 + 2D 2/1

leading to the total spin with S = 3/2 and 1/2 (integer)

50. Rotation operator for many spin systems

(a)

0 −1 ˆ ˆ A1 = −iσ y =   1 0 

i θ θ Rˆ θ )( = exp(− Aˆ θ) = cos 1ˆ − iσˆ sin y 2 1 2 y 2

When θ = π ,

π π 0 −1 ˆ ˆ ˆ ˆ ˆ Θ = Ry (θ = π ) = cos 1− iσ y sin = −iσ y ==   2 2 1 0 

Θˆ 2 = −1ˆ

79

______(b)

ˆ ˆ ˆ A2 = σˆ y ⊗1+1⊗σˆ y 0 − i − i 0     i 0 0 − i =  i 0 0 − i     0 i i 0 

0 0 0 1   i 0 0 −1 0 Θˆ = Rˆ π )( = exp(− Aˆ π ) = y 2 2 0 −1 0 0     1 0 0 0

Θˆ 2 =1ˆ ______(c)

ˆ ˆ ˆ ˆ ˆ ˆ ˆ A3 = σˆ y ⊗1⊗1+1⊗σˆ y ⊗1+1⊗1⊗σˆ y 0 − i − i 0 − i 0 0 0     i 0 0 − i 0 − i 0 0   i 0 0 − i 0 0 − i 0    0 i i 0 0 0 0 − i =    i 0 0 0 0 − i − i 0  0 i 0 0 i 0 0 − i   0 0 i 0 i 0 0 − i   0 0 0 i 0 i i 0 

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0 0 0 0 0 0 0 −1   0 0 0 0 0 0 1 0  0 0 0 0 0 1 0 0    ˆ ˆ i ˆ 0 0 0 0 −1 0 0 0  Θ = Ry π )( = exp(− A3π ) =   2 0 0 0 1 0 0 0 0  0 0 −1 0 0 0 0 0    0 −1 0 0 0 0 0 0    1 0 0 0 0 0 0 0 

Θˆ 2 = −1ˆ

51. Mathematica for the case of j = 3/2

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52. Spherical tensor under the time reversal Suppose that the operator Aˆ is either even or odd.

Θˆ AˆΘˆ −1 ±= Aˆ

So we see that we have

α~ Θˆ AˆΘˆ −1 α~ = α Aˆ α

α Aˆ α ±= α~ Aˆ α~

In an eigenstate of the angular momentum ( α j,, m , we have

ψ Aˆ ψ = ± ψ~ Aˆ ψ~ with

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ψ = α j,, m

m ψ~ = Θˆ α j,, m = im α j,, −m , ψ~ = (i* ) α j,, −m

Then we have

m α j,, m Aˆ α j,, m ±= im (i* ) α j,, −m Aˆ α j,, −m m ±= ()i 2 α j,, −m Aˆ α j,, −m ±= α j,, −m Aˆ α j,, −m

Let Aˆ be a component of a spherical tensor

ˆ ˆ k )( A = Tq=0

ˆ k)( which is Hermitian operator. Tq≠0 is not Hermitian operator. ______(( Note )) For example, the spherical components of rˆ are given by

1 1 T )1( −= (xˆ + yiˆ) , T )1( = zˆ , T )1( = (xˆ − yiˆ) 1 2 0 −1 2

+ 1 + + 1 T )1( −= (xˆ − yiˆ) , T )1( = zˆ , T )1( = (xˆ + yiˆ) 1 2 0 −1 2 where

1 Θˆ T )1( Θˆ −1 = − Θˆ (xˆ + yiˆ) 1 2 1 = − [Θˆ xˆ + Θ( yiˆ)] 2 1 = − (xˆ − yiˆ) 2 )1( = −T−1

83

ˆ )1( ˆ −1 ˆ ΘT0 Θ = Θzˆ = zˆ

1 Θˆ T )1( Θˆ −1 = Θˆ (xˆ − yiˆ) −1 2 1 = [Θˆ xˆ − Θ( yiˆ)] 2 1 = (xˆ + yiˆ) 2 )1( −= T1 ______k )( We define Tq=0 to be even (sign: +) or odd (sign: -) under the time reversal

ˆ k )( ˆ −1 k )( ΘTq=0Θ ±= Tq=0 .

Then we have

k )( k )( α j,, m Tq=0 α j,, m ±= α j,, −m Tq=0 α j,, −m (1)

The state α j,, −m is obtained by rotation

ˆ iϕ Ry (θ = π ) α j,, m = e α j,, −m

We also know the formula given by

k ˆ + ˆ k )( ˆ ˆ * ˆ k )( Ry θ)( Tq Ry θ)( = ∑ k,q Ry θ k q',)( Tq' q' −= k

So we get

k ˆ + ˆ k )( ˆ ˆ * ˆ k )( Ry π )( Tq=0 Ry π )( = ∑ k,q = 0 Ry π k q',)( Tq' q' −= k ˆ * ˆ k )( ˆ * ˆ k )( = k 0, Ry π k 0,)( T0 + ∑ k,q = 0 Ry π k q',)( Tq' q ≠0' k ˆ k ˆ * ˆ k = (− )1 T0 + ∑ k,q = 0 Ry π k q',)( Tq' q ≠0' where 84

ˆ k k 0, Ry π k 0,)( = Pk (cosπ ) = (− )1

We note that

ˆ k α j,, m Tq α j,, m = 0 for q ≠ 0 (from the m-selection rule)

So we get

ˆ + ˆ k )( ˆ k ˆ k α j,, m Ry π )( Tq=0Ry π )( α j,, m = (− )1 α j,, m T0 α j,, m or

ˆ k k ˆ + ˆ k )( ˆ α j,, m T0 α j,, m = (− )1 α j,, m Ry π )( Tq=0 Ry π )( α j,, m

k ˆ k )( = (− )1 α j,, −m Tq=0 α j,, −m or

ˆ k )( k ˆ k α j,, −m Tq=0 α j,, −m = (− )1 α j,, m T0 α j,, m (2) since

ˆ iϕ ˆ + −iϕ Ry (θ = π ) α j,, m = e α j,, −m , α j,, m Ry (θ = π ) = e α j,, −m

From Eqs.(1) and (2), we get

k )( k k )( α j,, m Tq=0 α j,, m = ±(− )1 α j,, m Tq=0 α j,, m

(( Note )) Wigner-Eckart theorem.

ˆ k )( ˆ k )( α j m',';' Tq α ,; mj = kj m,;, jkjq m',';, α j'' T αj with the selection rule,

m'= m + q , j'= j + , jk + k −1,...,| j − k | .

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53. Electric dipole moment of In the accepted time-reversal invariant theory of the electromagnetic interaction. The electric dipole moment, the electric dipole moment ( Dˆ = erˆ) with a charge e is even under the time reversal. This is consistent with the fact that no electric dipole moment has been observed for any particle.

)1( )1( α j,, m Tq=0 α j,, m = − α j,, m Tq=0 α j,, m

with

)1( ˆ Tq=0 = De z (which is even under the time-reversal)

Then we get

)1( ˆ α j,, m Tq=0 α j,, m = α j,, m De z α j,, m = 0

which implies that no electric dipole moment. At the present time very accurate measurements are being made on the neutron to see if even a very small electric dipole moment exists, Any such observation would indicate a breakdown of time-reversal symmetry in the theory. It would also indicate a breakdown of space inversion symmetry because accepted space inversion theory has the electric dipole moment as an odd parity operator. Thus if the particle has definite parity, the diagonal matrix elements of the dipole would also vanish for this reason.

(( Note )) The neutron electric dipole moment (EDM) is a measure for the distribution of positive and negative charge inside the neutron. A finite electric dipole moment can only exist if the centers of the negative and positive charge distribution inside the particle do not coincide. So far, no neutron EDM has been found. The current best upper limit amounts to

−26 dn < 9.2 ×10 e⋅cm.

______REFERENCES E.P. Wigner, Group Theory and its Application to the Quantum Mechanics of Atomic Spectra (Academic Press, 1959). P.H.E. Meijer and E. Bauer, Group Theory: The Application to Quantum Mechanics (North- Holland, 1962). K. Nishijima, Fundamental particles (W.A. Benjamin, 1963)

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M. Tinkham, Group Theory and Quantum Mechanics (McGraw-Hill, New York, 1964). P. Stehle, Quantum Mechanics (Holden-Day, Inc., 1966). L. Fonda and G.C. Ghirardi. Symmetry Principles in Quantum Physics (Marcel Dekker, 1970). F.J. Belinfante, Measurements and Time Reversal in Objective Quantum Theory (Pergamon, 1975). J.P. Elliott and P.G. Dawber, Symmetry in Physics vol. I and II (Macmillan, 1979). R.G. Sachs, The Physics of Time Reversal (University of Chicago, 1987). U. Fano and A.R.D. Rau, Symmetries in Quantum Physics (Academic Press, 1996). Eugen Merzbacher, Quantum Mechanics , third edition (John Wiley & Sons, New York, 1998). Albert Messiah, Quantum Mechanics , vol. I and vol. II (North-Holland, 1961, Dover 1999). L.E. Ballentine, Quantum Mechanics A Modern Development (World Scientific, 2000). K. Gottfried and T.-M. Yan, Quantum Mechanics , 2 nd edition (Springer, 2004). M.Le Bellac, Quantum Physics (Cambridge, 2006). N. Majlis, The Quantum Theory of magnetism , 2 nd edition (World Scientific, 2007). R.M. White, Quantum Theory of Magnetism , 3 rd edition (Springer, 2007). F. Scheck, Quantum Physics (Springer, 2007). M.S. Sozzi, Discrete Symmetries and CP Violation: From Experiment to Theory (Oxford, 2008). J.J. Sakurai and J. Napolitano, Modern Quantum Mechanics , second edition (Addison-Wesley, New York, 2011). T. Nagamiya, Theory of Magnetism, 3 rd edition (Yoshioka, 1992) [in Japanese]. ______APPENDIX Formula

ψ R = Θˆ ψ

1. Θˆ =UKˆˆ , Θˆ −1 = KUˆˆ + , (Uˆ ; unitary operator, Kˆ ; complex conjugate operator) Θˆ + = Kˆ +Uˆ + = KUˆˆ + . Θˆ −1 = Θˆ + . ΘΘˆˆ −1 = UKˆˆ 2Uˆ + = 1ˆ . 2. Kˆ 2 = 1 , Kˆ + Kˆ = 1, Kˆ + = Kˆ −1 = Kˆ . 3. ˆ KcK ˆ −1 = c* where c is a complex 4. KˆAˆKˆ −1 = Aˆ * where Aˆ is any operator. iπ 5. Θˆ =η exp(− Jˆ )Kˆ for spin j h y iπ 6. Θˆ = η exp(− σˆ )Kˆ = η(−iσˆ )Kˆ for spin 1/2 2 y y 7. Θˆ l,m = (− )1 m l,−m for orbital angular momentum 8. Θˆ 2 j,m = i−4mΘˆ j,m 9. Θˆ j,m = i2m j,−m

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ˆ N ˆ 10. Θ = (−i) σˆ1y ⊗σˆ2 y ⊗σˆ3y ⊗⋅⋅⋅⋅⊗ σˆ Ny K . ~ * 11. β α~ = α β = β α

~ * 12. β Θˆ AˆΘˆ −1 α~ = α Aˆ + β = β Aˆ α

13. α~ Θˆ AˆΘˆ −1 α~ = α Aˆ + α 14. α~ Θˆ AˆΘˆ −1 α~ = α Aˆ α ( Aˆ is Hermite operator) 15. For Θˆ AˆΘˆ −1 = Aˆ , α~ Aˆ α~ = α Aˆ α 16. For Θˆ AˆΘˆ −1 = −Aˆ , α~ Aˆ α~ = − α Aˆ α * 17. ψ R r t),( = r ψ R t)( = r ψ (−t) =ψ * r,( −t) 18. ψ R r )0,( = Θψ r )0,( =ψ * r )0,( 19. ψ R t)( = Θˆ ψ (−t) . ψ R )0( = Θˆ ψ )0(

~ * * 20. r φn = φn (−r) = − r φn

~ * * 21 k φn = φn (−k) = − k φn 22. Θˆ iΘˆ −1 = −i1ˆ 23. ΘHΘˆˆˆ −1 = Hˆ ( Hˆ : Hamiltonian) 24. Θˆ pˆΘˆ −1 −= pˆ 25. Θˆ rˆΘˆ −1 = rˆ 26. ΘJΘˆˆˆ −1 = −Jˆ ( Jˆ ; angular momentum 27. ΘRΘˆˆˆ −1 = Rˆ ( Rˆ , rotation operator) ˆˆ ˆ −1 ˆ ˆ 28. ΘTx a)( Θ = Tx a )( (Tx a )( ; translation operator) 29. Θˆ πˆΘˆ −1 = πˆ (πˆ ; parity operator) 30. Θˆ r' = r' 31. Θˆ p' = − p' ˆ k )( ˆ k )( 32. α j m',';' Tq α ,; mj = kj m,;, jkjq m',';, α j'' T αj ˆ ˆ ˆ ˆ 33. (A⊗ B)(ψ1 ⊗ ψ 2 ) = Aψ1 ⊗ Bψ 2

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