chapter 3 symmetry in quantum mechanics

Chapter 3

Symmetry in quantum mechanics

In this chapter we want to establish the importance of symmetry considerations in quan-tum mechanics and discuss the connection between

- symmetries;

- degeneracies;

- conservation laws.

3.1 Symmetries, conservation laws, degeneracies

3.1.1 Symmetries in classical physics

In classical mechanics, one usually considers the Lagrange formulation defined in terms ofthe Lagrangian L, which is a function of the generalized coordinates, qi, and the generalizedvelocities, qi, of the system.If, for instance, the Lagrangian L remains unchanged under displacements,

qi → qi + δqi,

then∂L

∂qi= 0,

which implies that:

dpidt

= 0 (3.1)

sinced

dt

(∂L

∂qi

)

− ∂L

∂qi= 0 and pi =

∂L

∂qi(canonical momentum).

Equation (3.1) means that pi is a conserved quantity.

In the Hamiltonian formulation,H(pi, qi),

65

66 CHAPTER 3. SYMMETRY IN QUANTUM MECHANICS

the Hamilton equations are given by:

pi = −∂H

∂qi, qi =

∂H

∂pi,

andd

dtpi = 0

whenever∂H

∂qi= 0.

In other words, if the Hamiltonian doesn’t explicitly depend on qi (which is equivalent tosaying that H is unchanged under qi → qi + δqi), we have a conserved quantity, in thiscase, the momentum pi.

3.1.2 Symmetries in quantum mechanics

In quantum mechanics:

(i) we associate a unitary operator O to a transformation that conserves probability.For instance, a rotation is described by a unitary operator. This operator is oftencalled a symmetry operator ;

(ii) we classify symmetries as continuous (rotation, translation) and discrete (parity,lattice translations, time reversal).

(ii) symmetry operations that differ infinitesimally from the identity transformation(continuous symmetries) are written as:

O = 1− iε

~G, (3.2)

where G is the hermitian generator of the symmetry operator we are describing.(Recall what we did in Chapter 2, where the angular momentum is the generator ofthe rotation operator. The same holds for the translation operator).

Translations in quantum mechanics. Translation operation

Let us assume that we start with a state which is well localized around ~r0. We define anoperation that changes this state into another well-localized state around ~r0 + d~r0 witheverything else unchanged. Such an operation is an infinitesimal translation by d~r0. Wedenote the corresponding operator by T (d~r0):

T (d~r0) |~r0〉 = |~r0 + d~r0〉 (3.3)

From expression (3.3), it is clear that |~r0〉 is not an eigenstate of T .

Properties of T (d~r0):

3.1. SYMMETRIES, CONSERVATION LAWS, DEGENERACIES 67

1. unitarity (since it has to fulfill probability conservation):

〈α|α〉 = 〈α| T †(d~r0)T (d~r0) |α〉

⇒ T †(d~r0)T (d~r0) = 1;

2. T (d~r1)T (d~r0) = T (d~r1 + d~r0);

3. a translation in the opposite direction must be the same as the inverse of the originaltranslation:

T (−d~r0) = T−1(d~r0);

4.

limd~r0→0

T (d~r0) = 1, identity operator;

5. a translation of a quantum state in position representation is:

⟨

~r|T (d~r0)|Ψ⟩

= 〈~r − d~r0|Ψ〉

6. the infinitesimal translation operator can be expressed as:

T (d~r0) = 1− iK · d~r0 , (3.4)

where (Kx, Ky, Kz) are hermitian operators. In order to show that, it is enough tosee that equation (3.4) fulfills the properties 1 to 4 (exercise).

Analogous to rotations and angular momenta as generators of rotations, K is the generatorof an infinitesimal translation. K is related to the momentum operator, ~p, as

K =~p

~.

(To see the proof: Sakurai “Modern Quantum Mechanics“, page 45).

G as a constant of motion

Now, let us go back to our general symmetry operator O (Eq. 3.2). We assume that theHamilton operator, H , describing the quantum system is invariant under O. In this case,

O†HO = H,

which is equivalent to[

G, H]

= 0.


In the Heisenberg representation, the corresponding equation of motion for G is (providedthat G doesn’t explicitly depend on time):

dG

dt=

1

i~

[

G, H]

→ dG

dt= 0,

what implies that G is a constant of motion. Then, for instance:

- if H is invariant under translation, then the momentum is a constant of motion;

- if H is invariant under rotation, then the angular momentum is a constant of motion.

In order to see that, let us consider that at t0 the system is in an eigenstate of G, |g, t0〉,with an eigenvalue g. The state at a later time, t, will be obtained by applying thetime-evolution operator, U(t, t0):

|g, t0; t〉 = U(t, t0) |g, t0〉 .

The new state, |g, t0; t〉, should be an eigenstate of G as well, with the same eigenvalue gsince [

G, H]

= 0

→[

G, U(t, t0)]

= 0

→ G [U(t, t0) |g〉] = U(t, t0)G |g〉 = g [U(t, t0) |g〉] .

Degeneracies

The question of degeneracy in quantum mechanics has important implications on theproperties of the quantum system.

Let us consider that the Hamiltonian of a quantum system commutes with the symmetryoperator O:

[

H, O]

= 0.

If |n〉 is an eigenstate of H ,

H |n〉 = En |n〉 ,then O |n〉 is an eigenstate of H as well since:

H(

O |n〉)

= O(

H |n〉)

= En

(

O |n〉)

.

Let us assume now that |n〉 and O |n〉 are different eigenstates with the same energy En.In this case, the two states are degenerate.

3.2. DISCRETE SYMMETRIES: PARITY 69

We consider the special case of rotations: O = R. If the Hamiltonian is rotationallyinvariant,

[

R, H]

= 0 →[

J, H]

= 0,[

J2, H]

= 0,

we can define a set of eigenstates of {H, J2, Jz} characterized by the quantum numbersn, j,m.From what we have seen above, all states of the form:

R |n; j,m〉

have the same energy. Since usually under rotation all m-values get mixed, R |n; j,m〉 isa linear combination of the 2j + 1 independent states:

R |n; j,m〉 =∑

m′

|n; j,m′〉C(j)m′m.

The degeneracy of R |n; j,m〉 is (2j + 1)-fold, which is equal to the number of possiblem-values. This can also be seen from the fact that [J±, H ] = 0 and therefore |n; j,m〉 form = −j, . . . , j have all the same energy.

Application of this degeneracy. As an illustration:Example 1: we consider an electron under the influence of a potential

V (r) + ε(r)~L · ~S,

with ~L · ~S being the spin-orbit coupling.

Both r and ~L · ~S (scalar product) are rotationally invariant. Therefore, we will have for

each atomic level, n, a (2j + 1)-fold degeneracy, with ~J = ~L + ~S. If we apply now anelectric field in the z-direction, the rotational symmetry becomes broken, and states withdifferent m-values acquire different energies.

Example 2: two spin 1/2 particles described by the Heisenberg interaction in thepresence of an external magnetic field,

H = J ~S1 · ~S2 +B∑

i=1,2

Siz.

→ Show the breaking of the triplet degeneracy for the interaction ~S1 · ~S2.

3.2 Discrete symmetries: parity

Rotations and translations are continuous symmetry operators, i.e., they describe oper-ations that can be obtained by applying successively infinitesimal operations. There arethough symmetry operations that cannot be obtained in this way, for instance:


- parity or space inversion

- lattice translation

- time reversal

They all are discrete symmetries.

Parity operation changes a right-handed (RH) system to a left handed (LH) system:We define the parity operator, Π (it can also be denoted as P ), in such a way that:

Π†~rΠ = −~r

or~rΠ = −Π~r.

Properties of Π:

1. it is unitary,ΠΠ† = 1,

and therefore Π and ~r anticommute:

{Π, ~r} = 0.

Note:{

A, B}

denotes anticommutator: AB + BA.

2. an eigenfunction of the position operator, |~r0〉, transforms under Π as:

Π |~r0〉 = eiδ |−~r0〉 , (3.5)

where eiδ is a phase factor (δ is real).

Proof : since~rΠ |~r0〉 = −Π~r |~r0〉 = (−~r0)Π |~r0〉 ,


Π |~r0〉 is an eigenstate of ~r with eigenvalue −~r0. This means that, up to a phasefactor, the state Π |~r0〉 should be the same as |−~r0〉. By convention, we take the

phase factor eiδ = 1 . With this choice of eiδ, eq. (3.5) becomes:

Π |~r0〉 = |−~r0〉 ,

Π2 |~r0〉 = |~r0〉

→ Π2 = 1 , identity operator.

3. since Π2 = 1, Π is also hermitian,

Π−1 = Π† = Π,

with eigenvalues +1 and −1.

4. the momentum operator, ~p, is also, as ~r, odd under parity:

Π†~pΠ = −~p .

Proof : the state |~p〉 in the position representation is:

|~p〉 = 1

(2π~)3/2

∫

d3r ei~p·~r/~ |~r〉 .

We act on |~p〉 by Π:

Π |~p〉 =1

(2π~)3/2

∫

d3r ei~p·~r/~Π |~r〉

=1

(2π~)3/2

∫

d3r ei~p·~r/~ |−~r〉

=1

(2π~)3/2

∫

d3r′ e−i~p·~r′/~ |~r ′〉 = |−~p〉 .

and therefore,

Π |~p〉 = |−~p〉 .

Then, for the operator ~p we have (in analogy to ~r):

Π†~pΠ = −~p; {Π, ~p} = 0.

This can also be seen from ~p = md~rdt.

5. the orbital angular momentum, ~L under parity:

[

Π, ~L]

= 0.


This can be shown by considering the definition of ~L, ~L = ~r× ~p, with ~r and ~p beingodd in parity, ~L is even in parity:

Π~L = ~LΠ.

or by considering the fact that ~L is the generator of spatial rotations. The 3 × 3orthogonal matrices, R(parity) and R(rotation), commute:

R(parity)R(rotation) = R(rotation)R(parity),

where R(parity) (or parity inversion) is

R(parity) =

−1 0 00 −1 00 0 −1

.

This commutation relation should be fulfilled by the corresponding operators, Πand R:

ΠR = RΠ,

with

R = 1− i

~dα~L · ~u.

Then,

Π

(

1− i

~dα~L · ~u

)

=

(

1− i

~dα~L · ~u

)

Π

→ Π~L = ~LΠ → [Π, ~L] = 0

or

Π†~LΠ = ~L.

This can be generalized to any angular momentum ~J . Both spin and ~L are evenoperators under parity:

Π†~SΠ = ~S .

Definitions :

-polar vectors : vector operators that are odd under parity, like

~r, ~p;

-axial vectors or pseudovectors: vector operators that are even under parity,like

~J ;


-scalar operators : under parity they transform as ordinary scalars, like

~L · ~S

→ Π−1~L · ~SΠ = ~L · ~S;-pseudoscalar operators: scalar operators which under parity transform as:

Π−1AΠ = −A.

e.g.,~S · ~r

→ Π−1~S · ~rΠ = −~S · ~r.

3.2.1 Wavefunctions under parity

→ 〈~r |Ψ〉 = Ψ(~r)

→ 〈~r|Π |Ψ〉 = 〈−~r |Ψ〉 = Ψ(−~r)

If |α〉 is an eigenstate of Π, thenΠ |α〉 = ± |α〉 .

In the position representation:

〈~r|Π |α〉 = ±〈~r |α〉and 〈~r|Π |α〉 = 〈−~r |α〉

}

〈~r |α〉 = ±〈−~r |α〉 .

Therefore, the eigenfunctions of parity can be classified as even and odd :

α(−~r) = α(~r) even

α(−~r) = −α(~r) odd

Example of wavefunctions with definite parity:

(1) Since[

~L,Π]

= 0, the eigenfunctions of ~L must have definite parity.

In spherical coordinates the transformation ~r → −~r is fulfilled by (see Fig. 3.1):

r → rθ → π − θφ → π + φ

(3.6)

Using relations (3.6) and the definition of Y ml (θ, φ),

Y ml (θ, φ) = (−1)m

√

(2l + 1)(l −m)!

4π(l +m)!Pml (cos θ)eimφ,

it can be shown that:Y ml (r) = (−1)lY m

l (−r) .


Figure 3.1: Transformation ~r → −~r

(2) Parity properties of energy eigenstates: if [H,Π] = 0, and Ψn is a non-degenerateeigenstate of H with eigenvalue En,

H |Ψn〉 = En |Ψn〉 ,

then |Ψn〉 is also an eigenstate of the parity operator.

For instance, in the case of the harmonic oscillator,

H = ~ω

(

n+1

2

)

, n = a†a,

we have that [

H,Π]

= 0 (shown in QM I)

andΨn(x) = (−1)nΨn(−x) .

Remark : for the previous statement to be valid, it is important that the eigenstatesof H are non-degenerate , otherwise is does not hold. For instance, consider thehydrogen atom in the non-relativistic limit,

H = T + V (r), [H,Π] = 0,

the eigenstates of H are defined by |n; lm〉. Considering the two states

2s : |2; 00〉

and2p−1 : |2; 1− 1〉 ,

one sees that an energy state cp |2p−1〉+ cs |2s〉 is not a parity eigenstate.


3.2.2 Example: double-well potential

Let us consider a symmetrical double-well potential (see Fig. 3.2):

Figure 3.2: Double-well potential

The solutions for a system in such a potential are given by sin and cos in the classicallyallowed region and sinh and cosh in the classically forbidden region and they are matchedat the position where the potential is discontinuous.

The two lowest-lying states can be described by a symmetrical state |S〉 and an antisym-

metrical state |A〉 (see Fig. 3.3):

(a) Symmetrical state |S〉 (b) Antisymmetrical state |A〉

Figure 3.3: Double-well solutions

Since [H,Π] = 0, these states are simultaneous eigenstates of the hamiltonian and theparity operator and the symmetrical state has lower energy:

EA > ES.

If the middle barrier gets thinner, the difference EA − ES gets smaller. We define theright-handed state as:

|R〉 = 1√2(|S〉+ |A〉) , (3.7)

and the left-handed state as:

|L〉 = 1√2(|S〉 − |A〉) (3.8)


The names that we have given to the above defined states reflect the fact that the state(3.7) concentrates mostly on the right-hand side well, and the state (3.8) concentratesmostly on the left-hand side well.

Note that |R〉 and |L〉 are NOT parity eigenstates. They are non-stationary states:

|R, t0 = 0; t〉 =1√2

(e−iESt/~ |S〉+ e−iEAt/~ |A〉

)

=1√2e−iESt/~

(|S〉+ e−i(EA−ES)t/~ |A〉

).

For t = T/2 ≡ 2π~2(EA−ES)

the system is in the left-handed state |L〉 and for t = T the

system is in the right-handed state |R〉. Thus, as a function of time, the system shows anoscillatory behavior between |R〉 and |L〉 with angular frequency:

ω =2π

T=

(EA − ES)

~.

A particle initially confined in the right-hand side well can tunnel through the classicallyforbidden region to the left-hand side well and back with angular frequency ω.

Let us now consider an infinitely high barrier (see Fig. 3.4)

Figure 3.4: Infinite barrier potential

In this case, there is no tunneling, and the |S〉 and |A〉 states are degenerate (see Fig.3.5).They are energy eigenstates, BUT they are not parity eigenstates; the |L〉 and |R〉 statesare also eigenstates of the Hamiltonian, but the system remains forever in |L〉 or in |R〉.

The ground state is asymmetrical even though the Hamiltonian is symmetrical underspace inversion. This is called broken symmetry related to degeneracy.

One example of broken symmetry in nature is ferromagnetism. The Hamiltonian for asystem of Fe atoms is rotationally invariant (H =

∑

ij J~Si~Sj), but the ferromagnet has

a definite direction in space. The ground state is degenerate. The infinite number ofdegenerate ground states of the ferromagnet are not rotationally invariant since the spinsare aligned along some definite direction (arbitrary).


Figure 3.5: Degenerate states

Ammonia molecule NH3

As another example of a system with broken symmetry, let us consider the ammoniamolecule NH3. In the ammonia molecule, the N atom can be either up or down as shownin Fig. 3.6:

Figure 3.6: Possible configurations for NH3

The three H atoms are in the corners of an equilateral triangle. The molecule rotatesalong the axis as shown in the figure.

The up/down positions of the NH3 molecule are analogous to the R/L states in thedouble-well potential. The system oscillates between the up and down positions, theparity and energy eigenstates, |S〉 and |A〉, are a superposition of the up/down states,and the oscillation frequency is

ω =(EA −ES)

~= 24000 MHz,

λ = 1 cm (microwave region).

Due to this two-state configuration, NH3 is used in maser (Microwave Amplification byStimulated Emission of Radiation) physics.

3.3. DISCRETE SYMMETRY: LATTICE TRANSLATION 79

which is a mathematical expression of the Laporte and Wigner rules allowing radiative

transitions to take place only between states of opposite parity. The electric dipole termin a multipole expansion is of the form ~E ·~r. If a Hamiltonian H is invariant under parity,the non-degenerate states cannot possess a permanent dipole moment,

〈Ψn|~r |Ψn〉 = 0.

Generally, any operator that is odd under parity (~p, ~r, ~S ·~r, . . .) has non-vanishing matrixelements only between states of opposite parity.

Weak interaction. The Hamiltonian that describes the weak interaction of elementaryparticles is not invariant under parity. Lee and Yang in 1956 proposed that parity, whichhad been believed to be an inherent symmetry in all important processes in physics,was not conserved in weak interactions. Namely, in decay processes the final states aresuperpositions of opposite parity states.

3.3 Discrete symmetry: lattice translation

We consider now a periodic potential in one dimension, such that

V (x± a) = V (x).

For instance, we have to deal with this type of potential if we are interested in thedescription of the state of an electron under the potential of a chain of regularly spacedpositive ions (see Fig. 3.7).

Figure 3.7: Periodic 1D potential

In the case that the barrier height between two adjacent lattice sites becomes infinite thenthe potential is as shown in Fig. 3.8.The Hamiltonian of the electron is, in general, not invariant under a translation with larbitrary:

T †(l)xT (l) = x+ l,

T (l) |x〉 = |x+ l〉 .However, if l = a, the lattice spacing, the situation changes:

T †(a)V (x)T (a) = V (x+ a) = V (x),


Figure 3.8: Periodic 1D potential with infinitely high barriers between adjacent sites

and since the kinetic term of the Hamiltonian is invariant under any translation then:

T †(a)HT (a) = H → [H, T (a)] = 0.

Properties

(i) The ground state of H for the case when the barrier between sites is infinitely highwill be a state in which the electron is localized at the nth site:

|Ψn〉 ,

withH |Ψn〉 = E0 |Ψn〉 ,

and in which the wavefunction 〈x |Ψn〉 = Ψn(x) is finite only in the nth site. BUT,any state localized at some other site has the same energy E0, and, therefore, thereare an infinite number of ground states Ψn, where n goes from −∞ to ∞.

(ii) |Ψn〉 is not an eigenstate of the translation operator T (a) since

T (a) |Ψn〉 = |Ψn+1〉 ,

even though[H, T (a)] = 0 and H |Ψn〉 = E0 |Ψn〉 .

This is so because we have an infinite-fold degeneracy, which means that the symme-try in which nature manifests itself not necessarily has to coincide with the symmetryof the energy eigenstates.

(iii) There must be though a basis set of states which are simultaneous eigenstates of Hand T (a). Let us consider the linear combination:

|Ψθ〉 ≡+∞∑

n=−∞

einθ |Ψn〉 , (3.9)

withθ ∈ R, −π ≤ θ ≤ π.


We will show that |Ψθ〉 is a simultaneous eigenstate of H and T (a). Acting with Hon eq. (3.9), we get:

H |Ψθ〉 =+∞∑

n=−∞

einθH |Ψn〉 = E0

+∞∑

n=−∞

einθ |Ψn〉 = E0 |Ψθ〉 ,

and then with T (a):

T (a) |Ψθ〉 =

+∞∑

n=−∞

einθT (a) |Ψn〉

=

+∞∑

n=−∞

einθ |Ψn+1〉 =|n′=n+1

+∞∑

n′=−∞

ei(n′−1)θ |Ψn′〉

= e−iθ |Ψθ〉 .

We see now that |Ψθ〉 is an eigenstate of H with eigenvalue E0, and an eigenstateof T (a) with eigenvalue e−iθ.

3.3.1 Tight-binding approximation

(iv) Let us consider the case of H when the barrier between two adjacent sites is notinfinite (more realistic situation) (see Fig. 3.7). In this case, one can also construct thelocalized state |Ψn〉 with T (a) |Ψn〉 = |Ψn+1〉. However, due to the quantum tunnelingeffect, the electron will have some probability to go into neighboring lattice sites, so that〈x |Ψn〉 has a tail extending to sites other than the nth site. Therefore, the diagonal matrixelements of H in the {|Ψn〉} basis are all equal (because of translational invariance),

〈Ψn|H |Ψn〉 = E0,

but H is not completely diagonal in this basis, i.e., there are non-zero off-diagonal matrixelements. In solid state, we make an approximation that the only important non-diagonalmatrix elements are those connecting nearest neighbors:

〈Ψn′ |H |Ψn〉 6= 0 only for

{n′ = n orn′ = n + 1,

and this approximation is called the tight-binding approximation.

With the definitions:〈Ψn±1|H |Ψn〉 = −∆

and〈Ψn |Ψn′〉 = δnn′,

we will have:H |Ψn〉 = E0 |Ψn〉 −∆ |Ψn+1〉 −∆ |Ψn−1〉


where uk(x) is a periodic function with period a uk(x − a) = uk(x). Then, substituting(3.11) in (3.10):

eik(x−a)uk(x− a) = eikxuk(x)e−ika,

we get the equation verified, which leads us to the following theorem.

Bloch’s theorem: the wavefunction |Ψθ〉, which is an eigenstate of T (a), can be written asa plane wave eikx times a periodic function with periodicity a:

〈x |Ψθ〉 = eikxuk(x).

This theorem holds even when the tight-binding approximation is not anymore valid.

Physical meaning of θ = ka

Bloch’s theorem states that the wave function of the electron is a plane wave characterizedby the propagation wave vector k modulated by a periodic function uk(x). As θ variesfrom −π to π, k varies from −π/a to π/a and the energy eigenvalue is a function of k:

E(k) = E0 − 2∆ cos ka .

This is valid within the tight-binding approximation.The wave vector k fulfills |k| ≤ π/a and describes the Brillouin zone. E(k) is the dis-persion relation for the electron states. Since the electron can tunnel, the infinite-folddegeneracy that we had for the case of infinite height wells is lifted, and the allowed en-ergy values form a continuous band between E0 − 2∆ and E0 + 2∆ (see Fig. 3.10).

Figure 3.10: Energy dispersion in the first Brillouin zone

Considering the case of many electrons moving in a periodic potential and neglectingCoulomb repulsion among them, we will have that, since the electrons, as fermionic par-ticles, satisfy the Pauli principle, they start filling the band (one electron per site). In thecase of a half-filled band, we will have a metallic state (see Fig. 3.11).


Figure 3.11: Metallic state

3.4 Discrete symmetry: time-reversal

Eugene Wigner (Physics Nobel Prize 1963, together with Maria Goeppert-Mayer and J.Hans D. Jensen) was the first to discuss this symmetry in 1932.

Let us consider a classical particle that at t = 0 stops and reverses its motion:

~p |t=0 → −~p |t=0

(a) at t = 0 the particle stops (b) reverses motion ~p |t=0 → −~p |t=0

Then, if ~r(t) is a solution of

m~r = −~∇V (~r),

~r(−t) is also a solution (as long as the force is non-dissipative), and it is difficult to tellwhich sequence, (1) or (2), is the correct one (see Fig. 3.12).

Figure 3.12: Possible parti-cle trajectory

This is not the case if we consider the trajectory of theparticle (in particular, if the particle is an electron) under

the influence of a magnetic field ~B. Microscopically, ~B isproduced by moving charges via an electric current; then,if we reverse the current that produces ~B, ~B will reversedirection.From the Maxwell equations (in SI unities),

~∇ · ~E = ρ/ǫ0 ~∇ · ~B = 0

~∇× ~B − ǫ0µ0∂ ~E∂t

= µ0~j ~∇× ~E = −∂ ~B

∂t,

3.4. DISCRETE SYMMETRY: TIME-REVERSAL 85

it follows that the Lorenz force,

~F = e[

~E +(

~v × ~B)]

,

is invariant under t → −t if

~E → ~E ~B → − ~B ρ → ρ ~j → −~j ~v → −~v

In quantum mechanics, if Ψ(~r, t) is a solution of the Schrodinger equation,

i~∂Ψ

∂t=

(

− ~2

2m∇2 + V

)

Ψ,

Ψ(~r,−t) is not a solution because of the first time derivation, but Ψ∗(~r,−t) is a solutionsince it fulfills the Schrodinger equation:

Ψ(~r, t) = Ψn(~r)e−iEnt/~,

Ψ∗(~r,−t) = Ψ∗n(~r)e

−iEnt/~.

Therefore, time reversal must be related with complex conjugation.

For a spinless system, at t = 0 the wave function is given by

Ψ(~r) = 〈~r|Ψ〉 ,

and the wave-function for the corresponding time-reversed state is

Ψ∗(~r) = 〈~r|Ψ〉∗ = 〈Ψ|~r〉 .

3.4.1 Antiunitary transformation

Rotations, translations, and parity are all unitary operators since they keep the scalarproduct of two states unchanged:

〈Ψ|Φ〉 = 〈Ψ|U †U |Φ〉 =⟨

Ψ|Φ⟩

,

where∣∣∣Ψ

⟩

and∣∣∣Φ

⟩

are the transformed states through U . For time reversal, one imposes

a weaker requirement:∣∣∣

⟨

Ψ|Φ⟩∣∣∣ = |〈Ψ|Φ〉| ,

which means that either ⟨

Ψ|Φ⟩

= 〈Ψ|Φ〉∗ = 〈Φ|Ψ〉or ⟨

Ψ|Φ⟩

= 〈Ψ|Φ〉


Definition: the transformation

|Ψ〉 →∣∣∣Ψ

⟩

= Θ |Ψ〉 , |Φ〉 →∣∣∣Φ

⟩

= Θ |Φ〉

is antiunitary if ⟨

Φ∣∣∣Ψ

⟩

= 〈Φ |Ψ〉∗

andΘ (c1 |Ψ〉+ c2 |Φ〉) = c∗1Θ |Ψ〉+ c∗2Θ |Φ〉 . (3.12)

Eq. (3.12) defines Θ as an antilinear operator as well.

An antiunitary operator can be written as

Θ = UK , (3.13)

the product of a unitary operator U and a complex-conjugate operator K defined as

Kc |Ψ〉 = c∗K |Ψ〉 , f.i., K

(1√2|Sz; +〉 ± i√

2|Sz;−〉

)

→ 1√2|Sz; +〉 ∓ i√

2|Sz;−〉 .

Note that the effect of K changes with the basis. For instance if the Sz eigenkets (|Sz; +〉,|Sz;−〉) are used as a base kets in order to represent the Sy states (see above) then K actsas above and changes the states. However, if we use the Sy eigenkets themselves (|Sy;±〉,they do not change under the action of K.To show that (3.12) fulfills (3.13):

Θ (c1 |Ψ〉+ c2 |Φ〉) = UK (c1 |Ψ〉+ c2 |Φ〉)= c∗1UK |Ψ〉+ c∗2UK |Φ〉= c∗1Θ |Ψ〉+ c∗2Θ |Φ〉 .

3.4.2 Time-reversal operator

We define the time-reversal operator Θ as an antiunitary operator, such that

|Ψ〉 → Θ |Ψ〉 ,

where Θ |Ψ〉 is the time-reversed state of |Ψ〉, f.i., Θ |~p〉 = |−~p〉 (up to a phase factor).

We consider a system described by |Ψ〉 at time t = 0. The system evolves under H , andat t = δt we have:

|Ψ, t0 = 0; t = δt〉 =(

1− iH

~δt

)

|Ψ〉 .

Now, let us first apply Θ at t = 0, and then let the system evolve under H . In this case,at t = δt we have: (

1− iH

~δt

)

Θ |Ψ〉 . (3.14)


If this evolution of the system is invariant under time reversal, (3.14) should be the sameas

Θ |Ψ, t0 = 0; t = −δt〉 ,i.e., when we first consider the state at t = −δt and then reverse ~p and ~J . In other words,

(1− iH

~δt)Θ |Ψ〉 = Θ

(1− iH

~(−δt)

)|Ψ〉

(1) (2)

Then,

−iHΘ |〉 = ΘiH |〉 , |〉: any ket. (3.15)

We see from eq. (3.15) that Θ has to be antiunitary. If Θ were unitary, we would havethe equation:

−HΘ = ΘH

andHΘ |Ψn〉 = −ΘH |Ψn〉 = −(En)Θ |Ψn〉 ,

which would imply that, if |Ψn〉 is an eigenstate of H with eigenvalue En, then Θ |Ψn〉 isan eigenstate of H with eigenvalue −En, what does not make sense.

For instance, if we consider a free particle with energy spectrum

E ∈ [0,+∞),

−E would ∈ (−∞, 0), but there is no state lower than a particle at rest, and, therefore,the energy spectrum (−∞, 0) does not make sense. From this we conclude that Θ shouldbe antiunitary,

ΘiH |〉 = −iΘH |〉 ,and eq. (3.15) is then

−iHΘ |〉 = −iΘH |〉 .After canceling the i’s we get:

HΘ = ΘH ; [H,Θ] = 0 ,


Note that:〈Ψ|Θ |Φ〉 = (〈Ψ|) (Θ |Φ〉)

and not(〈Ψ |Θ|) · |Φ〉 .

3.4.3 Transformation of operators under time reversal

Consider a linear operator A and states |Ψ〉 and |Φ〉,∣∣∣Ψ

⟩

= Θ |Ψ〉 ,∣∣∣Φ

⟩

= Θ |Φ〉 .

From the antiunitary nature of Θ it follows that

〈Φ| A |Ψ〉 =⟨

Ψ∣∣∣ΘA†Θ−1

∣∣∣Φ

⟩

. (3.16)

Proof : we define |γ〉 as|γ〉 ≡ A† |Φ〉 , 〈γ| = 〈Φ| A.

Then

〈Φ| A |Ψ〉 = 〈γ |Ψ〉 =⟨

Ψ |γ〉

=⟨

Ψ∣∣∣ΘA†

∣∣∣Φ

⟩

=⟨

Ψ∣∣∣ΘA†Θ−1Θ

∣∣∣Φ

⟩

=⟨

Ψ∣∣∣ΘA†Θ−1

∣∣∣ Φ

⟩

,

and we get eq. (3.16). If A = A† (hermitian), then:

〈Φ| A |Ψ〉 =⟨

Ψ∣∣∣ΘAΘ−1

∣∣∣Φ

⟩

(3.17)

Definitions :

ΘAΘ−1 = +A → observable A is even under time reversal symmetry,

ΘAΘ−1 = −A → observable A is odd under time reversal symmetry.

We substitute these definitions into eq. (3.17):

〈Φ| A |Ψ〉 = ±⟨

Ψ∣∣∣ A

∣∣∣Φ

⟩

.


For |Ψ〉 = |Φ〉,〈Ψ| A |Ψ〉 = ±

⟨

Ψ∣∣∣ A

∣∣∣Ψ

⟩

,

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

Momentum ~p: odd

By considering the corrspondence principle it is reasonable to assume that the ex-pectation value of ~p taken with respect to the time-reversed state will be of oppositesign:

〈Ψ| ~p |Ψ〉 = −⟨

Ψ∣∣∣ ~p

∣∣∣Ψ

⟩

,

therefore we take ~p to be an odd operator with respect to time-reversal symmetry:

Θ~pΘ−1 = −~p .

This has as a consequence,

~p (Θ |~p ′〉) = −Θ~p |~p ′〉 = (−~p ′)Θ |~p ′〉 ,

i.e., Θ |~p ′〉 is a momentum eigenstate of ~p with eigenvalue −~p ′.

Position ~r: even

Θ~rΘ1 = ~r

Θ |~r ′〉 = |~r ′〉

Angular momentum ~J : odd

To preserve

[Ji, Jj] = i~ǫijkJk,

the angular momentum must be odd under time reversal,

Θ ~JΘ−1 = − ~J,

which can immediately be seen when ~J = ~L = ~r × ~p.

3.4.4 Wavefunctions under time reversal

By acting on

|Ψ〉 =∫

d3r |~r〉〈~r |Ψ〉


with Θ:

Θ |Ψ〉 =

∫

d3rΘ |~r〉〈~r |Ψ〉∗

=

∫

d3r |~r〉〈~r |Ψ〉∗ (with the phase convention Θ |~r〉 = |~r〉),

one shows that under time reversal:

Ψ(~r) → Ψ∗(~r) .

For instance, the spherical harmonics:

Y ml (θ, φ) → Y m∗

l (θ, φ) = (−1)mY −ml (θ, φ);

therefore,

Θ |l, m〉 = (−1)m |l,−m〉 ,

which can be generalized for a state |j,m〉:

Θ |j,m〉 = i2m |j,−m〉 .

Note that if j if half-integer,

Θ2 |jm〉 = − |jm〉 ,

and if j is integer,

Θ2 |jm〉 = |jm〉 .

3.4.5 Kramers degeneracy

We consider a charged particle in an external electric field. For a static electric field, thepotential energy is:

V (~r) = eΦ(~r),

where Φ(~r) is the electrostatic potential. Since Φ(~r) is a real function of the time-reversaleven operator ~r, Θ and H commute:

[Θ, H ] = 0 . (3.18)

This relation doesn’t lead though to a conservation law since

ΘU(t, t0) 6= U(t, t0)Θ,

and there is no time-reversal quantum number, but it does lead to such a consequence asKramers degeneracy.


Kramers degeneracy

We assume that [Θ, H ] = 0 and H |Ψn〉 = En |Ψn〉. |Ψn〉 and Θ |Ψn〉 describe energyeigenstate and its time-reversed state, respectively, with the same energy eigenvalue since

ΘH |Ψn〉 = EnΘ |Ψn〉 = HΘ |Ψn〉 .We consider now that |Ψn〉 and Θ |Ψn〉 represent the same eigenstate. Then, |Ψn〉 andΘ |Ψn〉 can differ only by a phase factor:

Θ |Ψn〉 = eiδ |Ψn〉 .

In this case,

Θ2 |Ψn〉 = Θeiδ |Ψn〉 = e−iδΘ |Ψn〉 = e−iδeiδ |Ψn〉→ Θ2 |Ψn〉 = + |Ψn〉 .

This relation cannot be fulfilled by a system with half-integer j since for those systemsΘ2 = −1, and, therefore, |Ψn〉 and Θ |Ψn〉 must correspond to different states with thesame energy: they are degenerate.

Consequences : in a system composed of an odd number of electrons, in an external elec-tric field E each energy level must be at least two-fold degenerate, and this will haveimportant consequences in the distinct behavior between odd-electron and even-electronsystems.Kramers was the first to realize this degeneracy when studying the solutions of theSchrodinger equation, but it was Wigner who showed that this degeneracy is a conse-quence of time-reversal symmetry.

Magnetic field. If we set a charged particle in a magnetic field, the Hamiltonian willcontain terms of the form:

~S · ~B, ~p · ~A+ ~A · ~p (where ~B = ~∇× ~A)

since

H =1

2m

[

~p− e ~A]2

+ qV (~r)

(it will be worked out later on).

Now, since ~S and ~p are odd under time reversal,

ΘH 6= HΘ.

Thus, for instance, for a spin-1/2 system, the spin-up state |+〉 and its time-reversedstate |−〉 no longer have the same energy in the presence of an external magnetic field.Therefore, Kramers degeneracy in a system containing an odd number of electrons is liftedunder application of an external magnetic field.

Note: when ~B is an external field, it is not affected by the time reversal since time reversalis only applied to the closed quantum-mechanical system.

chapter 3 symmetry in quantum mechanics

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