Download Time reversal (reversal of motion)

Time reversal (reversal of motion)
The fact that the operator K does not change the base
states can be justified like:
The state |a0 i represented in the base {|a0 i} maps to the
column vector
 
0
 0 
 
then also x(−t) is a solution.
 .. 
 . 
At the moment t = 0 let there be a particle at the point
 
 0 
0
x(t = 0) with the momentum p(t = 0). Then a particle at

|a i 7→ 
 1 ,
the same point but with the momentum −p(t = 0)
 
 0 
follows the trajectory x(−t).
 
 . 
In the quantum mechanical Schrödinger equation
 .. 
0
h̄2 2
∂ψ
= −
∇ + V ψ,
ih̄
which is unaffected by the complex conjugation.
∂t
2m
Note The effect of the operator K depends thus on the
due to the first derivative with respect to the time,
choice of the basis states.
ψ(x, −t) is not a solution eventhough ψ(x, t) were, but
If U is a unitary operator then the operator θ = U K is
ψ ∗ (x, −t) is. In quantum mechanics the time reversal has antiunitary.
obviously something to do with the complex conjugation. Proof: Firstly
Let us consider the symmetry operation
θ(c1 |αi + c2 |βi) = U K(c1 |αi + c2 |βi)
|αi −→ |α̃i, |βi −→ |β̃i.
= (c∗1 U K|αi + c∗2 U K|βi)
= (c∗1 θ|αi + c∗2 θ|βi),
We require that the absolute value of the scalar product
is invariant under that operation:
so θ is antiliniear. Secondly, expanding the states |αi and
|βi in a complete basis {|a0 i} we get
|hβ̃|α̃i| = |hβ|αi|.
X
θ
ha0 |αi∗ U K|a0 i
|αi −→ |α̃i =
There are two possibilities to satisfy this condition:
The Newton equations of motion are invariant under the
transformation t −→ −t: if x(t) is a solution of the
equation
mẍ = −∇V (x)
a0
1. hβ̃|α̃i = hβ|αi, so the corresponding symmetry
operator is unitary, that is
hβ|αi −→ hβ|U U |αi = hβ|αi.
2. hβ̃|α̃i = hβ|αi∗ = hα|βi, so the symmetry operator
cannot be unitary.
We define the antiunitary operator θ so that
hβ̃|α̃i = hα|βi∗
θ(c1 |αi + c2 |βi) = c∗1 θ|αi + c∗2 θ|βi,
where
=
X
0
ha |αi∗ U |a0 i
a0
†
The symmetries treated earlier have obeyed this
condition.
=
X
hα|a0 iU |a0 i
a0
and
|β̃i =
X
ha0 |βi∗ U |a0 i ↔ hβ̃| =
a0
X
ha0 |βiha0 |U † .
a0
Thus the scalar product is
XX
ha00 |βiha00 |U † U |a0 ihα|a0 i
hβ̃|α̃i =
a00
=
X
a0
hα|a0 iha0 |βi = hα|βi
a0
= hβ|αi∗ .
|αi −→ |α̃i = θ|αi,
|βi −→ |β̃i = θ|βi
If the operator satisfies only the last condition it is called
antilinear.
We define the complex conjugation operator K so that
where Θ|αi is the time reversed (motion reversed) state.
If |αi is the momentum eigenstate |p0 i, we should have
Kc|αi = c∗ K|αi.
We present the state |αi in the base {|a0 i}. The effect of
the operator K is then
X
X
K
|αi =
|a0 iha0 |αi −→ |α̃i =
ha0 |αi∗ K|a0 i
a0
=
X
a0
a0
0
∗
0
ha |αi |a i.
The operator θ is thus indeed antiunitary.
Let Θ be the time reversal operator. We consider the
transformation
|αi −→ Θ|αi,
Θ|p0 i = eiϕ | − p0 i.
Let the system be at the moment t = 0 in the state |αi.
At a slightly later moment t = δt it is in the state
iH
|α, t0 = 0; t = δti = 1 −
δt |αi.
h̄
We apply now, at the moment t = 0, the time reversal
operator Θ and let the system evolve under the
Hamiltonian H. Then at the moment δt the system is in
the state
iH
δt Θ|αi.
1−
h̄
and
If the motion of the system is invariant under time
reversal this state should be the same as
In partcular, for a Hermitean observable A we have
Θ|α, t0 = 0; −δti,
i.e. we first look at the state at the earlier moment −δt
and then reverse the direction of the momentum p.
Mathematically this condition can be expressed as
iH
iH
1−
δt Θ|αi = Θ 1 −
(−δt) |αi.
h̄
h̄
Thus we must have
−iHΘ|i = ΘiH|i,
where |i stands for an arbitrary state vector.
If Θ were linear we would obtain the anticommutator
relation
HΘ = −ΘH.
If now |ni is an energy eigenstate corresponding to the
eigenvalue En then, according to the anticommutation
rule
HΘ|ni = −ΘH|ni = (−En )Θ|ni,
and the state Θ|ni is an energy eigenstate corresponding
to the eigenvalue −En . Thus most systems (those, whose
energy spectrum is not bounded) would not have any
ground state.
Thus the operator Θ must be antilinear, and, in order to
be a symmetry operator, it must be antiunitary. Using
the antilinearity for the right hand side of the condition
hβ| ⊗ |αi = hγ|αi = hα̃|γ̃i
= hα̃|Θ ⊗† |βi = hα̃|Θ ⊗† Θ−1 Θ|βi
= hα̃|Θ ⊗† Θ−1 |β̃i.
hβ|A|αi = hα̃|ΘAΘ−1 |β̃i.
We say that the observable A is even or odd under time
reversal depending on wheter in the equation
ΘAΘ−1 = ±A
the upper or the lower sign holds. This together with the
equation
hβ|A|αi = hα̃|ΘAΘ−1 |β̃i
imposes certain conditions on the phases of the matrix
elements of the operator A between the time reversed
states. Namely, they has to satisfy
hβ|A|αi = ±hβ̃|A|α̃i∗ .
In particular, the expectation value satisfies the condition
hα|A|αi = ±hα̃|A|α̃i.
Example The expectation value of the momentum
operator p.
We require that
hα|p|αi = −hα̃|p|α̃i,
so p is odd, or
ΘpΘ−1 = −p.
The momentum eigenstates satisfy
pΘ|p0 i = −ΘpΘ−1 Θ|p0 i
= (−p0 )Θ|p0 i,
−iHΘ|i = ΘiH|i
i.e. Θ|p0 i is the momentum eigenstates correponding to
the eigenvalue −p0 :
we can write it as
ΘiH|i = −iΘH|i.
So, we see that the operators commute:
ΘH = HΘ.
Θ|p0 i = eiϕ | − p0 i.
Similarly we can derive for the position operator x the
expressions
ΘxΘ−1 = x
Θ|x0 i = |x0 i
Note We have not defined the Hermitean conjugate of
the antiunitary operator θ nor have we defined the
meaning of the expression hβ|θ. That being, we let the
time reversal operator Θ to operate always on the right
and with the matrix element hβ|Θ|αi we mean the
expression (hβ|) · (Θ|αi).
Let ⊗ be an arbitrary linear operator. We define
when we impose the physically sensible condition
hα|x|αi = hα̃|x|α̃i.
We consider the basic commutation relations
|γi ≡ ⊗† |βi,
so that
[xi , pj ]|i = ih̄δij |i.
Now
hβ|⊗ = hγ|
Θ[xi , pj ]Θ−1 Θ|i = Θih̄δij |i,
from which, using the antilinearity and the time reversal
properties of the operators x and p we get
[xi , (−pj )]Θ|i = −ih̄δij Θ|i.
We see thus that the commutation rule
[xi , pj ]|i = ih̄δij |i
remains invariant under the time reversal.
Correspondingly, the requirement of the invariance of the
commutation rule
[Ji , Jj ] = ih̄ijk Jk
leads to the condition
ΘJ Θ−1 = −J .
This agrees with transformation properties of the orbital
angular momentum x × p.
Wave functions
so the states |ni and Θ|ni have the same energy. Because
the state |ni was supposed to be nondegenerate they
must represent the same state. The wave function of the
state |ni is hx0 |ni and the one of the state Θ|ni
correspondingly hx0 |ni∗ . These must be same (or more
accurately, they can differ only by a phase factor which
does not depend on the coordinate x0 ), i.e.
hx0 |ni = hx0 |ni∗
For example the wave function of a nondegenerate
groundstate is always real.
For a spinles particle in the state |αi we get
Z
Θ|αi = Θ dx0 hx0 |αi|x0 i
Z
=
dx0 hx0 |αi∗ |x0 i = K|αi,
Z
d3 x0 Θ|x0 ihx0 |αi∗
i.e. the time reversal is equivalent to the complex
conjugation.
On the other hand, in the momentum space we have
Z
Θ|αi =
d3 p0 | − p0 ihp0 |αi∗
Z
=
d3 p0 |p0 ih−p0 |αi∗ ,
Z
d3 x0 |x0 ihx0 |αi∗ ,
because
We expand the state |αi with the help of position
eigenstates:
Z
|αi = d3 x0 |x0 ihx0 |αi.
Now
Θ|αi =
=
HΘ|ni = ΘH|ni = En Θ|ni,
so under the time reversal the wave function
ψ(x0 ) = hx0 |αi
Θ|p0 i = | − p0 i.
The momentum space wave function transform thus
under time reversal like
φ(p0 ) −→ φ∗ (−p0 ).
transforms like
ψ(x0 ) −→ ψ ∗ (x0 ).
If in particular we have
ψ(x0 ) = R(r)Ylm (θ, φ),
We consider a spin 21 particle the spin of which is oriented
along n̂. The corresponding state is obtained by rotating
the state |Sz ; ↑i:
|n; ↑i = e−iSz α/h̄ e−iSy β/h̄ |Sz ; ↑i,
we see that
Ylm (θ, φ) −→ Ylm ∗ (θ, φ) = (−1)m Yl−m (θ, φ).
Because Ylm is the wave function belonging to the state
|lmi we must have
m
Θ|lmi = (−1) |l, −mi.
The probability current corresponding to the wave
function R(r)Ylm seems to turn clockwise when looked at
from the direction of the positive z-axis and m > 0. The
probability current of the corresponding time reversed
state on the other hand turns counterclockwise because m
changes its sign under the operation.
The spinles particles obey
Theorem 1 If the Hamiltonian H is invariant under the
time reversal and the energy eigenstate |ni nondegenerate
then the corresponding energy eigenfunction is real (or
more generally a real function times a phase factor
independent on the coordinate x0 ).
where α and β are the direction angles of the vector n̂.
Because
ΘJ Θ−1 = −J .
we see that
Θ|n; ↑i = e−iSz α/h̄ e−iSy β/h̄ Θ|Sz ; ↑i.
Furthermore, due to the oddity of the angular
momentum, it follows that
h̄
Jz Θ|Sz ; ↑i = − Θ|Sz ; ↑i,
2
so we must have
Θ|Sz ; ↑i = η|Sz ; ↓i,
where η is an arbitrary phase factor. So we get
Θ|n; ↑i = η|n; ↓i.
On the other hand we have
Thus we must have
|n; ↓i = e−iαSz /h̄ e−i(π+β)Sy /h̄ |Sz ; ↑i,
Θ2 = (−1)2j .
so
Often one chooses
−iSz α/h̄ −iSy β/h̄
η|n; ↓i = Θ|n; ↑i = e
e
Θ|Sz ; ↑i
−iαSz /h̄ −i(π+β)Sy /h̄
= ηe
e
|Sz ; ↑i.
Writing
Θ|jmi = i2m |j, −mi.
Spherical tensors
Θ = U K, U unitary
and recalling that the complex conjugation K has no
effect on the base states we see that
2Sy
K.
Θ = ηe−iπSy /h̄ K = −iη
h̄
Let us suppose that the operator A is either even or odd,
i.e.
ΘAΘ(−1) = ±A.
We saw that then we have
hα|A|αi = ±hα̃|A|α̃i.
Now
In an eigenstate of the angular momentum we have thus
e−iπSy /h̄ |Sz ; ↑i = +|Sz ; ↓i
e−iπSy /h̄ |Sz ; ↓i = −|Sz ; ↑i,
so the effect of the time reversal on a general spin
is
hα, jm|A|α, jmi = ±hα, j, −m|A|α, j, −mi.
1
2
state
Θ(c↑ |Sz ; ↑i + c↓ |Sz ; ↓i) = +ηc∗↑ |Sz ; ↓i − ηc∗↓ |Sz ; ↑i.
Applying the operator Θ once again we get
Θ2 (c↑ |Sz ; ↑i + c↓ |Sz ; ↓i)
= −|η|2 c↑ |Sz ; ↑i − |η|2 c↓ |Sz ; ↓i
= −(c↑ |Sz ; ↑i + c↓ |Sz ; ↓i),
Let now A be a component of a Hermitian spherical
tensor:
A = Tq(k) .
According to the Wigner-Eckart theorem it is sufficient to
consider only the component q = 0.
We define T (k) to be even/odd under the time reversal if
(k)
(k)
ΘTq=0 Θ−1 = ±Tq=0 .
Then we have
i.e. for an arbitrary spin orientation we have
(k)
(k)
hα, jm|T0 |α, jmi = ±hα, j, −m|T0 |α, j, −mi.
2
Θ = −1.
From the relation
Θ|lmi = (−1)m |l, −mi
we see that for spinles particles we have
The state |α, j, −mi is obtained by rotating the state
|α, jmi:
D(0, π, 0)|α, jmi = eiϕ |α, j, −mi.
On the other hand, due to the definition of the spherical
tensor
Θ2 = 1.
D
†
In general, one can show that
Θ2 |j half integeri = −|j half integeri
Θ2 |j integeri = +|j integeri.
(R)Tq(k) D(R)
=
k
X
(k)∗
(k)
Dqq0 (R)Tq0 ,
q 0 =−k
we get
(k)
D† (0, π, 0)T0 D(0, π, 0) =
X
Generally we can write
(k)
D0q (0, π, 0)Tq(k) .
q
Θ = ηe−iπJy /h̄ K.
Now
(k)
D00 (0, π, 0) = Pk (cos π) = (−1)k ,
Now
e−2iπJy /h̄ |jmi = (−1)2j |jmi,
so
so we have
(k)
D† (0, π, 0)T0 D(0, π, 0)
Θ2 |jmi =
Θ ηe−iπJy /h̄ |jmi
= |η|2 e−2iπJy /h̄ |jmi
= (−1)2j |jmi.
(k)
= (−1)k T0
+ (q 6= 0 components).
Furthermore
(k)
hα, jm|Tq6=0 |α, jmi = 0,
since the m selection rule would require m = m + q. So
we get
the Hamiltonian of an electron contains such terms as
S · B,
p · A + A · p.
(k)
hα, jm|T0 |α, jmi
(k)
= ±hα, jm|D† (0, π, 0)T0 D(0, π, 0)|α, jmi
(k)
= ±(−1)k hα, jm|T0 |α, jmi.
Note Unlike under other symmetries the invariance of
The magnetic field B is external, independent on the
system, so
[Θ, B] = 0 ja [Θ, A] = 0.
On the other hand, S and p are odd, or
the Hamiltonian under the time reversal
[Θ, H] = 0,
does not lead to any conservation laws. This is due to the
fact that the time evolution operator is not invariant:
ΘU (t, t0 ) 6= U (t, t0 )Θ.
Time reversal and degeneracy
Let us suppose that
[Θ, H] = 0.
Then the energy eigenstates obey
H|ni = En |ni
HΘ|ni = En Θ|ni.
If we now had
Θ|ni = eiδ |ni,
then, reapplying the time reversal we would obtain
Θ2 |ni = e−iδ Θ|ni = |ni,
or
Θ2 = 1.
This is, however, impossible if the system j is half integer,
because then Θ2 = −1. In systems of this kind |ni and
Θ|ni are degenerate.
Example Electon in electromagnetic field
If a particle is influenced by an external static electric
field
V (x) = eφ(x),
then clearly the Hamiltonian
H=
p2
+ V (x)
2m
is invariant under the time reversal:
[Θ, H] = 0.
If now there are odd number of electrons in the system
the total j is half integer. Thus, in a system of this kind
there is at least twofold degeneracy, so called Kramers’
degeneracy.
In the magnetic field
B =∇×A
ΘSΘ−1 = −S ja ΘpΘ−1 = −p,
so
[Θ, H] 6= 0.
We say that magnetic field breaks the time reversal
symmetry and lifts the Kramers degeneracy.