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Rotations, Boosts and Headaches

1 Rotations, Boosts and Headaches (marked by Richard)
In this exercise we are going to derive the commutation relationships between the generators
of rotations and boots.
(a) Prove that the result of applying an infinitesimal rotation of angle   1 around
the axis x1, then a boost of speed − along the axis x2, then their inverses (i.e. a
rotation of angle − around the axis x1 and a boost of speed  along the axis x2 in
that order) is equivalent to a single boost of speed 
2 along the axis x3. (Hint: Keep
in mind these are infinitesimal transformations:   1, so at some point you’ll need
to use that assumption)
(b) Using the result proved in part (a) and
e
iKˆµ e
iKˆν e
−iKˆµ e
−iKˆν = 1 + 
2
[Kˆ
ν, Kˆ
µ] + O


3

(1.1)
e
iω2
e
i
2Kˆ
λ = 1 + i2Kˆ
λ + iω21 + O(
3
) (1.2)
Prove that


1, Gˆ
2

= iGˆ
3 + ω121
where ω12 is a complex constant. Induce that (or compute) that the rest of the
commutators have the following form:


α, Gˆ
β

= iεαβγGˆ
γ + ωαβ1
(note that the last symbol εαβγ has nothing to do with )
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(c) Using the previous result, Jacobi identity with Jˆ
β, Jˆ
α and Gˆ
γ:

α, Jˆ
β

, Gˆ
γ

=

γ, Jˆ
β

, Jˆ
α

+

α, Gˆ
γ

, Jˆ
β

,
and the commutation relationships between Jˆ
α and Jˆ
β seen in class, prove that


α, Gˆ
α

= 0. By using the same identity prove also that ωαβ has to be of the form
ωαβ = iεαβγb
γ1.
with b
γ are real constants. Show that this term only produces an unphysical phase
factor that can be reabsorbed in the definition of Gˆ and therefore we obtain that


α, Gˆ
β

= iε
γ
αβ Gˆ
γ.
(Hint: the trick to solve this exercise is just to take appropriate particular values of the
labels α, β, γ in our particular Jacobi identity. As an intermediate step for the second
part of the exercise, it might be helpful to prove first that

α, Gˆ
β

= −


β, Gˆ
α

)
2 Zooming in and Running (Marked by Richard)
Galilean space is invariant under scaling (a rescaled treasure map is still a treasure map),
x → x
0 = e
cx where c is a real parameter. The corresponding unitary operator is e
−icDˆ
where Dˆ is the generator of dilations. Compute the commutator
D, ˆ Pˆ

between the generators of dilations and spatial translations.
(Hint: maybe equations (1.1) and (1.2) in the previous exercise can be helpful here. It can
also be helpful to prove, as an intermediate result, that applying an infinitesimal dilation 
along the axis x1, then an infinitesimal translation of  along the same axis and then their
inverses (a dilation of − and finally a translation of −) is equivalent to a translation of
magnitude −
2
)
3 Gauge invariance (Marked by Pipo)
It has been shown in the lectures that the most general Hamiltonian that describes the
motion of a particle compatible with Galilei invariance and an external force is
Hˆ =
(Pˆ − A(Xˆ ))2
2M
+ W(Xˆ ). (3.1)
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But that is not the end of the story. As you saw in question 4 of assignment 1, it
turns out that if we implement a unitary transformation in both observables and states
simultaneously, the physical predictions remain unchanged. That is, if for each state we
fix |ψ
0
i = Uˆ |ψi and Oˆ0 = UˆOˆUˆ†
, we get

0
|Oˆ0

0
i = hψ|Uˆ†UˆOˆUˆ†Uˆ|ψi = hψ|Oˆ|ψi. (3.2)
Coming back to the most general form of the Hamiltonian, these considerations make us
think whether given two pairs of different functions A, W and A0
, W0
they may describe the
same physical scenario. In other words, we want to know whether changes in A, W result
in different physics. You may guess that the answer is yes perhaps thinking that different
forces for sure will push the system differently (and you would be right!). However, and
interestingly, there changes in A, W that will not change physical predictions at all! This
is the object of this exercise.
In this exercise we will analyze transformations in A, W that generate unitary transformations in the observables, thereby generating unitarily equivalent representations and
not affecting the physical predictions. These are called gauge transformations, and are extremely important all the fundamental forces1
can be fully characterized by defining their
gauge transformations.
(a) First, given a unitary transformation in the state |ψ
0
i = Uˆ(t)|ψi dependent on time,
find the Hamiltonian Hˆ 0
that generates time translations. In other words, find the
new Schr¨odinger equation
i
d
dt

0
i = Hˆ 0

0
i. (3.3)
(b) Second, given a unitary transformation of the form Uˆ
χ = e
iχ(Xˆ ,t)
, where χ is a
function of the spatial coordinates (the position operator) and time, calculate Hˆ 0
explicitly.
Hint: you can use the following identity that you would get from Hadamard’s lemma
to BCH and a bit of spectral theorem juggling2
UˆPˆ
iUˆ† = Pˆ
i + i[χ(Xˆ , t), Pˆ
i
] = Pˆ
i + i∂iχ(Xˆ , t)[Xˆ
i
, Pˆ
i
] (3.4)
(c) Given the new Hamiltonian Hˆ 0
, discuss if it still respects the Galilei transformations
(i.e., does it have the same form as (3.1)) for some new A0
, W0
?) What are the new
A0 and W0
? How would an experimentalist know if the force they are measuring is
derived from A, W or from A0
, W0
.
1Yes! even gravity, if you do GR.
2You should totally be able to prove it.
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4 Discrete transformations (Marked by Erickson)
An inversion is a transformation that inverts spatial coordinates x → −x, i.e. taking
“mirror image”. A time-reversal is a transformation that inverts time direction, i.e.
t → −t. In quantum mechanics, inversion is implemented by the parity operator Π and ˆ
time-reversal is implemented by the time-reversal operator Tˆ. Note that both inversion
and time-reversal reverses classical momentum, since
p = m
dx
dt
Πˆ
−→ m
d(−x)
dt
= −p , (4.1)
p = m
dx
dt

−→ m
dx
d(−t)
= −p . (4.2)
(a) Show3
that for inversion,
Πˆ −1
◦ Xˆ ◦ Π =ˆ −Xˆ , Πˆ −1
◦ Pˆ ◦ Π =ˆ −Pˆ , (4.3)
while for time-reversal, we have
Tˆ−1
◦ Xˆ ◦ Tˆ = Xˆ , Tˆ−1
◦ Pˆ ◦ Tˆ = −Pˆ . (4.4)
(b) Show that if we require that Π preserves the commutation relation between ˆ Xˆ and
Pˆ, then Π must be a unitary operator. ˆ
(c) Show that if we require that Tˆ preserves the commutation relation s between Xˆ and
Pˆ, then Tˆ must be an anti-unitary operator.
Hint: saying Tˆ−1 ◦ i1 ◦ Tˆ = i1 is not correct. Think about it.
(d) An operator Oˆ represents a symmetry4
if [O, ˆ Hˆ ] = 0. Consider a family of states
{|ψ(t)i} which satisfies the Schr¨odinger equation
i
∂ |ψ(t)i
∂t = Hˆ |ψ(t)i . (4.5)
Show that to first order in t, we have
Tˆ−1
◦ iHˆ ◦ Tˆ = −iHˆ
3Note that we use the symbol ‘◦’ to denote composition of operators, for example [Xˆ , Pˆ] = Xˆ ◦Pˆ−Pˆ ◦Xˆ .
In quantum mechanics we don’t usually bother writing the composition explicitly as ◦ because the operators
are linear. However, the whole point of this exercise is that with discrete symmetries operators may not be
linear (could be anti-linear) and not every composition of operators is a ‘matrix multiplication’. Hence it
is important to write it explicitly for clarity.
4
In other words, Oˆ is a symmetry if it represents an observable that does not change with time (Hˆ
generates time translation).
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and hence prove that if Tˆ were unitary, it would imply that Hˆ = 0 (in other words,
Tˆ has to be anti-unitary if time-reversal is a valid symmetry of a system whose
nontrivial dynamics is governed by the Schr¨odinger equation).
5 Free particle: classical versus quantum
(or the multiple of identity in the structure constants)
(Marked by Erickson)
Let G be the Galilei group. We say that a group G acts on a vector space V if an element
gˆ ∈ G maps an element w ∈ V to another element w0 ∈ V . We write this compactly as
w0 = ˆgw. For example, when V = H is a Hilbert space of a quantum-mechanical free
particle, a spatial translation by a (which, as you know, is generated by Pˆ) is given by the
group element ˆg = e
−ia·Pˆ
. If we consider an element in H given by w = |xi, then the new
state reads5
|xi
0 = e
−ia·Pˆ
|xi = |x + ai =⇒ xˆ
0 = e
−ia·PˆXˆ e
ia·Pˆ
. (5.1)
We can do the same in classical mechanics, since V needs not be a Hilbert space of
quantum states. There are two common choices for V :
(1) Configuration space V1
∼= R
3
, where each point represents position6 x = (x1, x2, x3).
(2) Phase space V2
∼= R
6
, where each point is given by (x, p). Here x = (x1, x2, x3) is
position and p = (p1, p2, p3) is linear momentum.
It turns out that we can understand the difference in the physics of classical and quantummechanical free particle by looking at the choice of V on which the Galilei group G acts,
as this exercise will show.
(a) First, let us focus on configuration space. Consider translations and boosts in V1
which are given by
x → x + a , x → x + vt . (5.2)
Show that these are generated by Pˆ
α = i ∂
∂xα and Gˆ
α = −it

∂xα and hence they
commute7
, i.e. the commutator [Gˆ
α, Pˆ
β] = 0.
5Note that |xi is not normalizable and so you should think of H as rigged Hilbert space. In this question
we will ignore these technical, functional-analytic details which would not affect what follows.
6Let us ignore time in configuration space for simplicity in this exercise, since we will not do time shifts
and Galilei transformations do not mix time and space.
7Compare this with the lecture notes: you may notice that there is no multiple of identity here.
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Hint: The usual Taylor expansion will be helpful. However, note that a, vt are finite
translation and boost that do not need to be small, so your proof should not assume
truncation of any Taylor series expansion.
(b) On phase space V2, however, things work quite differently8 because we are not working
directly with the phase space elements (x, p). For a free particle, the Noether
charge9 Q(·) associated to the momentum and boost generators Pˆ
α, Gˆ
α are given by
Q(Pˆ
α) = p
α
, Q(Gˆ
α) = Mxα − p
α
t , α = 1, 2, 3 . (5.3)
Show that, at t = 0, these generators satisfy the Poisson bracket relation10
{Q(Gˆ
α), Q(Pˆ
β)} = δαβM , {f, g} := X
3
α=1
∂f
∂xα
∂g
∂pα

∂f
∂pα
∂g
∂xα
(5.4)
where M is a real number and {·, ·} is the Poisson bracket in classical mechanics.
Remark: this is where the ‘mass’ M that we saw in class appears in classical mechanics. The corresponding commutation relation in quantum mechanics can be obtained
by the canonical quantization procedure that maps
{·, ·} 7−→ i~[·, ·] , (5.5)
giving us the commutator [Gˆ
α, Pˆ
β] = i~δαβM1. We will take from now on ~ = 1.
(c) We have been saying that the ten generators {Jˆ
α, Gˆ
α, Pˆ
α, Hˆ } (α = 1, 2, 3) form
a Lie algebra11. However, any Lie algebra must be closed under Lie bracket,
i.e., if Kˆ
µ, Kˆ 0
µ are any two generators of the Lie algebra g, then their Lie bracket
[Kˆ
µ, Kˆ 0
ν
] ∈ g.
Using the closure property, explain why in quantum mechanics the ten generators
{Jˆ
α, Gˆ
α, Pˆ
α, Hˆ } (α = 1, 2, 3) associated to the Galilei group, whose commutators are
given in Block 2 notes, do not really form a Lie algebra. Find the “11th” generator
that extends the Galilei algebra so that it forms an 11-dimensional12 Lie algebra.
Hint: Find which operator (call it Aˆ for now) breaks the closure property and
show that the ‘extended’ algebra {Jˆ
α, Gˆ
α, Pˆ
α, H, ˆ Aˆ} is closed under the commutator.
8This is one of the technical aspects we ignore here, having to do with a concept called “lifting an
action”. The idea is that instead of the na¨ıve guess that we apply ˆg to (x, p), we need to work with the
commutator of the conserved charges associated to each generator.
9Recall from classical mechanics that a Noether charge is a conserved quantity associated to a symmetry.
10Note that the Poisson bracket is a Lie bracket in the sense of defining a Lie algebra associated to G.
Recall from Assignment 1 that a Lie bracket needs not be a commutator.
11See Assignment 1 Problem 3 for the notion of Lie algebra.
12Consequently, what we are using in the lecture notes is not what we would have called the Galilean
algebra (which is really 10-dimensional). This is why we can have [Gˆα, Pˆβ] = 0 in classical mechanics.
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Notice that Aˆ may include a real constant prefactor that does not depend on the
choice of the other ten generators.
(d) One interesting use of this eleven-dimensional (not ten!) algebra is the following: we
can argue that mass is really funny as a physical quantity, and it is not easy to think
of mass as a physical observable that has an associated self-adjoint operator.
Show that quantum-mechanical free particles of different masses M1, M2 are not
unitarily equilavent representations of the (extended) Galilei group.
Hint: two quantum free particles are unitarily equivalent representations if their
corresponding continuous symmetry generators are related by some unitary transformation. Note that an operator such as 1 generates precisely a phase transformation
(which is by the way, physically irrelevant for quantum states).
Remark: This means that we cannot na¨ıvely create a quantum state which is a
superposition of distinct masses, i.e.
|ψi = a |M1i + b |M2i |a|
2 + |b
2
| = 1 , (5.6)
where |Mj i are states corresponding to different mass Mj . For example, we cannot
prepare a state in a lab which is a superposition of proton and electron, since it would
mean that upon “mass measurement”, we may obtain proton or electron as outcomes.
This is relevant in phenomena such as the phenomenon of neutrino oscillation.
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