## 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

Jˆ

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:

Jˆ

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:

Jˆ

α, Jˆ

β

, Gˆ

γ

=

Gˆ

γ, Jˆ

β

, Jˆ

α

+

Jˆ

α, Gˆ

γ

, Jˆ

β

,

and the commutation relationships between Jˆ

α and Jˆ

β seen in class, prove that

Jˆ

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

Jˆ

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

Jˆ

α, Gˆ

β

= −

Jˆ

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

hψ

0

|Oˆ0

|ψ

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

|ψ

0

i = Hˆ 0

|ψ

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

Tˆ

−→ 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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