Research Progress

Srivatsan

February 22, 2014

1 Lorentz transformations

a) The definition of δ(4)(p) is:

δ(4)(p) =1

(2π)4

∫d4xeip·x. (1)

Here, k · x = xµpµ.

We could equally well choose to integrate with respect to xµ = Λµνxν .

δ(4)(p) =1

(2π)4

∫d4xeip·x =

1

(2π)4

∫d4xeipµΛµνx

ν

=1

(2π)4

∫d4xeip·x = δ(4)(p).

(2)

In the first line, we used the fact that the integration measure is invariant,d4x = d4x. In the second line we defined pν = pµΛµν .

b) We want to show that:

ω~kδ(3)(~k − ~k′) = ω

Λ~kδ(3)(Λ~k − Λ~k′) (3)

Integrating against an arbitrary function, we have:

F (~k′) =

∫d3kδ(3)(~k − ~k′)F (~k) =

∫d3k

ω~kω~kδ

(3)(~k − ~k′)F (~k)

=

∫d3k

ω~kω~kδ

(3)(~k − Λ~k′)F (Λ−1~k)

=

∫d3Λp

ωΛ~pωΛ~pδ

(3)(Λ~p− Λ~k′)F (~p)

=

∫d3k

ω~kω

Λ~kδ(3)(Λ~k − Λ~k′)F (~k).

(4)

1

As F was arbitrary this implies:

ω~kδ(3)(~k − ~k′) = ω

Λ~kδ(3)(Λ~k − Λ~k′). (5)

Note, in arriving at the last line, we used d3Λkω

Λ~k= d3k

ω~k, which we show in

part (d). Here, Λ~k ≡ Λiµkµ.

c) From part (b) we have:

[a

Λ~k, a†

Λ~k′

]= (2π)3δ(3)(Λ~k − Λ~k′) =

ω~kω

Λ~k

(2π)3δ(3)(~k − ~k′)

=ω~kω

Λ~k

[a~k, a

†~k′

].

(6)

We can then guess that:

aΛ~k

=

√ω~kω

Λ~k

a~k

a†Λ~k

=

√ω~kω

Λ~k

a†~k.

(7)

d)We want to show:∫

d3k

(2π)3

1

2ω~kf(k) =

∫d3k

(2π)3

1

2ω~kf(k) (8)

for any proper, orthocronos Lorentz transformation. In order to show this,first note that:∫

d3k

(2π)3

1

2ω~kf(k) =

∫d4k

(2π)4(2π)δ(k2 +m2)θ(k0)f(k). (9)

One can derive (??) by performing the k0 integral. Note that, as, k2 +m2 =−(k0)2 + ~k2 + m2, changing variables from k0 to u = (k0)2, changes the

measure as dk0 = 12k0du. Though the delta function k0 = ±

√~k2 +m2 has

two solutions, the theta function θ(k0) picks out only one.

2

With this relation we have:∫d3k

(2π)3

1

2ω~kf(k) =

∫d4k

(2π)4(2π)δ(k2 +m2)θ(k0)f(k)

=

∫d4k

(2π)4(2π)δ(k2 +m2)θ(k0)f(k)

=

∫d4k

(2π)4(2π)δ(k2 +m2)θ(k0)f(k) =

∫d3k

(2π)3

1

2ω~kf(k).

(10)

In getting to the last line, we used θ(k0) = θ(k0). This holds, as we are onlyconsidering the connected part of the Lorentz group, which does not changethe sign of k0.

2 Problem with relativistic quantum mechanics

a)Beginning with the Schrodinger equation, i∂tψ = − 1

2m~∇2ψ, and ρ =

|ψ|2, we have:

∂tρ = ψ∂tψ∗ + ψ∗∂tψ

= − i

2m

(ψ~∇2ψ∗ − ψ∗~∇2ψ

)=

i

2m~∇ ·(ψ∗~∇ψ − ψ~∇ψ∗

)= −~∇ · ~J

(11)

b) Proceeding as before, but now with ∂2t ψ = ~∇2ψ − m2ψ and ρ =

i2m (ψ∗∂tψ − ψ∂tψ∗) .

∂tρ =i

2m

(ψ∗∂2

t ψ − ψ∂2t ψ∗)

=i

2m

(ψ∗~∇2ψ − ψ~∇2ψ∗

)= −~∇ · ~J

(12)

c)The expression here for ρ, ρ = i

2m(ψ∗∂tψ − ψ∂tψ∗) = 1m Im(ψ∂tψ

∗), isnot positive definite, and so cannot be interpreted as a probability density.

3

3 Commutation relations of annihilation and cre-ation operators

a) The field and momentum at time t can be written in terms of the fieldand momentum at time 0 as: φ(~x, t) = eiHtφ(~x, 0)e−iHt and π(~x, t) =eiHtπ(~x, 0)e−iHt. With this the equal time commutation relations can bewritten as [

φ(~x, t), φ(~x′, t)]

= eiHt[φ(~x, 0), φ(~x′, 0)

]e−iHt[

π(~x, t), π(~x′, t)]

= eiHt[π(~x, 0), π(~x′, 0)

]e−iHt[

φ(~x, t), π(~x′, t)]

= eiHt[φ(~x, 0), π(~x′, 0)

]e−iHt.

(13)

As, eiHt0e−iHt = 0, and eiHtiδ3(~x − ~x′)e−iHt = iδ3(~x − ~x′), it is enough toimpose the commutation relations at t = 0.[

φ(~x, 0), φ(~x′, 0)]

= 0→[φ(~x, t), φ(~x′, t)

]= 0[

π(~x, 0), π(~x′, 0)]

= 0→[π(~x, t), π(~x′, t)

]= 0[

φ(~x, 0), π(~x′, 0)]

= iδ(3)(~x− ~x′)→[φ(~x, t), π(~x′, t)

]= iδ(3)(~x− ~x′)

(14)

b)

φ(~x, t) =

∫d3k

(2π)3

1√2ω~k

(a~ke

ik·x + a†~ke−ik·x

)π(~x, t) =

∫d3k

(2π)3− i√ω~k2

(a~ke

ik·x − a†~ke−ik·x

) (15)

Transforming expression (??), we have:

φ(~k) =1√2ω~k

(a~k + a†

−~k

)π(~k) = −i

√ω~k2

(a~k − a

†−~k

).

(16)

With this,

a~k =

√ω~k2φ(~k) + i

√1

2ω~kπ(~k)

a†~k=

√ω~k2φ(−~k)− i

√1

2ω~kπ(−~k).

(17)

4

It is also convenient to have the expressions:

φ(~k) =

√1

2ω~k

(a~k + a†

−~k

)π(~k) = −i

√ω~k2

(a~k − a

†−~k

).

(18)

c) Form the commutation relations, (??), we also have the momentum space,equal time, relations:[

φ(~k), φ(~k′)]

= 0[π(~k), π(~k′)

]= 0[

φ(~k), π(~k′)]

= i(2π)3δ(3)(~k + ~k′).

(19)

With these it is strait forward to calculate the commutators of creation andannihilation operators.[

a~k, a~k′]

=i

2

([φ(~k), π(~k′)

]+[π(~k), φ(~k′)

])= 0[

a~k, a~k′]

= 0[a~k, a

†~k′

]= − i

2

([φ(~k), π(−~k′)

]−[π(~k), φ(−~k′)

])= (2π)3δ(3)(~k − ~k′).

(20)

Here we have used the fact that [φ(~k), φ(~k′)] = [π(~k), π(~k′)] = 0.

4 Expressing Nother charges in terms of creationand annihilation operators

In analogy to the last problem set we have:

H =1

2

∫d3x

(π2 + ~∇φ2 +m2φ2

)P i = −

∫d3x

(π∂iφ

)Mµν =

∫d3x

(xµT 0ν − xνTµ0

) (21)

5

Note, the convention for P i differs by a sign from that on the last problemset. a) Beginning with H, we can rewrite (??) in momentum space.

H =1

2

∫d3k

(2π)3

(π(~k, t)π(−~k, t) + (~k2 +m2︸ ︷︷ ︸

ω2~k

)φ(~k, t)φ(−~k, t)

). (22)

Using expression (??), or equivalently, (??), we can write H in terms of aand a†.

H =

∫d3k

(2π)3

ω~k2

(a~ka†~k

+ a†~ka~k

)=

∫d3k

(2π)3ω~k a

†~ka~k︸︷︷︸

N~k

+

∫d3k

(2π)3(2π)3δ(3)(0)︸ ︷︷ ︸CH

=

∫d3k

(2π)3ω~kN~k + CH .

(23)

b)Again it is easiest to work in momentum space.

P i = −i∫

d3k

(2π)3π(~k, t)φ(−~k, t)ki

= −1

2

∫d3k

(2π)3

(a~k − a

†−~k

)(a−~k + a†~k

)ki

= −1

2

∫d3k

(2π)3

(a†~ka~k − a

†−~ka−~k + a~ka−~k − a

†−~ka†~k

+[a~k, a

†~k

])ki

= −∫

d3k

(2π)3ki a†~k

a~k︸︷︷︸N~k

− 1

2

∫d3k

(2π)3(2π)3δ(3)(0)ki︸ ︷︷ ︸CPi

= −∫

d3k

(2π)3kiN~k − CP i .

(24)

Here we used the fact that expressions such as,∫d3ka~ka−~kk

i = 0.c)To write Mµν in terms of creation and annihilation operators, it is useful

to write down the expression for the stress-energy tensor.

Tµν = ∂µφ∂νφ− 1

2ηµν

(∂ρφ∂

ρφ+m2φ2). (25)

6

For Mµν we need

T 0ν = −π∂νφ− 1

2η0ν(−π2 + ~∇φ · ~∇φ+m2φ2

)=(H(x), ~P(x)

).

(26)

With this we have:

M0i =

∫d3x

(tP i(x)− xiH(x)

)= tP i −

∫d3k

(2π)3

∂

∂ki(ω~kN~k),

M ij =

∫d3x

(xiPj(x)− xjP i(x)

)=

∫d3k

(2π)3

(kj

∂

∂ki− ki ∂

∂kj

)N~k.

(27)

Here we have dropped the normal ordering constants.

5 Second quantization

a) Varying the action L = iψ∗∂tψ − 12m~∇ψ∗ · ~∇ψ − V (~x)ψ∗ψ with respect

to ψ∗ and ψ yields:

i∂tψ +1

2m~∇2ψ − V (~x)ψ = 0

−i∂tψ∗ +1

2m~∇2ψ∗ − V (~x)ψ∗ = 0.

(28)

This is indeed the Schrodinger equation.b) The conjugate momentum to ψ is given as:

πψ =∂L∂∂tψ

= iψ∗. (29)

The Hamiltonian is:

H =

∫d3x [πψ∂tψ − L] =

∫d3x

[1

2m~∇ψ∗ · ~∇ψ + V (~x)ψ∗ψ

](30)

7

c) The canonical commutation relations are:[ψ(~x, t), πψ(~x′, t)

]= iδ(3)(~x− ~x′). (31)

This can also be written,[ψ(~x, t), ψ†(~x′, t)

]= δ(3)(~x− ~x′). (32)

d) Writing, ψ(~x, t) =∑

n ane−iEntun(~x), ψ†(~x, t) =

∑n a†neiEntun(~x). It is

enough to consider the commutator, (??), at t = 0.[ψ(~x, 0), ψ†(~x′, 0)

]=∑m,n

(ama

†n − a†nam

)um(~x)un(~x′) = δ(3)(~x− ~x′) (33)

Multiplying by um′(~x), un′(~x′), integrating d3x, d3x′ and dropping the primeson the indices, we have:[

am, a†n

]=

∫d3xd3x′um(~x)un(~x′)δ(3)(~x− ~x′) = δmn. (34)

Similarly, [ψ(~x, 0), ψ(~x′, 0)] = 0 implies [am, an] = [a†m, a†n] = 0.

e)Plugging the expression for ψ into (??) we have:

H =

∫d3xψ†(− 1

2m~∇2 + V (~x))ψ

=

∫d3x

∑m,n

a†mane−i(En−Em)tum(− 1

2m~∇2 + V (~x))un

=

∫d3x

∑m,n

a†mane−i(En−Em)tEnumun

=∑n

a†nan︸ ︷︷ ︸Nn

En.

(35)

In going from the second line to the third line, we used the fact that unsatisfies the time independent Schrodinger equation, (− 1

2m~∇2 + V (~x))un =

Enun. In getting the last line we used the orthonormality of un.f)

8

A generic state can be written as an arbitrary combination of creationoperators acting on the vacuum, with each creation operator aloud to actan arbitrary number of times.

|Ψ〉 ≡ |{mi}〉 = (a†1)m1(a†2)m2 . . . |0〉. (36)

Here, the {mi} range from 0 up to ∞. Note, do not confuse this notationwith that used in part (h).

g)A physical interpretation of the states |{mi}〉 is that this state has mi

indistinguishable particles in the energy level i. The reason for this inter-pretation comes from looking at the Hamiltonian. The energy level for thestate is just Etot =

∑iEimi. This is precisely the energy for many particles

interacting with an external potential with mi particles in energy level i.h) From the picture above the wave function would just be the two

particle wave function for one particle in state n1 and another in state n2.

Ψn1n2(~x1, ~x2, t) =1√2e−i(En1+En2 )t (un1(~x1)un2(~x2) + un2(~x1)un1(~x2)) .

(37)

We can also get this expression from the second quantized field.

1√2〈0|ψ(~x1, t)ψ(~x2, t)|n1, n2〉 =

1√2〈0|∑m,n

e−i(Em+En)tum(~x1)un(~x2)amana†n1a†n2|0〉

=1√2e−i(En1+En2 )t (un1(~x1)un2(~x2) + un2(~x1)un1(~x2)) .

(38)

9