Quantum Fields and ProbabilityHomework 1

Joona Oikarinen

January 23, 2018

On unbounded operators

Let H be a Hilbert space and D ⊂ H a dense subspace. An unbounded operator on H is alinear map D → H satisfying

supψ∈D ,‖ψ‖=1

‖Tψ‖ =∞ ,

or in other words there does not exist a constant C for which ‖Tψ‖ ≤ C‖ψ‖. The adjointT ∗ is defined in the following way. We say that ψ ∈ H belongs to the domain D(T ∗) of T ∗

if there exists χ ∈ H such that

〈ψ, Tφ〉 = 〈χ, φ〉 , ∀φ ∈ D .

Then we define T ∗ψ := χ. Note that D(T ∗) is not necessarily equal to D(T ) := D. We saythat T is self-adjoint if D(T ) = D(T ∗) and Tψ = T ∗ψ for all ψ ∈ D(T ).

For us the relevant operators are the position q̂, momentum p̂, creation a† and annihila-tion a. All of these operators are unbounded, and hence they operate on some subspace ofL2(R) instead of the whole space. The creation operator a† actually is the adjoint a∗ of theunbounded operator a, which we will prove below. Let us first consider a and a† as operatorsdefined on the space of Schwartz functions S (R) (smooth functions such that the functionand all its derivatives decay fast at infinity). Recall the definitions

a† =1√2

(√ωq − 1√

ω

d

dq

),

a =1√2

(√ωq +

1√ω

d

dq

).

Now for any ψ, φ ∈ S (R) we easily get the following result.

Lemma 0.1. It holds that S (R) ⊂ D(a∗) and a∗ψ = a†ψ for all ψ ∈ S (R).

1

Proof.

〈ψ, aφ〉 =

√ω√2

∫Rψ(q)qφ(q) dq − 1√

2ω

∫Rψ(q)φ′(q) dq

=

√ω√2

∫Rψ(q)qφ(q) dq +

1√2ω

∫Rψ′(q)φ(q) dq ,

where on the latter integral we used integration by parts1. Thus we have shown that

〈ψ, aφ〉 = 〈a†ψ, φ〉 .

So at least S (R) ⊂ D(a∗) and a∗ agrees with a† on this subspace.

Note that if we define

ψn =(a†)nψ0

‖(a†)nψ0‖,

then ψn ∈ S (R), since ψn is just a polynomial times a Gaussian function, and thus thedecay of ψn and its derivatives is of the type O(e−q

2).

To specify the full domain D(a∗) we need the following lemma.

Lemma 0.2. We have the relations

a†ψn =√n+ 1ψn+1 , aψn =

√nψn−1 .

Proof. It holds that

a(a†)nψ0 = (a†)a(a∗)n−1ψ0 + (a†)n−1ψ0

= (a†)2a(a†)n−2ψ0 + 2(a†)n−1ψ0

= . . .

= (a†)naψ0 + n(a†)n−1ψ0

= n(a†)n−1ψ0 ,

since aψ0 = 0 (essentially by definition, see the lecture note). Now, since (a†)nψ0 ∈ S (R)and a∗|S (R) = a†, we get

‖(a†)nψ0‖2 = 〈(a†)nψ0, (a†)nψ0〉 = 〈(a∗)nψ0, (a

†)nψ0〉= 〈ψ0, a

n(a†)nψ0〉= 〈ψ0, na

n−1(a†)n−1ψ0〉= 〈ψ0, (n− 1)nan−2(a†)n−2ψ0〉= . . .

= n!〈ψ0, ψ0〉= n! .

1To be precise, one has to split∫R ψφ

′ =∫B(0,R)

ψφ′ +∫B(0,R)c

ψφ′ and then do the integration by parts

on the first term. After this one has to argue that the boundary term and the B(0, R)c term go to zeroas R → ∞. This follows from the assumption that ψ and φ are Schwartz functions. The point is that theargument does not work for arbitrary ψ, φ ∈ L2(R).

2

Thus we have

ψn =(a†)nψ0√

n!,

which immediately implies that

a†ψn =(a†)n+1ψ0√

n!=√n+ 1ψn+1 .

The other relation is also immediate by the above computations:

aψn =a(a†)nψ0√

n!=n(a†)n−1ψ0√

n!=√nψn−1 .

Now we are ready to determine D(a∗). Since the sequence (ψn)∞n=0 forms an orthonormalbasis for L2(R), we can write each ψ ∈ L2(R) as

∑∞n=0 cnψn satisfying

∑∞n=0 |cn|2 < ∞.

Let’s now formally compute

a†ψ =∞∑n=0

cna†ψn =

∞∑n=0

cn√n+ 1ψn+1 .

This series converges if and only if∞∑n=0

|cn|2(n+ 1) <∞ ,

which holds if and only if∑∞

n=0 n|cn|2 <∞. Thus a† can be extended to have the domain

D(a†) = {ψ ∈ L2(R) | ψ =∑n

cnψn ,∑n

n|cn|2 <∞} .

Now we define the operator a† on this larger domain by2

a†ψ = a†

(∞∑n=0

cnψn

):=

∞∑n=0

cn√n+ 1ψn+1 .

Similar argument shows that the operator a can also be extended to this domain, i.e. wehave D(a†) = D(a) and for ψ ∈ D(a) we set

aψ :=∞∑n=0

cn√nψn−1 .

2For the reader who is wondering if this definition agrees with the earlier definition when operating onSchwartz-functions, an easy way out of this issue is to use in Lemma 0.1 the domain

Dfinite = {ψ | ψ =

k∑n=0

cnψn , k ∈ N}

instead of Schwartz-functions. This is still dense in L2(R) (and actually a subspace of S (R)) and Lemma0.1 holds also when we replace S (R) by this domain.

3

Theorem 0.3. We have D(a∗) = D(a†) (as defined above) and a∗ = a†.

Proof. The inclusion D(a†) ⊂ D(a∗) is easy since if ψ, φ ∈ D(a†), we write ψ =∑

n cnψnand φ =

∑n dnψn and get

〈a†ψ, φ〉 =∑n,m

cndm〈√n+ 1ψn+1, ψm〉

=∞∑n=0

cndn+1

√n+ 1 ,

〈ψ, aφ〉 =∑n,m

cndm〈ψn,√mψm−1〉

=∞∑n=0

cndn+1

√n+ 1

Thus ψ ∈ D(a∗) and a∗ψ = a†ψ for any ψ ∈ D(a†).Finally, we show that D(a∗) ⊂ D(a†). Let ψ ∈ D(a∗). This means that

〈a∗ψ, φ〉 = 〈ψ, aφ〉 , ∀φ ∈ D(a) . (0.1)

Thus we have

∞ > ‖a∗ψ‖2 = 〈a∗ψ, a∗ψ〉 =∑n,m

cncm〈a∗ψn, a∗ψm〉 =∑n,m

cncm〈ψn, aa∗ψm〉 .

The last equality is true because a∗ψm ∈ D(a) and (0.1). Now since a∗ψm = a†ψm, we get

∞ > ‖a∗ψ‖2 =∑n,m

cncm〈ψn, aa†ψm〉 =∑n,m

cncm〈ψn,mψm〉 =∑n

n|cn|2 .

Thus ψ ∈ D(a†). We conclude that D(a†) = D(a∗) and thus a† = a∗.

Some literature

1. Basics of unbounded operators and mathematical quantum mechanics can be found inthe book ”Quantum Theory for Mathematicians” by Brian Hall. The book is mathe-matically rigorous but does not cover quantum field theory. Mainly one particle quan-tum mechanics.

2. A more physical and less rigorous book is ”Quantum Field Theory: A Tourist Guidefor Mathematicians” by Gerald Folland. Good for learning actual physics.

3. ”Quantum Physics: A Functional Integral Point of View” by James Glimm and ArthurJaffe. A more advanced mathematics book on quantum field theory.

4. The 4 volumes of ”Methods of Modern Mathematical Physics” by Michael Reed andBarry Simon make life easier, especially Volumes 1 and 2.

4

1

Recall that the Hamiltonian H is given by

H = −1

2

d2

dq2+

1

2ω2q2

(in the units ~ = m = 1) and the creation and annihilation operators are defined by

a∗ =1√2

(√ωq − 1√

ω

d

dq

),

a =1√2

(√ωq +

1√ω

d

dq

).

These lead to the formula

H = ω

(a∗a+

1

2

).

1. First we prove that [a, a∗] = I. The operators a and a∗ are unbounded and thusdiscussing their commutator is a bit subtle. For this reason we will assume that weare working with states ψ ∈ H of the form

ψ =n∑k=0

ckψk

where ψk = (a∗)kψ0 as usual. These states belong to the domains of a, a∗, a∗a and aa∗

so we can consider commutation on this subset.

We have

[a, a∗]ψ(q) = aa∗ψ(q)− a∗aψ(q)

=1

2

(√ωq +

1√ω

d

dq

)(√ωqψ(q)− 1√

ωψ′(q)

)− 1

2

(√ωq − 1√

ω

d

dq

)(√ωqψ(q) +

1√ωψ′(q)

)=

d

dq(qψ(q))− qψ′(q)

= ψ(q) .

Thus [a, a∗] = I.

5

2. Next we show that [H, a] = −ωa and [H, a∗] = ωa∗. We have

(Ha− aH)ψ = ω

(a∗a+

1

2

)aψ − aω

(a∗a+

1

2

)ψ

= ω(a∗a2 − aa∗a)ψ

= ω[a∗, a]aψ

= −ωaψ ,(Ha∗ − a∗H)ψ = ωa∗[a, a∗]ψ

= ωa∗ψ .

3. Next we show that [H, an] = −ωnan and that [H, (a∗)n] = ωn(a∗)n. We proceed byinduction (the base case was done above). Hence assume that the relation holds forsome k. Then

[H, an] = Han − anH= (aH + [H, a])an−1 − anH= a[H, an−1]− ωan

= a(−ω(n− 1)an−1)− ωan

= −ωnan .

The proof for [H, (a∗)n] is almost identical. Alternatively, take the adjoint of the pre-vious result.

4. Next we show that eitHae−itH = e−itωa. First of all we need to consider the domainsof these operators. Since the states (ψn)∞n=0 form an orthonormal basis on L2(R), anyψ ∈ L2 can be written as

ψ =∞∑n=0

cnψn ,

where∑

n |cn|2 <∞. Note that we have a∗ψn =√n+ 1ψn+1. Thus for a∗ψ to belong

to L2 we need the further condition

‖a∗ψ‖ =∞∑n=0

(n+ 1)|cn|2 <∞,

i.e.∑

n n|cn|2 <∞. This fully determines the domain of a, so we have

D(a) = {ψ ∈ L2|ψ =∞∑n=0

cnψn,

∞∑n=0

n|cn|2 <∞} .

Since H is self-adjoint, the operator eitH is unitary and thus defined on the wholespace L2. Furthermore, eitHψn = eitω(n+1/2)ψn, so that eitHD(a) ⊂ D(a). Thereforethe domain of the operator eitHae−itH is the same as the domain of a.

6

For ψn = (a∗)nψ0/‖(a∗)nψ0‖ we have

e−itHψn =∞∑k=0

(−itH)k

k!ψn =

∞∑k=0

(−itω(n+ 1/2))k

k!ψn = e−itω(n+1/2)ψn .

Hence for all ψ ∈ L2 we have

e−itHψ =∞∑n=0

cne−itHψn =

∞∑n=0

cne−itω(n+1/2)ψn .

Thus, for any ψ ∈ D(a) we have

e−itHaeitHψ = e−itH∞∑n=0

cne−itω(n+1/2)

√nψn−1

=∞∑n=0

cne−itω(n+1/2)

√neitω(n−1+1/2)ψn−1

= e−itω∞∑n=0

cn√nψn−1

= e−itωaψ .

The computation for eitHa∗e−itH is similar (or, just take the adjoint of the previouscomputation).

7

2

By definition

q(t) = eitHqe−itH .

Using Exercise 1 and q = 1√2ω

(a+ a∗) we get

eitHqe−itH =1√2ω

(e−iωta+ eiωta∗) .

Thus

(ψ0, q(t1) . . . q(tn)ψ0) =1

(2ω)n/2(ψ0,

n∏j=1

(e−iωtja+ eiωtja∗)ψ0) .

At this point we already see that this is non zero only when n is even, since we must haveq(t1) . . . q(tn)ψ0 = Cψ0 (recall that the eigenstates are orthogonal) and this is possible onlywhen we get terms where we have equally many creation and annihilation operators actingon ψ0, i.e. when n is even. Thus from now on we assume that n is even.

Using 〈ψ0, a∗φ〉 = 〈aψ0, φ〉 = 0 and [a, q(t)] = 1√

2ωeiωt we get

(ψ0, q(t1) . . . q(tn)ψ0) =1√2ωe−iωt1(ψ0, aq(t2) . . . q(tn)ψ0)

=1√2ωe−iωt1(ψ0, q(t2)aq(t3) . . . ψ0) +

1

2ωeiω(t2−t1)(ψ0, q(t3) . . . q(tn)ψ0) .

We continue to commute a towards right in the first term

1√2ωe−iωt1(ψ0, q(t2)aq(t3) . . . ψ0) =

1√2ωe−iωt1(ψ0, q(t2)q(t3)aq(t4) . . . ψ)

+1

2ωeiω(t3−t1)(ψ0, q(t2)q(t3)q(t5) . . . q(tn)ψ0) .

Iterating this process we end up with

(ψ0, q(t1) . . . q(tn)ψ0) =1

2ω

n∑k=2

eiω(tk−t1)(ψ0,n∏

j=1,j /∈{1,k}

q(tj)ψ0) +1√2ωeiωt1(ψ0, q(t2) . . . q(tn)aψ0)

=1

2ω

n∑k=2

eiω(tk−t1)(ψ0,

n∏j=1,j /∈{1,k}

q(tj)ψ0) .

Now we proceed by induction. The claim for (ψ0, q(t1)q(t2)ψ0) is clear. Assume that theWick formula holds for (ψ0, q(s1) . . . q(sn−2)ψ0). Then the above formula implies

(ψ0, q(t1) . . . q(tn)ψ0) =n∑k=2

G(tk − t1)∑

p∈P ({1,...,n}\{1,k})

∏{a,b}∈p

G(ta − tb)

=∑

p∈P ({1,...,n})

∏{a,b}∈p

G(ta − tb) ,

where P (S) denotes the set of pair partitions of the set S.

8

3

For clarity, we denote the position operator by q̂. The operator eαq̂, α ∈ C, is defined by

(eαq̂ψ)(q) = eαqψ(q) =∞∑n=0

(αq)n

n!ψ(q) .

Thus the domain of eαq̂ consists of the functions ψ ∈ L2 for which∫|e2αq||ψ(q)|2 dq <∞.

The domains of the operators eαa∗

or eαa are a bit trickier. Since a and a∗ are notself-adjoint (not even normal), defining the exponential eαa can not be done using spectraltheory. Instead, we interpret eαa as the operator

(eαaψ)(q) :=∞∑n=0

(αa)n

n!ψ(q) .

One natural dense domain for this operator is

D(eαa) := {ψ ∈ L2(R) | ψ =k∑

n=0

cnψn , cn ∈ C} ,

i.e. finite linear combinations of the eigenstates of the Hamiltonian. For such ψ we have

‖eαaψ‖ =

∥∥∥∥∥∞∑n=0

(αa)n

n!ψ

∥∥∥∥∥ =

∥∥∥∥∥k∑

n=0

(αa)n

n!ψ

∥∥∥∥∥ <∞ ,

since anψk = 0 if n > k, i.e. the series converges in L2. Recall that ‖qnψ0‖ ≤ Cn√n! and

‖(a∗)nψ0‖ =√n!. These imply that the series we (implicitly) manipulate below actually

converge in L2.Let ψ ∈ D(eαa). Define a function fψ : C→ L2(R),

fψ(α) = e−αqeα√2ωa∗e

α√2ωaψ .

Then fψ(0) = ψ and3

f ′ψ(α) = −qfψ(α) + e−αqa∗√2ωe

α√2ωa∗e

α√2ωaψ + e−αqe

α√2ωa∗ a√

2ωe

α√2ωaψ .

3The derivative ddte

tAψ for an operator A acting on a Hilbert space means the limit

limh→0

e(t+h)Aψ − etAψh

in the Hilbert space. In our case all the exponentials can be represented as L2-convergent sums and thusthe differentiation ends up working ”as usual”, as long as we are operating on nice enough ψ.

9

Now by using the series form of e−αq and [qn, a∗] = n√2ωqn−1, we get

e−αqa∗√2ω

=a∗√2ωe−αq +

α

2ωe−αq .

Another straightforward commutation computation shows that

e−αqeα√2ωa∗ a√

2ω=

a√2ωfψ(α)

By combining the results we get

f ′ψ(α) = − α

2ωfψ(α) , fψ(0) = ψ .

Note that this is an L2(R)-valued differential equation. Since f ′ψ(α) = limh→0fψ(α+h)−fψ(α)

h

converges in L2, for any φ ∈ L2(R) we can write

〈f ′ψ(α), φ〉 = limh→0

〈fψ(α + h), φ〉 − 〈fψ(α), φ〉h

=d

dα〈fψ(α), φ〉 ,

which leads to the C-valued differential equation

d

dα〈fψ(α), φ〉 = − α

2ω〈fψ(α), φ〉 , 〈fψ(0), φ〉 = 〈ψ, φ〉 .

It is well known that this is solved by

〈fψ(α), φ〉 = e−α2

4ω 〈fψ(0), φ〉 = e−α2

4ω 〈ψ, φ〉 .

Since the above equation holds for any φ ∈ L2(R), we infer that

e−αqeα√2ωa∗e

α√2ωaψ = fψ(α) = e−

α2

4ωψ .

Note that the above computations heavily relied on the fact that we are working with a nicestate ψ, for example a finite linear combination of the eigenstates, since for these we havegood estimates for ‖qnψk‖ and ‖(a∗)nψk‖, leading to the convergence of all the series that(implicitly) are behind all our computations.

10

4

Using Exercise 3 we get

e−tHeipqψ0(q) = e−p2

4ω e−tHeipa∗√

2ω eipa√2ωψ0(q) = e−

p2

4ω e−tHeipa∗√

2ωψ0(q) ,

where the last formula holds because aψ0 = 0 (to prove this we can use the series represen-tation, which converges when operating on ψ0). Next we write (again, the series convergesbecause we are operating on ψ0)

e−tHeipa∗√

2ωψ0(q) = e−tH∞∑n=0

(ip/√

2ω)n

n!(a∗)nψ0(q) =

∞∑n=0

(ip/√

2ω)n

n!

1

‖(a∗)nψ0‖e−tHψn(q) .

Next we use e−tHψn(q) = e−tωnψn(q), which follows from Hψn(q) = ωnψn(q). Hence wehave

e−tHeipqψ0(q) = e−p2

4ω

∞∑n=0

(ip/√

2ω)n

n!e−tωn(a∗)nψ0(q)

= e−p2

4ω eip√2ωe−tωa∗

ψ0(q) .

Using Exercise 3 again we get

e−p2

4ω eip√2ωe−tωa∗

ψ0(q) = e−p2

4ω ep2

4ωe−2tω

eip√2ωe−tωq

e−ip√2ωe−tωa

ψ0(q)︸ ︷︷ ︸=ψ0(q)

.

Thus we have shown that

e−tHeipqψ0(q) =

∫Kt(q, q

′)eipq′ψ0(q

′) dq′ = e−p2

4ω ep2

4ωe−2tω

eip√2ωe−tωq

ψ0(q) .

Observe that the middle term is the inverse Fourier transform of the function fq(q′) =

Kt(q, q′)ψ0(q

′) evaluated at the point p. We take Fourier transform in p to get

Kt(q, x)ψ0(x) =1

2π

∫∫e−ipxKt(q, q

′)eipq′ψ0(q

′) dq′ dp

=1

2π

∫e−ipxe−

p2

4ω ep2

4ωe−2tω

eip√2ωe−tωq

ψ0(q) dp

=1

2πψ0(q)

∫exp

(p

(i√2ωe−tωq − ix

)+ p2

(e−2tω − 1

4ω

))dp

=1

2πψ0(q)

∫exp

(αp+ βp2

)dp

=1√2πψ0(q)

√1

−βeα2

−4β ,

11

with α = i( 1√2ωe−tωq − x) and β = e−2tω−1

4ω(we used the Gaussian integral formula). Thus

Kt(q, x) =1√2π

ψ0(q)

ψ0(x)

√2ω√

1− e−2tωe(x− 1√

2ωe−tωq)2 ω

1−e2tω

=

√ω√π

(1− e−2tω)−1/2 exp

(−q

2 − x2

2− (e−2tωq − x)2

1− e−2tω

).

12