The quantum harmonic oscillator is the single most important solvable system for QFT, because:
the free scalar field decomposes (in momentum space) into infinitely many decoupled oscillators , one for each p \mathbf{p} p
“particle number” in free QFT is literally an oscillator occupation number .
In QFT one often encounters quadratic Lagrangians of the form
L = 1 2 ϕ ˙ 2 − 1 2 ω 2 ϕ 2 , L=\frac12\dot\phi^2-\frac12\omega^2\phi^2, L = 2 1 ϕ ˙ 2 − 2 1 ω 2 ϕ 2 , which is just a harmonic oscillator with coordinate ϕ ( t ) \phi(t) ϕ ( t ) ω \omega ω ω \omega ω m m m ω p = p 2 + m 2 \omega_{\mathbf{p}}=\sqrt{\mathbf{p}^2+m^2} ω p = p 2 + m 2
The canonical momentum is
π ≡ ∂ L ∂ ϕ ˙ = ϕ ˙ . \pi \equiv \frac{\partial L}{\partial \dot\phi} = \dot\phi. π ≡ ∂ ϕ ˙ ∂ L = ϕ ˙ . The Hamiltonian is
H = π ϕ ˙ − L = 1 2 π 2 + 1 2 ω 2 ϕ 2 . H=\pi\dot\phi - L = \frac12\pi^2+\frac12\omega^2\phi^2. H = π ϕ ˙ − L = 2 1 π 2 + 2 1 ω 2 ϕ 2 . Canonical quantization promotes ϕ , π \phi,\pi ϕ , π
[ ϕ , π ] = i , [ ϕ , ϕ ] = [ π , π ] = 0. [\phi,\pi]=i,
\qquad
[\phi,\phi]=[\pi,\pi]=0. [ ϕ , π ] = i , [ ϕ , ϕ ] = [ π , π ] = 0.
This is the prototype of the QFT equal-time commutator[ ϕ ( t , x ) , π ( t , y ) ] = i δ ( 3 ) ( x − y ) [\phi(t,\mathbf{x}),\pi(t,\mathbf{y})]=i\delta^{(3)}(\mathbf{x}-\mathbf{y}) [ ϕ ( t , x ) , π ( t , y )] = i δ ( 3 ) ( x − y )
Define
a ≡ 1 2 ω ( π − i ω ϕ ) , a † ≡ 1 2 ω ( π + i ω ϕ ) . a \equiv \frac{1}{\sqrt{2\omega}}\left(\pi - i\omega\,\phi\right),
\qquad
a^\dagger \equiv \frac{1}{\sqrt{2\omega}}\left(\pi + i\omega\,\phi\right). a ≡ 2 ω 1 ( π − iω ϕ ) , a † ≡ 2 ω 1 ( π + iω ϕ ) . Using linearity and [ ϕ , π ] = i [\phi,\pi]=i [ ϕ , π ] = i
[ a , a † ] = 1 2 ω [ π − i ω ϕ , π + i ω ϕ ] = 1 2 ω ( i ω [ π , ϕ ] − i ω [ ϕ , π ] ) . [a,a^\dagger]
=
\frac{1}{2\omega}
\Bigl[\pi - i\omega\phi,\ \pi + i\omega\phi\Bigr]
=
\frac{1}{2\omega}\Bigl(i\omega[\pi,\phi] - i\omega[\phi,\pi]\Bigr). [ a , a † ] = 2 ω 1 [ π − iω ϕ , π + iω ϕ ] = 2 ω 1 ( iω [ π , ϕ ] − iω [ ϕ , π ] ) . Since [ π , ϕ ] = − [ ϕ , π ] = − i [\pi,\phi]=-[\phi,\pi]=-i [ π , ϕ ] = − [ ϕ , π ] = − i
[ a , a † ] = 1 2 ω ( i ω ( − i ) − i ω ( i ) ) = 1 2 ω ( ω + ω ) = 1. [a,a^\dagger]
=
\frac{1}{2\omega}\Bigl(i\omega(-i) - i\omega(i)\Bigr)
=
\frac{1}{2\omega}(\omega+\omega)=1. [ a , a † ] = 2 ω 1 ( iω ( − i ) − iω ( i ) ) = 2 ω 1 ( ω + ω ) = 1. Start from
H = 1 2 π 2 + 1 2 ω 2 ϕ 2 . H=\frac12\pi^2+\frac12\omega^2\phi^2. H = 2 1 π 2 + 2 1 ω 2 ϕ 2 . Invert the ladder-operator definitions:
ϕ = 1 2 ω ( a + a † ) , π = − i ω 2 ( a − a † ) . \phi = \frac{1}{\sqrt{2\omega}}(a+a^\dagger),
\qquad
\pi = -i\sqrt{\frac{\omega}{2}}(a-a^\dagger). ϕ = 2 ω 1 ( a + a † ) , π = − i 2 ω ( a − a † ) . Substitute into H H H
H = ω ( a † a + 1 2 ) . H=\omega\left(a^\dagger a+\frac12\right). H = ω ( a † a + 2 1 ) . The operator
N ≡ a † a N \equiv a^\dagger a N ≡ a † a is the number operator , and the 1 2 ω \frac12\omega 2 1 ω zero-point energy .
In field theory this becomes ∫ d 3 p 1 2 ω p \int d^3p\,\tfrac12\omega_{\mathbf{p}} ∫ d 3 p 2 1 ω p ( 2 π ) 3 (2\pi)^3 ( 2 π ) 3
From [ a , a † ] = 1 [a,a^\dagger]=1 [ a , a † ] = 1
[ N , a † ] = a † , [ N , a ] = − a . [N,a^\dagger]=a^\dagger,
\qquad
[N,a]=-a. [ N , a † ] = a † , [ N , a ] = − a . Proof (one line):
[ N , a † ] = [ a † a , a † ] = a † [ a , a † ] = a † . [N,a^\dagger]=[a^\dagger a,a^\dagger]=a^\dagger[a,a^\dagger]=a^\dagger. [ N , a † ] = [ a † a , a † ] = a † [ a , a † ] = a † . Similarly for [ N , a ] [N,a] [ N , a ]
Let ∣ n ⟩ |n\rangle ∣ n ⟩ N N N
N ∣ n ⟩ = n ∣ n ⟩ . N|n\rangle = n|n\rangle. N ∣ n ⟩ = n ∣ n ⟩ . Then
N ( a † ∣ n ⟩ ) = ( a † N + [ N , a † ] ) ∣ n ⟩ = a † n ∣ n ⟩ + a † ∣ n ⟩ = ( n + 1 ) ( a † ∣ n ⟩ ) , N(a^\dagger|n\rangle)
=
(a^\dagger N + [N,a^\dagger])|n\rangle
=
a^\dagger n|n\rangle + a^\dagger|n\rangle
=
(n+1)(a^\dagger|n\rangle), N ( a † ∣ n ⟩) = ( a † N + [ N , a † ]) ∣ n ⟩ = a † n ∣ n ⟩ + a † ∣ n ⟩ = ( n + 1 ) ( a † ∣ n ⟩) , so a † ∣ n ⟩ a^\dagger|n\rangle a † ∣ n ⟩ ∣ n + 1 ⟩ |n+1\rangle ∣ n + 1 ⟩
Similarly,
N ( a ∣ n ⟩ ) = ( n − 1 ) ( a ∣ n ⟩ ) , N(a|n\rangle)=(n-1)(a|n\rangle), N ( a ∣ n ⟩) = ( n − 1 ) ( a ∣ n ⟩) , so a ∣ n ⟩ ∝ ∣ n − 1 ⟩ a|n\rangle\propto |n-1\rangle a ∣ n ⟩ ∝ ∣ n − 1 ⟩
Write
a † ∣ n ⟩ = α n ∣ n + 1 ⟩ , a ∣ n + 1 ⟩ = α n ∗ ∣ n ⟩ . a^\dagger|n\rangle = \alpha_n |n+1\rangle,
\qquad
a|n+1\rangle = \alpha_n^* |n\rangle. a † ∣ n ⟩ = α n ∣ n + 1 ⟩ , a ∣ n + 1 ⟩ = α n ∗ ∣ n ⟩ . Then
⟨ n ∣ a a † ∣ n ⟩ = ∣ α n ∣ 2 . \langle n|a a^\dagger|n\rangle = |\alpha_n|^2. ⟨ n ∣ a a † ∣ n ⟩ = ∣ α n ∣ 2 . But a a † = a † a + 1 = N + 1 a a^\dagger = a^\dagger a + 1 = N+1 a a † = a † a + 1 = N + 1
∣ α n ∣ 2 = ⟨ n ∣ ( N + 1 ) ∣ n ⟩ = n + 1. |\alpha_n|^2 = \langle n|(N+1)|n\rangle = n+1. ∣ α n ∣ 2 = ⟨ n ∣ ( N + 1 ) ∣ n ⟩ = n + 1. Therefore one can choose phases so that α n = n + 1 \alpha_n=\sqrt{n+1} α n = n + 1
a † ∣ n ⟩ = n + 1 ∣ n + 1 ⟩ , a ∣ n ⟩ = n ∣ n − 1 ⟩ . a^\dagger|n\rangle=\sqrt{n+1}\,|n+1\rangle,
\qquad
a|n\rangle=\sqrt{n}\,|n-1\rangle. a † ∣ n ⟩ = n + 1 ∣ n + 1 ⟩ , a ∣ n ⟩ = n ∣ n − 1 ⟩ . Repeatedly lowering must eventually stop in a Hilbert space with positive norms . If there is a lowest state ∣ 0 ⟩ |0\rangle ∣0 ⟩
a ∣ 0 ⟩ = 0. a|0\rangle=0. a ∣0 ⟩ = 0. Then N ∣ 0 ⟩ = 0 N|0\rangle=0 N ∣0 ⟩ = 0 N N N nonnegative integers :
n = 0 , 1 , 2 , … n = 0,1,2,\dots n = 0 , 1 , 2 , … and the energy levels are
E n = ω ( n + 1 2 ) . E_n = \omega\left(n+\frac12\right). E n = ω ( n + 2 1 ) .
For finite matrices, t r ( A B ) = t r ( B A ) \mathrm{tr}(AB)=\mathrm{tr}(BA) tr ( A B ) = tr ( B A ) t r ( [ A , B ] ) = 0 \mathrm{tr}([A,B])=0 tr ([ A , B ]) = 0
If a , a † a,a^\dagger a , a †
t r ( [ a , a † ] ) = 0. \mathrm{tr}([a,a^\dagger]) = 0. tr ([ a , a † ]) = 0. But [ a , a † ] = 1 [a,a^\dagger]=\mathbf{1} [ a , a † ] = 1
t r ( [ a , a † ] ) = t r ( 1 ) = dim ( H ) ≠ 0 , \mathrm{tr}([a,a^\dagger])=\mathrm{tr}(\mathbf{1})=\dim(\mathcal{H})\neq 0, tr ([ a , a † ]) = tr ( 1 ) = dim ( H ) = 0 , a contradiction.
So the oscillator algebra forces an infinite-dimensional Hilbert space. This is one small hint of why QFT (with infinitely many modes) naturally lives in a very large Hilbert space.
Below is a simple ladder diagram that will later be reinterpreted as “particle number” levels in free QFT.
E
n = 0
n = 1
n = 2
n = 3
n = 4
a†
a
Eₙ = ω (n + 1/2)
A free relativistic scalar field (Klein–Gordon) decomposes into independent momentum modes. Each p \mathbf{p} p
ω p = p 2 + m 2 . \omega_{\mathbf{p}}=\sqrt{\mathbf{p}^2+m^2}. ω p = p 2 + m 2 . Schematically (suppressing ordering issues and constants),
H free ∼ ∫ d 3 p ( 2 π ) 3 ω p a p † a p + (zero-point) . H_{\text{free}}
\sim
\int \frac{d^3p}{(2\pi)^3}\,
\omega_{\mathbf{p}}\,
a^\dagger_{\mathbf{p}}a_{\mathbf{p}} \;+\; \text{(zero-point)}. H free ∼ ∫ ( 2 π ) 3 d 3 p ω p a p † a p + (zero-point) . This is the precise meaning of the “each p \mathbf{p} p QFT begins as an infinite collection of oscillators.
In the next chapters we will construct the field operator ϕ ( x ) \phi(x) ϕ ( x )
Starting from [ ϕ , π ] = i [\phi,\pi]=i [ ϕ , π ] = i [ a , a † ] = 1 [a,a^\dagger]=1 [ a , a † ] = 1
a = 1 2 ω ( π − i ω ϕ ) , a † = 1 2 ω ( π + i ω ϕ ) . a=\frac{1}{\sqrt{2\omega}}(\pi-i\omega\phi),
\quad
a^\dagger=\frac{1}{\sqrt{2\omega}}(\pi+i\omega\phi). a = 2 ω 1 ( π − iω ϕ ) , a † = 2 ω 1 ( π + iω ϕ ) .
Solution Compute directly:
[ a , a † ] = 1 2 ω [ π − i ω ϕ , π + i ω ϕ ] = 1 2 ω ( i ω [ π , ϕ ] − i ω [ ϕ , π ] ) . [a,a^\dagger]
=
\frac{1}{2\omega}[\pi-i\omega\phi,\ \pi+i\omega\phi]
=
\frac{1}{2\omega}(i\omega[\pi,\phi]-i\omega[\phi,\pi]). [ a , a † ] = 2 ω 1 [ π − iω ϕ , π + iω ϕ ] = 2 ω 1 ( iω [ π , ϕ ] − iω [ ϕ , π ]) . Use [ ϕ , π ] = i ⇒ [ π , ϕ ] = − i [\phi,\pi]=i\Rightarrow[\pi,\phi]=-i [ ϕ , π ] = i ⇒ [ π , ϕ ] = − i
[ a , a † ] = 1 2 ω ( i ω ( − i ) − i ω ( i ) ) = 1 2 ω ( ω + ω ) = 1. [a,a^\dagger]=\frac{1}{2\omega}\bigl(i\omega(-i)-i\omega(i)\bigr)=\frac{1}{2\omega}(\omega+\omega)=1. [ a , a † ] = 2 ω 1 ( iω ( − i ) − iω ( i ) ) = 2 ω 1 ( ω + ω ) = 1.
Using
ϕ = 1 2 ω ( a + a † ) , π = − i ω 2 ( a − a † ) , \phi = \frac{1}{\sqrt{2\omega}}(a+a^\dagger),
\qquad
\pi = -i\sqrt{\frac{\omega}{2}}(a-a^\dagger), ϕ = 2 ω 1 ( a + a † ) , π = − i 2 ω ( a − a † ) , show that
H = 1 2 π 2 + 1 2 ω 2 ϕ 2 = ω ( a † a + 1 2 ) . H=\frac12\pi^2+\frac12\omega^2\phi^2
=
\omega\left(a^\dagger a+\frac12\right). H = 2 1 π 2 + 2 1 ω 2 ϕ 2 = ω ( a † a + 2 1 ) .
Solution Insert the expressions:
Compute ω 2 ϕ 2 \omega^2\phi^2 ω 2 ϕ 2
ω 2 ϕ 2 = ω 2 ( a + a † ) 2 2 ω = ω 2 ( a + a † ) 2 . \omega^2\phi^2
=
\omega^2\frac{(a+a^\dagger)^2}{2\omega}
=
\frac{\omega}{2}(a+a^\dagger)^2. ω 2 ϕ 2 = ω 2 2 ω ( a + a † ) 2 = 2 ω ( a + a † ) 2 .
Compute π 2 \pi^2 π 2
π 2 = ( − i ω 2 ( a − a † ) ) 2 = − ω 2 ( a − a † ) 2 . \pi^2
=
\left(-i\sqrt{\frac{\omega}{2}}(a-a^\dagger)\right)^2
=
-\frac{\omega}{2}(a-a^\dagger)^2. π 2 = ( − i 2 ω ( a − a † ) ) 2 = − 2 ω ( a − a † ) 2 . Thus
H = 1 2 π 2 + 1 2 ω 2 ϕ 2 = − ω 4 ( a − a † ) 2 + ω 4 ( a + a † ) 2 . H=\frac12\pi^2+\frac12\omega^2\phi^2
=
-\frac{\omega}{4}(a-a^\dagger)^2+\frac{\omega}{4}(a+a^\dagger)^2. H = 2 1 π 2 + 2 1 ω 2 ϕ 2 = − 4 ω ( a − a † ) 2 + 4 ω ( a + a † ) 2 . Expand both squares and subtract. The mixed terms add:
( a + a † ) 2 − ( a − a † ) 2 = 4 a † a + 2 [ a , a † ] . (a+a^\dagger)^2-(a-a^\dagger)^2
=
4a^\dagger a + 2[a,a^\dagger]. ( a + a † ) 2 − ( a − a † ) 2 = 4 a † a + 2 [ a , a † ] . Using [ a , a † ] = 1 [a,a^\dagger]=1 [ a , a † ] = 1
H = ω 4 ( 4 a † a + 2 ) = ω ( a † a + 1 2 ) . H=\frac{\omega}{4}\Bigl(4a^\dagger a+2\Bigr)=\omega\left(a^\dagger a+\frac12\right). H = 4 ω ( 4 a † a + 2 ) = ω ( a † a + 2 1 ) .
Assume N ∣ n ⟩ = n ∣ n ⟩ N|n\rangle=n|n\rangle N ∣ n ⟩ = n ∣ n ⟩ ∣ n ⟩ |n\rangle ∣ n ⟩ ⟨ n ∣ n ⟩ = 1 \langle n|n\rangle=1 ⟨ n ∣ n ⟩ = 1
a ∣ n ⟩ = n ∣ n − 1 ⟩ , a † ∣ n ⟩ = n + 1 ∣ n + 1 ⟩ . a|n\rangle=\sqrt{n}\,|n-1\rangle,
\qquad
a^\dagger|n\rangle=\sqrt{n+1}\,|n+1\rangle. a ∣ n ⟩ = n ∣ n − 1 ⟩ , a † ∣ n ⟩ = n + 1 ∣ n + 1 ⟩ .
Solution From [ N , a † ] = a † [N,a^\dagger]=a^\dagger [ N , a † ] = a † a † ∣ n ⟩ ∝ ∣ n + 1 ⟩ a^\dagger|n\rangle\propto|n+1\rangle a † ∣ n ⟩ ∝ ∣ n + 1 ⟩ a † ∣ n ⟩ = α n ∣ n + 1 ⟩ a^\dagger|n\rangle=\alpha_n|n+1\rangle a † ∣ n ⟩ = α n ∣ n + 1 ⟩
∣ α n ∣ 2 = ⟨ n ∣ a a † ∣ n ⟩ = ⟨ n ∣ ( N + 1 ) ∣ n ⟩ = n + 1. |\alpha_n|^2
=
\langle n|a a^\dagger|n\rangle
=
\langle n|(N+1)|n\rangle
=
n+1. ∣ α n ∣ 2 = ⟨ n ∣ a a † ∣ n ⟩ = ⟨ n ∣ ( N + 1 ) ∣ n ⟩ = n + 1. Choose phases so α n = n + 1 \alpha_n=\sqrt{n+1} α n = n + 1
Show that no finite-dimensional matrices a , a † a,a^\dagger a , a † [ a , a † ] = 1 [a,a^\dagger]=\mathbf{1} [ a , a † ] = 1
Solution For finite-dimensional matrices, t r ( A B ) = t r ( B A ) \mathrm{tr}(AB)=\mathrm{tr}(BA) tr ( A B ) = tr ( B A ) t r ( [ A , B ] ) = 0 \mathrm{tr}([A,B])=0 tr ([ A , B ]) = 0 [ a , a † ] = 1 [a,a^\dagger]=\mathbf{1} [ a , a † ] = 1
0 = t r ( [ a , a † ] ) = t r ( 1 ) = dim ( H ) , 0=\mathrm{tr}([a,a^\dagger])=\mathrm{tr}(\mathbf{1})=\dim(\mathcal{H}), 0 = tr ([ a , a † ]) = tr ( 1 ) = dim ( H ) , a contradiction. Therefore the oscillator Hilbert space must be infinite-dimensional.
Show that in the number eigenstate ∣ n ⟩ |n\rangle ∣ n ⟩
⟨ n ∣ ϕ ∣ n ⟩ = 0 , ⟨ n ∣ ϕ 2 ∣ n ⟩ = n + 1 2 ω . \langle n|\phi|n\rangle=0,
\qquad
\langle n|\phi^2|n\rangle=\frac{n+\tfrac12}{\omega}. ⟨ n ∣ ϕ ∣ n ⟩ = 0 , ⟨ n ∣ ϕ 2 ∣ n ⟩ = ω n + 2 1 .
Solution Use ϕ = 1 2 ω ( a + a † ) \phi=\frac{1}{\sqrt{2\omega}}(a+a^\dagger) ϕ = 2 ω 1 ( a + a † ) a ∣ n ⟩ ∝ ∣ n − 1 ⟩ a|n\rangle\propto|n-1\rangle a ∣ n ⟩ ∝ ∣ n − 1 ⟩ a † ∣ n ⟩ ∝ ∣ n + 1 ⟩ a^\dagger|n\rangle\propto|n+1\rangle a † ∣ n ⟩ ∝ ∣ n + 1 ⟩ ⟨ n ∣ a ∣ n ⟩ = ⟨ n ∣ a † ∣ n ⟩ = 0 \langle n|a|n\rangle=\langle n|a^\dagger|n\rangle=0 ⟨ n ∣ a ∣ n ⟩ = ⟨ n ∣ a † ∣ n ⟩ = 0 ⟨ ϕ ⟩ = 0 \langle\phi\rangle=0 ⟨ ϕ ⟩ = 0
For ϕ 2 \phi^2 ϕ 2
ϕ 2 = 1 2 ω ( a 2 + a † 2 + a a † + a † a ) . \phi^2=\frac{1}{2\omega}(a^2+a^{\dagger 2}+aa^\dagger+a^\dagger a). ϕ 2 = 2 ω 1 ( a 2 + a † 2 + a a † + a † a ) . The terms a 2 a^2 a 2 a † 2 a^{\dagger 2} a † 2 n n n ± 2 \pm2 ± 2 ∣ n ⟩ |n\rangle ∣ n ⟩ a a † = N + 1 aa^\dagger=N+1 a a † = N + 1 a † a = N a^\dagger a=N a † a = N
⟨ n ∣ ϕ 2 ∣ n ⟩ = 1 2 ω ⟨ n ∣ ( N + 1 + N ) ∣ n ⟩ = 1 2 ω ( 2 n + 1 ) = n + 1 2 ω . \langle n|\phi^2|n\rangle=\frac{1}{2\omega}\langle n|(N+1+N)|n\rangle=\frac{1}{2\omega}(2n+1)=\frac{n+\tfrac12}{\omega}. ⟨ n ∣ ϕ 2 ∣ n ⟩ = 2 ω 1 ⟨ n ∣ ( N + 1 + N ) ∣ n ⟩ = 2 ω 1 ( 2 n + 1 ) = ω n + 2 1 . 