From Harmonic Oscillators to Fields

A free real scalar field contains independent real normal modes, each quantized as a harmonic oscillator. Its traveling-wave expansion uses complex exponentials, which can obscure how many oscillators are present. For a nonzero momentum pair $\{\mathbf k,-\mathbf k\}$, there are two independent real standing-wave oscillators, equivalently two independent traveling-wave annihilators. Field reality relates Fourier coordinates; it does not identify an annihilator at $-\mathbf k$ with a creator at $+\mathbf k$.

The oscillator algebra belongs to Quantum Harmonic Oscillator. The Reference oscillator-to-fields bridge derives the full normal-mode quantization. This page uses that result to resolve the mode-counting and regulator questions that arise in relativistic notation.

Required background. Quantum Harmonic Oscillator supplies the ladder algebra; Harmonic Oscillator to Fields provides the scalar normal-mode construction.

Helpful background. Why Fields Replace Wavefunctions separates states and fields; Plane-Wave Solutions fixes the relativistic frequency convention.

Two real modes for a nonzero momentum pair

Use $\hbar=c=1$ and a real scalar of mass $m>0$ in a periodic spatial box of volume $\mathcal V$. Choose a nonzero allowed momentum $\mathbf k$ and retain its pair with $-\mathbf k$. The orthonormal real spatial modes are

$$f_c(\mathbf x)=\sqrt{\frac2{\mathcal V}} \cos(\mathbf k\cdot\mathbf x),\qquad f_s(\mathbf x)=\sqrt{\frac2{\mathcal V}} \sin(\mathbf k\cdot\mathbf x).$$

They have the same frequency $\omega=\sqrt{\mathbf k^2+m^2}$. The contribution to the field is $\widehat\phi_{\rm pair}=f_cq_c+f_sq_s$, where the real-mode quantization gives

$$q_j(t)=\frac{a_je^{-i\omega t} +a_j^\dagger e^{i\omega t}}{\sqrt{2\omega}}, \qquad j=c,s.$$

The independent algebras are $[a_c,a_c^\dagger]=[a_s,a_s^\dagger]=1$, with all cross commutators zero. Their Hamiltonian is

$$H_{\rm pair} =\omega(a_c^\dagger a_c+a_s^\dagger a_s+1).$$

The final one is the sum of the two zero-point halves. There is no single real oscillator for the entire nonzero pair; that would discard either its cosine or sine coordinate.

Traveling annihilators are independent

Define a unitary change of oscillator basis,

$$a_+=\frac{a_c-ia_s}{\sqrt2},\qquad a_-=\frac{a_c+ia_s}{\sqrt2}.$$

Then

$$[a_+,a_+^\dagger]=[a_-,a_-^\dagger]=1, \qquad [a_+,a_-^\dagger]=0.$$

In particular the cross bracket is $\tfrac12(1+(-i)(-i))=0$. The number sum and Hamiltonian are unchanged:

$$\begin{aligned} N_c+N_s&=N_++N_-,\\ H_{\rm pair}&=\omega(N_++N_-+1). \end{aligned}$$

The traveling occupations carry momentum

$$\mathbf P_{\rm pair}=\mathbf k(N_+-N_-).$$

Thus $a_+^\dagger|0\rangle$ is a one-quantum state with momentum $+\mathbf k$, and $a_-^\dagger|0\rangle$ has momentum $-\mathbf k$. The reality of the scalar field does not require $N_+=N_-$ in every state. A real scalar is a neutral, self-conjugate species, but can carry nonzero momentum.

Imposing $a_-=a_+^\dagger$ would be inconsistent: it would give $[a_-,a_-^\dagger]=[a_+^\dagger,a_+]=-1$ instead of $+1$. It would also collapse two independent oscillators into one.

The Fourier reality condition acts on coordinates

Write the pair’s spatial Fourier expansion as

$$\widehat\phi_{\rm pair}(t,\mathbf x) =\frac1{\sqrt{\mathcal V}} \left[Q_{\mathbf k}(t)e^{i\mathbf k\cdot\mathbf x} +Q_{-\mathbf k}(t)e^{-i\mathbf k\cdot\mathbf x}\right].$$

The standing-wave relations give

$$Q_{\mathbf k}(t) =\frac{a_+e^{-i\omega t} +a_-^\dagger e^{i\omega t}}{\sqrt{2\omega}}, \qquad Q_{-\mathbf k}(t)=Q_{\mathbf k}(t)^\dagger.$$

This last relation is precisely Hermiticity of the real scalar field. Each Fourier coordinate contains an annihilator and the opposite traveling mode’s creator. Consequently, conjugating the coordinate does not conjugate only an annihilator.

Equivalently, the full field can be organized as a sum over all allowed momenta, with one annihilator for each momentum and its Hermitian-conjugate term. Or one can sum over one representative of each nonzero pair and retain both real standing modes. Both conventions have the same degrees of freedom. Summing over both signs and also inserting an independent cosine/sine pair for each sign would double count.

The zero momentum mode is different: it is its own opposite and has one real coordinate, not two. For $m>0$ its frequency is $m$ and it is an ordinary oscillator. In a massless periodic box, its frequency is zero; its free Hamiltonian is a kinetic term without a restoring potential. The usual oscillator vacuum and $1/\sqrt{2\omega}$ formulas do not apply to that mode without a separate prescription.

A standing-wave quantum is a momentum superposition

The unitary change of basis gives

$$a_c^\dagger|0\rangle =\frac{a_+^\dagger|0\rangle +a_-^\dagger|0\rangle}{\sqrt2}.$$

This is one quantum in a coherent superposition of opposite momenta, not two quanta moving in opposite directions. Its momentum expectation is zero. For the component parallel to $\mathbf k$, its variance is $|\mathbf k|^2$.

The state with one quantum in each traveling mode is instead

$$a_+^\dagger a_-^\dagger|0\rangle.$$

It has two quanta and exactly zero total momentum within this pair. The two states are distinguished by both energy and momentum statistics even though their mean momenta agree.

These examples also clarify the oscillator language: raising the motional level of a single trapped material particle does not create another such particle. The field-particle interpretation belongs to the normal modes of the quantized field, whose Hamiltonian and momentum define the excitations.

Cutoffs and the continuum boundary

A finite box makes momenta discrete. A momentum cutoff makes the number of retained oscillators finite. These regulate different limits: volume controls infrared spacing, while the cutoff controls short wavelengths. Increasing the number of modes without stating which limit changes does not by itself establish convergence.

With a finite cutoff, the equal-time field–conjugate-momentum commutator contains the corresponding truncated Fourier kernel instead of an exact continuum spatial delta function. One must control the cutoff removal before claiming the full continuum locality relation. The causality discussion states that continuum relation and its distinction from spacelike correlations.

Likewise, the regulated zero-point sum $\tfrac12\sum_{\mathbf k}\omega_{\mathbf k}$ can be subtracted relative to a free reference vacuum when defining normal-ordered excitation energies. That operation does not solve all interacting renormalization or gravitational vacuum-energy questions. Interactions also couple modes, so the free independent-oscillator solution is an input to their treatment, not their complete solution.

Exercises

One sine quantum. Express $a_s^\dagger|0\rangle$ in traveling states and compare its occupation probabilities with the cosine state.

Solution

$$a_s^\dagger|0\rangle =-\frac{i}{\sqrt2} \left(a_+^\dagger-a_-^\dagger\right)|0\rangle.$$

Both traveling momenta have probability $1/2$, as for the cosine state. Their relative phase differs, producing an orthogonal standing mode. Probabilities in one basis alone need not encode all the state’s coherence.

Count the oscillators. A momentum regulator retains zero momentum and three distinct nonzero opposite pairs. How many real oscillators and independent traveling annihilators are present for a massive real scalar?

Solution

There are $1+2(3)=7$ real oscillators and seven traveling annihilators. Reality relates conjugate Fourier coordinates but does not halve the latter count a second time.

Energy versus mean momentum. Compare the excitation energies, after subtracting the common pair vacuum energy, of the cosine one-quantum state and $a_+^\dagger a_-^\dagger|0\rangle$.

Solution

They have excitation energies $\omega$ and $2\omega$. Both have zero mean total momentum, but only the second has sharply zero total momentum in this two-mode sector.

References

• Schwartz, Matthew D. Quantum Field Theory and the Standard Model. Cambridge University Press (2014). doi:10.1017/9781139540940. Free scalar modes, state normalization, and field quantization.
• Tong, David. Quantum Field Theory. University of Cambridge lecture notes (2006), sections 2.1–2.4. Free fields. Oscillator quantization, the scalar vacuum, and particle modes.