Quantize one optical mode with explicit normalization
Connect a normalized transverse field mode to oscillator operators, photon energy, and vacuum quadrature noise.
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The question for today
What you’ll understand
- Normalize the spatial and polarization structure of a selected vacuum mode.
- Relate ladder operators to photon number and cavity energy.
- Declare a quadrature convention and compute its vacuum variance.
A cavity mode is a whole spatial field pattern with one oscillator amplitude, not a photon sitting at one chosen position. Quantization replaces that amplitude by an operator while retaining the mode’s classical boundary conditions and polarization. Its normalization fixes the electric field associated with one excitation. We will use a dimensionless annihilation operator, keep Planck’s constant in SI formulas, and declare quadratures whose vacuum variance is one half so that later noise comparisons have an unambiguous reference.
Normalize the spatial pattern before its amplitude
Choose one transverse vacuum normal mode with frequency omega and a vector profile u. Normalize the volume integral of its squared magnitude to one. The profile therefore has units of inverse length to the three-halves power and includes polarization. A uniform traveling-wave mode in a periodic box has unit polarization vector divided by square root of volume, with a spatial phase. A standing-wave cavity uses a different normalized profile satisfying its conducting boundaries.
With this convention the positive-frequency electric operator is proportional to u times the annihilation operator; its Hermitian conjugate supplies the negative-frequency part. An overall mode phase is conventional and can move the displayed factor of i into the operator. A separate factor of two for polarization must not be inserted into one selected mode. Other independent polarizations are additional modes, each with its own oscillator and normalization. Material dispersion or absorbing dielectrics require a more elaborate energy normalization.
Photons label oscillator excitations
The mode Hamiltonian is the harmonic-oscillator Hamiltonian with energy spacing hbar omega. The number operator has nonnegative integer eigenvalues, and the lowering operator removes one excitation with a square-root matrix element. The ground state has zero photons but nonzero zero-point energy. Adding one photon always increases energy by the same spacing in this ideal linear mode. A nonlinear cavity can have different spacings, so the simple oscillator description is a model assumption.
A number eigenstate has exact excitation number and zero mean field amplitude. It can nevertheless have substantial electric-field fluctuations and energy. Mean field, mean photon number, and a particular photon-count outcome are different quantities. A state can also be a superposition of number states, in which case repeated ideal number measurements produce a distribution. The amplitude coefficients and the chosen measurement determine that distribution; one cannot read a definite photon count from an oscillating expectation value alone.
Vacuum is the reference for quadrature noise
Define two Hermitian quadratures by real and imaginary combinations of the ladder operators divided by square root of two. Their commutator is i, and the product of their standard deviations is at least one half. The vacuum has equal variances of one half and saturates this bound. Another common convention divides by two and gives variance one quarter instead; both are valid, but mixing the formulas produces incorrect squeezing and detector calibrations.
For a number state, each quadrature variance is occupation plus one half, while both quadrature means vanish. The two-state lab can illustrate only the probabilities in a deliberately restricted zero-or-one-photon superposition. It does not implement the infinite oscillator ladder, field normalization, or a quadrature distribution. Its relative-phase playback preserves basis probabilities. Treat that as geometry within a two-dimensional slice, not as a cavity calculation or a photon-counting detector.
Work it through
Separate zero-point and photon energy
A selected optical mode has photon energy ℏω=2.0×10⁻¹⁹ J and is in the number state n=3. Find total mode energy, excitation energy above vacuum, energy added by one further photon, and both quadrature variances.
The total oscillator energy is (3+1/2)ℏω=7.0×10⁻¹⁹ J. The vacuum contribution is 1.0×10⁻¹⁹ J.
Subtract vacuum energy to obtain 3ℏω=6.0×10⁻¹⁹ J of excitation energy. Increasing n from three to four adds 2.0×10⁻¹⁹ J, independent of the original occupation.
Each dimensionless quadrature variance is n+1/2=3.5, and its mean is zero. Exact photon number therefore does not mean a noiseless electric-field quadrature.
The mode has total energy 7.0×10⁻¹⁹ J, excitation energy 6.0×10⁻¹⁹ J, and quadrature variances 3.5. One additional photon contributes 2.0×10⁻¹⁹ J.
The living notebook
Explore the model
Interactive lab · quantum
Inspect a zero-or-one-photon slice
For fresh runs Reset and set BOTH mixing angle 60°, relative phase 0°; then Reset and set mixing angle 120°, phase 0°. If basis states represent |0⟩ and |1⟩ only, predict one-photon probabilities 0.25 and 0.75. The pure two-state Bloch model has no higher Fock states or quadrature measurement.
- Probability |0⟩
- 71.1%
- Probability |1⟩
- 28.9%
- Bloch (x, y, z)
- 0.78, 0.45, 0.42
A normalized two-level pure state, not an interacting field simulation. Playback adds one radian per second to the relative phase; basis probabilities stay fixed. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.
Look forBasis probabilities can be studied in a truncated state space without mistaking it for a full bosonic field mode.
Pause & reason
Check your understanding
Your turn
Try it yourself
For the normalized state (|0⟩+|2⟩)/√2, find mean photon number, photon-number variance, and mean energy in units of ℏω.
Take a moment to make a prediction or work it out on paper. Open a hint when you need one.
Give me a hint
The two possible number outcomes have equal probabilities; compute the second moment separately.
Compare with the solution
Mean number is (0+2)/2=1 and its second moment is (0+4)/2=2, giving variance 2−1²=1. Mean energy is (1+1/2)ℏω=1.5ℏω. This does not imply a definite one-photon state: an ideal number measurement gives zero or two, each with probability one half.
Keep the idea
Take it with you
- Mode normalization fixes the field per oscillator excitation.
- Zero photons still leaves zero-point energy and quadrature fluctuations.
- Vacuum-noise numbers depend on the declared quadrature convention.
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