Phase in electromagnetic coupling.md

Phase in electomagnetic coupling

These notes build a clean toy-model Hamiltonian for two identical two-level systems (TLS) coupled to a single quantised electromagnetic (EM) mode, focusing on:

• How to write the interaction for electric and magnetic dipole transitions.
• Why and how phase factors like $e^{\pm i\phi}$ appear in cavity/waveguide Hamiltonians.
• The difference between:
  • spatial phase (from $e^{ikx}$ in a travelling wave), and
  • quadrature rotation (coupling to $X$ vs $P$ of the oscillator).
• A conceptual clarification of “phase” in EM waves (travelling vs standing waves).

Throughout, we use a single mode (one harmonic oscillator) and two TLS labelled donor $D$ and acceptor $A$.

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The basic ingredients

TLS Hamiltonian

For TLS $j\in{D,A}$ with transition frequency $\omega_j$,

$$ H_{\mathrm{TLS}}=\sum_{j=D,A}\frac{\hbar\omega_j}{2},\sigma_z^{(j)}. \label{eq:HTLS} $$

• $\sigma_z^{(j)}$ is the Pauli $z$ operator acting on TLS $j$.
• $\sigma_\pm^{(j)}$ are raising/lowering operators, with $\sigma_x^{(j)}=\sigma_+^{(j)}+\sigma_-^{(j)}$.

Single quantised mode Hamiltonian

One EM mode of frequency $\omega_c$ is described by bosonic operators $a,a^\dagger$:

$$ H_{\mathrm{mode}}=\hbar\omega_c,a^\dagger a. \label{eq:Hmode} $$

Total free Hamiltonian:

$$ H_0 = H_{\mathrm{mode}} + H_{\mathrm{TLS}}. \label{eq:H0} $$

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Dipole interactions: the starting point

A TLS transition can couple to:

• the electric field via an electric transition dipole $\hat{\mathbf d}$, and/or
• the magnetic field via a magnetic transition dipole $\hat{\boldsymbol\mu}$.

In the dipole approximation, at the TLS position $\mathbf r_j$:

$$ H_{\mathrm{int}}

-\sum_{j=D,A} \left( \hat{\mathbf d}^{(j)}\cdot\hat{\mathbf E}(\mathbf r_j) + \hat{\boldsymbol\mu}^{(j)}\cdot\hat{\mathbf B}(\mathbf r_j) \right). \label{eq:Hint_general} $$

Transition operators (minimal, aligned case)

To keep the algebra transparent, assume each transition dipole is aligned with the local field polarisation so dot products reduce to scalars:

• $\hat d^{(j)} = d,\sigma_x^{(j)}$
• $\hat\mu^{(j)} = \mu,\sigma_x^{(j)}$

Then Eq. $\ref{eq:Hint_general}$ becomes

$$ H_{\mathrm{int}}

-\sum_{j=D,A} \sigma_x^{(j)} \left( d,\hat E(\mathbf r_j) + \mu,\hat B(\mathbf r_j) \right). \label{eq:Hint_scalar} $$

(General vector/complex matrix-element versions give the same structure but with additional projection factors and complex conjugation; the key ideas below do not rely on the simplification.)

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Quantising the fields for a single mode

A single mode at position $\mathbf r$ can always be written as

$$ \hat{\mathbf E}(\mathbf r) = \mathbf E_{\mathrm{zpf}}(\mathbf r),a + \mathbf E_{\mathrm{zpf}}^_(\mathbf r),a^\dagger, \qquad \hat{\mathbf B}(\mathbf r) = \mathbf B_{\mathrm{zpf}}(\mathbf r),a + \mathbf B_{\mathrm{zpf}}^_(\mathbf r),a^\dagger. \label{eq:fields_singlemode} $$

• “zpf” means zero-point field amplitude (including polarisation and spatial mode function).
• The complex nature of $\mathbf E_{\mathrm{zpf}}(\mathbf r)$ or $\mathbf B_{\mathrm{zpf}}(\mathbf r)$ is exactly where phase factors come from.

A useful scalar shorthand at each TLS position is:

$$ \hat B(\mathbf r_j)= B_{\mathrm{zpf}}\big(u_j a + u_j^* a^\dagger\big), \qquad u_j \equiv \text{(complex mode factor at } \mathbf r_j\text{)}. \label{eq:Bmodefactor} $$

Write $u_j=|u_j|e^{i\theta_j}$:

$$ u_j a + u_j^* a^\dagger

|u_j|\left(e^{i\theta_j}a+e^{-i\theta_j}a^\dagger\right). \label{eq:phaseform} $$

This algebraic form is the origin of Hamiltonians containing $(e^{i\phi}a+e^{-i\phi}a^\dagger)$.

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Phase factors from spatial separation in a travelling wave (B-only)

Travelling-wave mode function

For a running (travelling) wave along $x$, a standard mode factor is

$$ u(x)=e^{ikx}. \label{eq:ux} $$

Then the magnetic field operator is

$$ \hat B(x)=B_{\mathrm{zpf}}\left(e^{ikx}a+e^{-ikx}a^\dagger\right). \label{eq:Btravel} $$

Two identical TLS at positions $x_D$ and $x_A$

Insert Eq. $\ref{eq:Btravel}$ into the magnetic-only part of Eq. $\ref{eq:Hint_scalar}$ (set $d=0$):

$$ H_{\mathrm{int}}^{(B)}

-\mu B_{\mathrm{zpf}} \sum_{j=D,A} \sigma_x^{(j)} \left(e^{ikx_j}a+e^{-ikx_j}a^\dagger\right). \label{eq:Hint_Btravel} $$

Define the couplings $V_j\equiv \mu B_{\mathrm{zpf}}$ (or include $|u(x_j)|$ if the mode has spatial envelope). Then

$$ H_{\mathrm{int}}^{(B)}

-\sum_{j=D,A} V_j,\sigma_x^{(j)} \left(e^{ikx_j}a+e^{-ikx_j}a^\dagger\right). \label{eq:Hint_Btravel2} $$

Only the relative phase matters

You can rephase the oscillator without changing $H_{\mathrm{mode}}$:

• Transform $a \rightarrow a,e^{-ikx_D}$ (and $a^\dagger \rightarrow a^\dagger e^{ikx_D}$).

This makes the donor term purely $(a+a^\dagger)$, and the acceptor term keeps the relative phase

$$ \Delta\phi = k(x_A-x_D). \label{eq:deltaphi} $$

So you obtain the familiar-looking structure

$$ H_{\mathrm{int}}^{(B)}

-(a+a^\dagger),V_D,\sigma_x^{(D)}

\left(e^{i\Delta\phi}a+e^{-i\Delta\phi}a^\dagger\right),V_A,\sigma_x^{(A)}. \label{eq:Hint_phasefactor} $$

Key point: in this B-only travelling-wave case, $\Delta\phi$ is simply the spatial phase sampled by the two TLS.

Special case: $\lambda/4$ separation

If $x_A-x_D=\lambda/4$, then $k(x_A-x_D)=2\pi(\lambda/4)/\lambda=\pi/2$, so $\Delta\phi=\pi/2$.

This is the cleanest route to a $\pi/2$ factor with B-only coupling: a travelling wave and a quarter-wavelength separation.

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Common confusion: is this “E vs B quadratures”?

The simple answer

• In the travelling-wave, B-only story above, the phase $\Delta\phi$ comes from $e^{ikx}$, i.e. different complex amplitudes at different positions.
• Writing $(e^{i\Delta\phi}a+e^{-i\Delta\phi}a^\dagger)$ as a “rotated quadrature” is just a re-expression of that same spatial phase.

There is no need to invoke electric coupling to explain the B-only relative phase.

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Quadratures: what the $e^{\pm i\phi}$ combination really means

Define oscillator quadratures

$$ X \equiv a+a^\dagger, \qquad P \equiv i(a^\dagger-a). \label{eq:XP} $$

Then the identity

$$ e^{i\phi}a+e^{-i\phi}a^\dagger

X\cos\phi + P\sin\phi \label{eq:rotated_quadrature} $$

shows that $(e^{i\phi}a+e^{-i\phi}a^\dagger)$ is a rotated quadrature.

What does $\phi=\pi/2$ mean?

Set $\phi=\pi/2$ in Eq. $\ref{eq:rotated_quadrature}$:

$$ e^{i\pi/2}a+e^{-i\pi/2}a^\dagger

P \times (\text{sign}). \label{eq:phi_pi_over_2} $$

So “$\pi/2$ in the Hamiltonian” is shorthand for “coupling to the orthogonal quadrature”.

In the travelling-wave case, that orthogonal-quadrature appearance is simply because sampling the mode at $x=\lambda/4$ rotates the combination of $a$ and $a^\dagger$.

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How coupling to both E and B rotates the quadrature at a single point

This is a different mechanism: it can produce an effective phase angle even without spatial separation.

Step 1: In one common convention, $E$ and $B$ correspond to different quadratures

A standard quantisation choice uses the vector potential $\hat{\mathbf A}\propto X$, then:

• $\hat{\mathbf B}=\nabla\times\hat{\mathbf A}\propto X$
• $\hat{\mathbf E}=-\partial_t\hat{\mathbf A}\propto P$

At a fixed position $\mathbf r_0$ (suppress vector details):

$$ \hat E(\mathbf r_0)=E_{\mathrm{zpf}},P, \qquad \hat B(\mathbf r_0)=B_{\mathrm{zpf}},X. \label{eq:EB_quadratures} $$

Step 2: Couple a single TLS to both

Use Eq. $\ref{eq:Hint_scalar}$ at $\mathbf r_0$:

$$ H_{\mathrm{int}}(\mathbf r_0)

-\sigma_x\left(d,\hat E(\mathbf r_0)+\mu,\hat B(\mathbf r_0)\right). \label{eq:Hint_singlepoint} $$

Insert Eq. $\ref{eq:EB_quadratures}$:

$$ H_{\mathrm{int}}(\mathbf r_0)

-\sigma_x\left(g_E,P + g_B,X\right), \label{eq:Hint_XP} $$

where

• $g_E \equiv d E_{\mathrm{zpf}}$
• $g_B \equiv \mu B_{\mathrm{zpf}}$

Step 3: Rewrite as a rotated quadrature

Define

$$ V=\sqrt{g_B^2+g_E^2}, \qquad \phi=\arctan\left(\frac{g_E}{g_B}\right). \label{eq:Vphi} $$

Then

$$ g_B X + g_E P = V\left(X\cos\phi + P\sin\phi\right). \label{eq:combine_XP} $$

Using Eq. $\ref{eq:rotated_quadrature}$:

$$ H_{\mathrm{int}}(\mathbf r_0)

• V,\sigma_x\left(e^{i\phi}a+e^{-i\phi}a^\dagger\right). \label{eq:Hint_rotated} $$

Interpretation: coupling to both $E$ and $B$ gives two independent linear couplings to the same oscillator, hence a rotated quadrature.

Important caveat (identical TLS at the same point)

If two TLS are identical and co-located (and oriented the same way), they have the same ratio $g_E/g_B$ and therefore the same $\phi$. In that case:

• there is no relative phase between them arising from this mechanism alone,
• you can rephase the mode to remove the common $\phi$ from both simultaneously.

To get a relative phase from this mechanism, something must make $g_E/g_B$ differ between the TLS (different position in a standing wave where $E$ and $B$ vary differently, different orientation, etc.).

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Conceptual interlude: are $E$ and $B$ “in phase” by Maxwell?

A major conceptual hurdle is distinguishing:

• travelling waves (propagating plane waves), and
• standing waves (superpositions of counter-propagating waves, e.g. cavity modes).

Travelling plane wave: $E$ and $B$ are in phase

A standard vacuum plane wave propagating in $+x$:

$$ E_y(x,t)=E_0\cos(kx-\omega t), \qquad B_z(x,t)=\frac{E_0}{c}\cos(kx-\omega t). \label{eq:plane_inphase} $$

At fixed $x$, the maxima/minima occur at the same times: no relative $\pi/2$ time shift.

A compact statement in complex-amplitude form is:

$$ \mathbf B_0=\frac{1}{\omega},\mathbf k\times \mathbf E_0, \label{eq:B0kE0} $$

so both share the same space–time factor $e^{i(\mathbf k\cdot\mathbf r-\omega t)}$.

Why the magnetic field “subtracts” in a standing-wave construction

A standing wave is built from two travelling waves of opposite propagation direction. For the same $E$ polarisation, reversing propagation flips the sign of $B$ because

• $\mathbf B \propto \hat{\mathbf k}\times \mathbf E$,
• $\hat{\mathbf k}\rightarrow -\hat{\mathbf k}$ implies $\mathbf B\rightarrow -\mathbf B$.

This is why, when you add counter-propagating waves, the $E$ fields add but the $B$ fields subtract.

Standing wave: $E$ and $B$ can be $\pi/2$ out of phase in time at a point

Adding the two waves yields (one common form):

$$ E_y(x,t)=2E_0\cos(kx)\cos(\omega t), \qquad B_z(x,t)=2\frac{E_0}{c}\sin(kx)\sin(\omega t). \label{eq:standing_quadrature} $$

At a fixed $x$ where both $\cos(kx)$ and $\sin(kx)$ are nonzero:

• $E\propto \cos(\omega t)$
• $B\propto \sin(\omega t)$

This is a genuine $\pi/2$ time shift.

Common confusion: “cos vs sin means $\pi/2$”

• Yes: $\sin(\omega t)=\cos(\omega t-\pi/2)$.
• But you must compare two signals with the same argument at the same location/time origin.

In a travelling wave, you can write both $E$ and $B$ using $\cos(kx-\omega t)$ (or both using $\sin$) consistently; switching $\cos \leftrightarrow \sin$ for one but not the other would correspond to shifting reference phase for one field only, which is not the physical plane-wave relation.

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Putting it all together: what “phase” means in the Hamiltonian

There are two distinct (but algebraically similar) appearances of phases:

1) Spatial phase (travelling wave, even with B-only)

• The mode function has $u(x)=e^{ikx}$.
• Different TLS positions sample different complex coefficients.
• This yields a relative phase $\Delta\phi=k(x_A-x_D)$ in the coupling, Eq. $\ref{eq:deltaphi}$.
• $\lambda/4$ separation gives $\Delta\phi=\pi/2$.

2) Quadrature rotation (coupling to both E and B at one point)

• In a common quantisation convention, $B\sim X$ and $E\sim P$.
• Coupling to both gives $g_BX+g_EP=V(X\cos\phi+P\sin\phi)$ with $\phi=\arctan(g_E/g_B)$.
• This produces an $e^{\pm i\phi}$ structure even without spatial separation, Eq. $\ref{eq:Hint_rotated}$.
• For identical TLS at the same point, this does not create a relative phase by itself.

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Takeaways

• A travelling-wave spatial mode $u(x)=e^{ikx}$ gives position-dependent complex coupling; the relative phase is $\Delta\phi=k\Delta x$ (Eq. $\ref{eq:deltaphi}$).
• A $\pi/2$ phase factor in $(e^{i\phi}a+e^{-i\phi}a^\dagger)$ corresponds to coupling to the orthogonal oscillator quadrature (Eq. $\ref{eq:rotated_quadrature}$).
• With B-only coupling, a $\pi/2$ relative phase arises cleanly from $\lambda/4$ spatial separation in a travelling wave (Eq. $\ref{eq:Hint_phasefactor}$).
• Coupling to both E and B at a single point yields a quadrature rotation with $\phi=\arctan(g_E/g_B)$ (Eq. $\ref{eq:Vphi}$), even without spatial separation.
• Maxwell’s equations allow $E$ and $B$ to be in phase in travelling waves (Eq. $\ref{eq:plane_inphase}$) but often $\pi/2$ out of phase in standing waves (Eq. $\ref{eq:standing_quadrature}$).