04 Jun 2026
Hamiltonian Formulation of Two Coupled Quantum Oscillators
This article explains the Hamiltonian for two coupled quantum harmonic oscillators in an external magnetic field. The construction has three parts: the energy of each oscillator, their mutual coupling, and the extra mixing produced by the magnetic field.
The starting Hamiltonian is
\[H=\omega_1\hat a^\dagger\hat a+\omega_2\hat b^\dagger\hat b +g^{\prime}(\hat a+\hat a^\dagger)(\hat b+\hat b^\dagger).\]| Symbol | Meaning | Role |
|---|---|---|
| $\omega_1,\omega_2$ | natural frequencies | set the two energy scales |
| $\hat a,\hat b$ | annihilation operators | remove one excitation quantum |
| $\hat a^\dagger,\hat b^\dagger$ | creation operators | add one excitation quantum |
| $g^{\prime}$ | coupling strength | controls interaction strength |
1. Oscillator Hamiltonian
A Hamiltonian is the energy operator of a quantum system and determines its time evolution:
\[i\hbar\frac{\partial}{\partial t}|\psi(t)\rangle=H|\psi(t)\rangle.\]For one oscillator, the classical energy is
\[H=\frac{p^2}{2m}+\frac{1}{2}m\omega^2x^2,\]where $x$ is position, $p$ is momentum, $m$ is mass, and $\omega$ is the natural frequency. In quantum mechanics, $x$ and $p$ become operators. The same oscillator can be described more compactly through ladder operators. For the two oscillators, the number operators are
\[\hat a^\dagger\hat a,\qquad \hat b^\dagger\hat b.\]Therefore the uncoupled part of the Hamiltonian is
\[\omega_1\hat a^\dagger\hat a+\omega_2\hat b^\dagger\hat b.\]2. Coupling Term
The interaction is
\[g^{\prime}(\hat a+\hat a^\dagger)(\hat b+\hat b^\dagger).\]Since $\hat a+\hat a^\dagger$ is proportional to the position of oscillator 1 and $\hat b+\hat b^\dagger$ is proportional to the position of oscillator 2, the coupling is essentially
\[\hat x_1\hat x_2.\]| Case | Meaning |
|---|---|
| $g^{\prime}=0$ | oscillators evolve independently |
| $g^{\prime}\neq0$ | the oscillators influence each other |
| larger $g^{\prime}$ | stronger correlation and energy exchange |
In position and momentum variables this interaction is written as
\[-gx_1x_2,\]where the rescaled coupling is
\[g=\frac{1}{2}g^{\prime}\sqrt{\omega_1\omega_2}.\]3. Magnetic Field
The magnetic field is introduced through a vector potential. The paper uses the symmetric gauge
\[\vec A=\frac{B}{2}(x_2-x_1),\]where $B$ is the magnetic-field strength. The field is related to the vector potential by
\[\vec B=\nabla\times\vec A.\]In quantum mechanics, the magnetic field modifies momentum through minimal coupling:
\[p\rightarrow p-\frac{e}{c}A,\]where $e$ is electric charge and $c$ is the speed of light. After this substitution and setting $m=1$, the Hamiltonian becomes
\[H= \frac{1}{2}\sum_{j=1}^{2} \left[p_j^2+(\omega_j^2+\omega_c^2)x_j^2\right] -gx_1x_2 +\omega_c(p_1x_2-p_2x_1),\]where
\[\omega_c=\frac{eB}{2c}\]is the cyclotron frequency. The terms have the following roles:
| Term | Meaning |
|---|---|
| $\frac{1}{2}\sum_j p_j^2$ | kinetic energy of both oscillators |
| $\frac{1}{2}\sum_j\omega_j^2x_j^2$ | ordinary harmonic potential |
| $\frac{1}{2}\sum_j\omega_c^2x_j^2$ | magnetic frequency shift |
| $-gx_1x_2$ | position-position coupling |
| $\omega_c(p_1x_2-p_2x_1)$ | magnetic position-momentum mixing |
The magnetic field therefore does two things: it shifts the oscillator frequencies and creates a cross-term that mixes position and momentum.
4. Rotating-Frame Transformation
The difficult term is
\[\omega_c(p_1x_2-p_2x_1)\]because it couples the momentum of one oscillator to the position of the other. Such a term is not like the usual kinetic energy $p^2$ or potential energy $x^2$. It mixes the two phase-space directions and makes the Hamiltonian look as if the system is rotating.
The idea is therefore to describe the same motion from a rotating frame. This is similar to ordinary mechanics: a particle moving in a magnetic field often has a rotational or circular character, so a rotating coordinate system can simplify the equations.
Start with the two position coordinates written as a vector:
\[\begin{pmatrix} x_1\\ x_2 \end{pmatrix}.\]Now rotate this vector by an angle $\phi(t)$ and introduce new coordinates $X_1,X_2$:
\[\begin{pmatrix} x_1\\ x_2 \end{pmatrix} = \begin{pmatrix} \cos\phi & \sin\phi\\ -\sin\phi & \cos\phi \end{pmatrix} \begin{pmatrix} X_1\\ X_2 \end{pmatrix}.\]The momenta must be rotated in the same way:
\[\begin{pmatrix} p_1\\ p_2 \end{pmatrix} = \begin{pmatrix} \cos\phi & \sin\phi\\ -\sin\phi & \cos\phi \end{pmatrix} \begin{pmatrix} P_1\\ P_2 \end{pmatrix}.\]This is a canonical transformation because it changes variables without changing the Hamiltonian structure of the problem. In other words, the new variables still behave like proper position and momentum pairs:
\[(x_j,p_j)\rightarrow(X_j,P_j).\]If the rotation angle were constant, we could simply substitute these relations into the Hamiltonian. But here $\phi$ depends on time, so the coordinate frame itself is moving. We must therefore see how the Hamiltonian action changes.
For one pair of variables, Hamiltonian mechanics is built from the combination
\[p\,dx-H\,dt.\]For two variables this becomes
\[p_1dx_1+p_2dx_2-H\,dt.\]It is cleaner to do the calculation in matrix notation. Define
\[q= \begin{pmatrix} x_1\\ x_2 \end{pmatrix}, \qquad Q= \begin{pmatrix} X_1\\ X_2 \end{pmatrix}, \qquad p= \begin{pmatrix} p_1\\ p_2 \end{pmatrix}, \qquad P= \begin{pmatrix} P_1\\ P_2 \end{pmatrix}.\]The rotation can be written as
\[q=R(\phi)Q,\qquad p=R(\phi)P,\]where
\[R(\phi)= \begin{pmatrix} \cos\phi & \sin\phi\\ -\sin\phi & \cos\phi \end{pmatrix}.\]Now differentiate $q=RQ$:
\[dq=d(RQ)=R\,dQ+(dR)Q.\]The phase-space one-form is
\[p^Tdq.\]Using $p=RP$, we have $p^T=P^TR^T$. Therefore
\[p^Tdq =P^TR^T\left(R\,dQ+(dR)Q\right).\]Since $R^TR=I$, the first part gives
\[P^TdQ.\]The second part is
\[P^TR^T(dR)Q.\]For this rotation matrix,
\[R^T(dR)= \begin{pmatrix} 0 & 1\\ -1 & 0 \end{pmatrix} d\phi.\]Hence
\[P^TR^T(dR)Q = \begin{pmatrix} P_1 & P_2 \end{pmatrix} \begin{pmatrix} 0 & 1\\ -1 & 0 \end{pmatrix} \begin{pmatrix} X_1\\ X_2 \end{pmatrix} d\phi.\]Multiplying the matrices gives
\[P^TR^T(dR)Q=(P_1X_2-P_2X_1)d\phi.\]Therefore, in ordinary notation,
\[p_1dx_1+p_2dx_2 =P_1dX_1+P_2dX_2 +(P_1X_2-P_2X_1)d\phi.\]Since $d\phi=\dot\phi(t)dt$, this becomes
\[p_1dx_1+p_2dx_2 =P_1dX_1+P_2dX_2 +\dot\phi(t)(P_1X_2-P_2X_1)dt.\]Substitute this into the action:
\[p_1dx_1+p_2dx_2-H\,dt =P_1dX_1+P_2dX_2 -\left[H-\dot\phi(t)(P_1X_2-P_2X_1)\right]dt.\]The term inside the square bracket is therefore the new Hamiltonian:
\[H'=H-\dot\phi(t)(P_1X_2-P_2X_1).\]The expression
\[P_1X_2-P_2X_1\]is the generator of rotations in this two-dimensional phase space. The extra term is not an assumption; it appears because the rotating coordinates themselves depend on time.
Now compare this correction with the magnetic cross-term. The original Hamiltonian contains a term proportional to
\[\omega_c(p_1x_2-p_2x_1).\]The rotating-frame correction contains the same kind of structure, but with coefficient $-\dot\phi(t)$. Therefore choosing the rotation rate
\[\dot\phi(t)=\omega_c\]removes the explicit magnetic position-momentum mixing. Integrating this equation gives
\[\phi(t)=\omega_ct+\theta,\]where $\theta$ is a constant initial angle. Thus the new frame rotates with angular speed $\omega_c$, exactly matching the magnetic rotation produced by the field.
The important point is this: the magnetic field has not disappeared. Its awkward position-momentum cross-term has been absorbed into the rotating frame. The field will reappear in the transformed Hamiltonian through time-dependent frequencies and coupling.
5. Rotating-Frame Hamiltonian
Now we continue directly from the previous result. The rotating frame was chosen so that
\[\dot\phi(t)=\omega_c.\]This choice cancels the magnetic position-momentum term. We now ask: what is left in the Hamiltonian after this cancellation?
Step 1: separate the remaining Hamiltonian
After the cross-term has been cancelled, the Hamiltonian still contains:
- the kinetic energy,
- the quadratic position terms,
- the position-position coupling.
The kinetic part is
\[\frac{1}{2}(p_1^2+p_2^2).\]The remaining position-dependent part is
\[\frac{1}{2}(\omega_1^2+\omega_c^2)x_1^2 +\frac{1}{2}(\omega_2^2+\omega_c^2)x_2^2 -gx_1x_2.\]This is the part that must be rewritten in terms of the rotated coordinates $X_1,X_2$.
Step 2: write the position part as a matrix
Define
\[q= \begin{pmatrix} x_1\\ x_2 \end{pmatrix}, \qquad Q= \begin{pmatrix} X_1\\ X_2 \end{pmatrix}, \qquad q=R(\phi)Q.\]The position-dependent part can be written compactly as
\[V(q)=\frac{1}{2}q^TKq,\]where
\[K= \begin{pmatrix} \omega_1^2+\omega_c^2 & -g\\ -g & \omega_2^2+\omega_c^2 \end{pmatrix}.\]To see that this matrix is correct, expand:
\[\frac{1}{2}q^TKq = \frac{1}{2}(\omega_1^2+\omega_c^2)x_1^2 +\frac{1}{2}(\omega_2^2+\omega_c^2)x_2^2 -gx_1x_2.\]The two off-diagonal entries are both $-g$. Since the matrix expression has a factor $\frac{1}{2}$, the two equal off-diagonal contributions combine to give exactly $-gx_1x_2$.
Step 3: rotate the quadratic form
Now use
\[q=RQ.\]Then
\[V(q)=\frac{1}{2}(RQ)^TK(RQ).\]Using $(RQ)^T=Q^TR^T$, this becomes
\[V(q)=\frac{1}{2}Q^T(R^TKR)Q.\]So the new stiffness matrix is
\[K'(t)=R^T(\phi)KR(\phi).\]The time dependence appears because $\phi=\phi(t)$.
Step 4: multiply the matrix entries
For compactness, write
\[c=\cos\phi(t),\qquad s=\sin\phi(t).\]Then
\[R= \begin{pmatrix} c & s\\ -s & c \end{pmatrix}, \qquad R^T= \begin{pmatrix} c & -s\\ s & c \end{pmatrix}.\]Also define
\[a=\omega_1^2+\omega_c^2, \qquad b=\omega_2^2+\omega_c^2.\]Then
\[K= \begin{pmatrix} a & -g\\ -g & b \end{pmatrix}.\]First calculate $KR$:
\[KR= \begin{pmatrix} a & -g\\ -g & b \end{pmatrix} \begin{pmatrix} c & s\\ -s & c \end{pmatrix} = \begin{pmatrix} ac+gs & as-gc\\ -gc-bs & -gs+bc \end{pmatrix}.\]Now multiply by $R^T$:
\[K'=R^TKR = \begin{pmatrix} c & -s\\ s & c \end{pmatrix} \begin{pmatrix} ac+gs & as-gc\\ -gc-bs & -gs+bc \end{pmatrix}.\]The $(1,1)$ entry is
\[K'_{11}=c(ac+gs)-s(-gc-bs).\]Therefore
\[K'_{11}=ac^2+2gcs+bs^2.\]Using $2cs=\sin2\phi(t)$, this becomes
\[K'_{11} =(\omega_1^2+\omega_c^2)c^2 +(\omega_2^2+\omega_c^2)s^2 +g\sin2\phi(t).\]Since $c^2+s^2=1$,
\[K'_{11} =\omega_1^2\cos^2\phi(t) +\omega_2^2\sin^2\phi(t) +\omega_c^2 +g\sin2\phi(t).\]This entry is called $\Omega_1^2(t)$:
\[\Omega_1^2(t)= \omega_1^2\cos^2\phi(t) +\omega_2^2\sin^2\phi(t) +\omega_c^2 +g\sin2\phi(t).\]The $(2,2)$ entry is
\[K'_{22}=s(as-gc)+c(-gs+bc).\]Therefore
\[K'_{22}=as^2-2gcs+bc^2.\]So
\[K'_{22} =\omega_1^2\sin^2\phi(t) +\omega_2^2\cos^2\phi(t) +\omega_c^2 -g\sin2\phi(t).\]This entry is called $\Omega_2^2(t)$:
\[\Omega_2^2(t)= \omega_1^2\sin^2\phi(t) +\omega_2^2\cos^2\phi(t) +\omega_c^2 -g\sin2\phi(t).\]The off-diagonal entry is
\[K'_{12}=c(as-gc)-s(-gs+bc).\]Thus
\[K'_{12}=(a-b)cs-g(c^2-s^2).\]Using
\[cs=\frac{1}{2}\sin2\phi(t), \qquad c^2-s^2=\cos2\phi(t),\]and $a-b=\omega_1^2-\omega_2^2$, we get
\[K'_{12} = \frac{\omega_1^2-\omega_2^2}{2}\sin2\phi(t) -g\cos2\phi(t).\]This off-diagonal entry is called $\Omega_{12}^2(t)$:
\[\Omega_{12}^2(t)= \frac{\omega_1^2-\omega_2^2}{2}\sin2\phi(t) -g\cos2\phi(t).\]Hence
\[K'(t)= \begin{pmatrix} \Omega_1^2(t) & \Omega_{12}^2(t)\\ \Omega_{12}^2(t) & \Omega_2^2(t) \end{pmatrix}.\]Step 5: write the transformed Hamiltonian
The potential energy is
\[\frac{1}{2}Q^TK'(t)Q = \frac{1}{2}\Omega_1^2(t)X_1^2 +\frac{1}{2}\Omega_2^2(t)X_2^2 +\Omega_{12}^2(t)X_1X_2.\]The kinetic energy is unchanged by the rotation because the rotation matrix is orthogonal:
\[p_1^2+p_2^2=P_1^2+P_2^2.\]Combining kinetic and potential parts gives
\[H'= \frac{1}{2}\sum_{j=1}^{2} \left[P_j^2+\Omega_j^2(t)X_j^2\right] +\Omega_{12}^2(t)X_1X_2.\]This is the rotating-frame Hamiltonian. It has no explicit term like
\[\omega_c(p_1x_2-p_2x_1),\]but the magnetic field remains present because
\[\phi(t)=\omega_ct+\theta.\]Thus $\omega_c$ enters the final Hamiltonian through the time dependence of $\Omega_1(t)$, $\Omega_2(t)$, and $\Omega_{12}(t)$.
| New quantity | Meaning | Physical role |
|---|---|---|
| $X_j,P_j$ | rotated position and momentum variables | describe the same system in the rotating frame |
| $\phi(t)$ | rotation angle controlled by $\omega_c$ | carries the magnetic-field dependence |
| $\Omega_1(t),\Omega_2(t)$ | effective time-dependent frequencies | determine the stiffness of each rotated oscillator |
| $\Omega_{12}(t)$ | effective time-dependent coupling | measures how strongly the rotated oscillators remain linked |
Thus Section 5 continues the result of Section 4: the rotating frame removes the magnetic position-momentum term, but the remaining quadratic position matrix becomes time dependent through $\phi(t)$. The system is simpler, but it is not generally decoupled.
6. Instantaneous Decoupling
From Section 5, the rotating-frame Hamiltonian is
\[H'= \frac{1}{2}\sum_{j=1}^{2} \left[P_j^2+\Omega_j^2(t)X_j^2\right] +\Omega_{12}^2(t)X_1X_2.\]The first part,
\[\frac{1}{2}\sum_{j=1}^{2} \left[P_j^2+\Omega_j^2(t)X_j^2\right],\]describes two separate oscillators in the rotated frame. The only term that still connects them is
\[\Omega_{12}^2(t)X_1X_2.\]Therefore, the two rotated oscillators are decoupled at an instant if this coefficient becomes zero:
\[\Omega_{12}^2(t)=0.\]This is called instantaneous decoupling because the condition may hold only at particular times.
Now use the expression obtained from the off-diagonal entry of the rotated stiffness matrix:
\[\Omega_{12}^2(t)= \frac{\omega_1^2-\omega_2^2}{2}\sin2\phi(t) -g\cos2\phi(t).\]Set this equal to zero:
\[\frac{\omega_1^2-\omega_2^2}{2}\sin2\phi(t) -g\cos2\phi(t)=0.\]Move the second term to the other side:
\[\frac{\omega_1^2-\omega_2^2}{2}\sin2\phi(t) =g\cos2\phi(t).\]Now divide by $\cos2\phi(t)$, assuming $\cos2\phi(t)\neq0$:
\[\frac{\omega_1^2-\omega_2^2}{2}\tan2\phi(t)=g.\]Hence
\[\tan2\phi(t)=\frac{2g}{\omega_1^2-\omega_2^2}.\]This is the decoupling condition.
| Quantity | Role |
|---|---|
| $g$ | coupling strength |
| $\omega_1^2-\omega_2^2$ | detuning between oscillators |
| $\phi(t)=\omega_ct+\theta$ | magnetic-field-controlled rotation angle |
The condition says that decoupling is possible when the rotation angle has just the right value to balance two effects:
- coupling coming from the original interaction strength $g$,
- coupling generated by rotating two oscillators with different frequencies.
Since
\[\phi(t)=\omega_ct+\theta,\]the angle changes with time when $\omega_c\neq0$. Thus the condition may be satisfied at isolated times. At such an instant, the Hamiltonian takes the temporarily separated form
\[H'= \frac{1}{2}\left[P_1^2+\Omega_1^2(t)X_1^2\right] +\frac{1}{2}\left[P_2^2+\Omega_2^2(t)X_2^2\right].\]There is then no $X_1X_2$ term at that instant, so the two rotated oscillators behave as independent modes. But as time passes, $\phi(t)$ changes, $\Omega_{12}^2(t)$ generally becomes nonzero again, and the modes become coupled once more.
7. Compact Picture
The model can be summarized as:
| Stage | Description |
|---|---|
| Start | two oscillators with frequencies $\omega_1,\omega_2$ |
| Add coupling | oscillators interact through $-gx_1x_2$ |
| Add magnetic field | frequencies shift and position-momentum mixing appears |
| Rotate frame | magnetic cross-term is removed |
| Final form | two oscillators with time-dependent frequencies and coupling |
Takeaway
The magnetic field does not merely shift oscillator frequencies. It also rotates the coupled system in phase space. In the rotating frame, the dynamics are controlled by $\Omega_1(t)$, $\Omega_2(t)$, and $\Omega_{12}(t)$, making the model useful for studying synchronization, mutual information, and quantum circuit complexity.
Discussion