# Proof of coherent state displacement operator solution

In $$3D$$ space I have two $$2\times2$$ non-Hermitian matrix operators, $$A$$ and $$A^\dagger$$, of the form: $$A=\begin{pmatrix} A_{11}(x_j,\partial_j) & A_{12}(x_j,\partial_j)\\ A_{21}(x_j,\partial_j) & A_{22}(x_j,\partial_j) \end{pmatrix}$$ That is, each component is a differential operator. These operators act on the states of the form: $$\psi(x_j)=\begin{pmatrix} \psi_\uparrow(x_j)\\ \psi_\downarrow(x_j) \end{pmatrix}$$

The displacement operator for the coherent states of these states is defined as: $$D(\alpha)=\exp(\alpha A^\dagger-\alpha^* A)$$

For $$\alpha\in\mathbb{C}$$. A coherent state can be defined as: $$\psi_\alpha=D(\alpha)\psi_0=\sum_{n=0}^\infty\frac{1}{n!}\left(\alpha A^\dagger-\alpha^* A\right)^n\psi_0$$

Through an educated guess using the symmetries of the system, I found the form of the state $$\psi_\alpha$$ for a specific $$\alpha\in\mathbb{R}$$, which I numerically confirmed after an expansion of over a dozen terms to an accuracy of over $$99\%$$.

Thus I have: $$\psi_\text{sol.}=D(\alpha)\psi_0$$ $$\Rightarrow \begin{pmatrix} \psi_{\text{sol.},\uparrow}(x_j)\\ \psi_{\text{sol.},\downarrow}(x_j) \end{pmatrix}=\sum_{n=0}^\infty\frac{\alpha^n}{n!}\left[\begin{pmatrix} A_{11}(x_j,\partial_j)^\dagger & A_{21}(x_j,\partial_j)^\dagger\\ A_{12}(x_j,\partial_j)^\dagger & A_{22}(x_j,\partial_j)^\dagger \end{pmatrix}-\begin{pmatrix} A_{11}(x_j,\partial_j) & A_{12}(x_j,\partial_j)\\ A_{21}(x_j,\partial_j) & A_{22}(x_j,\partial_j) \end{pmatrix}\right]^n\begin{pmatrix} \psi_{0,\uparrow}(x_j)\\ \psi_{0,\downarrow}(x_j) \end{pmatrix}$$

Where I know the starting state, $$\psi_0$$, the coherent state, $$\psi_\text{sol.}$$, the form of the differential operators, $$A_{ij}$$ and $$A_{ij}^\dagger$$, and the real number $$\alpha$$. I have confirmed this identity numerically, how can I prove it analytically? I can't see a way to start since the operators constantly couple the spin-up and down states together.

• Are your operators bosonic? Do they always satisfy $AA^\dagger-A^\dagger A=\mathbb{I}$? Commented Aug 4, 2023 at 18:21
• @QuantumMechanic no, they satisfy neither $AA^\dagger-A^\dagger A=\mathbb{I}$ nor $AA^\dagger+A^\dagger A=\mathbb{I}$ in general. Commented Aug 4, 2023 at 19:21
• Hm. Then I don't see why this should be the displacement operator per se. Is there any closure relation for $D(\alpha) D(\beta)$? Commented Aug 4, 2023 at 20:53
• @QuantumMechanic There is no closure relation that I can prove (i.e., anything of the form $D(\alpha)D(\beta)=D(\alpha\beta)$). There is a slightly modified version of these operators that are operators corresponding to the root vectors of an $SO(3,2)$ symmetry. These vectors are scalar multiples of $A$ and $A^\dagger$ similar to the scaling of the RL vector in the hydrogen atom (which includes the principle quantum number). They may satisfy the relation. I use the term "displacement operator" as I'm trying to find the "magnetic translation operators" of a system in 3D with a new gauge. Commented Aug 4, 2023 at 21:13
• @QuantumMechanic and when I say "They may satisfy the relation", I don't believe the state I've defined on the LHS of $\psi_\text{sol}=D(\alpha)\psi_0$ remains the solution if $A\rightarrow\tilde{A}$ (where $\tilde{A}$ is the scaled version that depends on one of the quantum numbers) Commented Aug 4, 2023 at 21:16