# How can an unknown Hamiltonian be solved for in an ODE?

Consider the following spinor (supposed to be a 2x1 matrix but formatting is not working correctly) $$\psi \equiv \times_{i}\left[\begin{array}{l}\chi_{i+}\\ \chi_{i-}\end{array}\right]$$ $$\hat{\mathcal{H}}=\frac{\hbar\omega_0}{2}(|r\rangle\langle r|-|g\rangle\langle g|)+\hbar\Omega e^{i\omega't}|g\rangle\langle r|+\hbar\Omega e^{-i\omega' t}|r\rangle\langle g|$$

The goal is to apply a time-dependent transformation (i.e. unitary time-evolution operator) such that the Hamiltonian under this transformation is given by $$\frac{\mathcal{H}}{\hbar}=c\mathbb{I}+h\sigma^x+h_{||}\sigma^x$$ where $$c,h,h_{||}$$ are all constants, and $$\sigma^x$$ is the Pauli matrix for $$x$$-oriented spin.

My current approach is to define $$\hat{U}\in\mathcal{L}(\mathscr{H})$$ such that $$\hat{U}(t,t_0=0)=e^{\frac{-it\mathcal{H}}{\hbar}}$$ and define another operator $$\hat{\tilde{U}}(t)\equiv \hat{A}^\dagger \hat{U}(t) \hat{A}(t)$$ $$\hat{\tilde{U}}(t)=e^{\frac{-it\tilde{\mathcal{H}}}{\hbar}}$$

Basically, I think that we need to identify an operator $$\hat{A}:\hat{A}^\dagger \mathcal{H}\hat{A}=\tilde{\mathcal{H}}$$

If we consider the time-dependent Schrodinger equation, we have $$i\hbar\partial_t|\psi\rangle = \mathcal{H}|\psi\rangle$$ We can make the ansatz $$|\psi\rangle = \hat{A}|\tilde{\psi}\rangle$$. This means that we have $$i\hbar\partial_t |\tilde{\psi}\rangle = \tilde{\mathcal{H}}|\tilde{\psi}\rangle$$ Here is where the confusion comes. There are two approaches I can think of: the first is to say $$\tilde{\mathcal{H}}=\langle \tilde{\psi}|i\hbar\partial_t \hat{A}|\tilde{\psi}\rangle$$ But writing the matrix of $$\hat{A}$$ as a linear combination of two arbitrary phase shifts $$e^{i\alpha t},e^{i\beta t}$$ such that the matrix is diagonal, this yields only two terms, while I should be getting three. Alternatively, we could try separation of variables and integrating, but I am not sure how I would integrate the Hamiltonian.

It is also worth noting that the original Hamiltonian of the system was stated to be (but not entirely relevant to this sub-problem) $$\frac{\mathcal{H}}{\hbar}=h\sum_i \sigma_i^x+J\sum_i \sigma_i^z\sigma_{i+1}^z$$ where $$i$$ indexes which spin the Pauli matrix acts on and $$J$$ is a parameter that determines ferromagnetic or antiferromagnetic properties. Any advice is appreciated.

• Why should you be getting three? It seems like the separation of the second two terms in your target form is arbitrary, since they are both proportional to $\sigma_x$. Feb 14 at 16:49

Your equation can solved completely. The easiest way is to make a classical analogy. You can map this general two state problem to a spin $$1/2$$ particle under an external magnetic field.
To make the analogy exact, the magnetic field is precessing at rate $$\omega$$ about an axis (say $$z$$). This gives you Bloch’s equation and you can solve it by going into the the moving frame. This physical insight immediately gives you the correct intermediate change of basis.
This is how you can find your appropriate $$\mathcal A$$ (notation is not standard). Here, the analogy with spin helps you find it directly from the conjugation relation using the properties of rotation matrices.