How to use the translation operator in order to find the eigenstates in a perturbed QM system?

Given a quantum mechanical system with Hamiltonian $$\hat{H_0}$$, introduce a perturbation $$\lambda \hat{H_1}$$ with $$\lambda$$ sufficiently small. Define now the spacial translation operator to be $$\hat{T}(x)=\exp(-i x \hat{p}/\hbar)$$. Assuming that the solutions of the eigenvalue equation of $$\hat{H_0}$$ are known, how does one compute the eigenstates of the perturbed Hamiltonian $$\hat{H_0}+\lambda\hat{H_1}$$ using taylor expansion of $$\hat{T}$$?

In particular, I'm trying to apply this to the problem of a linearly perturbed harmonic oscillator: Let $$\hat{H_0}=-\frac{\hbar^2}{2m}\frac{d^2}{dx^2}+\frac{1}{2}m\omega^2 x^2$$ and $$\hat{H_1}=x$$. Express $$\hat{T}$$ in terms of the ladder operators $$\hat{a}_{\pm}$$ and expand $$\hat{T}$$ up to the first order (for small deviations in $$x$$). Use this to compute the eigenstates of the perturbed Hamiltonian.

Hint: Let's first write the translation operator as $$T(a)=e^{-ipa/\hbar}\rightarrow e^{ad/dx}$$ For a given problem, $$H=\frac{P^2}{2m}+\frac{1}{2}m\omega^2X^2+\lambda X=\frac{P^2}{2m}+(\cdots)(X+\cdots)^2+\cdots$$ The $$\cdots$$ part in the end is just corresponds to a constant shift. The $$X+\cdots$$ part is what corresponds to translation. Find the solution for $$X$$ unperturb part and then translate the solution by amount $$\cdots$$ that will give you your desired result.
• Thank you, I have done as you suggested and obtained $\hat{T}(\lambda/(m\omega^2))|n\rangle=(\mathbb{1}-i\lambda\hat{p}/(m\hbar \omega^2))|n\rangle = |n^0 \rangle - \lambda |n^1 \rangle$ where $|n^k\rangle$ denotes the $k$-th order perturbation term for the $n$-th eigenstate (so $|n\rangle=|n^0\rangle + \lambda|n^1\rangle+...$). Shouldn't this be more something like $|n^0\rangle + \lambda|n^1\rangle$? Jul 8 at 12:10
• i.e. I would have to translate by amount $(-1)\cdot\ldots$ in order to obtain this Jul 8 at 12:13