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.


1 Answer 1


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.

  • $\begingroup$ 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 $|n^0\rangle + \lambda|n^1\rangle$? $\endgroup$
    – test123
    Jul 8, 2021 at 12:10
  • $\begingroup$ i.e. I would have to translate by amount $(-1)\cdot\ldots$ in order to obtain this $\endgroup$
    – test123
    Jul 8, 2021 at 12:13
  • $\begingroup$ You know the exact solution, You just need to act translation operator on it. You don't have to use perturbation theory for it. $\endgroup$ Jul 8, 2021 at 14:33

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