# Solving first order perturbation exactly in this situation

I have this homework problem in QM Perturbation Theory

The Hamiltonian of a system is given by $$H_0 = A L^2 + B L_z$$ where $A$ and $B$ are constants. If a perturbation $V = C L_y$ is added to the system (where $C \ll A, B$ , find the lowest order correction to the energy. Also solve the problem exactly

The lowest order correction to energy is

$$\langle l,m \rvert C L_y \lvert l, m \rangle = 0$$ as $L_y$ will either give me $\lvert l, m-1 \rangle$ or $\lvert l, m+1 \rangle$

But how do I solve it exactly? If I fix $l$, then $L_y$ is a $(2l+1) \times (2l+1)$, which is still more tedious than this assignment is supposed to be. Please help.

• Define $N = \sqrt{B^2+ C^2}$. Using a unitary transformation $U$ corresponding to a certain rotation, you have $U(AL^2 + BL_z+ CL_y)U^* = AL^2 + NL_z$. Unitary transformations do not change the eigenvalues. Therefore the exact eigevalues are $Al(l+1)+Nm$. The same eigenvalues as for $H_0$ with $B$ replaced for $N$. Commented Aug 31, 2016 at 19:40

(I assume all coefficients are real obviously.) Define $N=\sqrt{B^2+C^2}$. Using a unitary transformation $U$ representing in the Hilbert space a certain rotation, the one rotating $$\left(0,\frac{C}{N} ,\frac{B}{N}\right)$$ to $$(0,0,1)$$ which therefore leaves $L^2$ invariant, you have $$U(AL^2+BL_z+CL_y)U^∗=AL^2+NL_z\:.$$ Unitary transformations do not change the eigenvalues so that eigenvalues of $AL^2+NL_z = U(AL^2+BL_z+CL_y)U^∗$ are the same as of $AL^2+BL_z+CL_y$.
From this we conclude that the exact eigevalues of $AL^2+BL_z+CL_y$ are the ones of $AL^2+NL_z$: $$Al(l+1)+Nm\:, \quad l=0,1,2,\ldots \quad m = \pm l$$ the same eigenvalues as for $H_0$ with $B$ replaced for $N$.
Define $N=B^2+C^2$. Using a unitary transformation $U$ corresponding to a certain rotation, you have $U(AL^2+BL_z+CL_y)U^∗=AL^2+NL_z$. Unitary transformations do not change the eigenvalues. Therefore the exact eigevalues are $Al(l+1)+Nm$: the same eigenvalues as for $H_0$ with $B$ replaced for $N$.