1
$\begingroup$

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.

$\endgroup$
1
  • 2
    $\begingroup$ 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 $N$. $\endgroup$ Commented Aug 31, 2016 at 19:40

2 Answers 2

1
$\begingroup$

(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 $N$.

$\endgroup$
0
$\begingroup$

As Valter Moretti answered in the comments, adding this here for archive purposes.

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 $N$.

$\endgroup$
1
  • $\begingroup$ I have written my comment into an answer a bit re-elaborating it. You may cancel yours. Yesterday I was in a hurry... $\endgroup$ Commented Sep 1, 2016 at 6:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.