Degenerate Perturbation Theory: Zero Diagonal Does Not Mean No Splitting
A particle of mass m moves in a two-dimensional isotropic harmonic oscillator H0 = p²/(2m) + (1/2)mω²(x²+y²). The first excited level (n_x+n_y=1) is two-fold degenerate, spanned by |1,0> and |0,1>. A perturbation H' = λxy is switched on, with λ small. A student computes the diagonal matrix elements <1,0|H'|1,0> = <0,1|H'|0,1> = 0 and concludes that the first-order shift vanishes and the degeneracy is not lifted. You are asked to check this with degenerate perturbation theory. Use ħω = 0.10 eV and choose λ so that the off-diagonal matrix element has magnitude 0.020 eV. Ignore coupling to other levels unless you are explicitly discussing second order.
1. Explain why the nondegenerate formula is insufficient and write the 2×2 perturbation matrix in the degenerate subspace.
Read the worked answer
Because the two states have the same unperturbed energy, any linear combination is an equally good zeroth-order state. The nondegenerate formula applies only when the unperturbed level is isolated. One must therefore diagonalize H' inside the two-dimensional subspace spanned by |1,0> and |0,1>. Writing x and y in ladder operators,
x = sqrt(ħ/(2mω))(a_x + a_x†), y = sqrt(ħ/(2mω))(a_y + a_y†).
The diagonal elements vanish because xy changes n_x and n_y by one unit and cannot return to the same state at first order. The off-diagonal element is
<1,0|λxy|0,1> = λ [ħ/(2mω)] = V.
By symmetry the other off-diagonal element is also V. Thus the matrix is [[0, V], [V, 0]], with V = 0.020 eV by choice of λ.
2. Diagonalize the matrix and give the first-order energies and eigenstates.
Read the worked answer
The matrix [[0,V],[V,0]] has eigenvalues +V and -V. Therefore the first-order energies are
E± = 2ħω ± V = 0.20 eV ± 0.020 eV,
so the splitting is 2V = 0.040 eV. The corresponding eigenstates are
|+> = (|1,0> + |0,1>)/sqrt(2), |-> = (|1,0> - |0,1>)/sqrt(2).
The degeneracy is lifted linearly in λ. The student correctly found zero diagonal elements but stopped too early; the off-diagonal coupling within the degenerate subspace is exactly what produces the splitting.
3. What is wrong with the student's reasoning, and would second-order perturbation change the qualitative conclusion?
Read the worked answer
The student confused a zero diagonal matrix element with a zero perturbation. In degenerate perturbation theory, the correct criterion is whether H' has nonzero matrix elements anywhere inside the degenerate subspace, not just on its diagonal. Here the off-diagonal element V = 0.020 eV is nonzero, so the degeneracy is lifted at first order. Second-order corrections do exist: H' couples |1,0> and |0,1> to states such as |0,0> and |2,0>, producing shifts of order V²/(ħω) ≈ (0.020)²/0.10 = 0.004 eV. These are an order of magnitude smaller than the first-order splitting and shift both levels in the same direction; they do not cancel the ±V splitting. The qualitative conclusion of a split doublet is therefore robust as long as V is nonzero and small compared with ħω.