The Stark shift of a diatomic molecule

1. The Stark shift of a diatomic molecule

2. Contributions to the Stark shift Hamiltonian for the Stark interaction: Energy shift by perturbation theory, to second order... Zero Small, when i and j differ only by rotation The Stark interaction of heteronuclear diatomic molecules is dominated by the rotational contributions.

3. The Stark shift of a rigid rotor E m B

4. The Stark shift of a rigid rotor...by perturbation theory To complete the calculation we need to evaluate E H0 H’ m B The unperturbed eigenstates are the angular momentum eigenstates ,|J,M>. The unperturbed eigenvalues are WJM = B J(J+1) For a small field, the approximate correction to the energy can be found by perturbation theory:

5. Evaluating the matrix element For convenience, define the renormalized spherical harmonic: (1) Use the Wigner-Eckart theorem: Remember your spherical harmonics. In particular: So, our matrix element can be written (2) Find the reduced matrix element – ours is a standard result in angular momentum theory: N.B. Only non-zero if M’ = M and J’ = J ± 1

6. Returning to our perturbation calculation... Start with the ground state, (J,M)=(0,0). The only non-zero matrix element is with (J’,M’)=(1,0). The energy shift is negative, and goes quadratically with E Next, consider the state (J,M)=(1,0). It has non-zero matrix elements with (0,0) and (2,0) The energy shift is positive, goes quadratically with E, and is smaller than the ground state shift

7. Returning to our perturbation calculation... Now, the (J,M) = (1,1) state. It’s only non-zero matrix element is with (2,1). The energy shift is negative, goes quadratically with E, and is smaller than the other two

8. The perturbation theory results... (1,0) (1,±1) (0,0) N.B. This is the energy shift. The J=1 states have their unperturbed energies up at 2B

9. Higher fields – a more complete treatment Solve the eigenvalue problem with Use the angular momentum eigenstates |J,M> as a basis. Construct the matrix for the complete Hamiltonian in this basis and diagonalize it to find the eigenvalues. We already know all of the matrix elements of H in the |J,M> basis: A problem: To get the exact solution we need an infinitely large basis. Solution: Truncate the basis. Far away states don’t matter much. To illustrate the procedure, choose to truncate at J=1. Also, since H doesn’t couple different M’s, we can do each value of M separately. Then, the only states in our basis are |0,0> and |1,0>. In this case, we just have a 2x2 matrix:

10. The eigenvalues of this 2x2matrix The eigenvalues, l, of a 2x2 matrix, are found by setting with the result So we get the eigenvalues of our problem: Sanity check: are they correct at zero field? For small values of E we can expand the square root... The result for the ground state Stark shift is the same as we got from perturbation theory.

11. For higher accuracy, truncate at higher J... Truncation at J=3. M=0 only: Truncation at J=2. M=0 only: Truncation at J=8. M=0 only:

12. Truncation at J=3. All M:

13. The answer Diagonalize the matrix (numerically) to get the eigenvaluesto any degree of accuracy you like. Bigger basis gives higher accuracy. N.B. Your computer is better at this than you are! (2,0) (2,±1) (2,±2) (1,0) (1,±1) (0,0) We found this answer by considering a rigid-rotor model. This is an excellent approximation for all heteronuclear diatomic molecules in S states. For non-S states the answer is a bit different (we won’t go into that!)

14. More of the answer • At small fields the Stark shift is quadratic in E, while at high fields it is linear. • Some states shift down, others up. • The down-shifters are called high-field seeking states. • The up-shifters are called weak-field seeking states. • All the weak-field seeking states turn over and become strong-field seeking at high field. • The Stark shift is big – big enough to control molecular motion.