- #1
Malamala
- 345
- 28
Hello! I am asking this question from a molecular physics perspective (i.e. diatomic molecule placed in an external electric field), but it's quite general in terms of the formulation. I have a 2 level system (call the levels ##\ket{0}## and ##\ket{1}##) that can be connected by an electric field through the Stark interaction. The Hamiltonian of the system is (I call the energies of the 2 levels in the absence of electric field ##E_1## and ##E_2##):
$$
\begin{pmatrix}
E_0 & -D_{01}E \\
-D_{01}E & E_1
\end{pmatrix}
$$
with ##D_{01} = \braket{0|D|1}##. If I diagonalize this, I get (say for the lowest energy state), something of the form: ##\ket{\tilde{0}} = a\ket{0}+b\ket{1}##, with ##a^2+b^2=1##. Then, the polarization in the ground state is given by:
$$p = \frac{\braket{\tilde{0}|D|\tilde{0}}}{D_{01}}$$
Thus, for a very small external field, ##b## is very small and the polarization is close to zero, while for a very large field, I have ##a=b=\sqrt{2}/2## and thus the polarization is 1, as expected i.e. the molecule is fully polarized in a large external field. However, if I add a 3rd level, connected to ##\ket{1}## by a dipole interaction, the Hamiltonian becomes:
$$
\begin{pmatrix}
E_0 & -D_{01}E & 0\\
-D_{01}E & E_1 & -D_{12}E \\
0 & -D_{12}E & E_2
\end{pmatrix}
$$
with ##D_{12} = \braket{1|D|2}## (which doesn't need to be equal to ##D_{01}## in general and I also assume that the dipoles are real numbers). Now, the ground state wavefunction is ##\ket{\tilde{0}} = a\ket{0}+b\ket{1}+c\ket{2}##, but I am not sure how to define the polarization anymore. I know I still need to take the expectation value: ##\braket{\tilde{0}|D|\tilde{0}} = 2abD_{01}+2bcD_{12}## but I am not sure what to divide this by (the same way I divided by ##D_{01}## before), or if I need a different definition. I know I need to have the same behaviour i.e. close to zero poalrization at low field and close to 1 at high field, but I am not sure how to get this in the case of multiple levels. Thank you!
$$
\begin{pmatrix}
E_0 & -D_{01}E \\
-D_{01}E & E_1
\end{pmatrix}
$$
with ##D_{01} = \braket{0|D|1}##. If I diagonalize this, I get (say for the lowest energy state), something of the form: ##\ket{\tilde{0}} = a\ket{0}+b\ket{1}##, with ##a^2+b^2=1##. Then, the polarization in the ground state is given by:
$$p = \frac{\braket{\tilde{0}|D|\tilde{0}}}{D_{01}}$$
Thus, for a very small external field, ##b## is very small and the polarization is close to zero, while for a very large field, I have ##a=b=\sqrt{2}/2## and thus the polarization is 1, as expected i.e. the molecule is fully polarized in a large external field. However, if I add a 3rd level, connected to ##\ket{1}## by a dipole interaction, the Hamiltonian becomes:
$$
\begin{pmatrix}
E_0 & -D_{01}E & 0\\
-D_{01}E & E_1 & -D_{12}E \\
0 & -D_{12}E & E_2
\end{pmatrix}
$$
with ##D_{12} = \braket{1|D|2}## (which doesn't need to be equal to ##D_{01}## in general and I also assume that the dipoles are real numbers). Now, the ground state wavefunction is ##\ket{\tilde{0}} = a\ket{0}+b\ket{1}+c\ket{2}##, but I am not sure how to define the polarization anymore. I know I still need to take the expectation value: ##\braket{\tilde{0}|D|\tilde{0}} = 2abD_{01}+2bcD_{12}## but I am not sure what to divide this by (the same way I divided by ##D_{01}## before), or if I need a different definition. I know I need to have the same behaviour i.e. close to zero poalrization at low field and close to 1 at high field, but I am not sure how to get this in the case of multiple levels. Thank you!
