- #1
kelly0303
- 573
- 33
Hello! If we have a 2 level system (say an atom) with energies ##0## and ##\Delta## and opposite parities, and we add an electric field, E, the hamiltonian would now look like ##H = H_0 +H_{Stark}## (where ##H_0## is the unperturbed Hamiltonian), which is equal to ##H=H_0-e\vec{r}\vec{E}##, where (assume we are having an electron in an atom) e is the electric charge and r is the position operator. If we assume that the field is in the z direction we get in the end ##H=H_0-ezE_z##. Now, if we calculate the expectation value of this Stark Hamiltonian in one of the 2 states we would have something like ##<+|-ezE_z|+>=-eE_z<+|z|+>=0##, where ##|+>## is the state of positive parity and we get zero because z is an odd parity operator. On the other hand the diagonal matrix element is not zero and we have ##<+|-ezE_z|->=d_zE_z##, where we define ##d_z = <+|-ez|->##. So in matrix form we get $$H =
\begin{pmatrix}
0 & d_zE_z \\ d_zE_z & \Delta
\end{pmatrix}
$$
And from here we can diagonalize and get the new eigenstates and energy levels. Is this correct? Now my next question is, assume we have an electric field that depends on the position, such as ##E = E_0z##, pointing in the z direction. Now the hamiltonian would be ##H = H_0 - eE_0z^2##. But given that we have ##z^2## instead of ##z##, the parity changes, so in this case the Stark hamiltonian has even parity which means that ##<+|-eE_0z^2|->=0## and ##<+|-eE_0z^2|+>=E_0d_z^+## and ##<-|-eE_0z^2|->=E_0d_z^-##. So in matrix form we get
$$H =
\begin{pmatrix}
E_0d_z^- & 0 \\ 0 & \Delta + E_0d_z^+
\end{pmatrix}
$$
Is my derivation correct? So if we have a uniform electric field, we get off-diagonal term, but if we have a field varying linearly with z (or at least with an odd power of z), we get diagonal term, and the off-diagonals are unchanged. Is my conclusion correct? Thank you!
\begin{pmatrix}
0 & d_zE_z \\ d_zE_z & \Delta
\end{pmatrix}
$$
And from here we can diagonalize and get the new eigenstates and energy levels. Is this correct? Now my next question is, assume we have an electric field that depends on the position, such as ##E = E_0z##, pointing in the z direction. Now the hamiltonian would be ##H = H_0 - eE_0z^2##. But given that we have ##z^2## instead of ##z##, the parity changes, so in this case the Stark hamiltonian has even parity which means that ##<+|-eE_0z^2|->=0## and ##<+|-eE_0z^2|+>=E_0d_z^+## and ##<-|-eE_0z^2|->=E_0d_z^-##. So in matrix form we get
$$H =
\begin{pmatrix}
E_0d_z^- & 0 \\ 0 & \Delta + E_0d_z^+
\end{pmatrix}
$$
Is my derivation correct? So if we have a uniform electric field, we get off-diagonal term, but if we have a field varying linearly with z (or at least with an odd power of z), we get diagonal term, and the off-diagonals are unchanged. Is my conclusion correct? Thank you!