The Hydrogen Atom in a Static Electric Field

We address the problem of the hydrogen atom in a static electric field, ${\vec {\mathcal {E}}}={\mathcal {E}}{\hat {z}}$. The hamiltonian for this problem can be written

$H=H_{0}+H^{\prime }=H_{0}+e{\mathcal {E}}z$

where $H_{0}$ is the "unperturbed" hamiltonian for the hydrogen atom.

We chose to solve this via matrix methods. The first step is to write down the matrix elements for the hamiltonian is a basis of our choosing. Let's try with the $|nlm\rangle$ basis kets, the eigenkets of $H_{0}$. So, $H_{0}$ only contributes diagonal elements to the matrix, $E_{n}$. As $e$ and $E$ are scalars, not operators, we need only consider the effect of $z$. First, $z$ is a parity odd operator, connecting only states of different parity. Thus $H^{\prime }$ contributes nothing to the diagonal entries nor to any entries with the same angular momentum, $l$. States of the same parity but whose angular momentum differ by more than $\Delta l=\pm 1$ also result in zero because ... Finally, $H^{\prime }$ also only connects states of the same $m$. One can see this by noting that

$z=Y_{1}^{0}r{\sqrt {\frac {4\pi }{3}}}=r\cos \theta$

which is an even function in $\theta$. Any states differing by $\Delta m=\pm 1$ would then result in an integral of two even functions (one of those being the $\cos \theta$ originating from the $z$) and an odd function in $\theta$ which is zero. This resulta can also be seen directly by noting a result of the Wigner-Eckhart theorem that $\langle njm|z|njm^{\prime }\rangle =\alpha (n,j)\langle njm|J_{z}|njm^{\prime }\rangle$ where $\alpha$ is just a number. Thus, we produce the "selection rules" for the ${\vec {E}}=E{\hat {z}}$ operator,

$\Delta m=0~~~~\ l=\pm 1$

NOTE that this strictly applies only to this specific operator. If $E$ were pointing in some other direction then things might (and do) change.

The matrix for the our hamiltonian reads then

${\begin{pmatrix}E_{1}&0(e/o)&0(e/o)&eE\langle z\rangle &0(p)&...\\~&E_{2}&0(p)&0(p)&0(p)&~\\~&~&E_{2}&0(p)&0(e/o)&~\\~&~&~&E_{2}&eE\langle z\rangle &~\\~&~&~&~&E_{2}&~\end{pmatrix}}$

where the entries arranged in $|100\rangle ,|211\rangle ,|21-1\rangle ,|210\rangle ,|200\rangle$ order. The 0's are designated with an indication of "why" those particular entries in the matrix are zero, $e/o$ meaning even/odd ($\Delta m=0$ selection rule) and $p$ meaning parity ($\Delta l=\pm 1$ selection rule). The $H^{\prime }$ contribution to the diagonal elements is zero due to parity. Because the $n=2$ states are degenerate, degenerate perturbation theory must be used to solve the problem. Of course we know that in reality the problem is more complex than this. Both fine, hyperfine and the Lamb shift have been neglected. Solving the problem taking this into account would indicate the use of second order perturbation theory.

To see how this all shakes out, let's go ahead and apply perturbation theory directly.

${\begin{array}{rcl}E_{n}^{(1)}&=&\langle n|H^{\prime }|n\rangle =0~by~parity\\E_{n}^{(2)}&=&\Sigma _{m,m\neq n}{\frac {|\langle n|H^{\prime }|m\rangle |^{2}}{E_{n}-E_{m}}}=e^{2}{\mathcal {E}}^{2}\Sigma _{m,m\neq n}{\frac {|\langle n|z|m\rangle |^{2}}{E_{n}-E_{m}}}\\|n^{(1)}\rangle &=&\Sigma _{m,m\neq n}|m\rangle {\frac {|\langle n|H^{\prime }|m\rangle |^{2}}{E_{n}-E_{m}}}\end{array}}$

This simple perturbation theory does not work in the case of degenerate states (leading to $E_{m}-E_{n}=0$ in the denominator). Then one has to diagonalize the Hamiltonian in relation to $H^{\prime }$. If you do that for the case of $n=2$ you find that the eigenstates are

${\begin{array}{rcl}|n=2,l=1,m=1\rangle &~~~&E_{2}\\|2,1,-1\rangle &~~~&E_{2}\\{\frac {1}{\sqrt {2}}}(|2,1,0\rangle +|2,0,0\rangle )&~~~&E_{2}+cE\\{\frac {1}{\sqrt {2}}}(|2,1,0\rangle -|2,0,0\rangle )&~~~&E_{2}-cE\end{array}}$

where $c$ is a constant. The last two states have a linear response to the electric field, or a linear Stark effect. Even is there were a small splitting between the different states in the $n=2$ manifold, if the field interaction were higher that the splitting then there would be also be a linear Stark effect. At lower fields the interaction would be second order (second order pertubation theory would be called for) and the response would be quadratic in the applied electric field. Notice the the new eigenstates are a mixture of states of different parity. This mixture allows for a dipole to be formed and it is the interaction of the electic field with this dipole that gives rise to a linear response to the field. It is this dipole that is talked about by chemists when they say that a molecule "has a dipole moment". Molecules "have dipole moments" because they have closely lying states of opposite parity so small fields put them in the linear Stark regime. But make no mistake, at low enough fields, the response would be quadratic, just as it is in the case of atoms.

Now, all of this has been talked about under the (essentially correct) assumption that $[H,\pi ]=0$ and, therefore, that the eigenstates of the H atom are also parity eigenstates. But what if $[H,\pi ]\neq 0$ ? This occurs when the weak force is involved and will likely be present in nature and described, eventually, by extensions to the Standard Model. Such mechanisms can lead to the presence of permanent electic dipole moments of elementary particles.