Atoms in strong electric fields

Beyond the quadratic Stark effect

It should be obvious from the previous discussion that the Stark effect for a state of $g$ is quadratic only when

${\begin{aligned}\ \epsilon <<{\frac {E_{i}-E_{g}}{e|\langle i|{\bf {r}}|g\rangle |}}\end{aligned}}$

(EQ_ beyondone)

when $i$ is the nearest state of opposite parity to $g$.

If $g$ is the ground state, we can expect $E_{i}=E_{g}\sim 0.5$ $E_{g},~E_{g}\sim 0.3$ Hartree and $|<r>|^{-1}\approx |<r^{-1}>|=2E_{g}/e^{2}$ (virial theorem). Hence the Stark shift should be quadratic if the field is well below the critical value

${\begin{aligned}\ \epsilon _{crit}={\frac {0.5\times 2(0.3)^{2}m^{2}e^{8}}{e^{3}\hbar ^{4}}}\approx 0.1{\frac {e}{a_{0}^{2}}}\end{aligned}}$

(EQ_ beyondtwo)

[$e/a_{0}^{2}$ is atomic unit of field] $\approx 5\times 10^{8}V/{\rm {cm}}$ —a field three orders of magnitude in excess of what can be produced in a laboratory except in a vanishingly small volume.

If $g$ is an excited state, say $|g\rangle =|n\ell \rangle$, this situation changes <it> dramatically </it>. In general, the matrix element $<n,\ell +1|{\bf {r}}|n,\ell >\sim n^{2}a_{0}^{2}$ and $\Delta E$ to the next level of opposite parity depends on the quantum defect:

${\begin{aligned}\ \Delta E=E_{n,\ell +1}-E_{n,\ell }={\frac {-R_{H}}{(n-\delta _{\ell +1)^{2}}}}-{\frac {-R_{H}}{(n-\delta _{\ell })^{2}}}\approx 2R_{H}(\delta _{\ell +1}-\delta _{\ell })/n^{*3}\end{aligned}}$

(EQ_ beyondthree)

Thus the critical field is lowered to

$\epsilon _{\rm {crit}}={\frac {\Delta E}{e<|{\bf {r}}|>}}={\frac {me^{4}}{\hbar ^{2}}}{\frac {1}{ea_{0}}}{\frac {\delta _{\ell +1}-\delta _{\ell }}{n^{*5}}}$

${\begin{aligned}\ ={\frac {e}{a_{0}^{2}}}{\frac {\delta _{\ell +1}-\delta _{\ell }}{n^{5}}}\approx 5\times 10^{9}{\frac {\delta _{\ell -1}-\delta _{\ell }}{n^{*5}}}{\frac {\rm {volts}}{\rm {cm}}}\end{aligned}}$

(EQ_ beyondfour)

Considering that quantum defects are typically $\leq 10^{-5}$ when $\ell \geq \ell _{\rm {core}}+2\ (\ell _{\rm {core}}$ is the largest $\ell$ of an electron in the core), it is clear that even 1 V/cm fields will exceed $\epsilon _{\rm {crit}}$ for higher $\ell$ levels if $n>7$. Large laboratory fields ($10^{5}$ V/cm) can exceed $\epsilon _{\rm {crit}}$ even for $S$ states if $n^{*}\geq 5$.

When the electric field exceeds $\epsilon _{\rm {crit}}$ states with different $\ell$ but the same $n$ are degenerate to the extent that their quantum defects are small. Once $\ell$ exceeds the number of core electrons, these states will easily become completely mixed by the field and they must be diagonalized exactly. The result is eigenstates possessing apparently permanent electric dipoles with a resulting linear Stark shift (see following figure). As the field increases, these states spread out in energy. First they run into states with the same $n$ but different quantum defects; then the groups of states with different $n$ begin to overlap. At this point a matrix containing all $n,\ell$ states with $\ell$ greater or equal to $m_{\ell }$ must be diagonalized. The only saving grace is that the lowest $n$ states do not partake in this strong mixing; however, the $n$ states near the continuum always do if there is an $\epsilon$-field present.\

The situation described above differs qualitatively for hydrogen since it has no quantum defects and the energies are degenerate. In this case the zero-field problem may be solved using a basis which diagonalizes the Hamiltonian both for the atom above and also in the presence of an electric field. This approach corresponds to solving the H atom in parabolic–ellipsoidal coordinates and results in the presence of an integral quantum number which replaces $\ell$. The resulting states possess permanent dipole moments which vary with this quantum number and therefore have linear Stark effects even in infinitesimal fields. Moreover the matrix elements which mix states from different $n$ manifolds vanish at all fields, so the upper energy levels from one manifold cross the lower energy levels from the manifold above without interacting with them.\

The following example shows the high field stark effect for Li. Only the $S$ term in Li has an appreciable quantum defect, and it has been suppressed by selecting final states with $m_{\ell }=1$.

The dramatic difference between the physical properties of atoms with $n>10$ and the properties of the same atoms in their ground state, coupled with the fact that these properties are largely independent of the type of atom which is excited, justifies the application of the name Rydberg atoms to highly excited atoms in general.\

File:06-E-FIELD/Stark pattern.eps

Stark effect and field ionization in Li for levels with $m=1$. Each vertical line represents a measurement at that field of the number of atoms excited (from the $3s$ state) by radiation whose energy falls the indicated amount below the ionization limit. Thus the patterns made by absorption peaks at successive field strengths represent the behavior of the energy levels with increasing field. At zero field the levels group according to the principal quantum number $n$; at intermediate field the levels display a roughly linear Stark effect, and at high fields they disappear owing to field ionization. The solid line is the classically predicted ionization field (see next section). Figure taken from Littman, Kash and Kleppner.

Field ionization

If an atom is placed in a sufficiently high electric field it will be ionized, a process called <it> field ionization </it>. An excellent order of magnitude estimate of the field $\epsilon _{\rm {ion}}$, required to ionize an atom which is initially in a level bound by energy $E$ can be obtained by the following purely classical argument: the presence of the field adds the term $U(z)=e\epsilon z$ to the potential energy of the atom. This produces a potential with a maximum $U_{\rm {max}}<0$ and the atom will ionize if $U_{\rm {max}}<-E$.

File:06-E-FIELD/Field ion.eps

Potential diagram for field ionization.

The figure shows the combined potential as well as $U_{\rm {atom}}$ and $U_{\rm {field}}$

${\begin{aligned}\ U_{\rm {total}}(z)=U_{\rm {atom}}(z)+U_{\rm {field}}(z)={\frac {-e^{2}}{|z|}}+e\epsilon z\end{aligned}}$

(EQ_ fieldionone)

The appropriate maximum occurs at

${\begin{aligned}\ z_{\rm {max}}=-{\left({\frac {e}{\epsilon }}\right)^{1/2}}\end{aligned}}$

(EQ_ fieldiontwo)

as determined from $dU/dz=0$. Equating $U_{\rm {max}}$ and $-E$ gives

${\begin{aligned}\ \epsilon _{ion}={\frac {E^{2}}{4e^{2}}}={\frac {1}{16n^{*4}}}=3.2\times 10^{8}(n^{*})^{-4}\ {\rm {V/cm}}\end{aligned}}$

(EQ_ fieldionthree)

for level with energy $-E$ and quantum number $n^{*}$.

The predictions of this formula for $E_{\rm {ion}}$ is usually accurate within 20% in spite of its neglect of both quantum tunneling and the change in $E$ produced by the field. [This latter deficiency is remedied in the comparison with Li data shown in the preceding part of this section because the eye naturally uses the ionization field appropriate to the perturbed energy of the state rather than its zero-field energy.] Tunneling manifests itself as a finite decay rate for states which classically lie lower than the barrier. The increase of the ionization rate with field is so dramatic, however, that the details of the experiment do not influence the field at which ionization occurs very much: calculations [u'BHR65'] show the ionization rate increasing from $10^{5}$/sec to $10^{10}$/sec for a 30% increase in the field.\

Oddly enough the classical prediction works worse for H than for any other atom. This is a reflection of the fact that certain matrix elements necessary to mix the $n,\ell$ states (so the wave function samples the region near $U_{\rm {min}}$) are rigorously zero in H, as discussed in the preceding part of this section. Hence the orbital ellipse of the electron does not precess and can remain on the side of the nucleus. There its energy will increase with $\epsilon$, but it will not spill over the lip of the potential and ionize.