Atoms in electric fields

This section deals with how atoms behave in static electric fields. The method is straightforward, involving second order perturbation theory. The treatment describes the effects of symmetry on the basic interaction, polarizability, and the concept of oscillator strength.

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► Category:Atoms in electric fields

  Atoms

  Atoms in electric fields

  Atoms in Magnetic Fields

  Coherence

  Effects of the Nucleus on Atomic Structure

  Fine Structure and Lamb Shift

  Higher-Order Radiation Processes

  Interaction of an atom with an electromagnetic field

  Old Coherence

  Resonances

  Two-Photon Excitation

The Hydrogen Atom in a Static Electric Field

We first 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.

Perturbation Theory of Polarizability

We will find the energy and polarizability of an atom in a static field along the +z direction. We apply perturbation theory taking $H_{0}$ to describe the unperturbed atomic system and

$H^{\prime }=-d\cdot {\hat {z}}{\mathcal {E}}=ez{\mathcal {E}}$

Non-degenerate eigenstates have to be eigenstates of parity. Since $H^{\prime }$ is odd under parity operation, parity requires that $H_{mm}^{\prime }=0$. So the first order perturbation vanishes. To second order, the energy is given by

$E_{n}=E_{n}^{(0)}-e^{2}{\mathcal {E}}^{2}\sum '_{m}{\frac {|\langle m|z|n\rangle |^{2}}{E_{m}-E_{n}}}$

If we define now the polarizability in state $n$ as

{EQ_polarsix}

$\alpha _{n}=2e^{2}\sum '_{m}{\frac {|\langle m|z|n{\rangle }|^{2}}{E_{m}-E_{n}}}$

we obtain $E_{n}-E_{n}^{(0)}=\alpha {\mathcal {E}}^{2}/2$

The dipole moment is the expectation value of the dipole operator, using the first order perturbed state vector.

$d_{nm}=(\langle n^{(0)}|+\langle n^{(1)}|)\;{\bf {d}}\;(|n^{(0)}\rangle +|n^{(1)}\rangle )$

$=2{\rm {Re}}[\langle n^{(0)}|d|n^{(1)}\rangle ]=2e^{2}{\rm {Re}}{\left[\sum _{s,m}{\frac {\langle n^{(0)}|s|m\rangle \langle m|z|n^{(0)}\rangle }{E_{m}-E_{n}}}\right]}{\hat {s}}\cdot {\hat {z}}{\mathcal {E}}$

where the sum is over $s=x,y,z.$ Only the term $s=z$ will contribute, and we can express the induced dipole moment by the polarizability:

$d=\alpha {\mathcal {E}}{\hat {z}}$

Note that the Stark shift is $E_{n}-E_{n}^{(0)}=-\langle d\rangle {\mathcal {E}}/2$ and not equal to $\langle H'\rangle =-\langle d\rangle {\mathcal {E}}$. $\langle H'\rangle$ is the expectation value for the electrostatic potential energy of the dipole moment, but the total energy change is only one half of this since energy is needed to admix excited states into the ground state.

Note that polarizability has the dimensions of length$^{3}$, i.e. volume. As an example, for the ground state of hydrogen we can obtain a lower limit for the polarizability by considering only the contribution to the sum of the $2P$ state. Values for the various moments in hydrogen are given in Bethe and Salpeter, Section 63. Using $|\langle 2P|r|1S\rangle |^{2}$ = 1.666, and $E_{2p}-E_{1S}=3/8$, we obtain $\alpha =2.96$ atomic units (i.e. $2.96\;a_{0}^{3}$).

The polarizability of the ground state of hydrogen can be calculated exactly. It turns out that the $2P$ state makes the major contribution, and that the higher bound states contribute relatively little. However, the continuum makes a significant contributions. The exact value is 4.5.

To put the above result for the polarizability in perspective, note that the potential of a conducting sphere of radius $R$ in a uniform electric field ${\mathcal {E}}$ is given by

$V(r,\theta )=-{\mathcal {E}}\cos \theta \left(r-{\frac {R^{3}}{r^{2}}}\right)\ (r\geq R)$

The induced dipole moment is $R^{3}{\mathcal {E}}$, so that the polarizability is $R^{3}$. For the ground state of hydrogen, ${\bar {r}}^{3}=2.75$, so to a crude approximation, in an electric field hydrogen behaves like a conducting sphere.

Polarizability may be approximated easily, though not accurately, using Unsold's approximation in which the energy term in the denominator of Eq. \ref{EQ_polarsix} is replaced by an average energy interval ${\overline {E_{m}}}-E_{n}$. The sum can then be evaluated using the closure rule $\sum _{m}|m\rangle \langle m|=1$. (Note that the term $m=n$ does not need to be excluded from the sum, since $\langle n|z|n\rangle =0$.). With this approximation,

$a_{n}={\frac {2e^{2}}{{\overline {E_{m}}}-E_{n}}}\sum _{m}\langle n|z|m\rangle \langle m|z|n{\rangle }={\frac {2e^{2}\langle n|z^{2}|n{\rangle }}{{\overline {E_{m}}}-E_{n}}}$

For hydrogen in the ground state, ${\overline {z^{2}}}=1$. If we take the average excitation energy to be ${\overline {E_{m}}}=0$, the result is $\alpha =4$.

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.

Atoms in an Oscillating Electric Field

There is a close connection between the behavior of an atom in a static electric field and its response to an oscillating field, i.e. a connection between the Stark effect and radiation processes. In the former case, the field induces a static dipole moment; in the latter case, it induces an oscillating moment. An oscillating moment creates an oscillating macroscopic polarization and leads to the absorption and emission of radiation. We shall calculate the response of an atom to an oscillating field

${\mathcal {E}}(\omega ,t){\hat {e}}={\mathcal {E}}{\hat {e}}\cos \omega t$

where ${\hat {e}}$ is the polarization vector for the field. For a weak field the time varying state of this system can be found from first order time dependent perturbation theory. We shall write the electric dipole operator as D = -er. (This is a change of notation. Previously the symbol was d.) The Hamiltonian naturally separates into two parts, $H=H_{0}+H^{\prime }(t)$, where $H_{0}$ is the unperturbed Hamiltonian and

$H^{\prime }=-D\cdot {\hat {e}}{\mathcal {E}}\cos \omega t=-{\frac {1}{2}}(e^{i\omega t}+e^{-i\omega t}){\mathcal {E}}{\hat {e}}\cdot D$

We shall express the solution of the time dependent Schroedinger equation in terms of the eigenstates of $H_{0},|n\rangle$.

$|\psi \rangle =\sum _{n}a_{n}e^{-i\omega _{n}t}|n\rangle ~~~H_{0}|\psi \rangle =\hbar \sum _{n}a_{n}\omega _{n}e^{-i\omega _{n}t}|n\rangle$

where $\omega _{n}=E_{n}/\hbar$. Because of the perturbation $H^{\prime }(t)$, the $a_{n}$'s become time dependent, and we have

$i\hbar {\frac {|\partial \psi \rangle }{\partial t}}={\left(H_{0}+H^{\prime }\right)}\sum _{n}a_{n}e^{-i\omega _{n}t}|n\rangle =\sum _{n}\hbar {\left(a_{n}\omega _{n}+{\dot {a}}_{n}\right)}|n\rangle e^{-i\omega _{n}t}$

Left multiplying the final two expressions by $\langle k|$ to project out the $k$-th terms yields

${\dot {a}}_{k}=(i\hbar )^{-1}\sum _{n}\langle k|H^{\prime }(t)|n\rangle a_{n}e^{i\omega _{kn}t}$

where $\omega _{kn}=\omega _{k}-\omega _{n}$. In perturbation theory, this set of equations is solved by a set of approximations to $a_{k}$ labeled $a_{k}^{(i)}(t)$. Starting with

$a_{n}^{(0)}(t)=a_{n}(0)$

one sets

${\dot {a}}_{k}^{(i+1)}(t)=(i\hbar )^{-1}\sum _{n}\langle k|H^{\prime }(t)|n\rangle a_{n}^{(i)}(t)e^{i\omega _{kn}t}$

and solves for the successive approximations by integration.

We now apply this to the problem of an atom which is in its ground state $g$ at $t=0$, and which is subject to the interaction of Eq.\ \ref{EQ_atomoef2}. Consequently $a_{g}(0)=1$, $a_{n\not =g}(0)=0$. Substituting in Eq.\ \ref{EQ_atomoef7} and integrating from $t^{\prime }=0$ to $t$ gives

${\begin{array}{rcl}a_{k}^{(1)}(t)&=&(i\hbar )^{-1}\int _{0}^{t}dt^{\prime }\langle k|H^{\prime }(t^{\prime })|g\rangle e^{i\omega _{kg}t^{\prime }}\\&=&-(i\hbar )^{-1}\langle k|{\hat {e}}\cdot D|g\rangle {\frac {\mathcal {E}}{2}}\int _{0}^{t}dt^{\prime }{\left[e^{i(\omega _{kg}+\omega )t^{\prime }}+e^{i(\omega _{kg}-\omega )t^{\prime }}\right]}\\&=&{\frac {\mathcal {E}}{2\hbar }}\langle k|{\hat {e}}\cdot D|g\rangle {\left[{\frac {e^{i(\omega _{kg}+\omega )t}-1}{\omega _{kg}+\omega }}+{\frac {e^{i(\omega _{kg}-\omega )t}-1}{\omega _{kg}-\omega }}\right]}\end{array}}$

The -1 terms in the square bracketed term arises because it is assumed that the field was turned on instantaneously at $t=0$. They represent transients that rapidly damp and can be neglected.

The term with $\omega _{kg}+\omega$, in the denominator is the counter-rotating term. It can be neglected if one is considering cases where $\omega \approx \omega _{kg}$ (i.e. near resonance), but we shall retain both terms and calculate the expectation value of the first order time dependent dipole operator $\langle D(\omega ,t)\rangle$

${\begin{array}{rcl}\langle D(\omega ,t)\rangle &=&2{\rm {Re}}{\left\{\langle g|{\bf {D}}|\sum _{k}a_{k}^{(1)}(t)e^{-i\omega _{kg}}|k\rangle \right\}}\\&=&{\mathcal {E}}{\rm {Re}}{\left[\sum _{k}{\frac {\langle g|D|k\rangle \langle k|{\hat {e}}\cdot D|g\rangle }{\hbar }}{\left\{{\frac {e^{i\omega t}}{\omega _{kg}+\omega }}+{\frac {e^{-i\omega t}}{\omega _{kg}-\omega }}\right\}}\right]}\end{array}}$

If we consider the case of linearly polarized light $({\hat {e}}={\hat {z}})$, then

$d_{z}(\omega ,t)={\frac {2e^{2}}{\hbar }}\sum _{k}{\frac {\omega _{kg}|\langle k|z|g\rangle |^{2}}{\omega _{kg}^{2}-\omega ^{2}}}{\mathcal {E}}\cos \omega t$

We can write $d_{z}$ in terms of a polarizability $\alpha (\omega )$:

$\alpha (\omega )={\frac {2e^{2}}{\hbar }}\sum _{k}{\frac {\omega _{kg}|\langle k|z|g\rangle |^{2}}{\omega _{kg}^{2}-\omega ^{2}}}$

This result diverges if $\omega \rightarrow \omega _{kg}$. Later, when we introduce radiative damping, the divergence will be avoided in the usual way.

Oscillator Strength

Eq.\ \ref{EQ_atomoef11} resemble the oscillating dipole moment of a system of classical oscillators. Consider a set of oscillators having charge $q_{k}$, mass $m$, and natural frequency $\omega _{k}$, driven by the field ${\mathcal {E}}\cos \omega t$. The amplitude of the motion is given by

$z_{k}={\frac {q_{k}}{m(\omega _{k}^{2}-\omega ^{2})}}{\mathcal {E}}\cos \omega t$

If we have a set of such oscillators, then the total oscillating moment is given by

$d_{z}(\omega ,t)={\frac {1}{m}}\sum _{k}{\frac {q_{k}^{2}}{(\omega _{k}^{2}-\omega ^{2})}}{\mathcal {E}}\cos \omega t$

This is strongly reminiscent of Eq.\ \ref{EQ_atomoef10}. It is useful to introduce the concept of oscillator strength, a dimensionless quantity defined as

$f_{kj}={\frac {2m}{\hbar }}\omega _{kj}|\langle k|z|j\rangle |^{2}$

where $k$ and $j$ are any two eigenstates. Note that $f_{kj}$ is positive if $E_{k}>E_{j}$, i.e. for absoprtion, and negative if $E_{k}<E_{j}$ Then, Eq.\ \ref{EQ_atomoef10} becomes

$d_{z}(t)=\sum _{k}f_{kg}{\frac {e^{2}}{m(\omega _{kg}^{2}-\omega ^{2})}}{\mathcal {E}}\cos \omega t$

Comparing this with Eq.\ \ref{EQ_ostre2}, we see that the behavior of an atom in an oscillating field mimics a set of classical oscillators with the same frequencies as the eigenfrequencies of the atom, but having effective charge strengths $q_{k}^{2}=f_{kg}e^{2}$.\\

The oscillator strength is useful for characterizing radiative interactions and also the susceptibiltiy of atoms. It satisfies an important sum rule, the Thomas-Reiche-Kuhn sum rule:

$\sum _{k}f_{kg}=1$

We prove by considering the general Hamiltonian

$H={\frac {1}{2}}\sum _{j}p_{j}^{2}+V(r_{1},r_{2}\cdots ).$

Using the commutator relation

$[A,B^{2}]=[A,B]B+B[A,B],$

and the relation $[r_{j},p_{k}]=i\hbar \delta _{jk}$, we have

$[r,H]={\frac {i\hbar }{m}}p$

where $r=\sum r_{j}$, and $p=\sum p_{j}$. However,

$\langle j|[r,H]|k\rangle =(E_{k}-E_{n})\langle j|r|k\rangle$

Consequently,

$\langle j|r|k\rangle ={\frac {i}{m}}{\frac {\langle j|p|k\rangle }{\omega _{kj}}}$

where $\omega _{kj}=(E_{k}-E_{j})/\hbar$. Thus, we can write Eq.\ \ref{EQ_ostre3} in either of two forms:

$f_{kj}={\frac {2i}{\hbar }}\langle j|p_{z}|k\rangle \langle |k|z|j\rangle =-{\frac {2i}{\hbar }}\langle k|p_{z}|j\rangle \langle |j|z|k\rangle$

Taking half the sum of these equations and using the closure relation $\sum _{k}|k\rangle \langle k|=1$, we have

$\sum _{k}f_{kj}={\frac {i}{\hbar }}{\left[\langle j|p_{z}z-zp_{z}|j\rangle \right]}=1$

We have calculated this for a one-electron atom, but the application to a Z-electron atom is straightforward because the Hamiltonian in Eq.\ \ref{EQ_ostre6} is quite general. In this case

$\sum _{k}f_{kj}=Z.$

Here $j$ is some eigenstate of the system, and the index $k$ describes all the eigenstates of all the electrons -- including continuum states. In cases where only a single electron will be excited, however, for instance in the optical regime of a "single-electron" atom where the inner core electrons are essentially unaffected by the radiation, the atom behaves as if it were a single electron system with $Z=1$.

Note that $f_{kj}$ is positive if $\omega _{kj}>0$, i.e. if the final state lies above the initial state. Such a transition corresponds to absorption of a photon. Since $f_{jk}=-f_{jk}$, the oscillator strength for emission of a photon is negative.

Our definition of oscillator strength, Eq.\ \ref{EQ_ostre3}, singles out a particular axis, the ${\hat {z}}$-axis, fixed by the polarization of the light. Consequently, it depends on the orientation of the atom in the initial state and final states. It is convenient to introduce the average oscillator strength (often simply called the oscillator strength), by letting $|z_{kj}|^{2}\rightarrow |r_{kj}|^{2}/3$, summing over the initial $m$ state and averaging over the final state.\\

${\overline {f_{kj}}}={\frac {2}{3}}{\frac {m}{\hbar }}\omega _{kj}{\frac {1}{2J_{j}+1}}\sum _{m,m^{\prime }}|\langle j,J_{j},m^{\prime }|r|k,J_{k},m\rangle |^{2}$

(This is the conversion followed by Sobelman.) It is evident that

${\overline {f_{jk}}}=-{\frac {2J_{j}+1}{2J_{k}+1}}{\overline {f_{kj}}}=-{\frac {g_{j}}{g_{k}}}{\overline {f_{kj}}},$

where $g_{j}$ is the multiplicity factor for state $j$. An extensive discussion of the sum rules and their applications to oscillator strengths and transition momentums can be found in Bethe and Salpeter, section 6.1. Among the interesting features they point out is that transitions from an initial state $|n,\ell \rangle$ to a final state $|n^{\prime },\ell ^{\prime }\rangle$ on the average have stronger oscillator strengths for absorption if $\ell ^{\prime }>\ell$, and stronger oscillator strengths for emission if $\ell ^{\prime }<\ell$. In other words, atoms "like" to increase their angular momentum on absorption of a photon, and decrease it on emission. The following page gives a table of oscillator strengths for hydrogen in which this tendency can be readily identified. (Taken from {\it The Quantum Mechanics of One- and Two-Electron Atoms}, H.A. Bethe and E.E. Salpeter, Academic Press (1957).)

\caption{ Oscillator strengths for hydrogen. From Mechanics of One- and Two-Electron Atoms}

Index of refraction

(in preparation)