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3.3: Example of degenerate perturbation theory - Stark Effect in Hydrogen

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    The change in energy levels in an atom due to an external electric field is known as the Stark effect. The perturbing potential is thus \(\hat{V} = eEz = eEr \cos \theta\). Ignoring spin, we examine this effect on the fourfold degenerate n=2 levels. We will label these by their appropriate quantum number: \(|l, m \rangle\).

    \[u_{00} = (8 \pi a^3_0 )^{−1/2} (1 − r/2a_0)e^{−r/2a_0} ; \quad u_{10} = (8 \pi a^3_0 )^{−1/2} (r/2a_0) \cos \theta e^{−r/2a_0} \nonumber\]

    \[u_{11} = ( \pi a^3_0 )^{−1/2} (r/8a_0) \sin \theta e^{i \phi} e^{ −r/2a_0} \quad u_{1−1} = ( \pi a^3_0 )^{−1/2} (r/8a_0) \sin \theta e^{−i \phi} e^{−r/2a_0} \nonumber\]

    From the analysis above, we need to calculate the matrix elements.

    \[V_{lm,l'm'} = \langle l, m|eEz|l' , m' \rangle = eE \int \int \int u^*_{lm} (r \cos \theta ) u_{l' m'} r^2 \sin \theta d \theta d \phi dr \nonumber\]

    It turns out that many of these are zero, since if any of the integrals are zero their product will be. Looking first at parity, it is clear that \(eEz\) has odd parity \((eE(r) \cos( \pi − \theta ) = −eEr \cos \theta )\), \(u_{00}\) has even parity and \(u_{1m}\) have odd parity. Since the integral over all space of any odd function is zero, \(V_{00,00} = V_{1m,1m0} = 0\). Secondly, \(\int^{2\pi}_0 e^{ \pm i \phi} d \phi = 0\), so \(V_{00,11} = V_{00,1−1} = V_{11,00} = V_{1−1,00} = 0\).

    Since the perturbation is real, \(V_{00,10} = V_{10,00}\) and the only remaining non-zero matrix element is:

    \[\langle 00|eEr \cos \theta |10 \rangle = (8 \pi a^3_0 )^{−1} \int^{2 \pi}_0 d \phi \int^{\pi}_0 \cos^2 \theta \sin \theta d \theta \int^{\infty}_0 (1 − r/2a_0)e^{−r/a_0} r^4 /2a_0dr = −3eEa_0 \nonumber\]

    This is best solved as a matrix problem, the determinantal equation is then:

    \[\begin{vmatrix} − \Delta E & −3eEa_0 & 0 & 0 \\ −3eEa_0 & − \Delta E & 0 & 0 \\ 0 & 0 & − \Delta E & 0 \\ 0 & 0 & 0 & − \Delta E \end{vmatrix} = ( \Delta E)^4 − ( \Delta E)^2 (3eEa_0)^2 = 0 \nonumber\]

    The solutions to this are \(\Delta E = \pm 3eEa_0\), 0, 0. The degeneracy of the states \(u_{11}\) and \(u_{1−1}\) is not lifted, but the new non-degenerate eigenstates corresponding to \(\Delta E_n = \pm 3eEa_0\) are mixtures, \((u_{00} \mp u_{10})/ \sqrt{2}\). Consequently, the spectral line corresponding to the \(n = 2 \rightarrow n = 1\) Lyman-\(\alpha\) transition is split into three if the hydrogen atom is in an electric field.

    A curious aspect of these eigenstates is that they are not eigenstates of \({\bf L^2}\), although they are eigenstates of \(L_z\). Nor do they have definite parity. In an electric field, therefore, the total angular momentum is not a good quantum number. Note that this effect is specific to hydrogen, since in other elements the \(s\) and \(p\) levels are not degenerate.

    Experimental results confirm this theory beautifully - the splitting of levels in hydrogen varies linearly with the applied field strength, while in all other atoms it varies quadratically: the first order perturbation is zero.

    Looking at the electrostatics: the energy of a spherically symmetric charge density in a uniform field is clearly independent of orientation. To have any orientation dependence the object must have a dipole moment. The combination of 2s and 2p wavefunctions achieves this.

    This page titled 3.3: Example of degenerate perturbation theory - Stark Effect in Hydrogen is shared under a CC BY 4.0 license and was authored, remixed, and/or curated by Graeme Ackland via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.

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