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## 物理代写|量子力学代写Quantum mechanics代考|PHY402LEC Time-Independent Perturbation Theory: The Degenerate Case

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## 物理代写|量子力学代写Quantum mechanics代考|Time-Independent Perturbation Theory: The Degenerate Case

The perturbation method we developed in the previous section fails when the unperturbed energy eigenkets are degenerate. The method of the previous section assumes that there is a unique and well-defined unperturbed ket of energy $E_{n}^{(0)}$ which the perturbed ket approaches as $\lambda \rightarrow 0$. With degeneracy present, however, any linear combination of unperturbed kets has the same unperturbed energy; in such a case it is not a priori obvious to what linear combination of the unperturbed kets the perturbed ket is reduced in the limit $\lambda \rightarrow 0$. Here specifying just the energy eigenvalue is not enough; some other observable is needed to complete the picture. To be more specific, with degeneracy we can take as our base kets simultaneous eigenkets of $H_{0}$ and some other observable $A$, and we can continue labeling the unperturbed energy eigenket by $\left|k^{(0)}\right\rangle$, where $k$ now symbolizes a collective index that stands for both the energy eigenvalue and the $A$ eigenvalue. When the perturbation operator $V$ does not commute with $A$, the zeroth-order eigenkets for $H$ (including the perturbation) are in fact not $A$ eigenkets.

From a more practical point of view, a blind application of formulas like (5.42) and (5.44) obviously runs into difficulty because
$$\frac{V_{n k}}{E_{n}^{(0)}-E_{k}^{(0)}}$$
becomes singular if $V_{n k}$ is nonvanishing and $E_{n}^{(0)}$ and $E_{k}^{(0)}$ are equal. We must modify the method of the previous section to accommodate such a situation.

Whenever there is degeneracy we are free to choose our base set of unperturbed kets. We should, by all means, exploit this freedom. Intuitively we suspect that the catastrophe of vanishing denominators may be avoided by choosing our base kets in such a way that $V$ has no off-diagonal matrix elements [such as $V_{n k}=0$ in (5.74)]. In other words, we should use the linear combinations of the degenerate unperturbed kets that diagonalize $H$ in the subspace spanned by the degenerate unperturbed kets. This is indeed the correct procedure to use.

## 物理代写|量子力学代写Quantum mechanics代考|Linear Stark Effect

As an example of degenerate perturbation theory, let us study the effect of a uniform electric field on excited states of the hydrogen atom. As is well known, in the Schrödinger theory with a pure Coulomb potential with no spin dependence, the bound-state energy of the hydrogen atom depends only on the principal quantum number $n$. This leads to degeneracy for all but the ground state because the allowed values of $l$ for a given $n$ satisfy
$$0 \leq l<n .$$

To be specific, for the $n=2$ level, there is an $l=0$ state called $2 s$ and three $l=1(m=\pm 1,0)$ states called $2 p$, all with the same energy, $-e^{2} / 8 a_{0}$. As we apply a uniform electric field in the $z$-direction, the appropriate perturbation operator is given by
$$V=-e z|\mathbf{E}|,$$
which we must now diagonalize. Before we evaluate the matrix elements in detail using the usual $(\mathrm{nlm})$ basis, let us note that the perturbation (5.90) has nonvanishing matrix elements only between states of opposite parity, that is, between $l=1$ and $l=0$ in our case. Furthermore, in order for the matrix element to be nonvanishing, the $m$-values must be the same because $z$ behaves like a spherical tensor of rank one with spherical component (magnetic quantum number) zero. So the only nonvanishing matrix elements are between $2 s$ ( $m=0$ necessarily) and $2 p$ with $m=0$. Thus
Explicitly,
\begin{aligned} \langle 2 s|V| 2 p, m=0\rangle &=\langle 2 p, m=0|V| 2 s\rangle \ &=3 e a_{0}|\mathbf{E}| \end{aligned}

## 物理代写|量子力学代写Quantum mechanics代考|Time-Independent Perturbation Theory: The Degenerate Case

$$\frac{V_{n k}}{E_{n}^{(0)}-E_{k}^{(0)}}$$

## 物理代写|量子力学代写Quantum mechanics代考|Linear Stark Effect

$$0 \leq l<n .$$

$$V=-e z|\mathbf{E}|$$

$$\langle 2 s|V| 2 p, m=0\rangle=\langle 2 p, m=0|V| 2 s\rangle \quad=3 e a_{0}|\mathbf{E}|$$

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