Approximation Methods

---

4.1 Time-Independent Perturbation Theory

All the perturbation theory methods are based on the idea of this kind of hamiltonian: \[ H(\lambda) = H^{(0)} + \lambda \delta H, \lambda \in [0, 1] \]

why \(\lambda\)?

• \(\lambda = 0\): \(H^{(0)}\)
• \(\lambda = 1\): \(H^{(0)} + \delta H\)

so we can mark the rank of the perturbation as \(\lambda\)

Non-degenerate Perturbation Theory

Formal Expansion

Assume the original Schrodinger equation has a solution of the form: \[ H^{(0)} \ket{n^{(0)}} = E_n^{(0)} \ket{n^{(0)}} \] where \(n = 1, 2, 3, \ldots\) represents the energy eigenstates and eigenvalues of the unperturbed system.

So we have the perturbed Schrodinger equation: \[ (H_0 + \lambda V) \ket{n}_\lambda = E_n^{(\lambda)} \ket{n}_\lambda \]

Define the scalar \(\Delta_n \equiv E_n - E_n^{(0)}\), so we have \[ (E_n^{(0)} - H_0) \ket{n} = (E_n^{(0)} - (H_0 + \lambda V) + \lambda V) \ket{n} = (E_n^{(0)} + \lambda V - E_n) \ket{n} = (\lambda V - \Delta_n) \ket{n} \]

Because of non-degeneracy, we have \[ \bra{n^{(0)}} (\lambda V - \Delta_n) \ket{n} = 0 \]

So we have: \[ \boxed{\Delta_n = \lambda \bra{n^{(0)}} V \ket{n}} \]

Define the complementary projection operator: \[ \phi_n \equiv \mathbf{1} - \ket{n^{(0)}}\bra{n^{(0)}} = \sum_{k \ne n} \ket{k^{(0)}}\bra{k^{(0)}} \] where \[ \frac{1}{E_n^{(0)} - H_0} \phi_n = \sum_{k \ne n} \frac{\ket{k^{(0)}}\bra{k^{(0)}}}{E_n^{(0)} - E_k^{(0)}} \]

So we have: \[ \phi_n (\lambda V - \Delta_n) \ket{n} = (\lambda V - \Delta_n) \ket{n} \]

Substituting the expression for \(\frac{1}{E_n^{(0)} - H_0} \phi_n\): \[ \ket{n} \sim{} \frac{1}{E_n^{(0)} - H_0} \phi_n (\lambda V - \Delta_n) \ket{n} \]

Add special solution \(\ket{n^{(0)}}\): \[ \boxed{\ket{n} = \ket{n^{(0)}} + \frac{1}{E_n^{(0)} - H_0} \phi_n (\lambda V - \Delta_n) \ket{n}} \]

Method of Undetermined Coefficients

Given \[ H^{(0)}, \ket{k^{0}}, k = 0, 1, \cdots, \bra{k^{0}}\ket{l^{0}} = \delta_{kl} \] \[ H^{(0)} \ket{k^{0}} = E_{k}^{0} \ket{k^{0}}, E_0^{(0)} < E_1^{(0)} < \cdots \]

Non-degeneracy means that each energy level is distinct and there is no degeneracy in the unperturbed system.

Assume that \(\ket{n^{0}}\) is not degenerate, i.e. \(E_{n-1}^{(0)} < E_{n}^{(0)} < E_{n+1}^{(0)}\).

We want to solve: \[ H(\lambda) \ket{n}_\lambda = E_n(\lambda) \ket{n}_\lambda \]

Assume an expansion (method of undetermined coefficients): \[ E_n(\lambda) = E_n^{(0)} + \lambda E_n^{(1)} + \lambda^2 E_n^{(2)} + \cdots \] \[ \ket{n}_\lambda = \ket{n^{(0)}} + \lambda \ket{n^{(1)}} + \lambda^2 \ket{n^{(2)}} + \cdots \]

Substitute into the original Schrodinger equation: \[ \left( H^{(0)} + \lambda \delta H - E_n(\lambda) \right) \ket{n}_\lambda = 0 \]

Expand the left hand side by \(\lambda\): \[ \left\{ \begin{aligned} \lambda^0 : & \left( H^{(0)} - E_n^{(0)} \right) \ket{n^{(0)}} = 0 \\ \lambda^1 : & \left( H^{(0)} - E_n^{(0)} \right) \ket{n^{(1)}} = \left( E_n^{(1)} - \delta H \right) \ket{n^{(0)}} \\ \lambda^2 : & \left( H^{(0)} - E_n^{(0)} \right) \ket{n^{(2)}} = \left( E_n^{(1)} - \delta H \right) \ket{n^{(1)}} + E_n^{(2)} \ket{n^{(0)}} \\ \vdots \\ \lambda^k : & \left( H^{(0)} - E_n^{(0)} \right) \ket{n^{(k)}} = \left( E_n^{(1)} - \delta H \right) \ket{n^{(k-1)}} + E_n^{(2)} \ket{n^{(k-2)}} + \cdots + E_n^{(k)} \ket{n^{(0)}} \end{aligned} \right. \]

1st Order Non-degenerate Case

From the first-order equation: \[ \left( H^{(0)} - E_n^{(0)} \right) \ket{n^{(1)}} = \left( E_n^{(1)} - \delta H \right) \ket{n^{(0)}} \]

Left multiply by \(\bra{n^{(0)}}\): \[ \bra{n^{(0)}} \left( H^{(0)} - E_n^{(0)} \right) \ket{n^{(1)}} = \bra{n^{(0)}} \left( E_n^{(1)} - \delta H \right) \ket{n^{(0)}} \]

Because of \((H^{(0)} - E_n^{(0)})\ket{n^{(0)}} = 0\), We get the first-order energy correction: \[ \boxed{ E_n^{(1)} = E_n^{(1)} \bra{n^{(0)}} \ket{n^{(0)}} = \bra{n^{(0)}} E_n^{(1)} \ket{n^{(0)}} = \bra{n^{(0)}} \delta H \ket{n^{(0)}} } \]

Substituting back into the first-order equation:

\[ \ket{n^{(1)}} = \frac{E_n^{(1)} - \delta H}{H^{(0)} - E_n^{(0)}} \ket{n^{(0)}} \]

We can’t directly get the fraction term because it involves the inverse operator \((H^{(0)} - E_n^{(0)})^{-1}\), which is not easily computable in general. So we need to expand by \(\ket{k^{(0)}}\) basis:

\[ \begin{aligned} \ket{n^{(1)}} &= \mathbf{1} \cdot \frac{E_n^{(1)} - \delta H}{H^{(0)} - E_n^{(0)}} \ket{n^{(0)}} \\ &= \sum_{k = 1, 2, ...} \ket{k^{(0)}} \bra{k^{(0)}} \cdot \frac{E_n^{(1)} - \delta H}{H^{(0)} - E_n^{(0)}} \ket{n^{(0)}} \end{aligned} \]

Calculate the inner product term: \[ \bra{k^{(0)}} \frac{E_n^{(1)} - \delta H}{H^{(0)} - E_n^{(0)}} \ket{n^{(0)}} = \begin{cases} 0 & \text{if } k = n \\ \frac{\bra{k^{(0)}} \delta H \ket{n^{(0)}}}{E_n^{(0)} - E_k^{(0)}} & \text{if } k \neq n \end{cases} \]

The derivation here is not rigorous. We cannot directly divide through; instead, we should first left-multiply by \(\bra{k^{(0)}}\) to obtain the inner product expression.

Finally we derive the full expression for the first-order state correction:

\[ \boxed{ \ket{n^{(1)}} = \sum_{k \neq n} \frac{\bra{k^{(0)}} \delta H \ket{n^{(0)}}}{E_n^{(0)} - E_k^{(0)}} \ket{k^{(0)}} } \]

2nd Order Non-degenerate Case

At first we calculate the energy correction \(E_n^{(2)}\). By

\[ \left( H^{(0)} - E_n^{(0)} \right) \ket{n^{(2)}} = \left( E_n^{(1)} - \delta H \right) \ket{n^{(1)}} + E_n^{(2)} \ket{n^{(0)}} \]

We get: \[ \begin{aligned} E_n^{(2)} &= \bra{n^{(0)}} \delta H \ket{n^{(1)}} \\ &= \sum_{k \neq n} \frac{\bra{n^{(0)}} \delta H \ket{k^{(0)}} \bra{k^{(0)}} \delta H \ket{n^{(0)}}}{E_n^{(0)} - E_k^{(0)}} \\ &= \sum_{k \neq n} \frac{|\bra{n^{(0)}} \delta H \ket{k^{(0)}}|^2}{E_n^{(0)} - E_k^{(0)}} \end{aligned} \]

NoteDiscussion about Non-degenerate Perturbation Theory

1. Base Energy Correction: The first order \(\bra{n^{(0)}} \delta H \ket{n^{(0)}}\) always increases the energy, while the second order correction \(-\sum_{k \neq 0} \frac{|\bra{0^{(0)}} \delta H \ket{k^{(0)}}|^2}{E_0^{(0)} - E_k^{(0)}}\) always decreases the energy.

2. Level Repulsion (also called avoided crossing or level anti-crossing):

   Consider two unperturbed levels \(E_n^{(0)}\) and \(E_k^{(0)}\) with \(E_n^{(0)} < E_k^{(0)}\). The second-order correction to \(E_n\) from the coupling to \(\ket{k^{(0)}}\) is:

   \[ E_n^{(2)} \supset \frac{|\bra{n^{(0)}} \delta H \ket{k^{(0)}}|^2}{E_n^{(0)} - E_k^{(0)}} < 0 \quad (\text{pushes } E_n \text{ down}) \]

   while the correction to \(E_k\) from coupling to \(\ket{n^{(0)}}\) is:

   \[ E_k^{(2)} \supset \frac{|\bra{k^{(0)}} \delta H \ket{n^{(0)}}|^2}{E_k^{(0)} - E_n^{(0)}} > 0 \quad (\text{pushes } E_k \text{ up}) \]

   The lower level is pushed further down, and the upper level is pushed further up — the two levels repel each other. The closer the unperturbed levels are, the stronger the repulsion (due to the energy denominator). In the exact solution, two levels connected by a perturbation matrix element can never cross as the perturbation strength varies — this is the von Neumann–Wigner non-crossing rule.

Note2nd Order Stark Effect

The Stark effect is the shift of atomic energy levels in an external electric field \(\vec{\mathcal{E}} = \mathcal{E}\hat{z}\). The perturbation is:

\[ \delta H = -e\mathcal{E}z = e\mathcal{E}r\cos\theta \]

First-order correction: For hydrogen eigenstates \(\ket{n,l,m}\), parity gives \(\ket{n,l,m} \to (-1)^l \ket{n,l,m}\), while \(z\) is odd under parity. Therefore:

\[ E_n^{(1)} = \bra{n,l,m} e\mathcal{E}z \ket{n,l,m} = 0 \]

for all non-degenerate states (states with definite \(l\)). The first-order Stark effect vanishes for hydrogen.

Second-order correction (for the ground state \(\ket{1,0,0}\)):

\[ E_1^{(2)} = \sum_{n \neq 1} \frac{|\bra{n,l,m} e\mathcal{E}z \ket{1,0,0}|^2}{E_1^{(0)} - E_n^{(0)}} \]

Selection rules from \(z = r\cos\theta \propto r Y_1^0\) require \(\Delta l = \pm 1\), \(\Delta m = 0\). So only \(\ket{n,1,0}\) states contribute. The sum can be evaluated by closure or explicitly:

\[ E_1^{(2)} = -\frac{9}{4} \frac{e^2 \mathcal{E}^2 a_0^3}{E_1^{(0)}} = -\frac{9}{4} a_0^3 \mathcal{E}^2 \cdot 4\pi\epsilon_0 \]

where \(a_0\) is the Bohr radius. This gives the quadratic Stark effect:

\[ \boxed{\Delta E = -\frac{1}{2}\alpha_d \mathcal{E}^2, \quad \alpha_d = \frac{9}{2}\cdot 4\pi\epsilon_0 a_0^3} \]

where \(\alpha_d\) is the static electric polarizability of the hydrogen ground state. The energy shift is always negative (the atom is attracted to regions of stronger field), consistent with level repulsion.

For the \(n = 2\) level, which has a 4-fold degeneracy (\(l = 0, 1\)), one must use degenerate perturbation theory. The states \(\ket{2,0,0}\) and \(\ket{2,1,0}\) are mixed by \(\delta H\), giving a linear Stark effect \(\Delta E^{(1)} = \pm 3 e a_0 \mathcal{E}\).

NoteTwo Level System

Consider a general two-level Hamiltonian:

\[ H = \begin{pmatrix} E_1 & 0 \\ 0 & E_2 \end{pmatrix} + \lambda \begin{pmatrix} V_{11} & V_{12} \\ V_{21} & V_{22} \end{pmatrix} \]

where \(V_{21} = V_{12}^*\) (Hermiticity). For simplicity, set \(V_{11} = V_{22} = 0\) and \(V_{12} = V\) (real). Then \(H = \begin{pmatrix} E_1 & \lambda V \\ \lambda V & E_2 \end{pmatrix}\).

Exact solution: The eigenvalues are

\[ E_\pm = \frac{E_1 + E_2}{2} \pm \sqrt{\left(\frac{E_1 - E_2}{2}\right)^2 + \lambda^2 V^2} \]

Defining \(\Delta = E_1 - E_2\) and expanding for small \(\lambda V / \Delta\):

\[ E_\pm = \frac{E_1 + E_2}{2} \pm \frac{|\Delta|}{2}\sqrt{1 + \frac{4\lambda^2 V^2}{\Delta^2}} \approx \frac{E_1 + E_2}{2} \pm \frac{|\Delta|}{2}\left(1 + \frac{2\lambda^2 V^2}{\Delta^2} + \cdots\right) \]

Perturbation theory results: With \(E_n^{(1)} = V_{nn} = 0\) and second-order corrections:

\[ E_1^{(2)} = \frac{|V|^2}{E_1 - E_2} = \frac{V^2}{\Delta}, \quad E_2^{(2)} = \frac{|V|^2}{E_2 - E_1} = -\frac{V^2}{\Delta} \]

So perturbation theory gives \(E_1 \approx E_1 + \lambda^2 V^2 / \Delta\) and \(E_2 \approx E_2 - \lambda^2 V^2 / \Delta\), which matches the Taylor expansion of the exact result.

Key observations:

• The exact eigenvalues show avoided crossing: as \(\Delta \to 0\), the gap between \(E_+\) and \(E_-\) approaches \(2|\lambda V|\) rather than zero.
• Perturbation theory breaks down when \(\lambda V \sim \Delta\) (the coupling becomes comparable to the level spacing). In this regime the exact solution must be used.
• This model illustrates level repulsion in the simplest possible setting.

Degenerate Perturbation Theory

Formal Theory

1. 引言与问题设定 (Introduction and Setup)

在量子力学中，当非微扰哈密顿量 \(H_0\) 的某个能级对应多个线性独立的本征态时，我们称该能级是兼并的 (Degenerate)。标准的非兼并微扰理论在这种情况下会失效，因为其能量分母会出现零点发散。因此，我们需要通过在兼并子空间内对角化微扰矩阵来处理这一问题。

设定总哈密顿量为： \[ H = H_0 + \lambda V \] 其中 \(H_0\) 是精确可解的非微扰哈密顿量，\(V\) 是微扰项，\(\lambda\) 是无量纲的小参数（在此形式推导中，最后可令 \(\lambda \to 1\)）。

1.1 兼并子空间

假设 \(H_0\) 的第 \(n\) 个能级 \(E_n^{(0)}\) 是 \(g\) 重兼并的。对应的本征子空间 \(\mathcal{D}_n\) 由一组正交归一的基底向量张成： \[ \{ | \phi_{n\alpha}^{(0)} \rangle \}_{\alpha=1}^g \]

满足： \[ H_0 | \phi_{n\alpha}^{(0)} \rangle = E_n^{(0)} | \phi_{n\alpha}^{(0)} \rangle, \quad \alpha = 1, \dots, g \]

且正交归一条件为 \(\langle \phi_{n\alpha}^{(0)} | \phi_{n\beta}^{(0)} \rangle = \delta_{\alpha\beta}\)。

对于该能级之外的非兼并态，我们记为 \(| k^{(0)} \rangle\)，其能量为 \(E_k^{(0)} \neq E_n^{(0)}\)。

2. 微扰展开 (Perturbation Expansion)

当微扰 \(\lambda V\) 开启时，兼并能级 \(E_n^{(0)}\) 通常会发生分裂。我们将受扰动的能量本征值 \(E_{n}\) 和本征态 \(| \psi_{n} \rangle\) 展开为 \(\lambda\) 的幂级数： \[ \begin{aligned} E_{n} &= E_n^{(0)} + \lambda E_n^{(1)} + \lambda^2 E_n^{(2)} + \dots \\ | \psi_{n} \rangle &= | \psi_{n}^{(0)} \rangle + \lambda | \psi_{n}^{(1)} \rangle + \lambda^2 | \psi_{n}^{(2)} \rangle + \dots \end{aligned} \]

2.1 零级波函数的选择

这是兼并微扰理论最关键的一步。当 \(\lambda \to 0\) 时，\(| \psi_{n} \rangle\) 并不一定会收敛到某个特定的基底 \(| \phi_{n\alpha}^{(0)} \rangle\)，而是收敛到兼并子空间 \(\mathcal{D}_n\) 中的某个特定的线性组合。我们将这个”正确的”零级波函数记为： \[ | \psi_{n}^{(0)} \rangle = \sum_{\beta=1}^g c_\beta | \phi_{n\beta}^{(0)} \rangle \]

我们的目标之一就是确定系数 \(c_\beta\)。

3. 薛定谔方程的级数解 (Series Solution)

将展开式代入定态薛定谔方程 \((H_0 + \lambda V) | \psi_{n} \rangle = E_{n} | \psi_{n} \rangle\)： \[ (H_0 + \lambda V) \left( \sum_{j=0}^\infty \lambda^j | \psi_{n}^{(j)} \rangle \right) = \left( \sum_{i=0}^\infty \lambda^i E_n^{(i)} \right) \left( \sum_{j=0}^\infty \lambda^j | \psi_{n}^{(j)} \rangle \right) \]

按 \(\lambda\) 的幂次归类，得到一系列方程：

\(\lambda^0\) 阶: \[ H_0 | \psi_{n}^{(0)} \rangle = E_n^{(0)} | \psi_{n}^{(0)} \rangle \] 这验证了 \(| \psi_{n}^{(0)} \rangle\) 确实属于 \(E_n^{(0)}\) 的本征子空间。

\(\lambda^1\) 阶: \[ H_0 | \psi_{n}^{(1)} \rangle + V | \psi_{n}^{(0)} \rangle = E_n^{(0)} | \psi_{n}^{(1)} \rangle + E_n^{(1)} | \psi_{n}^{(0)} \rangle \]

移项整理得： \[ (H_0 - E_n^{(0)}) | \psi_{n}^{(1)} \rangle = (E_n^{(1)} - V) | \psi_{n}^{(0)} \rangle \quad (*) \tag{1}\]

4. 一阶能量修正与久期方程 (First-Order Energy and Secular Equation)

为了求解 \(E_n^{(1)}\) 和系数 \(c_\beta\)，我们将方程 \((*)\) 投影到兼并子空间 \(\mathcal{D}_n\) 上。

左乘兼并基底中的任意一个态 \(\langle \phi_{n\alpha}^{(0)} |\) (其中 \(\alpha = 1, \dots, g\))： \[ \langle \phi_{n\alpha}^{(0)} | (H_0 - E_n^{(0)}) | \psi_{n}^{(1)} \rangle = \langle \phi_{n\alpha}^{(0)} | (E_n^{(1)} - V) | \psi_{n}^{(0)} \rangle \]

利用 \(H_0\) 的厄米性，\(H_0\) 作用在左边的 \(\langle \phi_{n\alpha}^{(0)} |\) 上给出 \(E_n^{(0)}\)，因此方程左边恒为零： \[ \langle \phi_{n\alpha}^{(0)} | H_0 - E_n^{(0)} | \psi_{n}^{(1)} \rangle = (E_n^{(0)} - E_n^{(0)}) \langle \phi_{n\alpha}^{(0)} | \psi_{n}^{(1)} \rangle = 0 \]

于是我们得到： \[ 0 = E_n^{(1)} \langle \phi_{n\alpha}^{(0)} | \psi_{n}^{(0)} \rangle - \langle \phi_{n\alpha}^{(0)} | V | \psi_{n}^{(0)} \rangle \]

将 \(| \psi_{n}^{(0)} \rangle = \sum_{\beta=1}^g c_\beta | \phi_{n\beta}^{(0)} \rangle\) 代入上式： \[ E_n^{(1)} \sum_{\beta=1}^g c_\beta \underbrace{\langle \phi_{n\alpha}^{(0)} | \phi_{n\beta}^{(0)} \rangle}_{\delta_{\alpha\beta}} = \sum_{\beta=1}^g c_\beta \langle \phi_{n\alpha}^{(0)} | V | \phi_{n\beta}^{(0)} \rangle \]

定义微扰矩阵元 \(V_{\alpha\beta} = \langle \phi_{n\alpha}^{(0)} | V | \phi_{n\beta}^{(0)} \rangle\)，上式简化为： \[ E_n^{(1)} c_\alpha = \sum_{\beta=1}^g V_{\alpha\beta} c_\beta \]

或者写成矩阵形式： \[ \sum_{\beta=1}^g (V_{\alpha\beta} - E_n^{(1)} \delta_{\alpha\beta}) c_\beta = 0 \]

4.1 久期方程 (Secular Equation)

这是一个关于系数向量 \(\mathbf{c} = (c_1, \dots, c_g)^T\) 的齐次线性方程组。为了使方程有非零解，系数矩阵的行列式必须为零： \[ \det (V_{\alpha\beta} - E_n^{(1)} \delta_{\alpha\beta}) = 0 \]

或者更直观地写成： \[ \begin{vmatrix} V_{11} - E_n^{(1)} & V_{12} & \dots & V_{1g} \\ V_{21} & V_{22} - E_n^{(1)} & \dots & V_{2g} \\ \vdots & \vdots & \ddots & \vdots \\ V_{g1} & V_{g2} & \dots & V_{gg} - E_n^{(1)} \end{vmatrix} = 0 \]

求解这个 \(g\) 次代数方程，我们可以得到 \(g\) 个根 \(E_{n,1}^{(1)}, E_{n,2}^{(1)}, \dots, E_{n,g}^{(1)}\)。这些根就是兼并能级 \(E_n^{(0)}\) 在一阶微扰下的能量修正。

• 如果所有 \(g\) 个根都互不相同，则兼并完全消除。
• 如果仍有重根，则兼并只是部分消除，需要更高阶微扰理论处理剩余的兼并。

求出每个特征值 \(E_{n,r}^{(1)}\) 后，代回线性方程组即可求出对应的特征向量 \(\mathbf{c}^{(r)}\)，从而确定零级波函数 \(| \psi_{n,r}^{(0)} \rangle\)。

5. 一阶波函数修正 (First-Order Wavefunction Correction)

现在我们需要求 \(| \psi_{n}^{(1)} \rangle\)。其可以展开为全空间的基底： \[ | \psi_{n}^{(1)} \rangle = \sum_{k \notin \mathcal{D}_n} a_k | k^{(0)} \rangle + \sum_{\beta=1}^g b_\beta | \phi_{n\beta}^{(0)} \rangle \]

通常我们选择中间归一化条件 (Intermediate Normalization)：\(\langle \psi_{n}^{(0)} | \psi_{n} \rangle = 1\)。这蕴含了 \(\langle \psi_{n}^{(0)} | \psi_{n}^{(1)} \rangle = 0\)。如果微扰完全解除了兼并，我们可以令 \(| \psi_{n}^{(1)} \rangle\) 与整个兼并子空间正交（即 \(b_\beta = 0\)）。

回到方程 \((*)\)： \[ (H_0 - E_n^{(0)}) | \psi_{n}^{(1)} \rangle = (E_n^{(1)} - V) | \psi_{n}^{(0)} \rangle \]

左乘兼并子空间之外的态 \(\langle k^{(0)}|\) (其中 \(E_k^{(0)} \neq E_n^{(0)}\))： \[ \langle k^{(0)} | (H_0 - E_n^{(0)}) | \psi_{n}^{(1)} \rangle = \langle k^{(0)} | (E_n^{(1)} - V) | \psi_{n}^{(0)} \rangle \]

左边项利用 \(H_0\) 的厄米性： \[ (E_k^{(0)} - E_n^{(0)}) \langle k^{(0)} | \psi_{n}^{(1)} \rangle = E_n^{(1)} \underbrace{\langle k^{(0)} | \psi_{n}^{(0)} \rangle}_{0} - \langle k^{(0)} | V | \psi_{n}^{(0)} \rangle \]

因为 \(| k^{(0)} \rangle\) 与兼并子空间正交，所以 \(\langle k^{(0)} | \psi_{n}^{(0)} \rangle = 0\)。

解出展开系数 \(\langle k^{(0)} | \psi_{n}^{(1)} \rangle\)： \[ \langle k^{(0)} | \psi_{n}^{(1)} \rangle = \frac{\langle k^{(0)} | V | \psi_{n}^{(0)} \rangle}{E_n^{(0)} - E_k^{(0)}} \]

因此，波函数的一阶修正（在兼并子空间正交补上的部分）为： \[ | \psi_{n}^{(1)} \rangle = \sum_{k \notin \mathcal{D}_n} \frac{\langle k^{(0)} | V | \psi_{n}^{(0)} \rangle}{E_n^{(0)} - E_k^{(0)}} | k^{(0)} \rangle \]

NoteTwo-fold Degenerate Case

这里给出一个完整的二重简并微扰例子：自旋 \(1/2\) 粒子在零磁场时的二重简并，在弱磁场作用下被劈裂。这个例子直接体现简并微扰的标准步骤：在简并子空间内写出扰动矩阵并对角化。

设 \[ H = H_0 + \lambda V,\qquad \lambda\ll 1. \]

1. 零级问题与二重简并

取 \[ H_0 = 0. \] 则能量 \(E^{(0)}=0\) 对整个自旋空间成立，所以这是一个二重简并能级。先选取 \(S_z\) 的本征基作为简并子空间的一组基底： \[ |1\rangle \equiv |\uparrow z\rangle,\qquad |2\rangle \equiv |\downarrow z\rangle. \]

2. 加入扰动：磁场沿 \(x\) 方向

施加一个沿 \(x\) 方向的弱磁场，取扰动为 \[ V = -\gamma B\,S_x = -\gamma B\,\frac{\hbar}{2}\,\sigma_x. \] 在 \(\{|\uparrow z\rangle,|\downarrow z\rangle\}\) 基底下 \[ \sigma_x= \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}. \] 因此简并子空间上的扰动矩阵（也记作 \(W\)）为 \[ W_{ij}=\langle i|V|j\rangle =-\gamma B\frac{\hbar}{2} \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}. \] 注意这里 \(W\) 有非对角元：这意味着你原先选的 \(|\uparrow z\rangle,|\downarrow z\rangle\) 并不是“正确的零级态”。如果硬套非简并的一阶公式，会遇到简并导致的分母为零问题。

3. 对角化 \(W\)：得到一阶能量修正

简并微扰论告诉我们：\(W\) 的本征值就是一阶能量修正 \(E^{(1)}\)，即 \[ \det(W-E^{(1)}I)=0. \] 对本例直接解得 \[ E^{(1)}_\pm = \pm\gamma B\frac{\hbar}{2}. \] 因此能量到一阶为 \[ E_\pm \approx E^{(0)}+\lambda E^{(1)}_\pm = \lambda\left(\pm\gamma B\frac{\hbar}{2}\right). \]

4. “正确的零级态”是简并态的线性组合

对应的本征向量满足 \((W-E^{(1)}_\pm I)\mathbf c^{(\pm)}=0\)，解得 \[ \mathbf c^{(+)}\propto\begin{pmatrix}1\\-1\end{pmatrix},\qquad \mathbf c^{(-)}\propto\begin{pmatrix}1\\1\end{pmatrix}. \] 所以被扰动选出的零级态（在简并子空间内对角化 \(V\) 的那组基）为 \[ |\psi^{(0)}_+\rangle=\frac{1}{\sqrt2}\left(|\uparrow z\rangle-|\downarrow z\rangle\right)=|{-}x\rangle, \] \[ |\psi^{(0)}_-\rangle=\frac{1}{\sqrt2}\left(|\uparrow z\rangle+|\downarrow z\rangle\right)=|{+}x\rangle. \]

一旦用这组 \(|\psi^{(0)}_\pm\rangle\) 作为简并能级被“分裂后”的零级态，后续一阶波函数修正就可以用你上面推导的公式，并且求和只对简并子空间外的 \(k\notin\mathcal D\) 进行。

Application of Perturbation Theory

Fine Structure of Hydrogen

The fine structure of hydrogen consists of two relativistic corrections to the non-relativistic Coulomb Hamiltonian \(H_0 = \frac{p^2}{2m} - \frac{e^2}{4\pi\epsilon_0 r}\), both of order \(\alpha^2\) relative to \(E_n^{(0)}\) (where \(\alpha = e^2/(4\pi\epsilon_0\hbar c) \approx 1/137\) is the fine-structure constant).

1. Relativistic kinetic energy correction:

From the relativistic energy \(E = \sqrt{p^2c^2 + m^2c^4}\), expanding to next order beyond \(p^2/2m\):

\[ T = \sqrt{p^2c^2 + m^2c^4} - mc^2 \approx \frac{p^2}{2m} - \frac{p^4}{8m^3c^2} + \cdots \]

The perturbation is:

\[ H'_{\text{rel}} = -\frac{p^4}{8m^3c^2} = -\frac{1}{2mc^2}\left(\frac{p^2}{2m}\right)^2 \]

Using \(\frac{p^2}{2m} = E_n + \frac{e^2}{4\pi\epsilon_0 r}\) to express \(p^4\) in terms of expectation values of \(1/r\) and \(1/r^2\):

\[ E_{\text{rel}}^{(1)} = -\frac{1}{2mc^2}\left[E_n^2 + 2E_n \frac{e^2}{4\pi\epsilon_0}\langle r^{-1}\rangle_{nl} + \left(\frac{e^2}{4\pi\epsilon_0}\right)^2 \langle r^{-2}\rangle_{nl}\right] \]

With the known hydrogen expectation values \(\langle r^{-1}\rangle = 1/(n^2 a_0)\) and \(\langle r^{-2}\rangle = 1/[n^3 a_0^2 (l+1/2)]\), this gives:

\[ E_{\text{rel}}^{(1)} = -\frac{E_n^{(0)^2}}{2mc^2}\left(\frac{4n}{l+1/2} - 3\right) \]

2. Spin-Orbit Coupling:

In the electron’s rest frame, the proton orbits the electron, creating a magnetic field that interacts with the electron’s spin magnetic moment. The correct result (including the Thomas precession factor of \(1/2\)) is:

\[ H'_{SO} = \frac{e^2}{8\pi\epsilon_0 m^2 c^2}\frac{1}{r^3}\vec{L}\cdot\vec{S} \]

To evaluate \(\langle\vec{L}\cdot\vec{S}\rangle\), use \(\vec{J} = \vec{L} + \vec{S}\), so:

\[ \vec{L}\cdot\vec{S} = \frac{1}{2}(J^2 - L^2 - S^2) \quad \Rightarrow \quad \langle\vec{L}\cdot\vec{S}\rangle = \frac{\hbar^2}{2}[j(j+1) - l(l+1) - s(s+1)] \]

where \(j = l \pm 1/2\) (for \(l \geq 1\)). With \(\langle r^{-3}\rangle = 1/[n^3 a_0^3 l(l+1/2)(l+1)]\):

\[ E_{SO}^{(1)} = \frac{E_n^{(0)^2}}{mc^2}\frac{n[j(j+1) - l(l+1) - 3/4]}{l(l+1/2)(l+1)} \quad (l \neq 0) \]

Combined fine structure: Remarkably, the sum \(E_{\text{rel}}^{(1)} + E_{SO}^{(1)}\) depends only on \(j\) (not separately on \(l\)):

\[ \boxed{ E_{\text{fs}} = -\frac{E_n^{(0)^2}}{2mc^2}\left(\frac{4n}{j+1/2} - 3\right) = \frac{13.6\,\text{eV}}{n^2}\cdot\frac{\alpha^2}{n^2}\left(\frac{n}{j+1/2} - \frac{3}{4}\right) } \]

This explains why states with the same \(n\) and \(j\) but different \(l\) (e.g., \(2S_{1/2}\) and \(2P_{1/2}\)) remain degenerate at this order — the degeneracy is only lifted by the Lamb shift (a QED effect).

Zeeman Effect

An external magnetic field \(\vec{B} = B\hat{z}\) adds the perturbation:

\[ H'_Z = -(\vec{\mu}_L + \vec{\mu}_S)\cdot\vec{B} = \frac{eB}{2m}(L_z + 2S_z) = \frac{eB}{2m}(J_z + S_z) \]

Weak-field (anomalous) Zeeman effect (\(H'_Z \ll H'_{\text{fs}}\)): Fine structure is the dominant perturbation, so good quantum numbers are \(\{n, l, j, m_j\}\). Using the projection theorem:

\[ \langle J_z + S_z \rangle = \langle J_z\rangle + \langle S_z\rangle = m_j\hbar + \langle S_z\rangle \]

The projection theorem gives \(\langle\vec{S}\rangle = \frac{\langle\vec{J}\cdot\vec{S}\rangle}{\langle J^2\rangle}\langle\vec{J}\rangle\), so:

\[ \langle S_z \rangle = \frac{j(j+1) + s(s+1) - l(l+1)}{2j(j+1)} m_j\hbar \]

Defining the Landé g-factor:

\[ g_J = 1 + \frac{j(j+1) + s(s+1) - l(l+1)}{2j(j+1)} \]

The Zeeman energy shift is:

\[ \boxed{E_Z^{(1)} = g_J m_j \mu_B B} \]

where \(\mu_B = e\hbar/(2m)\) is the Bohr magneton. Each fine-structure level \(j\) splits into \(2j+1\) equally spaced sub-levels.

Strong-field (Paschen-Back) effect (\(H'_Z \gg H'_{\text{fs}}\)): The magnetic field dominates, so good quantum numbers are \(\{n, l, m_l, m_s\}\). The Zeeman energy is \(E_Z = (m_l + 2m_s)\mu_B B\), and fine structure is treated as a perturbation on top.

Van der Waals Interaction

Consider two neutral hydrogen atoms separated by distance \(R\) (with \(R \gg a_0\)). Each atom has a nucleus and electron; label them \((1)\) and \((2)\) with position vectors \(\vec{r}_1\) and \(\vec{r}_2\) relative to their respective nuclei.

The unperturbed Hamiltonian is \(H_0 = H_{\text{atom}}^{(1)} + H_{\text{atom}}^{(2)}\), with ground state \(\ket{0} = \ket{1s}_1 \ket{1s}_2\) and \(E_0 = 2E_{1s}\).

The perturbation is the dipole-dipole interaction (leading order in \(a_0/R\)):

\[ V = \frac{e^2}{4\pi\epsilon_0 R^3}\left(x_1 x_2 + y_1 y_2 - 2z_1 z_2\right) + \mathcal{O}(R^{-4}) \]

First-order shift: \(E^{(1)} = \bra{0}V\ket{0} = 0\), because \(\langle 1s|x|1s\rangle = 0\) (the ground state has no permanent dipole moment).

Second-order shift (Van der Waals energy):

\[ E^{(2)} = \sum_{n_1, n_2 \neq 0} \frac{|\bra{n_1, n_2} V \ket{0}|^2}{E_0 - E_{n_1} - E_{n_2}} < 0 \]

Since each excited state energy is at least \(E_{2p}\), and using dimensional analysis:

\[ \boxed{E_{\text{vdW}} = -\frac{C_6}{R^6}} \]

where \(C_6 > 0\) is a constant that can be computed exactly: \(C_6 = 6.499\,e^2 a_0^5 / (4\pi\epsilon_0)\). The \(R^{-6}\) scaling is characteristic of the van der Waals (London dispersion) interaction between two non-polar atoms. It is always attractive and arises from correlated quantum fluctuations of the dipole moments.

4.2 Time-Dependent Perturbation Theory

Interaction (Dirac) Picture

The Schrödinger and Heisenberg pictures are two extremes: in the former, all time dependence is in states; in the latter, all is in operators. The interaction picture is a hybrid that is particularly convenient for perturbation theory.

Split the Hamiltonian: \(H(t) = H_0 + V(t)\). Define interaction-picture states and operators:

\[ \ket{\psi_I(t)} = e^{iH_0 t/\hbar}\ket{\psi_S(t)}, \quad \hat{A}_I(t) = e^{iH_0 t/\hbar}\hat{A}_S e^{-iH_0 t/\hbar} \]

The equation of motion for the state is:

\[ \begin{aligned} i\hbar\frac{d}{dt}\ket{\psi_I(t)} &= i\hbar\left(\frac{i}{\hbar}H_0 e^{iH_0 t/\hbar}\ket{\psi_S} + e^{iH_0 t/\hbar}\frac{d}{dt}\ket{\psi_S}\right) \\ &= -H_0\ket{\psi_I} + e^{iH_0 t/\hbar}(H_0 + V)\ket{\psi_S} \\ &= e^{iH_0 t/\hbar}V(t) e^{-iH_0 t/\hbar}\ket{\psi_I} \end{aligned} \]

Defining \(V_I(t) = e^{iH_0 t/\hbar}V(t)e^{-iH_0 t/\hbar}\):

\[ \boxed{i\hbar\frac{d}{dt}\ket{\psi_I(t)} = V_I(t)\ket{\psi_I(t)}} \]

Picture	State evolution	Operator evolution
Schrödinger	\(i\hbar\dot{\ket{\psi}} = H\ket{\psi}\)	\(\hat{A}\) fixed
Heisenberg	\(\ket{\psi}\) fixed	\(i\hbar\dot{\hat{A}} = [\hat{A}, H]\)
Interaction	\(i\hbar\dot{\ket{\psi_I}} = V_I\ket{\psi_I}\)	\(i\hbar\dot{\hat{A}_I} = [\hat{A}_I, H_0]\)

Transition Probability

Expand the state in the eigenbasis of \(H_0\): \(\ket{\psi_I(t)} = \sum_n c_n(t)\ket{n}\), with \(H_0\ket{n} = E_n\ket{n}\). The equation becomes:

\[ i\hbar\dot{c}_f(t) = \sum_n \bra{f}V_I(t)\ket{n}\,c_n(t) = \sum_n V_{fn}(t)\,e^{i\omega_{fn}t}\,c_n(t) \]

where \(\omega_{fn} = (E_f - E_n)/\hbar\) and \(V_{fn}(t) = \bra{f}V(t)\ket{n}\).

First-order approximation: If the system starts in \(\ket{i}\) at \(t = 0\), then \(c_n(0) = \delta_{ni}\). To first order, substitute \(c_n(t) \approx \delta_{ni}\) on the RHS:

\[ c_f^{(1)}(t) = -\frac{i}{\hbar}\int_0^t V_{fi}(t')\,e^{i\omega_{fi}t'}\,dt' \quad (f \neq i) \]

The transition probability from \(\ket{i}\) to \(\ket{f}\) is:

\[ \boxed{P_{i\to f}(t) = |c_f^{(1)}(t)|^2 = \frac{1}{\hbar^2}\left|\int_0^t V_{fi}(t')\,e^{i\omega_{fi}t'}\,dt'\right|^2} \]

Constant Perturbation Turned On at \(t = 0\)

If \(V(t) = V\theta(t)\) (constant for \(t > 0\)), then:

\[ c_f^{(1)}(t) = -\frac{i}{\hbar}V_{fi}\int_0^t e^{i\omega_{fi}t'}\,dt' = -\frac{V_{fi}}{\hbar\omega_{fi}}\left(e^{i\omega_{fi}t} - 1\right) \]

\[ P_{i\to f}(t) = \frac{|V_{fi}|^2}{\hbar^2}\cdot\frac{4\sin^2(\omega_{fi}t/2)}{\omega_{fi}^2} = \frac{|V_{fi}|^2}{\hbar^2}\,t^2\,\text{sinc}^2\!\left(\frac{\omega_{fi}t}{2}\right) \]

The function \(\frac{\sin^2(\omega t/2)}{\omega^2}\) is sharply peaked around \(\omega = 0\) with width \(\sim 2\pi/t\) and height \(\sim t^2/4\). As \(t \to \infty\):

\[ \frac{\sin^2(\omega t/2)}{\omega^2/4} \to 2\pi t\,\delta(\omega) \]

Fermi’s Golden Rule

For transitions to a continuum of final states with density of states \(\rho(E_f)\), sum over all final states near energy \(E_f \approx E_i\):

\[ \Gamma_{i\to f} = \frac{d}{dt}P_{i\to f} = \frac{2\pi}{\hbar}|V_{fi}|^2\,\rho(E_f)\bigg|_{E_f = E_i} \]

\[ \boxed{\Gamma_{i\to f} = \frac{2\pi}{\hbar}\left|\bra{f}V\ket{i}\right|^2 \rho(E_i)} \]

This is Fermi’s golden rule: the transition rate is proportional to (1) the square of the matrix element and (2) the density of final states at the energy of the initial state. Energy conservation (\(E_f = E_i\)) emerges naturally from the \(\delta\)-function in the long-time limit.

NoteHarmonic Perturbation

For a sinusoidal perturbation \(V(t) = V e^{-i\omega t} + V^\dagger e^{i\omega t}\) (e.g., electromagnetic radiation), the first-order amplitude becomes:

\[ c_f^{(1)}(t) = -\frac{i}{\hbar}\left[V_{fi}\frac{e^{i(\omega_{fi}-\omega)t}-1}{\omega_{fi}-\omega} + V_{fi}^*\frac{e^{i(\omega_{fi}+\omega)t}-1}{\omega_{fi}+\omega}\right] \]

The first term peaks at \(\omega_{fi} \approx \omega\) (absorption: \(E_f \approx E_i + \hbar\omega\)), and the second at \(\omega_{fi} \approx -\omega\) (stimulated emission: \(E_f \approx E_i - \hbar\omega\)). Applying the same long-time limit gives Fermi’s golden rule with an energy-conserving \(\delta\)-function:

\[ \Gamma_{\text{abs}} = \frac{2\pi}{\hbar}|V_{fi}|^2\,\rho(E_i + \hbar\omega), \quad \Gamma_{\text{em}} = \frac{2\pi}{\hbar}|V_{fi}|^2\,\rho(E_i - \hbar\omega) \]

4.3 Variational Method

The Variational Principle (Rayleigh-Ritz)

Theorem: For any normalized trial state \(\ket{\psi_{\text{trial}}}\), the expectation value of the Hamiltonian provides an upper bound to the ground state energy:

\[ \boxed{E_0 \leq \langle \hat{H} \rangle_{\text{trial}} = \bra{\psi_{\text{trial}}} H \ket{\psi_{\text{trial}}}} \]

Proof: Expand the trial state in the exact eigenbasis \(\{|n\rangle\}\) of \(H\):

\[ \ket{\psi_{\text{trial}}} = \sum_n c_n \ket{n}, \quad H\ket{n} = E_n\ket{n}, \quad E_0 \leq E_1 \leq E_2 \leq \cdots \]

Then:

\[ \langle H \rangle = \sum_n |c_n|^2 E_n \geq E_0 \sum_n |c_n|^2 = E_0 \]

Equality holds if and only if \(\ket{\psi_{\text{trial}}} = \ket{0}\) (the exact ground state). \(\blacksquare\)

Variational Method in Practice

1. Choose a family of trial wavefunctions \(\psi_\alpha(\vec{r})\) parameterized by one or more variational parameters \(\alpha\).
2. Compute \(E(\alpha) = \langle \psi_\alpha | H | \psi_\alpha \rangle\).
3. Minimize: \(\frac{\partial E}{\partial \alpha} = 0\) gives the optimal \(\alpha^*\) and the best variational estimate \(E(\alpha^*) \geq E_0\).

NoteExample: Helium Ground State Energy

The helium atom has two electrons orbiting a \(Z = 2\) nucleus. The Hamiltonian is (in atomic units where \(\hbar = m_e = e = 4\pi\epsilon_0 = 1\)):

\[ H = -\frac{1}{2}\nabla_1^2 - \frac{1}{2}\nabla_2^2 - \frac{Z}{r_1} - \frac{Z}{r_2} + \frac{1}{r_{12}} \]

The electron-electron repulsion \(1/r_{12}\) makes this problem exactly unsolvable. Treat it variationally.

Trial wavefunction: Use a product of hydrogen-like \(1s\) orbitals with an effective nuclear charge \(Z^*\) as the variational parameter:

\[ \psi(r_1, r_2) = \frac{(Z^*)^3}{\pi}\,e^{-Z^*(r_1 + r_2)} \]

Kinetic + nuclear energy: Each electron contributes (from the hydrogen-atom result with nuclear charge \(Z^*\)):

\[ \langle T + V_{\text{nuc}}\rangle = 2\left[\frac{(Z^*)^2}{2} - Z\cdot Z^*\right] = (Z^*)^2 - 2Z\cdot Z^* \]

Electron-electron repulsion: The integral \(\langle 1/r_{12}\rangle\) for two \(1s\) orbitals with effective charge \(Z^*\) gives:

\[ \langle 1/r_{12}\rangle = \frac{5}{8}Z^* \]

Total energy:

\[ E(Z^*) = (Z^*)^2 - 2Z\cdot Z^* + \frac{5}{8}Z^* = (Z^*)^2 - \left(2Z - \frac{5}{8}\right)Z^* \]

Minimizing \(\frac{dE}{dZ^*} = 2Z^* - 2Z + \frac{5}{8} = 0\):

\[ Z^* = Z - \frac{5}{16} = 2 - 0.3125 = 1.6875 \]

The optimal energy is:

\[ E = -\left(Z - \frac{5}{16}\right)^2 = -(1.6875)^2 = -2.848\,\text{a.u.} = -77.5\,\text{eV} \]

The experimental value is \(-79.0\,\text{eV}\), so this simple one-parameter variational estimate achieves \(\sim 98\%\) accuracy. The effective charge \(Z^* = 1.69 < 2\) reflects the partial screening of the nuclear charge by the other electron.

NoteExcited States and Generalization

The variational principle can be extended to excited states:

1. Orthogonality constraint: If the trial state is constrained to be orthogonal to the exact ground state (\(\langle\psi_{\text{trial}}|0\rangle = 0\)), then \(\langle H\rangle \geq E_1\).
2. Linear variational method: Express the trial state as \(\ket{\psi} = \sum_i c_i \ket{\phi_i}\) in a finite basis \(\{\ket{\phi_i}\}\). Minimizing \(E\) with respect to \(\{c_i\}\) leads to the generalized eigenvalue problem: \(\mathbf{H}\vec{c} = E\,\mathbf{S}\vec{c}\), where \(H_{ij} = \bra{\phi_i}H\ket{\phi_j}\) and \(S_{ij} = \braket{\phi_i|\phi_j}\). This yields upper bounds to the first \(N\) energy levels (Hylleraas-Undheim-MacDonald theorem).

4.4 Scattering Theory

This part is based on Sakurai’s textbook.

Difference between scattering states and bound states: wave functions of bound states can be normalized, while those of scattering states cannot. So we always use ratio of wave functions to define transition amplitudes.

Format Theory

Now in this section we define the following rules:

1. Input wave functions: \(\psi^{in}(\vec r) = e^{ikz}\), with \((-\frac{\hbar^2}{2m} \nabla^2 - E) \psi^{in}(\vec r) = 0\)
2. Local field around the interaction region: WE DON’T CARE!
3. Remote field: \(\psi(r, \theta, \psi) = \sum_{lm} c_{lm} \frac{u_l(r)}{r} Y_{lm}(\theta, \phi)\), where \(u_l(r)\) are radial functions and \(Y_{lm}(\theta, \phi)\) are spherical harmonics. By \(\frac{d^2}{dr^2}u_l(r) + k^2 u_l(r) = 0 \to u_l(r) = e^{\pm i k r}\)

After all, we obtain the asymptotic form of the wave function: \[ \boxed{ \psi(\vec r) = \bra{\vec r} \ket{\psi} = A [e^{ikz} + f(\theta, \phi) \frac{e^{ikr}}{r}] } \] where \(f(\theta, \phi)\) is the scattering amplitude.

if the scattering potential is symmetric around z-axis, then \(m\) is good quantum number, which means \(f(\theta, \phi) = f(\theta)\) iff input wave \(m=0\).

The total scattering cross section is given by

\[ \sigma_{tol} = \int d \Omega \left| f(\theta, \phi) \right|^2 = \int_0^{2\pi} d\phi \int_0^\pi \sin\theta d\theta \left| f(\theta, \phi) \right|^2 \]

Green Function Method

Finally we have:

\[ \boxed{ f(\theta, \phi) = - \frac{(2\pi)^{3/2}}{4\pi} \frac{2m}{\hbar^2} \int d^3 r' e^{-i\vec k_f \cdot \vec r'} V(r') \psi_p^+ (\vec r') } \]

Because

\[ \bra{\vec r} \ket{\vec k_f} = \frac{1}{(2\pi)^{3/2}} e^{i\vec k_f\cdot \vec r} \]

So we have:

\[ \boxed{ f(\theta, \phi) = - \frac{4\pi^2 m}{\hbar^2} \bra{\vec k_f} V \ket{\psi_{k_i}^+} } \]

Born Approximation and Optical Theorem

Because of \[ \psi_p^+ (\vec r') = \frac{1}{(2\pi)^{3/2}} \left[ e^{i \vec k_i \cdot \vec r'} + (2\pi)^{3/2} \int d^3 r'' G_0(\vec r' - \vec r'') V(\vec r'') \psi_p^+ (\vec r'')\right] \]

So we have: \[ \ket{\psi_{k_i}^+} = \ket{k_i} + \hat G_0^+ \hat V \ket{\psi_{k_i}^+} \]

Define \(\hat T\) Operator \[ \hat T^\pm \ket{k_i} \equiv \hat V \ket{\psi_{k_i}^\pm} \]

So Born Expansion is \[ \hat T^\pm = \hat V + \hat V \hat G_0^\pm \hat V + \hat V \hat G_0^\pm \hat V \hat G_0^\pm \hat V + \cdots \]