Nhiễu loạn yếu và các không gian con suy biến
Đã giới thiệu| English | Ý nghĩa |
|---|---|
| perturbation/pətəˈbeɪʃn/ | A small extra term added to a solvable Hamiltonian. |
| expectation value/ekspɪkˈteɪʃn ˈvæljuː/ | The average result of measuring an observable over many identical preparations. |
| degenerate/dɪˈdʒenəreɪt/ | Sharing the same energy with another independent state. |
Zero shift, changed state
- A weak tilt $\gamma(x - L/2)$ across a symmetric well gives every state a zero first-order energy shift — yet the wavefunctions do change, and second-order shifts appear.
- GRE perturbation 微扰 theory tests first-order shifts as expectation values 期望值, symmetry arguments, second order and degenerate 简并的 levels.
First order
- $H = H_0 + \lambda W$: $E_n^{(1)} = \lambda\langle n|W|n\rangle = \lambda\int|\psi_n|^2W\,dx$ — a weighted average over the whole state.
- Infinite well $0 < x < L$: $\langle x\rangle = L/2$, so $W = \gamma x$ shifts every level by $\gamma L/2$. 2 nm well, $\gamma = 0.05$ eV/nm: 0.05 eV.
A 3 nm infinite well gets $W = \gamma x$ with $\gamma = 0.1$ eV/nm. First-order shift in eV?
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The first-order shift is $\langle\gamma x\rangle = \gamma L/2 = 0.1 \times 3/2 = 0.15$ eV, since the probability is symmetric about the centre.
Symmetry
- $|\psi_n|^2$ is symmetric about $L/2$, so odd perturbations like $\gamma(x - L/2)$ give zero first-order shift.
- $\eta(x - L/2)^2$ gives $\eta L^2\left[\tfrac1{12} - \tfrac{1}{2\pi^2n^2}\right] > 0$, depending on $n$.
Zero at first order?
Use symmetry of the infinite-well density to predict first-order shifts.
For an infinite-well state, $W = \gamma(x - L/2)^3$ gives a first-order shift of
Độ khó 1/5 · 100 xu + một phần thưởng nhỏ cho lần hoàn thành đầu tiên trong ngày
$|\psi|^2$ is symmetric about $L/2$ while $(x - L/2)^3$ is odd about it, so the expectation value is zero.
Second order
- $E_n^{(2)} = \sum_{m \ne n}\dfrac{|\lambda W_{mn}|^2}{E_n^0 - E_m^0}$: always $\le 0$ for the ground state.
- Valid when couplings are small compared with level gaps; near-degeneracy breaks it.
The second-order shift of a nondegenerate ground state is
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Every denominator is negative.
Degenerate levels
- Write $W$ as a matrix in the degenerate subspace and diagonalise: eigenvalues are the first-order shifts.
$\varepsilon\begin{pmatrix} 2&1\\1&2 \end{pmatrix}$: shifts $3\varepsilon$ (symmetric combination) and $\varepsilon$ (antisymmetric). Reading the diagonal would wrongly give $2\varepsilon$ twice. A multiple of $I$ never splits a degeneracy.
A degenerate pair has perturbation matrix $\varepsilon\begin{pmatrix} 1&2\\2&1 \end{pmatrix}$ with $\varepsilon = 0.002$ eV. Splitting in eV?
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The matrix has eigenvalues $3\varepsilon$ and $-\varepsilon$, so the splitting is $4\varepsilon = 0.008$ eV.
A perturbation equal to a constant times the identity shifts a degenerate level without splitting it.
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True. Every state in the level gets the same shift, so the degeneracy remains.
Recap
- $E^{(1)} = \langle W\rangle$ in the unperturbed state.
- Odd perturbation, symmetric density: zero at first order.
- Ground-state $E^{(2)} \le 0$; degenerate levels: diagonalise the block.
- perturbation: A small extra term added to a solvable Hamiltonian.
- expectation value: The average result of measuring an observable over many identical preparations.
- degenerate: Sharing the same energy with another independent state.