Cơ học lượng tử và vật lý nguyên tử
Đã giới thiệu| English | Ý nghĩa |
|---|---|
| normalisation/ˌnɔːməlaɪˈzeɪʃn/ | Chuẩn hóa |
| commutator/ˈkɒmjuːteɪtə/ | The operator AB minus BA. |
| Hermitian/hɜːˈmɪʃn/ | Equal to its own conjugate transpose, so with real eigenvalues. |
A box has no resting particle
- Confine an electron to a box and its lowest energy is not zero: the wavefunction must vanish at the walls, so it must curve.
- GRE quantum questions test normalisation 归一化, measurement probabilities, the infinite well and commutators 对易子.
Probabilities from coefficients
- $\int |\psi|^2 = 1$. For orthonormal $\phi_j$, $\psi = \sum c_j\phi_j$ has $\sum |c_j|^2 = 1$; measuring gives $E_j$ with probability $|c_j|^2$.
- $\psi = (\phi_1 + 2\phi_2)/\sqrt5$: $P(E_1) = \tfrac15$, $P(E_2) = \tfrac45$. Relative phases affect position densities, not these probabilities.
$\psi = (\phi_1 + 2\phi_2)/\sqrt5$. What is $\langle E\rangle$ in units of $E_1$ for an infinite well?
Độ khó 2/5 · 140 xu + một phần thưởng nhỏ cho lần hoàn thành đầu tiên trong ngày
$\tfrac15 \cdot 1 + \tfrac45 \cdot 4$.
For an equal superposition of 4 orthonormal states, each coefficient has magnitude 1/____.
Độ khó 1/5 · 100 xu + một phần thưởng nhỏ cho lần hoàn thành đầu tiên trong ngày
Normalisation needs $4|c|^2 = 1$, so $|c| = 1/\sqrt4 = 1/2$.
Observables and averages
- Observables are Hermitian 厄米的 operators: real eigenvalues. $\langle A\rangle = \int \psi^* A\psi\,dx$ is an average over many preparations, not one result.
- If $A^4 = I$ and $A$ is Hermitian, its eigenvalues are $\pm1$ — the complex fourth roots are excluded.
The infinite well
- $0 < x < L$: $\psi_n = \sqrt{2/L}\sin(n\pi x/L)$, $E_n = \dfrac{n^2\pi^2\hbar^2}{2mL^2} = \dfrac{n^2h^2}{8mL^2}$, $n = 1, 2, \dots$
- $E_n = n^2E_1$. Doubling $L$ divides every level by 4. $n = 0$ is not a state ($\psi \equiv 0$).
$E_2 = 4E_1$, $E_3 = 9E_1$. The transition $3 \to 2$ emits a photon of energy $5E_1$.
Do not swap $h$ and $\hbar$: $n^2h^2/8mL^2$ and $n^2\pi^2\hbar^2/2mL^2$ are the same number, but mixing the forms is off by $4\pi^2$.
Photon energy of the $4 \to 2$ transition in an infinite well, in units of $E_1$?
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Infinite-well energies go as $n^2E_1$: $E_4 - E_2 = (16 - 4)E_1 = 12E_1$.
The well width is tripled. The ground-state energy is multiplied by
Độ khó 1/5 · 100 xu + một phần thưởng nhỏ cho lần hoàn thành đầu tiên trong ngày
Since $E_n \propto 1/L^2$, tripling $L$ divides every level by $3^2 = 9$.
Commutators
- $[A, B] = AB - BA$. Commuting observables share an eigenbasis. $[x, p] = i\hbar$.
- $p$ commutes with $p^2/2m$ but not with $V(x)$. $[J_x, J_y] = i\hbar J_z$; $J^2$ commutes with each component.
Do they commute?
Recognise compatible and incompatible observables.
$H = p^2/2m + V(x)$ commutes with $p$ only when $V$ is constant.
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True. $[H, p] = [V(x), p] = i\hbar V'(x)$, which vanishes only if $V' = 0$; then momentum is conserved.
Recap
- Probabilities are $|c_j|^2$; normalise first.
- Infinite well: $E_n = n^2E_1$, $E \propto 1/L^2$, ground state above zero.
- Commuting operators can be measured together; $x, p$ and $J_x, J_y$ cannot.
- commutator: The operator AB minus BA.
- Hermitian: Equal to its own conjugate transpose, so with real eigenvalues.