量子力学および原子物理学
導入済み| English | 意味 |
|---|---|
| operator/ˈɒpəreɪtə/ | 演算子 |
| normalisation/ˌnɔːməlaɪˈzeɪʃn/ | 正規化 |
A decision before an answer
- A particle in a box cannot have zero ground-state kinetic energy. Its confinement changes what states are allowed.
- Your goal: Use wavefunctions, operators and measurement probabilities.
Read the relationship
- A state wavefunction must satisfy the model’s boundary conditions and have total probability one: integrate |ψ(x)|² over the allowed domain. For orthonormal states φ_j, a superposition ψ=Σc_jφ_j has norm squared Σ|c_j|² because the cross terms integrate to zero. Thus an equal superposition of N orthonormal states has coefficient magnitude 1/sqrt(N). The probability of finding one constituent energy is |c_j|², not c_j. Relative phases may affect position probabilities even when energy probabilities are unchanged.
- Solve standard bound-state models and tunnelling reasoning.
In an infinite well, E3/E1 is:
難易度 1/5 · 100 コイン + 今日初回完了時に小ボーナス付与
Energy scales as n².
Use the defining rule
- An observable is represented by a Hermitian operator. Its eigenvalues are real, and a normalised state gives expectation value ⟨A⟩=∫ψ* Aψ dx. The expectation is an average over repeated preparations, not necessarily the result of one measurement. If A satisfies a polynomial identity, apply it to an eigenstate: for A⁴=I, a real eigenvalue must be +1 or −1. A complex fourth root is excluded by Hermiticity even though it solves the polynomial.
- Apply angular momentum, spin and atomic spectral principles.
The probability density is:
難易度 1/5 · 100 コイン + 今日初回完了時に小ボーナス付与
The modulus squared gives density.
Check the conditions
- For an infinite well 0<x<L, ψ(0)=ψ(L)=0 allows sine modes sin(nπx/L), with n=1,2,… . Their energies are E_n=n²π²ℏ²/(2mL²), so the ground state is not zero and doubling L divides every energy by four. The equivalent h expression is n²h²/(8mL²); never substitute h for ℏ without its 2π factor. A finite barrier instead permits exponential tails, and a travelling wave in a classically allowed region has oscillatory rather than exponentially decaying spatial dependence.
- Apply angular momentum, spin and atomic spectral principles.
For an infinite well, E2=4E1 and E3=9E1. A transition from n=3 to n=2 releases 5E1. Doubling L lowers every energy by factor four. n=0 would give the zero wavefunction, not a physical state.
If E1=2 eV, E2=____ eV.
難易度 1/5 · 100 コイン + 今日初回完了時に小ボーナス付与
E2=4E1=8 eV.
Apply the task format
- Operators commute when [A,B]=AB−BA=0; compatible observables can have a common eigenbasis. Since kinetic energy p²/(2m) is a function of p, it commutes with p. A position-dependent potential generally makes H=p²/(2m)+V(x) fail to commute with p. Angular momentum obeys [J_x,J_y]=iℏJ_z and cyclic permutations, while J² commutes with each component. Atomic transitions exchange positive photon energy equal to a level difference; angular momentum and mechanism-specific selection rules restrict which transitions occur.
- Apply angular momentum, spin and atomic spectral principles.
A stationary state can have a nonzero energy even though its probability density is time-independent.
このケースに当てはまる答えはどれか？
波動関数、演算子および測定確率を用いる
A normalised quantum state has total probability one.
難易度 1/5 · 100 コイン + 今日初回完了時に小ボーナス付与
Integrating its probability density over the full domain gives one.
Keep the distinctions
- normalisation 归一化 — Making total probability equal to one.
- operator 算符 — A mathematical action representing an observable.
- Use wavefunctions, operators and measurement probabilities.
- Solve standard bound-state models and tunnelling reasoning.
- Apply angular momentum, spin and atomic spectral principles.
Match each term with its precise meaning in this lesson.
難易度 2/5 · 140 コイン + 今日初回完了時に小ボーナス付与
Keep the distinctions stated in the teaching example.
Put this lesson’s reasoning or event sequence in order.
難易度 3/5 · 200 コイン + 今日初回完了時に小ボーナス付与
The order follows the stated process; check each stage before the next.