Quantum mechanics and atomic physics · Mecânica quântica e física atômica
| English | Português |
|---|---|
| operator/ˈɒpəreɪtə/ | operador |
| normalisation/ˌnɔːməlaɪˈzeɪʃn/ | normalização |
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:
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:
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.
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.
Which answer fits this case?
Use wavefunctions, operators and measurement probabilities
A normalised quantum state has total probability one.
Integrating its probability density over the full domain gives one.
Keep the distinctions
- normalisation 归一化 — Making total probability equal to one.
- operator · operador 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.
Keep the distinctions stated in the teaching example.
Put this lesson’s reasoning or event sequence in order.
The order follows the stated process; check each stage before the next.