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GRE Subject Test · GRE Physics Chủ đề 5 6:11 Giọng đọc tiếng Anh · phụ đề tiếng Anh + 中文 được ghi trực tiếp
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A particle in a box cannot have zero kinetic energy, even in its ground state: confining it changes which states are allowed.
盒中粒子即使处于基态，动能也不可能为零：束缚改变了允许存在的状态。
This topic uses wavefunctions, operators and measurement probabilities; solves the standard bound-state models and tunnelling; and applies angular momentum, spin and atomic spectra.
本主题运用波函数、算符和测量概率；求解标准束缚态模型和隧穿问题；并应用角动量、自旋和原子光谱。
A state wavefunction must satisfy the model's boundary conditions and have total probability one.
态波函数必须满足模型的边界条件，且总概率为一。
Integrate the absolute value of psi of x, squared, over the allowed domain.
在允许的定义域上对 |ψ(x)|² 进行积分。
For orthonormal states phi sub j, a superposition psi equals the sum of c sub j times phi sub j has norm squared equal to the sum of the absolute value of c sub j, squared, because the cross terms integrate to zero.
对于正交归一态 φ_j，叠加态 ψ = Σc_jφ_j 的模平方为 Σ|c_j|²，因为交叉项积分为零。
Thus an equal superposition of N orthonormal states has coefficient magnitude equal to 1 divided by the square root of N.
因此，N 个正交归一态的等幅叠加中，系数的模长为 1/√N。
The probability of finding one constituent energy is the absolute value of c sub j, squared, not c sub j.
测得某一组成能量本征值的概率是 |c_j|²，而非 c_j。
Relative phases may affect position probabilities even when energy probabilities are unchanged.
即使能量概率不变，相对相位仍可能影响位置概率。
An observable is represented by a Hermitian operator.
可观测物理量由厄米算符表示。
Its eigenvalues are real, and a normalised state gives expectation value angle bracket A angle bracket equals the integral of psi star A psi dx.
其特征值为实数，归一化态给出的期望值⟨A⟩等于ψ* A ψ对x的积分。
The expectation is an average over repeated preparations, not necessarily the result of one measurement.
期望值是多次重复制备的平均结果，不一定是一次测量的结果。
If A satisfies a polynomial identity, apply it to an eigenstate. For A to the fourth power equals I, a real eigenvalue must be plus 1 or minus 1.
若A满足多项式恒等式，则将其作用于本征态：对于A的四次方等于I的情况，实特征值必为+1或−1。
A complex fourth root is excluded by Hermiticity even though it solves the polynomial.
尽管复四次根满足该多项式，但会被厄米性排除。
For an infinite well between zero and L, the boundary conditions psi of zero equals psi of L equals zero allow sine modes sin of n pi x over L, with n equal to one, two, and so on.
对于无限深势阱，在零到L之间，边界条件ψ(0) = ψ(L) = 0允许正弦模式sin(nπx/L)，其中n等于1、2等。
Their energies are E sub n equals n squared pi squared h-bar squared divided by two m L squared, so the ground state is not zero and doubling L divides every energy by four.
它们的能量为E_n = n²π²ℏ²/(2mL²)，因此基态不为零，且将L加倍会使所有能量除以四。
The equivalent h expression is n squared h squared divided by eight m L squared; never substitute h for h-bar without its 2 pi factor.
等效的h表达式为n²h²/(8mL²)；切勿在没有乘以2π因子的情况下用h代替ℏ。
A finite barrier instead permits exponential tails, and a traveling wave in a classically allowed region has oscillatory rather than exponentially decaying spatial dependence.
有限势垒则允许指数尾部存在，而在经典允许区域中的行波具有振荡而非指数衰减的空间依赖关系。
Operators commute when the commutator of A and B equals A times B minus B times A, which equals zero. Compatible observables can have a common eigenbasis.
算符满足 = AB−BA = 0 时彼此对易；相容可观测量可拥有共同本征基。
Since kinetic energy p squared over two m is a function of p, it commutes with p.
由于动能 p²/(2m) 是 p 的函数，它与 p 对易。
A position-dependent potential generally makes H equal to p squared over two m plus V of x fail to commute with p.
位置相关的势通常导致 H = p²/(2m)+V(x) 与 p 不对易。
Angular momentum obeys the commutator of J sub x and J sub y equals i h-bar J sub z and cyclic permutations, while J squared commutes with each component.
角动量满足 = iℏJ_z 及其循环置换关系，而 J² 与各分量均对易。
Atomic transitions exchange positive photon energy equal to a level difference. Angular momentum and mechanism-specific selection rules restrict which transitions occur.
原子跃迁交换的正光子能量等于能级差；角动量和机制特定的选择定则限制了哪些跃迁可以发生。
In an infinite well from zero to L, the wavefunction must vanish at both walls.
在从 0 到 L 的无限深势阱中，波函数在两壁处必须为零。
That allows only the sine modes, root two over L times sin of n pi x over L, with energies n squared pi squared h-bar squared over two m L squared.
因此只允许正弦模式：根号下 2/L 乘以 sin(nπx/L)，能量为 n²π²ħ² 除以 2mL²。
Prepare psi equals psi sub 1 plus i psi sub 2, all over the square root of 2.
准备波函数 ψ = (ψ₁ + iψ₂) / √2。
Orthogonality makes its norm one.
正交性使其归一化为 1。
Energy measurements give E sub 1 or 4 E sub 1 with equal probability.
能量测量结果为 E₁ 或 4E₁，且概率相等。
Each outcome has probability one half, so the mean energy is half of E one plus half of four E one: five halves E one.
每个结果的概率都是二分之一，所以平均能量等于 E₁ 的一半加 4E₁ 的一半，即 2.5 倍 E₁。
This average is not an allowed single energy result.
这个平均值不是允许的单一能量结果。
The relative phase changes in time, so a superposition of different energies is not a stationary state.
相对相位随时间变化，因此不同能量的叠加态不是定态。
In the same well, a normalized state is psi one plus two psi three, all over the square root of five.
在同一势阱中，一个归一化态为 (ψ₁ + 2ψ₃) / √5。
Find its possible measured energies and probabilities, mean energy, and energy variance.
求其可能的测量能量及概率、平均能量和能量方差。
Is the mean an allowed result of one measurement?
平均值是否可能为单次测量的结果？
The results are E 1 with probability one fifth and 9 E 1 with probability four fifths.
测量结果为 E1，概率为五分之一；以及 9 E1，概率为五分之四。
The expectation value of E equals 37 E 1 over 5.
E 的期望值等于 37 E1 除以 5。
The expectation value of E squared equals 65 E 1 squared.
E 平方的期望值等于 65 乘以 E1 的平方。
The variance of E equals 65 E 1 squared minus the square of 37 E 1 over 5, which simplifies to 256 E 1 squared over 25.
方差等于 65 乘以 E1 的平方，减去 37 E1 除以 5 的平方，化简为 256 乘以 E1 的平方再除以 25。
The mean, 7.4 E 1, is not an energy eigenvalue of the well.
均值 7.4 E1 并不是势阱的能量本征值。
Award results, probabilities, mean, second moment, variance calculation, and interpretation.
评分要点包括：结果、概率、均值、二阶矩、方差计算及解释。
A stationary state can have a nonzero energy even though its probability density is time-independent.
定态可以具有非零能量，即使其概率密度不随时间变化。
normalisation: Making total probability equal to one.
归一化：使总概率等于一。
operator: A mathematical action representing an observable.
算符：代表可观测量的数学作用。
Retrieve the meanings without reading the definitions, then change a parameter in the original transfer and test the limiting case.
在不阅读定义的情况下回忆含义，然后在原始传递中更改一个参数并检验极限情况。