Quantum occupations and limits of equipartition
GRE Subject Test · GRE Physics Topic 36 6:41 English narration · English + 中文 subtitles burned in
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A fermion energy level can hold several particles when it contains several distinct states: the exclusion rule applies to each complete state, not to the level.
当一个费米子能级包含若干个不同的态时，它可以容纳多个粒子：不相容原理作用于每一个完整的量子态，而不是整个能级。
This topic computes fermion and boson occupations, counts identical-particle arrangements, and tests where equipartition fails.
本主题计算费米子和玻色子的占据数，计算全同粒子的排列数，并检验能均分定理在哪里失效。
For noninteracting particles in thermal and particle exchange equilibrium, use chemical potential mu and x equals epsilon minus mu divided by k B T.
对于处于热平衡和粒子交换平衡中的非相互作用粒子，使用化学势μ以及x等于(ε−μ)除以kBT。
Mean occupation of one complete state is n sub F equals 1 over e to the x plus 1 for fermions and n sub B equals 1 over e to the x minus 1 for bosons.
单个完整态的平均占据数为：费米子为n_F等于1除以(e^x+1)，玻色子为n_B等于1除以(e^x−1)。
Fermionic occupation of that state is zero or one; its mean lies between them.
该态的费米子占据数只能是0或1，其平均值介于两者之间。
Bosonic occupation can exceed one.
玻色子的占据数可以超过1。
A level of degeneracy g has total mean g times the single-state occupation when its states share the same energy.
当简并度为g且各态能量相同时，总平均占据数为g乘以单态占据数。
Count spin as part of a complete state.
请将自旋视为完整态的一部分。
The Bose denominator requires epsilon greater than mu for the ordinary finite expression, with the ground-state limit treated separately.
普通有限表达式要求玻色分母中ε大于μ，基态极限需单独处理。
For equilibrium photons mu equals 0 because photon number is not conserved; do not set mu equals 0 for every material particle gas.
对于平衡态光子，由于光子数不守恒，故μ等于0；切勿对所有实物粒子气体都设定μ等于0。
Count occupation patterns rather than labelling identical particles.
要统计占据模式，而不是给全同粒子编号。
Two identical fermions distributed among four distinct complete states have 4 choose 2, which is 6, allowed patterns.
两个全同费米子分布在四个不同的完整态中，有从 4 个中选 2 个，即 6 种允许的模式。
Two identical bosons among those states have 5 choose 2, which is 10, patterns because both may share one state.
两个全同玻色子在这些态中有从 5 个中选 2 个，即 10 种模式，因为两者可以占据同一个态。
Two labelled distinguishable particles would have 4 squared, which is 16, assignments.
两个可编号的可区分粒子则有 4 的平方，即 16 种分配。
These are different counting models, not three interchangeable answers to the same specification.
这些是不同的计数模型，而不是同一题设下三个可互换的答案。
If a question supplies spin degeneracy, first decide whether its stated number counts complete states or just orbital levels.
若题目给出自旋简并度，先判断所给数字计的是完整态还是仅仅轨道能级。
A Pauli prohibition on two identical complete states does not prohibit opposite-spin fermions in one spatial orbital.
泡利原理禁止两个全同粒子处于同一个完整态，但不禁止自旋相反的费米子占据同一个空间轨道。
When x is large and positive, occupation is small and both denominators are dominated by e to the minus x.
当x为较大的正数时，占据数很小，且两个分母均由e的负x次方主导。
So n sub F is approximately equal to n sub B, which is approximately equal to e to the minus x.
因此，费米子占据数n_F约等于玻色子占据数n_B，均约等于e的负x次方。
This is the Maxwell-Boltzmann dilute limit.
这是麦克斯韦-玻尔兹曼稀薄极限。
At x equals natural log of 4, the mean is one fifth for fermions, one third for bosons, and one quarter in the classical approximation. The difference is still significant.
在x等于ln 4时，费米子的平均值为1/5，玻色子为1/3，经典近似下为1/4，差异仍然显著。
A large total particle number alone does not justify classical statistics.
仅总粒子数大并不能证明使用经典统计是合理的。
Density, temperature, and accessible states control occupation.
密度、温度和可及态共同决定占据情况。
For fermions at low T, states below mu become nearly occupied and those above become nearly empty.
对于低温下的费米子，能量低于μ的能级几乎被占满，而高于μ的能级几乎为空。
At epsilon equals mu, a fermionic state has a mean of one half. Inserting that value into the ordinary Bose formula would instead produce a divergent denominator and requires different limiting treatment.
在ε = μ处，费米子态的平均占据数为1/2；若将此值代入普通玻色公式，会导致分母发散，需采用不同的极限处理方法。
Classical equipartition assigns k B T over 2 of mean energy to each independent quadratic term in an equilibrated Hamiltonian.
经典能量均分定理为平衡哈密顿量中的每个独立二次项分配 k_B T/2 的平均能量。
A monatomic ideal gas has three translational terms, giving U equals 3 N k B T over 2 and C sub V equals 3 N k B over 2.
单原子理想气体具有三个平动项，给出 U = 3Nk_BT/2 和 C_V = 3Nk_B/2。
One classical one-dimensional harmonic oscillator has kinetic and potential terms, giving mean k B T and C equals k B.
一个经典的一维谐振子具有动能和势能项，给出平均 k_BT 和 C = k_B。
For a quantum oscillator with fixed spacing epsilon equals h-bar omega, the thermal energy above its temperature-independent zero point is epsilon over e to the minus x minus 1, where x equals epsilon over k B T, and C over k B equals x squared times e to the x, all over e to the x minus 1, squared.
对于具有固定间距 ε = ℏω 的量子谐振子，其高于与温度无关的零点的热能为 ε/(e^x - 1)，其中 x = ε/(k_BT)，且 C/k_B = x²e^x/(e^x - 1)²。
At high T it approaches the classical value; at low T excitation freezes out and C goes to 0.
在高温下它趋近于经典值；在低温下激发冻结，C → 0。
A zero-point energy epsilon over 2 shifts U but not C.
零点能 ε/2 改变 U 但不改变 C。
Molecular rotational or vibrational contributions likewise need their energy scales checked before assigning classical quadratic terms.
分子转动或振动贡献同样需要在分配经典二次项之前检查其能量尺度。
Occupation number belongs to one complete state.
占据数属于一个完整的量子态。
At x equal to ln 4, the fermion occupation is one over 4 plus 1, one fifth, and the boson occupation is one over 4 minus 1, one third.
当 x 等于 ln 4 时，费米子占据数为 1 除以 4 加 1，即五分之一；玻色子占据数为 1 除以 4 减 1，即三分之一。
The classical estimate, e to the minus x, gives one quarter, between the two: not exact here.
经典估计 e 的负 x 次方给出四分之一，介于两者之间：这里它并不精确。
A level with g states holds g times this mean, so a fermion level can hold more than one particle while each state still obeys exclusion.
一个含 g 个态的能级平均容纳 g 倍的占据数，所以费米子能级可以容纳多于一个粒子，而每个态仍然遵守不相容原理。
Three identical particles access four complete states with no energy restriction.
三个全同粒子进入四个完整态，无能量限制。
Count occupation patterns for fermions and bosons, and compare with three labelled particles.
分别计算费米子和玻色子的占据模式，并与三个可标记粒子的情况进行比较。
Explain each counting rule.
解释每种计数规则。
Fermions select three of four states, which is 4 choose 3 equals 4.
费米子从四个态中选择三个，即 C(4,3) = 4。
Bosons distribute three indistinguishable particles among four states, which is 3 plus 4 minus 1 choose 3, equals 20, by stars and bars.
玻色子将三个不可区分粒子分布在四个态中，根据隔板法，C(3+4-1, 3) = 20。
Labelled particles each choose any state, so 4 to the power of 3 equals 64.
标记粒子每个都可以选择任意态，所以 4^3 = 64。
Award each count and exclusion or indistinguishability reasoning.
请给出每个计数以及排除或不可区分性的推理。
Spin is already included in complete states.
自旋已经包含在完整态中。
Exclude two fermions from one complete state, not from an entire degenerate energy level.
从单个完整态中排除两个费米子，而非整个简并能级。
State which x is used; chemical-potential occupations and fixed oscillator excitation formulas are different models.
需明确所使用的是哪个x。 化学势占据数公式与固定振子激发公式属于不同模型。
mean occupation: Ensemble average number of particles in one complete quantum state or a specified group of states.
平均占据数：一个完整量子态或指定状态组中粒子的系综平均值。
equipartition: Classical equilibrium rule assigning kBT/2 to each independent quadratic Hamiltonian term.
能量均分：经典平衡规则，为每个独立的二次哈密顿项分配 kBT/2。
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其含义，然后更改原始传递中的一个参数并检验极限情况。