量子力学期中real
量子力学期中考试试题
量子力学期中考试试题物理常数：光速：812.99810c m s -=⨯⋅；普朗克常数：346.62610h J s -=⨯⋅；玻尔兹曼常数：231.38110/B k J K -=⨯；电子质量：319.10910e m kg -=⨯；碳原子质量：2612 2.00710C m u kg -==⨯；电子电荷：191.60210e C -=⨯一、填空题：1、 量子力学的基本特征是 。
2、 波函数的性质是 。
3、1924年，德布洛意提出物质波概念，认为任何实物粒子，如电子、质子等，也具有波动性，对于具有一定动量p 的自由粒子，满足德布洛意关系： ； 假设电子由静止被150伏电压加速，求加速后电子的的物质波波长： （保留1位有效数字）；对宏观物体而言，其对应的德布洛意波波长极短，所以宏观物体的波动性很难被我们观察到，但最近发现介观系统（纳米尺度下的大分子）在低温下会显示出波动性。
计算1K 时，60C 团簇（由60个C 原子构成的足球状分子）热运动所对应的物质波波长：_______________（保留2位有效数字）。
4.一粒子用波函数Φ(,)rt 描写,则在某个区域dV 内找到粒子的几率为 。
5、线性谐振子的零点能为 。
6、厄密算符的本征值必为 。
7、氢原子能级n =5的简并度为 。
8、完全确定三维空间的自由粒子状态需要三个力学量，它们是 。
9、测不准关系反映了微观粒子的 。
10. 等人的实验验证了德布罗意波的存在。
11. 通常把 称为束缚态。
12. 波函数满足的三个基本条件是： 。
13.一维线性谐振子的本征能量与相应的本征函数分别为： 14．两力学量对易的说明： 。
15. 坐标与动量的不确定关系是： 。
16. 氢原子的本征函数一般可以写为： 。
17. 何谓定态： 。
1. 束缚态、非束缚态及相应能级的特点。
2. 简并、简并度。
3. 用球坐标表示，粒子波函数表为 ()ϕθψ,,r ，写出粒子在立体角Ωd 中被测到的几率。
量子力学基本概念总结
量子力学基本概念总结量子力学是一门描述微观粒子行为的物理学分支，它提供了一种理论框架，用于解释和预测原子、分子和基本粒子的现象。
以下是一些量子力学的基本概念的总结。
1. 波粒二象性（Wave-particle duality）量子力学中的一个重要概念是波粒二象性，即微观粒子既可以表现出粒子特性也可以表现出波动特性。
例如，电子可以像波一样传播，但也可以被当作是粒子来计算。
2. 不确定性原理（Heisenberg's Uncertainty Principle）不确定性原理是由波粒二象性导致的。
它表明在粒子的位置和动量之间存在一种固有的不确定性。
换句话说，我们无法同时准确知道一个粒子的位置和动量，只能知道它们之间的不确定性。
3. 玻尔模型（Bohr model）玻尔模型是描述原子结构的经典模型之一。
它基于量子力学中能级的概念，认为电子围绕着原子核在不同的能级轨道上运动。
这个模型解释了原子光谱、电离能和跃迁等现象。
4. 波函数（Wave function）波函数是量子力学中用来描述粒子状态的数学函数。
它包含了所有关于粒子位置、动量和能量等信息。
根据波函数，我们可以计算出粒子的一些物理性质。
5. 测量与观测（Measurement and Observation）量子力学强调测量和观测对系统产生影响。
在测量时，波函数将塌缩到某个确定的状态，并给出对应的测量结果。
这种波函数塌缩导致了一系列奇特的现象，如量子纠缠和量子隐形。
6. 量子纠缠（Quantum Entanglement）量子纠缠是量子力学中的一个非常奇特的现象。
当两个或更多粒子处于纠缠状态时，它们的态无法独立地描述，而必须考虑整个系统的态。
当一个粒子的状态发生改变时，纠缠粒子的状态也会瞬间发生变化，即使它们之间的距离很远。
7. 施特恩-盖拉赫实验（Stern-Gerlach Experiment）施特恩-盖拉赫实验是证明电子具有自旋的经典实验之一。
epx函数范文
epx函数范文def epx(z):real_part = z.realimag_part = z.imagexp_real = math.exp(real_part)result = exp_real * exp_imagreturn result其中，z是一个复数，real_part代表复数的实部，imag_part代表复数的虚部。
函数使用math库中的exp函数计算实部的指数形式，并使用math库中的cos和sin函数计算虚部的指数形式。
最后，将实部和虚部的指数形式相乘，得到最终的结果。
epx函数的应用广泛，特别是在电路分析、信号处理和量子力学等领域。
在电路分析中，复数常用于表示电压、电流或阻抗等。
通过epx函数，可以进行复数的指数运算，如计算复数的平方、立方等。
在信号处理中，复数常用于表示频域信号，通过epx函数可以进行傅里叶变换等运算。
在量子力学中，复数表示量子态，通过epx函数可以计算量子态的时间演化等。
举例来说，假设有一个复数z=3+4i，我们可以使用epx函数计算其指数形式。
import mathdef epx(z):real_part = z.realimag_part = z.imagexp_real = math.exp(real_part)result = exp_real * exp_imagreturn resultresult = epx(z)print(result)运行结果为：因此，epx函数是一个非常实用的数学函数，可以方便地进行复数的指数运算，丰富了数学计算的工具箱。
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II. T he Machinery of Quantum MechanicsBased on the results of the experiments described in the previous section, we recognize that real experiments do not behave quite as we expect. This section presents a mathematical framework that reproduces all of the above experimental observations. I am not going to go into detail about how this framework was developed. Historically, the mathematical development of QM was somewhat awkward; it was only years after the initial work that a truly rigorous (but also truly esoteric) foundation was put forth by Von Neumann. At this point, we will take the mathematical rules of QM as a hypothesis that is consistent with all the experimental results we have encountered.Now, there is no physics or chemistry in what we are about to discuss; the physics always arises from the experiments. However, just as Shakespeare had to learn proper spelling and grammar before he could write Hamlet, so we must understand the mathematics of QM before we can really start using it to make interesting predictions. This is both the beauty and the burden of physical chemistry; the beauty because once you understand these tools you can answer any experimental question without having to ask a more experienced colleague; the burden because the questions are very hard to answer.A. Measurements Happen in Hilbert SpaceAll the math of QM takes place in an abstract space that called Hilbert Space. The important point to realize is that Hilbert Space has no connection with the ordinary three dimensional space that we live in. For example, a Hilbert Space can (and usually does) have an infinite number of dimensions. These dimensions do not correspond in any way to the length, width and height we are used to. However, QM gives us a set of rules that connect operations in Hilbert Space to measurements in real space. Given a particular experiment, one constructs the appropriate Hilbert Space, and then uses the rules of QM within that space to make predictions.1. Hilbert Space Operators Correspond to Observables The first rule of QM is: all observables are associated with operators in Hilbert Space. We have already encountered this rule, we just didn’t know the operators lived in Hilbert space. Now, for mostintents and purposes, Hilbert Space operators behave like variables: you can add them, subtract them, multiply them, etc. and many of thefamiliar rules of algebra hold, for example (Z Y Xˆ,ˆ,ˆare arbitrary operators):Addition Commutes:X Y Y Xˆˆˆˆ+=+ Addition is Associative: ()()Z Y X Z Y Xˆˆˆˆˆˆ++=++ Multiplication is Associative:()()Z Y X Z Y X ˆˆˆˆˆˆ= However, the multiplication of operators does not commute :Multiplication does not commute:X Y Y Xˆˆˆˆ≠ We already knew that this was true; in the case of the polarizationoperators we showed that x Pˆ and 'ˆx P do not commute: y x x y P P P Pˆˆˆˆ''≠ Thus, the association of observables with operators allows us to describe the first quantum effect we discovered in the experiments: non-commuting observations . Also, note that uncertainty comes solely from the fact that the order of measurements matters; hence we can’t know the result of both measurements simultaneously.Now, deciding that operators have all the above features (e.g. associative multiplication, commutative addition) may seem rather arbitrary at first. For example, why does operator multiplication need to be associative? The deep result that motivates this is a theorem that asserts that if a set of operators satisfies the above relations (together with a few other benign conditions) guarantees that operators in Hilbert space can always be represented bymatrices . Hence a better way to remember how to multiply and add operators is to remember that they work just like matrices; any relation that is true for two arbitrary matrices is also true for two arbitrary operators.2. The System is Described by a State VectorIn Hilbert Space, the system is represented by a state . Again, we already knew this, but the fact that the states live in Hilbert space lets us know some new facts. First, we note that there are three simpleoperations one can execute on a state. First, one can multiply it by aconstant to obtain a new state:c c ψψ=In general, this constant can be complex . It does not matter which side the constant appears on. The second thing one can do is to addtwo states together to make a new state:21ψψψ+=As we have seen before, ψ is a superposition of the two states 1ψ and 2ψ. Finally, there is one new operation we need to introduce, called Hermitian conjugation . By definition, the Hermitian conjugate (dentoed by ‘†’) is given by:()**2211†2211c c c c ψψψψ+=+()2211†2211**ψψψψc c c c +=+Where ‘*’ denotes complex conjugation. Further, Thus, the Hermitian conjugate takes kets to bras (and vice versa) and takes the complex conjugate of any constant. Hermitian conjugation in Hilbert space isanalogous to the transpose in a traditional vector space. Thus:()T =⇔=†ψψ To be precise, we will ultimately find that Hermitian conjugation is the same as taking the transpose and complex conjugate simultaneously. Finally, we note one important fact about a Hilbert space. There always exists a basis of states, {}αφ, such that any other state canbe written as a linear combination of the basis states:=αααφψcWe have as yet said nothing about the number of these states. In general, the basis for a Hilbert space involves an infinite number of states. The definition above assumes they are denumerable (i.e. we can assign them numbers i=1,2,3,4…) In some situations, the basis will be continuous . In these situations, we can replace the sum byan integral:()αφαψαd c =.3. Bra-Ket Gives ProbabilityNow, in order to make predictions, we need to understand a few properties of the bra-ket product. To be mathematically precise, bar and ket states are dual to one another. The illustration in terms of vectors is invaluable in understanding what this means, because column vectors and row vectors are also dual to one another. Thus, essentially all the properties of row and column vectors can betransferred over to bra and ket states. Most notably, one can define an overlap (or inner product) analogous to the dot product forordinary vectors.()⇔.ψχ The overlap between a bra and a ket has all the same intuitivecontent as the dot product: it tells you how similar the two states are. If the overlap is zero, the two states are orthogonal . We can also define the norm of a state by:ψψψ=2One of the properties of the bracket product in Hilbert space is that the norm of a state is always greater than or equal to zero and it can only be zero for the trivial state that corresponds to the origin. It turns out that the norm of the state has no physical relevance; any value between 0 and gives the same physical answer. In practice it is often easiest to multiply the wavefunction by a normalizationconstant, 2/1−=ψψc , that makes the norm 1. This does not affectour predictions but often makes the expressions simpler. If two states are both orthogonal to one another and normalized, they are said to be orthonorma l .As mentioned above, operators can be associated with matrices. It is therefore natural to associate an operator acting on a ket state with amatrix-vector product:× ⇔ψO ˆ This allows us to define the Hermitian Conjugate (HC) of an operatorby forcing the HC ψOˆ to be the HC of ψ times the HC of O ˆ:()††ˆˆOO ψψ≡ This defines †ˆO, the HC of O ˆ. This is also called the adjoint of the operator Oˆ. If an operator is equal to its adjoint, it is hermitian . This is analogous to a symmetric matrix.It is important to notice that the order of operations is crucial at this point. Operators will always appear to the left of a ket state and to the right of a bra state. The expressionsOand O ˆˆψψ are not incorrect; they are simply useless in describing reality. This might be clearer if we write the associated matrix expressions:()and One can give meaning to these expressions (in terms of a tensor product) but the result is not useful.We are now in a position to restate the third rule of QM: for a system, is given by:ψψψψO O ˆˆ=. Note that this equation simplifies if ψis normalized, in which caseψψO Oˆˆ=. 4. Operators and EigenvaluesOne important fact is that operators in Hilbert Space are always linear , which means:()2121ˆˆˆψψψψO O O+=+ This is another one of the traits that allows operators to berepresented in terms of a matrix algebra (they call it linear algebra for a reason).Now, one can associate a set of eigenvalues, αo , and eigenstates,αψ, with any linear operator, Oˆ, by finding all of the solutions of the eigenvalue equation:αααψψo O=ˆ This allows us to state the final two rules of QM: when measuring the value of the observable O , the only possible outcomes are the eigenvalues of Oˆ. If the spectrum of eigenvalues of O ˆ is discrete, this immediately implies that the resulting experimental results will be quantized , as we know is quite often the case. If the spectrum ofeigenvalues of Oˆ is continuous, then this rule gives us little information. And, finally, after O has been observed and found to have a valueothen the wavefunction of the system collapses into α.5. Some Interesting Facts Before moving on to describe the experiments from the previous section in terms of our newly proposed rules, it is useful to define a few concepts. The first is the idea of an outer product. Just as we can write the inner product as (bra)x(ket), we can write the outer product as (ket)x(bra). This is in strict analogy to the case of vectors where the outer product is a column vector times a row vector:() ⇔ψχ As we have seen in the polarization experiments, the outer product is an operator; if we act on a state with it, we get another state back: ()()φψχφψχφψχ≡==c c This is, again, in direct analogy with vector algebra, where the outer product of two vectors is a matrix. One interesting operator is the outer product of a ket with its own bra, which is called the density operator :ψψψ=P ˆ If ψ is normalized, this operator happens to be equal to its own square:ψψψψψψP P P ˆˆˆ===1This property is called idempotency . Hence, we see that the density operator for any quantum state is idempotent. Further, we see that ψP ˆ acting on any state gives back the state ψ times a constant: ()()φψψφψψφψψ≡==c cBy this token, density operators are also called projectionoperators , because they project out the part of a given wavefunction that is proportional to ψ.One very important fact about Hilbert space is that there is always a complete orthonormal basis , {i φ, of ket states. As the nameimplies, these states are orthonormal (the overlap between different states is zero and each state is normalized) and the form a basis (any state ψ can be written as a linear combination of these states). We can write the orthonormality condition in shorthand asij j i δφφ=Where we have defined the Kroenecker delta - a symbol that is one if i=j and zero otherwise.The first important results we will prove concern Hermitian operators.Given a Hermitian operator, Hˆ, it turns out that 1) the eigenvalues of Hˆ are always real, and 2) the eigenstates can be made to form a complete orthonormal basis. Both these facts are extremely important. First, recall that we know experimental results (which correspond to eigenvalues) are always real numbers; thus, it is not surprising that every observable we deal with in this course will be associated with a Hermitian operator. Also, note that up to now we have appealed to the existence of an orthonormal basis, but gave no hints about how such a basis was to be constructed. We now see that every Hermitian operator associated with an observation naturally defines its own orthonormal basis!As with nearly all theorems in chemistry, the most important part of this is the result and not how it is obtained. However, we will outline the proof of this theorem, mostly to get a little more practice with ins and outs of Dirac notation.______________________________________________________1) Consider the eigenvalue equation and its Hermitian conjugate:*ˆˆααααααψψψψh H h H Conjugate Hermitian = → = Now we apply one of our tricks and take the inner product of the left equation with αψ and the inner product of the right equation with αψ:ααααααααααψψψψψψψψ*ˆˆh H h H ==We see that the left hand sides (l.h.s.) of both equations are the same, so we subtract them to obtain: ()ααααψψ*0h h −= . In order to have the right hand side (r.h.s) be zero, either:()*0ααh h −= or ααψψ=0Since we defined our states so that their norms were not zero , we conclude that()*0ααh h −= Which implies that αh is real2) Here, we need to prove that the eigenstates are a) normalized, b) orthogonal and c) form a complete basis. We will take these points in turn.a) The eigenstates can be trivially normalized, since if αψ isan eigenstate of H ˆ, then so is αψc : ()()αααααααψψψψψc h ch H c c H c H====ˆˆˆ So given an unnormalized eigenstate, we can always normalize it without affecting the eigenvalueb) Consider the ket eigenvalue equation for one value of α and the bra equation for 'α'''ˆˆααααααψψψψh Hh H == where we have already made use of the fact that *''ααh h =.Now, take the inner product of the first equation with 'αψ andthe second with αψ. Then:ααααααααααψψψψψψψψ'''''ˆˆh Hh H == Once again, the l.h.s. of the equations are equal and subtracting gives: ()ααααψψ''0h h −=Thus, either:()'0ααh h −= or ααψψ'0=Now, recall that we are dealing with two different eigenstates (i.e. 'αα≠). If the eigenvalules are not degenerate (i.e.'ααh h ≠), then the first equation cannot be satisfied and theeigenvectors must be orthogonal. In the case of degeneracy, however, we appear to be out of luck; the first equality issatisfied and we can draw no conclusions about theorthogonality of the eigenvectors. What is going on? Notice that, if h h h ≡='αα, then any linear combination of the twodegenerate eigenstates, 'ααψψb a +, is also an eigenstate with the same eigenvalue :()()''''ˆˆˆααααααααψψψψψψψψb a h bh ah H b H a b a H+=+=+=+ So, when we have a degenerate eigenvalue, the definition of the eigenstates that correspond to that eigenvalue are notunique, and not all of these combinations are orthogonal to one another. However, there is a theorem due to Gram andSchmidt – which we will not prove – that asserts that at least one of the possible choices forms an orthonormal set . The difficult part in proving this is that there may be two, three, four… different degenerate states. So, for non-degenerate eigenvalues, the states must be orthogonal, while for adegenerate eigenvalue, the states are not necessarily orthogonal, we are free to choose them to be orthogonalc) The final thing we need to prove is that the eigenstates form a complete basis. Abstractly, this means that we can write any other state as a linear combination of the eigenstates:=αααψχcThis turns out to be difficult to prove, and so we simply defer to our math colleagues and assert that it can be proven_______________________________________________________Finally, it is also useful to define the commutator of two operators:[]A B B A B Aˆˆˆˆˆ,ˆ−≡ If two operators commute, then the order in which they appear does not matter and the commutator vanishes. Meanwhile, if the operatorsdo not commute, then the commutator measures “how much” the order matters.。
量子力学中的自由粒子和动量本征态
量子力学中的自由粒子和动量本征态量子力学是现代物理学中最重要的分支之一，它描述了微观世界中粒子的行为。
在量子力学中，自由粒子是一种特殊的粒子，它没有受到外部力的作用，可以自由运动。
而动量本征态则是描述自由粒子的状态。
本文将详细介绍量子力学中的自由粒子和动量本征态。
首先，让我们来了解一下自由粒子的概念。
自由粒子是指在没有外部力作用下的粒子，它可以自由地运动。
在经典物理学中，自由粒子的运动可以由牛顿的运动定律描述，但在量子力学中，自由粒子的行为却具有一些奇特的特性。
在量子力学中，自由粒子的运动状态可以用波函数来描述。
波函数是一个复数函数，它包含了粒子的位置和动量信息。
对于自由粒子来说，其波函数可以用平面波函数表示。
平面波函数是一种特殊的波函数形式，它具有均匀的振幅和相位分布。
自由粒子的波函数可以写成如下形式：ψ(x,t) = A * exp(i(kx - ωt))其中，ψ(x,t)是波函数，A是振幅，k是波矢，x是位置，ω是角频率，t是时间。
这个波函数描述了自由粒子在空间中的分布和随时间的演化。
接下来，让我们来介绍动量本征态。
动量本征态是指具有确定动量的粒子的状态。
在量子力学中，动量是一个算符，它作用在波函数上可以得到一个复数，表示粒子的动量。
动量算符的本征态就是动量本征态，它们满足如下的本征方程：P |p⟩ = p |p⟩其中，P是动量算符，|p⟩是动量本征态，p是动量的本征值。
动量本征态具有特殊的性质，它们是正交归一的。
这意味着不同动量本征态之间的内积为零，同一动量本征态的内积为一。
动量本征态可以用平面波函数表示。
对于自由粒子来说，动量本征态的波函数可以写成如下形式：ψp(x,t) = N * exp(i(px - Et)/ħ)其中，ψp(x,t)是动量本征态的波函数，N是归一化常数，p是动量，E是能量，ħ是普朗克常数除以2π。
这个波函数描述了具有确定动量的自由粒子的空间分布和随时间的演化。
动量本征态具有一些重要的性质。
量子力学中的粒子自旋与统计
量子力学中的粒子自旋与统计量子力学是研究微观世界的一门重要学科，而粒子自旋与统计则是其中的一个核心概念。
在量子力学中，我们了解到粒子的自旋是一个与传统的旋转概念不同的性质，而粒子的统计则是描述粒子在量子态上的行为规律。
本文将深入探讨粒子自旋与统计的相关内容，帮助读者更好地理解这一复杂而又神秘的领域。
在传统的经典物理学中，我们通常将物体的旋转描述为一个具有确定方向和大小的量。
然而，在量子力学中，粒子的自旋并不是描述粒子旋转的角动量，而是与粒子的内禀性质有关的一个量。
粒子的自旋可以有两个可能的取值，分别为正自旋和负自旋，通常用1/2和-1/2来表示。
粒子的自旋是量子力学中一项非常重要的概念，它决定了粒子在外加磁场中的行为。
根据物理学家斯特恩和格拉赫的实验结果，我们得知自旋1/2的粒子具有和指针磁场发生相互作用的能力，并且可以出现两个不同的测量结果：自旋向上和自旋向下。
这种特殊的性质使得粒子的自旋在某种程度上不同于经典物理学中的旋转概念。
除了自旋的量子性质外，粒子的统计也是量子力学中的重要概念。
根据粒子的统计性质，它们可以分为两类：玻色子和费米子。
玻色子是一种具有整数自旋的粒子，如光子，声子等。
费米子则是指自旋为半整数的粒子，如电子，质子等。
这两类粒子的统计行为截然不同。
根据泡利不相容原理，费米子遵循的是反对称的统计规则，即多个费米子不能处于同一个量子态；而玻色子则遵循的是对称的统计规则，多个玻色子可以同时处于同一个量子态。
这就是为什么电子组成的原子电子壳层中同一能级上只能有两个电子，而光子的波函数可以有无限个。
通过粒子自旋与统计的研究，我们深入理解了量子力学中的一些基本原理。
例如，弗恩海曼微观理论中的费米子路径积分和玻色子路径积分等都建立在对粒子自旋与统计的理解之上。
这些基本理论为粒子的行为提供了定量的描述和解释。
此外，粒子自旋与统计对于量子信息科学的发展也起到了重要作用。
粒子自旋的量子态可以用作量子比特的载体，而粒子的统计则决定了量子比特之间的可交换性。
关于量子力学的知识点总结
关于量子力学的知识点总结量子力学是现代物理学的一个重要分支，研究微观世界的行为规律。
它涉及到很多的知识点，下面将对其中的一些重要知识点进行总结。
1. 波粒二象性：量子力学中的基本粒子既可以表现出粒子的性质，又可以表现出波动的性质。
例如，电子、光子等粒子既可以像粒子一样具有位置和动量，又可以像波动一样具有频率和波长。
2. 不确定性原理：由于波粒二象性的存在，无法同时准确测量粒子的位置和动量，因为测量其中一个属性会对另一个属性造成不确定性。
这是因为波粒二象性使得微观粒子的位置和动量不能同时具有确定值。
3. 波函数：在量子力学中，波函数描述了一个量子系统的状态，其平方表示在不同位置寻找粒子的概率。
波函数形式为ψ(x)，其中x代表位置。
4. 叠加原理：当两个或多个波函数重叠时，它们可以相互叠加形成新的波函数。
这种叠加可以导致干涉现象，即波的相位相加或相减，形成波纹增强或波纹消除的现象。
5. 薛定谔方程：薛定谔方程是描述量子系统随时间演化的基本方程。
它能够确定系统的波函数随时间的变化，并给出粒子的能量以及其他物理量。
6. 量子态与态矢量：量子力学描述粒子的态称为量子态，用态矢量表示。
一个粒子的量子态是一个复数的线性组合，它确定了粒子在不同物理量上的测量结果的概率。
7. 纠缠：当两个或多个粒子通过量子力学的相互作用使得它们的量子态互相关联时，就产生了纠缠现象。
纠缠态的特点是不能将其视为单个粒子的状态，而必须将其作为整个系统的态来描述。
8. 可观测量与算符：在量子力学中，物理量的观测结果用可观测量表示。
每个可观测量都有对应的算符，通过作用于波函数求得其期望值。
例如，位置可观测量对应位置算符，动量可观测量对应动量算符。
9. 自旋：自旋是粒子特有的内禀角动量，与其自身特性相关。
自旋可能采取离散值，如电子的自旋即为1/2。
10. 荷质比：荷质比是粒子带电性质与其质量的比值。
根据量子力学理论，荷质比具有量子化的性质。
量子力学-斯坦福大学-Chapter03-08
k ' k
d (k k ')dk ' 1
if
k k
Copyright – Michael D. Fayer, 2007
Therefore
2 if P ' P 1 P' P 2 c 0 if P ' P
P
c2 c
are orthogonal and the normalization constant is
Function representing state of system in a particular representation and coordinate system called wave function.
Copyright – Michael D. Fayer, 2007
P i
classical rock in interstellar space Know momentum, p, and position, x, can can predict exact location at any subsequent time. Solve Newton's equations of motion. p = mV What is the quantum mechanical description? Should be able to describe photons, electrons, and rocks.
Adjust c to make equal to 1.
Evaluate
i ( k k ') x e dx
cos(k k ') x i sin(k k ') x dx
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上海师范大学标准试卷科目：量子力学期中考试专业本科 年级 班 姓名 学号我承诺，遵守《上海师范大学考场规则》，诚信考试。
签名：__________ 1. 请简述态叠加原理。
态叠加原理：如果1ψ和2ψ是体系可能的状态，那么，它们的线性迭加1122c c ψψψ=+ （c 1，c 2是复数）也是这个体系的一个可能状态。
2. 具有完备的共同本征函数系的两个力学量算符有什么特征？球谐函数),(ϕθlm Y 是哪两个算符的共同本征函数？具有完备的共同本征函数系的两个力学量算符对易。
球谐函数),(ϕθlm Y 是L 2和L z的共同本征函数3. 请简述波函数的统计解释及波函数的标准条件。
波函数的统计解释：波函数在空间某一点的强度（振幅绝对值的平方）和在该点找到粒子的几率成正比。
波函数的标准条件：单值性，有限性，连续性一、问答题（本大题满分20分，每小题答对得5分）4. 一维空间两粒子体系的归一化波函数为),(21x x ψ，写出下列概率：发现粒1的位置介于x 和dx x +之间。
（不对粒子2进行观测）概率dx x x dx p ]),([222ψ⎰+∞∞-=1．（15分）氦原子的动能是kT E 23=，k 是玻耳兹曼常数，求K T 1=时，氦原子的德布罗意波长。
已知普朗克常数秒焦耳⋅⨯=-3410626.6h ，玻耳兹曼常数1231038.1--⋅⨯=K k 焦耳，质子质量kg .m p 2710671-⨯=，氦原子的质量近似取为质子质量的四倍。
解：根据 eV .K k 410862501-⨯=⋅ ┈┈ 4分 知本题氦原子的动能为： eV .kT E 41029375123-⨯==┈┈ 4分显然远小于氦原子的静止能量2mc 。
这样，便有λ=m .910261-⨯= ┈┈ 7分2．（15分）假设一平面转子角速度为ω，转动惯量为I ，试用波尔-索莫非条件求其能量可能值解：根据波尔-索莫非条件202pdq I d I nh πωϕπω===⎰⎰（2分）(1,2,3......)2nh nhn I Iωπ=== （2分） 平面转子能量212E I ω= （2分）二、计算题（本大题满分60分，请写出解题的主要步骤，否则不得分）将ω代入能量公式，得能量量子化公式：222122n n E I I ⎛⎫== ⎪⎝⎭ （2分）3. （15分）一粒子在一维无限深势阱⎪⎩⎪⎨⎧>∞≤≤<∞=a x a x x x U ,,0,0,0,)（ 中运动，求粒子的能级和相应的归一化波函数。
解：t x U 与)(无关，是定态问题。
其定态SE)()()()(2222x E x x U x dx d m ψψψ=+-┈┈ 2分在各区域的具体形式为Ⅰ： )()()()(20111222x E x x U x dx d m x ψψψ=+-< ① Ⅱ： )()(2 0 22222x E x dx d m a x ψψ=-≤≤ ② Ⅲ： )()()()(2 333222x E x x U x dxd m a x ψψψ=+-> ③ ┈┈ 3分由于(1)、(3)方程中，由于∞=)(x U ，要等式成立，必须 0)(1=x ψ 3()0x ψ= 即粒子不能运动到势阱以外的地方去。
方程(2)可变为0)(2)(22222=+x mEdx x d ψψ 令222 mE k =，得0)()(22222=+x k dxx d ψψ其解为 kx B kx A x cos sin )(2+=ψ ④┈┈ 3分根据波函数的标准条件确定系数A ，B ，由连续性条件，得)0()0(12ψψ=⑤)()(32a a ψψ=⑥⑤ 0=⇒B0sin =⇒ka A),3 ,2 ,1( 0sin 0==⇒=∴≠n n ka ka A π∴x an A x πψsin )(2= ┈┈ 2分 由归一化条件1)(2=⎰∞dx x ψ得 1sin 022=⎰axdx an Aπ由mn abaxdx a n x a m δππ⎰=*2sin sinx a n a x aA πψsin 2)(22=∴=⇒┈┈ 2分222 mE k =),3,2,1( 22222 ==⇒n n ma E n π可见E 是量子化的。
┈┈ 1分对应于n E 的归一化的定态波函数为⎪⎩⎪⎨⎧><≤≤=-a x a x a x xe an a t x tE in n , ,0 0 ,sin 2),( πψ┈┈ 2分3. 设氢原子处于归一化状态211021111(,,)()(,)()(,)2r R r Y R r Y ψθϕθϕθϕ-=求其能量、角动量平方及角动量Z 分量的可能值，这些可能值出现的几率和这些力学量的平均值。
（15分）解：由波函数211021111(,,)()(,)()(,)2r R r Y R r Y ψθϕθϕθϕ-=可得 该状态所对应各量子数取值为：n =2，l =1，m =0，-1 （3分） 能量可能值为：442222228s s n e e E nμμ==-=-（2分）出现的几率：221122⎛⎛⎫+= ⎪ ⎝⎭⎝⎭（1分） 能量平均值：428s e E μ=-（1分）角动量平方可能取值： ()2212l l += （2分）出现的几率为：22112⎛⎫+= ⎪⎝⎭⎝⎭（1分） 平均值：2ˆ2L= （1分） 角动量Z 分量可能取值：m ，故可取0，- （2分）出现的几率分别为0：21124⎛⎫= ⎪⎝⎭；-：234=⎝⎭ （1分） 平均值：133ˆ0()444zL =⨯+-⨯=- （1分）1、（10分）利用坐标和动量算符的对易关系，证明轨道角动量算符的对易关系：证明：zyxLiLLˆ]ˆ,ˆ[=]ˆˆ,ˆˆ[]ˆ,ˆ[zxyzyxp xp zp zp yLL--=]ˆˆ,ˆ[]ˆˆ,ˆ[zxyzxzp xp zp zp xp zp y---=]ˆ,ˆ[]ˆ,ˆ[]ˆ,ˆ[]ˆ,ˆ[zyxyzzxzp xp zp zp zp xp yp zp y+--=]ˆ,ˆ[]ˆ,ˆ[zyxzp xp zp zp y+=yzzyzxxzpp xzp xpzpp zyp zpyˆ]ˆ,[]ˆ,ˆ[ˆ]ˆ,[]ˆ,ˆ[+++=yzxzpp xzp zpyˆ]ˆ,[]ˆ,ˆ[+=yzyzxzxzppxzppzxpzpyppyzˆˆ],[ˆ]ˆ,[ˆ],ˆ[]ˆ,ˆ[+++=yxpixpiyˆ)(ˆ)(+-=]ˆˆ[xyp yp xi-=zLiˆ=三、证明题（本大题满分10分）指出下列算符哪个是厄米算符，说明其理由。
224 dxd dx d i dx d ，， 不是厄米算符，，当解：dxddxdx ddx dx ddx dx d dx dx d x dx dx d dx dxd*)( *)( * *0 * * *-∴≠-=-=∴→→±∞→-=⎰⎰⎰⎰⎰⎰∞∞-∞∞-∞∞-∞∞-∞∞-∞∞∞∞-φψφψφψφψφψφψφψφψ是厄米算符dxdi dx dxdi dx dx d i dx dx d i i dx dx di∴=-=-=⎰⎰⎰⎰∞∞-∞∞-∞∞-∞∞∞∞- *)( *)( * * * -φψφψφψφψφψ四、分析题（本大题满分10分）。