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CHAPTER 6 Quantum Mechanics IIThepsatri Rajabhat University

Quantum mechanicsAbhaykumar vishwakarma

Concepts and Problems in Quantum Mechanics, Lecture-II By Manmohan DashManmohan Dash

Quantum physics the bottom up approachSpringer

Problems and solutions statistical physics 1Alberto de Mesquita

Ph 101-9 QUANTUM MACHANICSChandan Singh

- ADVANCE QUANTUM MECHANICS Instructor: Prof. Amir Al-Q’rawee E- mail: University of Karbala M.Sc. Course 2016-2017 1
- Course Description This course for graduate students. The students are assumed undergraduate-level quantum mechanics and basic mathematics background. It aims to help students gain advanced foundation in quantum mechanics. Text book: Quantum Mechanics L.I. SCHIFF Quantum Mechanics J. POWELL Modern Quantum Mechanics J. J. Sakurai 2
- THE COURSE WILL COVER THE FOLLOWING SUBJECTS An advanced undergraduate level introduction to the principles, formalism and results of quantum mechanics; including: historical background, Schrödinger equations, particle in box, harmonic oscillator, hydrogen atom, angular momentum, Hilbert space, Dirac notation, introduction to Approximation methods for stationary states, and scattering theory. Objectives: 1- To achieve an understanding of the theory of quantum mechanics, and an ability to apply that theory correctly to important physical systems. 2- To became aware of the necessity for quantum methods in the analysis of physical systems of atomic and nuclear physics. 3
- The Wave Function •The Schrödinger Equation •The Statistical Interpretation •Probability •Normalization •Momentum •The Uncertainty Principle 4
- Wave function One of the most profound and mysterious principles in all of physics is the Born Rule, named after Max Born. In quantum mechanics, particles don’t have classical properties like “position” or “momentum”; rather, there is a wave function that assigns a (complex) number, called the “amplitude,” to each possible measurement outcome. Q.M says that every object in the universe is associated with a mathematical expression that encodes in it every property (its charge, mass, location, energy...) that it is possible to know about the object. This math expression is called the object’s wave function (psi) ψ. 5
- The Schrödinger has two "forms", Time dependent Schrödinger wave equation: Time independent Schrödinger equation: Born's Statistical Interpretation We have seen that matter must be consider to have wave- like properties in order to explain experimental data, but what is the nature of these waves? What is "waving"? Born postulated that the wave function, , that describes a particle's behavior is related to the probability of finding the particle by: The Schrödinger Equation 6
- Where 2 is the complex square, or * and * is the complex conjugate of the (perhaps complex) wave function. This tells us that the wave function itself does not represent the probability, but a probability amplitude, and that the information contained in only represents the probability that one would measure a certain dynamical quantity, but cannot give pre- determined results in the same way that the deterministic (if you know) the initial conditions, you can say exactly where the particle will be at a later point in time), quantum mechanics only tells you statistical information about what the possible measurements will be. This interpretation, although since born out by much experimentation, caused much debate in the history of quantum mechanics, and continues to prick at our intuition. b) and a between t at time particle the finding of y probabilit ( ) , ( Ψ 2 b a dx t x 7
- What exactly is the wave function. Born came up with a statistical interpretation of the wave function, which says that gives the probability of finding the particle at point x, at time t, or more precisely, The particle would be relatively likely to be found near A, and unlikely to be found near B. The shaded area represents the probability of finding the particle in the range dx. t at time ) ( and between particle the finding of y probabilit ) , ( Ψ 2 dx x x dx t x 8
- The probability P(r,t)dV to find a particle associated with the wave function Ψ(r,t) within a small volume dV around a point in space with coordinate r at some instant t is P(r,t) is the probability density For one-dimensional case dV t dV t P 2 ) , ( ) , ( r r dx t x dV t x P 2 ) , ( ) , ( Here |Ψ(r,t)|2 = Ψ*(r,t)Ψ(r,t) Probability Y Z X r dV 9
- The wave function contains information about where the particle is located, its square being a probability density. A wave function must be "well behaved", in other words it should be defined and continuous everywhere. In addition it must be square-integrable, meaning: dx t x 2 ) , ( Ψ 10
- Properties of a valid wave function: Boundary conditions 1) In order to avoid infinite probabilities, the wave function must be finite everywhere. 2) In order to avoid multiple values of the probability, the wave function must be single valued. 3) For finite potentials, the wave function and its derivative must be continuous. This is required because the second-order derivative term in the wave equation must be single valued. (There are exceptions to this rule when V is infinite.) 4) In order to normalize the wave functions, they must approach zero as x approaches infinity. 11
- The wave function itself is complex, but 2 = (where is the complex conjugate of ) is real and non negative – as a probability must be. * = ||2 = “Probability distribution function” ||2 dV = For a stationary state, • * is independent of time • * = |(x,y,z)|2 probability of finding a particle near a given point x,y,z at a time t A wave function must: Be single valued Be continuous Be differentiable Be integrable 12
- dx t x P dx t x P everywhere b a ab 2 2 ) , ( ) , ( For one-dimensional case, the probability of funding the particle in the arbitrary interval a ≤ x ≤ b is Example: Let two functions and be defined for 0 ≤ 𝑥 ≤ ∞. Explain why (𝑥) = 𝑥 can not be a wave function but could be a valid wave function. Solution: Both functions are continuous and defined on the interval of interest. They are both single valued and differentiable. However, consider the integral of 𝑥: 2 ) ( Φ x e x 0 0 3 3 0 2 3 ) ( x dx x dx x ψ 13
- Given that, (𝑥) = 𝑥 is not square integrable over this range it cannot be a valid wav function, On the other hand: Therefore, is square integrable so we can say it’s well behaved over the interval of interest. This makes it a valid candidate wave function. 0 2 2 0 2 8 ) ( Φ π dx e dx x x 2 ) ( Φ x e x Example: Suppose that a certain probability distribution is given for 1 ≤ 𝑥 ≤ 3. Find the probability that 5/2 ≤ 𝑥 ≤ 3. Solution: If this were a wave function, there would be about a 5.5 percent chance of finding the particle between 5/2 ≤ 𝑥 ≤ 3. 3 1 4 9 ) ( x x p 3 2 5 3 2 5 2 3 055 . 0 200 11 25 4 9 1 8 9 ) 2 ( 4 9 1 4 9 x dx x p 14
- 3 1 4 9 ) ( x x p 3 2 5 3 2 5 2 3 055 . 0 200 11 25 4 9 1 8 9 ) 2 ( 4 9 1 4 9 x dx x p Example: Suppose that a certain probability distribution is given for 1 ≤ 𝑥 ≤ 3. Find the probability that 5/2 ≤ 𝑥 ≤ 3. Solution: If this were a wave function, there would be about a 5.5 percent chance of finding the particle between 5/2 ≤ 𝑥 ≤ 3. Since ̋ Ψ*(𝑥,𝒕) Ψ(𝑥,𝒕) 𝑑𝑥 ̋ is a probability, then the sum of all probability must equal one, Thus This is the Normalization Condition and is very useful in solving problems. 1 ) , , , ( Ψ ) , , , ( Ψ or 1 ) , ( Ψ ) , ( Ψ dV t z y x t z y x dx t x t x Normalizing the wave function means you find the exact form of ψ 15
- Expectation values Although particle is never in a definite location, it is more likely to be in one location than others. We have seen that is the probability density of a measurement of a particle's displacement yielding the value at time . Suppose that we made a large number of independent measurements of the displacement on an equally large number of identical systems. In general, measurements made on different systems will yield different results. However, from the definition of probability, the mean of all these results is simply 16
- The expectation value of x2 is given by And the stander deviation, or uncertainty, in x is given by Although particle is never in a definite location, it is more likely to be in one location than others, if any potential V is active. The definition of a weighted average position: space all dx x p x x 2 2 ) ( 2 2 Δ x x x : ) ( ) ( ) ( e Us ) ( ) ( x ψ x ψ x p dx x p dx x xp x 17
- ) ( ) ( ) ( ) ( dx x ψ x ψ dx x ψ x x ψ x This is the “expectation value of 𝑥” By convention, place 𝑥 between ’s If has been normalized, this denominator is. Examples: Lets assume tha wave function of QM particle is of an observable ̋𝑥 ̋ is given by: We can normalize Ψ to get the constant A: otherwise 0 1 1 - ) , ( Ψ x Ax t x 18
- x x dx x A dx t x t x 1 1 2 2 1 or ) , ( Ψ ) , ( Ψ 1 Then we can calculate the expectation value of 𝑥, <𝑥> as: Note: That the probability Ψ*Ψ of observing the QM particle is 0 at 𝑥 = 0 but multiple measurements will average to net weighted average measurement of 0. 1 1 2 3 2 3 dx x x x x 2 3 3 2 3 1 2 1 1 3 2 A A x A 0 x 4 1 2 3 ) ( 2 3 1 1 4 1 1 3 x dx x x 19
- As the previous case illustrates, sometimes the “average value” of an observable is not all we need to know. Particularly, since the integrand of our <𝑥> equation was odd, the integrand was 0. (Looking at whether a function is evev or odd can often save calculation effort). We may like to also know the “stander deviation” or the variance (stander deviation squared) of the observable around the average value. In statistics, the standard deviation is: And since <𝑥>, all we need is: 2 2 x x σ 1 1 2 2 2 3 or ) , ( Ψ ) , ( Ψ xdx xx dx t x t x x 5 3 10 3 1 1 5 2 x x 775 . 0 5 3 0 5 3 2 2 x x σ 20
- Expectation Value for Momentum of a Free Particle dx x x x i p dx x x i x dx x p x p ) ( ) ( ) ( ) ( ) ( ˆ ) ( * * * Generally p k dx Ae Ae k dx Ae ik i Ae p dx Ae x i Ae p A dx Ae Ae dx x Ae x ikx ikx ikx ikx ikx ikx ikx ikx ikx * * * * 2 n integratio of limits as 0 where , 1 ) ( with ) ( Free Particle Similarly, we could also calculate other higher order moments of distributions such as skewness, curtosis, etc … 21
- Q) A particle is described by the wave function calculate the expectation values of momentum and energy. Solution: The normalization condition Determines N up to an arbitrary phase factor. We choose N to be real. Then constant 0 ; 1 ) , ( 2 b e e N t r t i br 1 ) , ( 2 3 dV t r R 0 0 2 0 4 2 2 3 2 2 sin dr d d e r e dV e e e e N br t i br R t i br 0 3 4 2 8b 4 dr e r br 22
- Operators: Every classical obtained dynamical variable can be replaced by an "Operator" that "acts on the wave function". An operator is merely the mathematical rule used to describe a certain mathematical operation. For example, the "X " derivative operator" is defined as "d/dx". The wave function is said to be the operators, operand, i.e. what is being acted on. The following table lists the corresponding of the classical dynamical variables and their corresponding quantum mechanical operator: An operator, , operates on a wave function, ψ , to produce an observable, O An operator, , is a mathematical entity which transforms one function into another. 23
- The operator x represents position, and the operator represents momentum, in quantum mechanics; to calculate expectation values, we sandwich the appropriate operator between and , and integrate. The fact is, all such quantities can be written in terms of position and momentum. Kinetic energy, for example, is m p mv T 2 2 1 2 2 24
- 2 ) )( ( x px 1. The limitations imposed by the uncertainty principle have nothing to do with quality of the experimental equipment 2. The uncertainty principle does imply that one cannot determine the position or the momentum with arbitrary accuracy It refers to the impossibility of precise knowledge about both: e.g. if Δx = 0, then Δ px is infinity, and vice versa 3. The uncertainty principle is confirmed by experiment, and is a direct consequence of the de Broglie’s hypothesis The Uncertainty Principles 25
- Energy and time, for instance, form a pair of complementary variable. Their simultaneous measurement must obey the time-energy uncertainty relation Estimate the uncertainty in the position of a) a neutron moving at (5 x 106 m/s) and b) a 50Kg person moving at (2 m/s). This distance is comparable to the size of a nucleus. An uncertainty of this magnitude is beyond human detection, therefore can be neglected. So the position and momentum uncertainty are important for microscopic systems, but negligible for macroscopic systems. 2 t E 26
- Probability current density is conserved If a particle is not being created or destroyed it's integrated probability always remains constant (=1 for a normalized wave function). However, if the particle is moving, we can define a "probability current density" and a probability continuity equation that describes the particles movement through a Gaussian surface (analogous to electromagnetic). Probability current density equation 0 J t electric current density electric charge density 27
- J → probability current density and ρ is the probability density ρ = ψ* ψ (∂ρ ∕ ∂t) → electric charge density electric current density We need an analogous expression to describe probability density ρProb and probability current JProb which can flow in space but remain conserved. Assume ρProb and Jprob involve ψ somehow, but in an unspecified function. Plan: 1. Use the only equation we have for ψ: the Schrödinger Equation 2. Manipulate (معالجة ) it to get the form J 28
- Convert 1− d Schrödinger Eq to 3 - d : Form [ψ *• (Sch Eq)] - [ψ • (Sch Eq)*]⇒ ÷ by iћ and collect terms : real) is V if (0 2 0 2 2 2 mi t i V V mi t t V is real because, Imaginary potentials do cause probability not to be conserved. 29
- Prob Prob 0 2 J mi t Q) A particle is described by the wave function Calculate the current density when r2 = x2 + y2 +z2. How does the probability current density behave for large r. constant 0 ; ) ( a r e r r a ik 30
- Mathematical Structure of Quantum Mechanics CHAPTER TWO The Hilbert Space Dirac Notation Operators Hermitian Adjoint Commutator Algebra 31
- Hilbert space is the mathematical foundation used for quantum mechanics. This formalism is based on the basic ideas of vector analysis, with functions taking the role of vectors and each vector can have finite or infinite number of elements. A Hilbert space ℋ consists of a set of vectors ψ, ϕ, χ,….. and a set of scalars a, b, c,….. which satisfy the following four properties: (a) ℋ is a linear space The properties of a linear space. A linear vector space consists of two sets of elements and two algebraic rules: A vector space is a set of objects called vectors ( |A〉, |B〉, |C〉,...) and a set of numbers called scalars (a, b, c,...) along with a rule for vector addition and a rule for scalar multiplication. The Hilbert Space 32
- If the scalars are real, we have a real vector space; if the scalars are complex, we have a complex vector space. The set must be closed under vector addition and scalar multiplication. Vector addition must have these properties: • The sum of any 2 vectors is a vector: |A〉 + |B〉 = |C〉 • Vector addition is commutative and associative: |A〉 + |B〉 = |B〉 + |A〉 and |A〉 + (|B〉 + |C〉) = (|A〉 +| B〉) + C〉 • There exists a zero vector |0〉 such that : |A〉 + |0〉 = |A〉 for any vector |A〉 • For every vector |A〉 there is an inverse vector |-A〉 such that |A〉 + |-A〉 = |0〉 33
- Scalar multiplication must have these properties: • The product of a scalar and a vector is another vector: b |A〉 = |C〉 • It is distributive with respect to vector addition and scalar addition: a (|A〉 + |B〉) = a |B〉 + a |A〉 and (a + b) |A〉 = a |A〉 + b |A〉 • It is associative with respect to ordinary scalar multiplication: a (b |A〉) = (a b) |A〉 • Multiplication by the scalars 0 and 1 yields the expected: 0 |A〉 = |0〉 and 1 |A〉 = |A〉 34
- The scalar product of an element ψ with another element ϕ is in general a complex number, denoted by (ψ,ϕ), where (ψ,ϕ) = complex number. Because the scalar product is a complex number, the quantity (ψ,ϕ), is generally not equal to (ϕ,ψ), : (ψ,ϕ) = ψ*ϕ while (ϕ,ψ) = ϕ*ψ, The scalar product satisfies the following properties: The scalar product of ψ with ϕ is equal to the complex conjugate of the scalar product of ϕ with ψ: , , The scalar product of ϕ with ψ is linear with respect to the second factor if ψ =aψ1 + bψ2: 2 1 2 1 , , , b a b a (b) ℋ has a defined scalar product that is strictly positive 35
- and antilinear with respect to the first factor if ϕ = aϕ1 + bϕ2 (aϕ1 + bϕ2,ψ) = a*(ϕ1,ψ) + b*(ϕ2,ψ) . The scalar product of a vector ψ with itself is positive real number: where the equality holds only for ψ = 0. (c) ℋ is separable: (d) ℋ is complete : , 0 , 2 36
- Question: Check whether the following sets of functions are linearly independent or dependent on the real x-axis. x e x h x x g x f a 2 2 ) ( , ) ( , 4 ) ( ) 3 2 ) ( , ) ( , ) ( ) x x h x x g x x f b Dirac Notation In QM, the state of a system is completely characterized by a state vector. Dirac developed general notation to describe these state vectors which was independent of basis, today this is called Dirac notation. 37
- Dirac bra/ket notation d 2 1 Ket: ψ always denotes a column vector, e.g. Bra: ψ always denotes a row vector that is the conjugate transpose of ψ, e.g. [ * 1 * 2 * d ] Bra-ket Dirac denoted the scalar (inner) product by the symbol 〈 ∣ 〉, which he called a a bra-ket. For instance, the scalar product (ψ,ϕ) is denoted by the bra-ket 〈 ψ ∣ ϕ 〉: Note: When a ket (or bra) is multiplied by a complex number, we also get a ket (or bra). 38
- 〈 ψ ∣ ϕ 〉 = the inner product The inner product satisfies 〈 ψ ∣ ϕ 〉 = 〈 ϕ ∣ ψ 〉* ∣ψ〉〈ϕ∣ = the outer product Properties of kets, bras, and bra-kets Every ket has a corresponding bra To every ket ∣ψ〉, there corresponds a unique bra 〈ψ∣ and vice versa: ∣ψ〉 ⟷ 〈ψ∣ a∣ψ〉 + b∣ϕ〉 ⟷ a*〈ψ∣ + b*〈ϕ∣ Where a and b are complex numbers. The following common notation: ∣aψ〉 = a∣ψ〉, 〈aψ∣ = a*〈ψ∣ 39
- Properties of the scalar product In quantum mechanics, since the scalar product is a complex number. We must be careful to distinguish a scalar product from its complex conjugate; 〈ψ∣ϕ〉 is not the same thing as 〈ϕ∣ψ〉 : 〈ϕ∣ψ〉* = 〈ψ∣ϕ〉 the scalar product 〈ϕ∣ψ〉 is given by 3 , , dr t r t r This property becomes clearer : 〈ϕ∣ψ〉* = 〈ψ∣ϕ〉 r d t r t r r d t r t r 3 3 , , , , 40
- additional properties of the scalar product: 〈ψ∣𝑎1ψ1+𝑎2ψ2〉 = 𝑎1〈ψ∣ψ1〉 + 𝑎2〈ψ∣ψ2〉 , 〈𝑎1ϕ1 + 𝑎2ϕ2∣ψ〉 = 𝑎1*〈ϕ1∣ψ〉 + 𝑎2*〈ϕ2∣ψ〉 , 〈𝑎1ϕ1 + 𝑎2ϕ2∣b1ψ1+b2ψ2〉 = 𝑎1*b1〈ϕ1∣ψ1〉 + 𝑎1*b2〈ϕ1∣ψ2〉 + 𝑎2*b1〈ϕ2∣ψ1〉 + 𝑎2*b2〈ϕ2∣ψ2〉 Orthogonal states Two kets, ∣ψ〉 and ∣ϕ〉, are said to be orthogonal if they have vanishing scalar product Bra and Ket orthogonal if scalar product equals zero. 0 B A 41
- 2 Bras or 2 Kets orthogonal; and orthogonal if 0 B A B A and orthogonal if 0 B A B A Two states of a dynamical system are orthogonal if the vectors representing them are orthogonal. Orthonormal states Two kets, ∣ψ〉 and ∣ϕ〉, are said to be orthonormal if they are orthogonal and if each one of them has a unit norm: 〈ψ∣ϕ〉 = 0 , 〈ψ∣ψ〉 = 1 , 〈ϕ∣ϕ〉 = 1 42
- dx x x dx x x ) ( ) ( The integration ψ*(x) ψ(x) over all values of x will yield 1 if ψ(x) is normalized. Forbidden quantities If ∣ψ〉 and ∣ϕ〉 belong to the same vector (Hilbert) space, products of the type ∣ψ〉 ∣ϕ〉 and 〈ψ∣〈ϕ∣ are forbidden. They are nonsensical. 43
- Operators Definition of an operator: An operator Â is a mathematical rule that when applied to a ket ∣ψ〉 transforms it into another ket ∣ψ'〉 of the same space and when it acts on a bra 〈ϕ∣ transforms it into another bra 〈ϕ’∣ : Â∣ψ〉 = ∣ψ'〉 , 〈ϕ∣Â = 〈ϕ'∣ Operators stand to the left of kets and to the right of bras. A similar definition applies to wave functions: ), ( ) ( ˆ r r A ) ( ˆ ) ( r A r consider some simple operators: The Identity Operator:- The simplest operator of all is the identity operator, which does nothing to a ket or Unity operator I ˆ 44
- Outer Product The outer product between a ket and a bra is written as follows: ∣ψ〉〈ϕ∣ (∣ψ〉〈ϕ∣)∣χ〉 = ∣ψ〉〈ϕ∣χ〉 = 𝛂∣ψ〉 inner product(complex number) Products of operators: The product of two operators ,the operators do not commute. The product of operators is associative: A B B A ˆ ˆ ˆ ˆ C B A C B A C B A ˆ ) ˆ ˆ ( ) ˆ ˆ ( ˆ ˆ ˆ ˆ 45
- Linear operators An operator that obeys the distributive law That is, an operator Â is linear if, for any vectors ∣ψ1〉 and ∣ψ2〉 and any complex numbers 𝑎1 and 𝑎2, we have: Â(𝑎1∣ψ1〉 + 𝑎2∣ψ2〉) = 𝑎1Â∣ψ1〉 + 𝑎2Â∣ψ2〉 , and (〈ψ1∣𝑎1 + 〈ψ2∣𝑎2)Â = 𝑎1〈ψ1∣Â + 𝑎2〈ψ2∣Â . The expectation or mean value Â of an operator Â with respect to a state ∣ψ〉 is defined by: A A ˆ ˆ A A ˆ ˆ 46
- Hermitian Adjoint Hermitian Conjugate or adjoint of an operator Â to be A†. Since the inner product is just a complex number, we can form the complex conjugate which is given by relation 〈ψ∣ϕ〉 = 〈ϕ∣ψ〉*. When an operator is present inside the inner product this becomes 〈ψ∣Â†∣ϕ〉 = 〈ϕ∣Â∣ψ〉* DEFINITION: Forming the Adjoint of a General Expression 1. Replace any constants by their complex conjugates 2. Replace kets by their associated bras, and bras by their associated kets (∣ψ〉)† = 〈ψ∣ , (〈ψ∣)† = ∣ψ〉 3. Replace each operator by its adjoint 4. Reverse the order of all factors in the expression 47
- The adjoint operation also has the following properties: (Â†)† = Â (𝑎Â)†=𝑎*Â† (Âⁿ)†=(Â†)ⁿ 48
- The expectation or mean value Â of an operator Â with respect to a state |ψ is defined by The quantity |ψ| is a linear operators in Dirac's notation Since is a complex number. 49
- Example: (a) Discuss the hermiticity of the operators (Â + Â†),i((Â + Â†), and (Â - Â†). (b) Find the Hermitian adjoint of f(Â) =(1 + i Â + 3 Â2)(1-2i Â - 9 Â2)/(5 + 7 Â). (c) Show that the expectation value of a Hermitian operator is real and that of an anti-Hermitian operator is imaginary. Solution: (a)The operator † is Hermitian regardless of whether or not Â is Hermitian, Since Similarly the operator i(Â - Â†) is also Hermitian; but i(Â + Â†) is anti-Hermitian, since [i(Â + Â†)]† = -i(Â + Â†). A A B ˆ ˆ ˆ 50
- (b) Since the Hermitian adjoint of an operator function f(Â) is given by f †(Â) = f *(Â†), We can write (c) From Â = Â† or 〈ψ∣Â†∣ϕ〉 = 〈ϕ∣Â∣ψ〉* we immediately infer that the expectation value of a Hermitian operator is real, for it satisfies the following property: 〈ψ∣Â†∣ϕ〉 = 〈ϕ∣Â∣ψ〉* that is, if Â† = Â then 〈ϕ∣Â∣ψ〉 is real. Similarly, for anti- Hermitian operator, we have which means that is purely imaginary number. 51
- Find the Hermitian Conjugate of x 52
- Commutator Algebra 3a – 5b + 7a original (given) statement 3a + 7a – 5b Commutative Property (3a + 7a) – 5b Associative Property a(3 + 7) – 5b Distributive Property a(10) – 5b simplification (3 + 7 = 10) 10a – 5b Commutative Property 53
- Commutative Laws: a + b = b + a a × b = b × a Associative Laws: (a + b) + c = a + (b + c) (a × b) × c = a × (b × c) Distributive Law: a × (b + c) = a × b + a × c 54
- The commutator of two operator Â and , denoted by , is defined by And anticommutator , is defined by Two operators are said to commute if their commutator is equal to zero and hence Any operator commutes with itself: 55
- A B B A B A ˆ ˆ ˆ ˆ ] ˆ , ˆ [ 0 ] ˆ , ˆ [ ] ˆ , ˆ [ A B B A 0 ] ˆ , ˆ [ A A ] ˆ , ˆ [ ] ˆ , ˆ [ ] ˆ ˆ , ˆ [ C A B A C B A ] ˆ , ˆ [ ] ˆ , ˆ [ ] ˆ , ˆ ˆ [ C B C A C B A ] ˆ , ˆ [ ˆ ˆ ] ˆ , ˆ [ ] ˆ ˆ , ˆ [ C A B C B A C B A ] ˆ , ˆ [ ˆ ˆ ] ˆ , ˆ [ ] ˆ , ˆ ˆ [ C B A B C A C B A Consider the following properties: ] ˆ , ˆ [ ] ˆ , ˆ [ A B B A Distributive Antisymmetric linearity 56
- If two operators are Hermitian and their product is also Hermitian, these operators commute: Operator commute with scalar: an operator Â commute with any scalar b: [Â,b⦌=0 57
- Example: (a) Show that the commutator of two Hermitian operators is anti-Hermitian. (b) Evaluate the commutator Solution: (a) If Â and are Hermitian, we can write that is, the commutator of Â and is anti-Hermitian ] ˆ ] ˆ , ˆ [ , ˆ [ D C B A 58
- (b) Using the distributivity relation, we have D C B A D A C B D C B A ˆ ]] ˆ , ˆ [ , ˆ [ ] ˆ , ˆ ][ ˆ , ˆ [ ] ˆ ] ˆ , ˆ [ , ˆ [ 59
- CHAPTER THREE The Time-Independent Schrodinger Equation Stationary States The Infinite Square Well The Free Particle The Finite Square Well The Harmonic Oscillator The Delta-Function Potential 60
- The idea of "stationary states" was first introduced by Bohr as a name given to those states of hydrogen atom for which the orbits that the electron occupied were stable, i.e. the electron remained in the same orbital state for all time. This idea is now no longer appropriate, the notation of a stationary state remains a valid one. It is a term that is now used to identify those states of a quantum system that do not change in time. This is not to say that stationary state is one for which "nothing happens" – there is still a rich collection of dynamical physics to be associated with such a state-but a stationary state turns out to be one for which the probabilities of outcomes of a measurement of any property of the system is the same no matter at what time the measurement is made. stationary states A stationary state is called stationary because the system remains in the same state as time elapses. The Hamiltonian is unchanging in time 61
- The starting point is to let the time-dependent Schrodinger equation for a particle of mass m moving in a time-dependent potential V(r,t) can be written as follows: Now, let us consider the particular case of time-independent potentials: V(r,t) = V(r). In this case the Hamiltonian operator will also be time independent, and hence the Schrodinger equation will have solutions that are separable, i.e. solutions that consist of a product of two functions, one depending only on r and the other only on time: For separable solutions we have ) ( ) ( , t f x t x , dt df t 2 2 2 2 dx d x Schrodinger Equation 62
- And the Schrodinger equation Dividing by f Now the left side is a function of t alone, and the right side is a function of x alone. The only way this can possibly be true is if both sides are in fact constant; this constant, which we denote by E , has the dimensions of energy. f V f dx d m dt df i 2 2 2 2 V dx d m dt df f i 2 2 2 1 2 1 ) ( ) ( t Ef dt t df i 63
- and or Separation of variables has turned a potential differential equation into two ordinary differential equations. The first of these is easy to solve The second term in the S.E is called the time-independent Schrödinger equation: , f iE dt df E V dx d m 2 2 2 1 2 E V dx d m 2 2 2 2 ) ( ) ( ˆ x E x H If a Hamiltonian does not depend on time, then the solution of the Schrödinger equation can be expressed in terms of eigenfunctions of the Hamiltonian Stationary wave function 64
- This particular solution of the Schrodinger equation for time- independent potential is called a stationary state. Why is this is called stationary? The reason is obvious: the probability density is stationary, i.e., it does not depend on time: Note that such a state a precise value for the energy, E = ħ. In summary, stationary states, which are given by the solutions of exist only for time-independent potentials. The set of energy levels that are solutions to this equation are called the energy spectrum of the system. The states corresponding to discrete and continuous spectra are called bound and unbound states, respectively. E V dx d m 2 2 2 2 65
- One-Dimensional Problems Properties of One-Dimensional Motion To study the dynamical properties of a single particle moving in a one-dimensional potential, let consider a potential V(x). One such potential is display in Fig.(1); it is finite at x = , V(-) = V1 and V(+) = V2 with V1 smaller than V2, and it has a minimum, Vmin. Fig.(1) Shape of general potential 66
- Discrete Spectrum(Bound States) Bound states occur whenever the particle cannot move to infinity. That is, the particle is confined or bound at all energies to move within a finite and limited region of space. The Schrodinger equation in this region admits only solutions that are discrete. The infinite square well potential and harmonic oscillator are typical examples that display bound states. In the potential of Fig.(1), the motion of the particle is bounded between the classical turning points x1 and x2 when the particle's energy lies between Vmin and V1: The states corresponding to this energy range are called bound states. They are defined as states whose wave functions are finite (or zero) at x . 67
- In a one-dimensional problem the energy levels of a bound state system are discrete and not degenerate. The wave function n(x) of a one-dimensional bound state system has n nodes if n = 0 corresponds to the ground state and (n-1) nodes if n = 1 corresponds to the ground state. Continuous Spectrum (Unbound States) Unbound states occur in those cases where the motion of the system is not confined; a typical example is the free particle. For the potential displayed in Fig.(1) there are two energy ranges where the particle's motion is finite: V1 E V2 and E V2. 68
- Mixed Spectrum Potentials that confined the particle for only some energies give rise to mixed spectra; the motion of the particle for such potentials is confined for some energy values only. For instance, for the potential display in Fig.(1), if the energy of the particle is between Vmin E V1, the motion of the particle is confined (bound) and its spectrum is discrete, but if E V2, the particle's motion is unbounded and its spectrum is continuous (if V1 E V2 , the motion is unbounded only along the x = - direction). Other typical examples where mixed spectra are encountered are the finite square well potential. Nondegenerate spectrum Since the eigenstates of the parity operator have definite parity, the bound eigenstates of a particle moving in a one-dimensional symmetric potential have definite parity; they are either even or odd: (-x) = (x) 69
- Degenerate spectrum If the spectrum of the Hamiltonian corresponding to a symmetric potential is degenerate, the eigenstates are expressed only in terms of even and odd states. That is, the eigenstates do not have definite parity. The Free Particle: Continuous States This is the simplest one-dimensional problems; it corresponds to V(x)=0 for any value of x. In this case the Schrodinger equation is given by Where k2 = 2mE/ ħ2 , k being the wave number The most general solution is a combination of two linearly independent plane waves 0 ) ( ) ( ) ( 2 2 2 2 2 2 2 x k dx d x E dx x d m ikx ikx k e A e A x ) ( 70
- Where A+ and A- are two arbitrary constants. The complete wave function is thus given by the stationary state. Since = E/ħ = ħk2/2m. The first term represents a wave traveling to the right, while the second term represents a wave traveling to the left. The intensities of these waves are given by A+2 and A-2, respectively. We should note that the waves +(x,t) and -(x,t) are associated, respectively, with a free particle traveling to the right and to the left with well-defined momenta and energy: p = ħk, E=ħ2k2/2m. We will comment on the physical implications of this in a moment. Since there are no boundary conditions, there are no restriction on k or on E ; all values yield solutions to the equation. ) ( ) ( ) , ( t kx i t kx i k e A e A t x ) 2 ( ) 2 ( 2 2 m t k kx i m t k kx i e A e A 71
- There are three Physical subtleties: First, the probability densities corresponding to either solutions p(x,t) = (x,t)2 = A2 are constant, for they depend neither on x nor on t. This is due to the complete loss of information about the position and time for a state with definite values of momentum, p = ħk, E=ħ2k2/2m. This is a consequence of Heisenberg’s uncertainty principle: when the momentum and energy of a particle are known exactly, ∆p = 0 and ∆E=0, there must be total uncertainty about its position and time: ∆x and ∆t. The second subtlety pertains to an apparent discrepancy between the speed of the wave and the speed of the particle it is supposed to represent. The speed of the plane waves (x,t) is given by 72
- On the other hand, the classical speed of the particle is given by Third, the wave function is not normalizable: The solutions (x,t) are thus unphysical; physical wave function must be square integrable. The problem can be traced to this: a free particle cannot have sharply defined momenta and energy. According to above subtleties the Schrodinger equation that are physically acceptable cannot be plane waves. Can construct physical solutions by means of a linear superposition of plane waves. The answer is provided by wave packets. 73
- Where (k), the amplitude of the wave packet, is given by the Fourier transform of (x,0) as initial wave function ψ (x,0) at time t = 0 The wave packet solution cures and avoids all subtleties raised above. First, the momentum, the position and the energy of the particle are no longer known exactly; only probabilistic outcomes are possible. Second, the wave packet and particle travel with the same speed vg=p/m, called the group speed or the speed of the whole packet. Third, the wave packet is normalizable. Group velocity phase velocity 74
- Example 1. A free particle moves along x- direction. Set up its Schrödinger equation. Solution: The motion of free particle has kinetic energy only, so we have: 0 2 2 1 2 1 2 2 2 2 p x x k E m P m mv m mv E Substituting x i Px into above equation, we get 2 2 2 2 2 2 ˆ dx d x H 75
- ) ( ) ( ˆ r E r H Substituting the above Hamiltonian operator into the following equation The schrödinger equation could be obtained 0 2 2 2 2 2 2 2 2 mE dx d E dx d m 0 ) ( 2 2 2 2 x V E m dx d 76
- The Potential Step: Unbounded states Another simple problem consists of a particle that is free everywhere, but beyond a particular point, say x = 0, the potential increases sharply(i.e. it becomes repulsive or attractive). A potential of this type is called a potential step as shown in Fig.(2): Two cases: Case E V0 The particles are free for x 0. Consider a particle of energy E moving in region in which the potential energy is the step function. What happened when a particle moving from left to right encounters the step? The classical answer is simple: All particles with E < Vo will be reflected back. All particles with E > Vo will be pass through into zone II. An unbound state occurs when the energy is sufficient to take the particle to infinity, E > V(). 77
- 78
- the particle moves with a speed v = √2E/m. As the particles enter the region x 0, where the potential now is V = V0, they slowdown to momentum 2m(E-V0) ; they will then conserve this momentum as they travel to the right. as example of potential step is a charged particle moves along the axis of cylindrical electrodes held at different voltages. If the total energy E of the electron is GREATER than the work function of the metal, V0, when the electron reaches the end of the wire, it will… either be reflected or transmitted with some probability. Wave function splits into two pieces – transmitted part and reflected part. e 79
- Fig.(2) Potential step and propagation directions of the incident, reflected, and transmitted waves, plus their probability densities (x)2 when E V0. V(x)= 0 V(x)= V0 Since the particles have sufficient energy to penetrate into the region x 0, there will be total transmission: all the particles will emerge to the right with a smaller kinetic energy E – V0. This is then a simple scattering problem in one dimension. 80
- The probabilities of reflection and transmission can be calculated by solving the Schrödinger equation in each region of space and comparing the amplitudes of transmitted waves and reflected waves with that of the incident wave. where and . The most general solutions to these two equations are waves: (x < 0) (x ≥ 0) 2 2 1 / 2 mE k 2 0 2 2 / ) ( 2 V E m k x ik x ik x ik x ik De Ce x Be Ae x 2 2 1 1 ) ( ) ( 2 1 81
- Since no wave is reflected from the region x 0 to the left, then D = 0 and the complete wave function is thus given by: x < 0 x ≥ 0 Using the boundary conditions of the wave function at x = 0 to determine the constants. Since both the wave function and its first derivative are continuous at x = 0. ) ( 2 ) ( ) ( 1 2 1 1 ) ( ) ( ) , ( t x k i t i t x k i t x k i t i Ce e x Be Ae e x t x is continuous: 1(0) = 2(0) ⟹ A + B = C is continuous: ⟹ A – B = x x x ) 0 ( ) 0 ( 2 1 C k k 1 2 82
- The reflection and transmission coefficient, R and T, is defined by: To calculate R and T 2 1 incident ) ( ) ( ) ( ) ( 2 A m k dx x ψ d x ψ dx x ψ d x ψ m i J i i i i 2 2 d transmitte 2 1 reflected , C m k J B m k J 2 2 1 2 2 2 , A C k k T A B R 83
- , 2 , ; ) ( , 2 1 1 2 1 2 1 2 1 A k k k C A k k k k B C k B A k C B A 2 2 2 1 2 1 2 2 2 2 1 2 2 1 1 4 4 , ) 1 ( ) 1 ( ) ( ) ( k k k k T k k k k R where = k2/k1 = √1-V0/E . The sum of R and T is equal to 1. When E gets smaller and smaller, T also gets smaller and smaller so that when E = V0 the transmission coefficient T becomes zero and R = 1. On the other hand, when E » V0, we have κ = 1-V0/E 1; hence R = 0 and T = 1. 84
- Case E V0 The Schrodinger equation is given by: .(2) Potential step and propagation directions of the incident, reflected, and transmitted waves, plus their probability densities (x)2 when E V0. (x ≥ 0) ) ( 2 2 2 2 2 x k dx d 85
- 2 0 2 2 / ) ( 2 E V m k Where . This equation’s solution is (x ≥ 0) Since the wave function must be finite everywhere, and since the term diverges when x ⟹ , the constant D has to be zero. Thus, the complete wave function is x < 0, x≥ 0. x k x k De Ce x 2 2 ) ( 2 x k e 2 ) ( ) ( ) ( 2 1 1 ) , ( t x k i t x k i t x k i Ce Be Ae t x 86
- Thus , the reflected coefficient is given by 1 2 2 2 1 2 2 2 1 2 2 k k k k A B R The Potential Barrier and Well Consider a beam of a particles of mass m and energy E approaching a potential barrier of height V0 and the potential everywhere else is zero. . , 0 , 0 0, , 0 ) ( 0 a x a x V x x V 87
- Case E V0 unbound states(continuum states) We will first consider the case when the energy is greater than the potential barrier (figure 3). The wave function will consist of an incident wave, a reflected wave, and a transmitted wave. The potentials and the Schrödinger wave equation for the three regions are as follows: 0 2 0 ) Region( 0 ) ( 2 ) Region(0 0 2 0 0) Region( 3 2 2 3 2 2 0 2 2 2 2 0 1 2 2 1 2 E m dx d V a x V E m dx d V V a x E m dx d V x 88
- , , ) ( , 0 , ) ( , 0 , ) ( ) ( 1 2 2 1 1 3 2 1 a x Ee x a x De Ce x x Be Ae x x x ik x ik x ik x ik x ik The corresponding solutions are: E V(x) V0 Particle x 0 ɑ Fig.(3) Potential barrier and propagation directions of the incident, reflected, and transmitted waves, plus their probability densities (x)2 when E V0 89
- where and .The constant B, C, D, and E can be obtained in terms of A from the boundary conditions: (x) and d/dx must be continuous at x = 0 and x = ɑ, respectively: These equation yield Solving for E, we obtain 2 1 / 2 mE k 2 0 2 / ) ( 2 V E m k . ) ( ) ( ), ( ) ( , ) 0 ( ) 0 ( (0), (0) 3 2 3 2 2 1 2 1 dx a d dx a d a a dx d dx d ), ( ) ( 2 1 D C ik B A ik D C B A a ik a ik a ik a ik a ik a ik Ee ik De Ce ik Ee De Ce 1 2 2 1 2 2 1 2 ) ( , 1 2 2 1 2 2 1 2 1 ] ) ( ) [( 4 2 2 1 a ik a ik a ik e k k e k k Ae k k E . )] sin( ) ( 2 ) cos( 4 [ 4 1 2 2 2 2 1 2 2 1 2 1 1 a k k k i a k k k Ae k k a ik 90
- 91
- The Case E V0: Tunneling: bound states Now we consider the situation where classically the particle does not have enough energy to surmount the potential barrier (figure3). Classically, we would expect total reflection: every particle that arrives at the barrier (x = 0) will be reflected back; no particle can penetrate the barrier, where it would have a negative kinetic energy. Quantum mechanical predictions differ sharply from their classical counterparts, for the wave function is not zero beyond the barrier. The solutions of the Schrödinger equation in the three regions yield expression that are similar to , , ) ( , 0 , ) ( , 0 , ) ( ) ( 1 2 2 1 1 3 2 1 a x Ee x a x De Ce x x Be Ae x x x ik x ik x ik x ik x ik 92
- E V(x) V0 Particle x 0 ɑ + Fig.(3) Potential barrier and propagation directions of the incident, reflected, and transmitted waves, plus their probability densities (x)2 when E V0 Except that should be replaced with x ik x ik De Ce x 2 2 ) ( 2 x k x k De Ce x 2 2 ) ( 2 93
- , , ) ( , 0 , ) ( , 0 , ) ( ) ( 1 2 2 1 1 3 2 1 a x Ee x a x De Ce x x Be Ae x x x ik x k x k x ik x ik where and The behavior of the probability density corresponding to this wave function is expected, as display in Fig.(3), to be oscillatory in the regions x 0 and x a, and exponentially for 0 x a. To find the reflection and transmission coefficients, 2 2 1 / 2 mE k . / ) ( 2 2 0 2 2 E V m k 2 2 2 2 , A E T A B R 94
- we need only to calculate B and E in terms of A. The continuity conditions of the wave function and its derivative at x = 0 and x = a yield The last two equations lead to the following expressions for C and D: Inserting these two expression into the two above equations and dividing by A, ), ( ) ( 2 1 D C k B A ik D C B A a ik a k a k a ik a k a k Ee ik De Ce k Ee De Ce 1 2 2 1 2 2 1 2 ) ( , a k ik a k ik e k k i E D e k k i E C 2 1 2 1 2 1 2 1 1 2 , 1 2 95
- Solving these two equations for B/A and E/A, we obtain: 96
- Thus, the coefficients R and T become Rewrite R in terms of T as 97
- Since cosh2(k2a) = 1 + sinh2(k2a) can reduce T to Note T is finite. This means that the probability of the transmission of the particle into the region x a is not zero. 98
- The Infinite Square Well Potential The Asymmetric Square Well Consider a particle of mass m confined to move inside an infinitely deep asymmetric potential well V(x) is infinite outside the region 0 ≤ x ≤ a, the wave function of the particle must be zero outside the boundary. The solutions only inside the well. 99
- V V V(x) 0 a 0 V The solutions are, The wave function vanishes at the walls, ψ(0) = ψ(a) = 0, the condition ψ(0) = 0 gives B = 0, while ψ(a) = Asin(ka) = 0 gives kna = n (n = 1,2,3,……) This condition determines the energy ) cos( ) sin( ) ( ) ( kx B kx A x e B e A x ikx ikx ,.....) 3 , 2 , 1 ( 2 2 2 2 2 2 2 2 n n ma k m E n n 100
- Normalisation of the wave functions gives: So that: Example: A typical diameter of a nucleus is about 10-14m. Use the infinite square well potential to calculate the transition energy from the first excited state to the ground state for a proton confined to the nucleus. Solution: The first excited state energy is found to be E2 = 4E1 = 8MeV, and the transition energy is ΔE =E2 – E1 = 6MeV. This is a reasonable value for proton in the nucleus. a n n n n n a A dx x k A dx 0 2 2 2 , 1 ) ( sin ) sin( 2 ) ( x k a x n n MeV c ma c E n ma En 0 . 2 2 , 2 2 2 2 2 2 1 2 2 2 2 101
- A particle in an infinite square-well potential has ground-state energy 4.3eV. (a) Calculate and sketch the energies of the next three levels, and (b) sketch the wave functions on top of the energy levels. a 102
- A particle in a box according to Newton's laws of classical mechanics (A), and according to the Schrödinger equation of quantum mechanics (B-F). In (B-F), the horizontal axis is position, and the vertical axis is the real part (blue) and imaginary part (red) of the wavefunction 103
- Why there is no state with zero energy for square well potential, if the particle has zero energy, it will be at rest inside the well, and this violates Heisenberg's uncertainty principle. Remember Heisenberg uncertainty relation Particle is confined in box, so x ~ a. Must be an uncertainty in momentum Since momentum cannot be zero, minimum energy must be of order E p m ma min ( ) 2 2 2 2 2 Generally, the lowest energy in a quantum system is called the zero point energy. Zero-point energy 104
- Wave function Probability distribution function a a 105
- Example (zero point): 106
- 107
- The Symmetric Potential Well What happens if the potential become symmetric? First, the energy spectrum to remain unaffected by this translation, since the Hamiltonian is invariant under spatial translation, as it contains only a kinetic part, it commutes with the particle's momentum, [H,P] = 0. The energy spectrum is discrete and nondegenerate. Second, for symmetric potentials, V(-x) = V(x), the wave function of bound states must be either even or odd. The wave function corresponding for above potential can be written as follows: 108
- That is, the wave functions corresponding to odd quantum numbers n = 1,3,5,….. are symmetric, ψ(-x) = ψ(x), and those corresponding to even numbers n = 2,4,6,…. are symmetric, ψ(-x) = -ψ(x). cos (-x) = cos(x) and sin(-x) = -sin(x) n odd n even 109
- The Finite Square Well The infinite potential is an oversimplification which can never be realised. A more realistic finite square well. Consider a particle of mass m moving in the following symmetric potential: The two physically interesting cases are E V0 and E V0. We expect the solutions to yield a continuous doubly- degenerate energy spectrum for E V0 and discrete nondegenerate spectrum for 0 E V0. 110
- Fig.(4) Finite square well potential and propagation directions of the incident, reflected and transmitted waves when E V0 and 0 E V0. 111
- The Scattering Solutions (E > V0) Classically, if the particle is initially incident from left with constant momentum √2m(E-V0) it will speed up to √2mE between –a/2≤ x ≤ a/2 and then slow down to its initial momentum in the region x > a. All the particles that come from the left will be transmitted, non will be reflected back; therefore T = 1 and R = 0. Quantum mechanically, and as we did for step and barrier potentials, we can verify that we get a finite reflection coefficient. The solution is straightforward to obtain. The wave function has an oscillating pattern in all three regions. 112
- The Bound state Solution (0 < E < V0) Classically, when E < V0 the particle is completely confined to the region -a/2 ≤ x ≤ a/2; it will bounce back and fourth between x = -a/2 and x = a/2 with constant momentum p = √2mE. Quantum mechanically, the solution are particularly interesting for they are expected to yield a discrete energy spectrum and wave functions that decay in the two regions x < -a/2 and x > a/2, but oscillate in -a/2 ≤ x ≤ a/2. In these three regions, the Schrödinger equation can be written as: 113
- , 2 1 0 ) ( , 2 1 2 1 - 0 ) ( , 2 1 0 ) ( 3 2 1 2 2 2 2 2 2 2 1 2 1 2 2 a x x k dx d a x a x k dx d a x x k dx d where k1 = √2m (V0-E)/ ħ and k1 = √2mE/ ħ. Eliminating the physically unacceptable solutions which grow exponentially for large values of ∣x∣, we can write the solution to this Schrödinger equation in the regions x < -a/2 and x> a/2 as follows: 114
- ). 2 1 ( ) ( ), 2 1 ( ) ( 1 1 3 1 a x De x a x Ae x x k x k Since the bound state eigenfunctions of symmetric one- dimensional Hamiltonian are either even or odd under space inversion, the solutions are then either antisymmetric (odd) , 2 / , 2 / 2 / ), sin( 2 / , ) ( 1 1 2 a x De a x a x k C a x Ae x x k x k a 115
- or symmetric (even) To determine the eigenvalues, we need to use the continuity conditions at x = ± a/2. The continuity of the logarithmic derivative, (1/a(x) )da(x)/dx, of a(x) at x = ± a/2 yields , 2 / , 2 / 2 / ), cos( 2 / , ) ( 1 1 2 a x De a x a x k B a x Ae x x k x k s 116
- Solve these equations to obtain: function parity odd . 2 cot 1 2 2 k a k k Similarly, the continuity of (1/s(𝑥))ds(𝑥)/d𝑥 at 𝑥 = ± 𝒂/2 gives function parity even . 2 tan 1 2 2 k a k k ..) (1,2,3,... 2 π n αn Since αn 2 = m𝒂2En / (2h2) we see that we recover the energy expression for the infinite well 2 2 2 2 2 2 n ma π E π n α n n 117
- ) 2 /( and ) 2 /( where , cot and tan with of on intersecti by the given are they : potential well sequare finit for the solutions Graphical Fig.(5) 2 2 2 2 n 2 2 n 2 2 o n n n n n V ma R E ma R 118
- In the limiting case V0→∞, the circle’s radius R also becomes infinite, and hence the function infinite become cot and tan both when because , 2 / asymptotes at the , cot and tan cross will n 2 2 n n o n n n n n V n R ...) 1,2,3,.... (n cot ..) 0,1,2,3,.. (n 2 1 2 tan n n n n n n states. bound other two yields cot on with intersecti its and 0,2, n states, bound two yields tan with the circle larger the of on intersecti the whereas 0, n boundstate one only yields tan curve with the circle small the of on intersecti The n n n n n n 119
- When is. small how matter no on) intersecti one (i.e. state bound one least at always is e that ther Note states. bound of number the greater the hence and R, of value larger the the well, e broader th and deeper The . ) 2 /( since well, the of width the and depth on the depends in turn which R, of size on the depends solutions of number The 2 2 o o o V V ma R a V 2 0 2 2 0 or 2 0 ma V R there is only one bound state corresponding to n = 0 as shown in Fig. ; this state-the ground state-is even. Then, and when 2 2 2 0 2 2 or 2 ma V R there are two bound states; an even state (the ground state) corresponding to n = 0 and the first odd state corresponding to n = 1. Now, if 120
- 2 2 2 0 2 2 2 2 2 3 2 or 2 3 ma V ma R there exist three bound states; an even state (the ground state) corresponding to n = 0, the first excited state (odd state), corresponding to n = 1, and the second excited state (even state), which corresponds to n = 2. In general, the well width at which n states are allowed is given by 2 2 2 2 0 2 2 or 2 n ma V n R 121
- Example: Find the number of bound states and the corresponding energies for the finite square well when: (a) R=1, and (b) R = 2 Solution: equations ing correspond the of solutions numerical The ly. respective 1, n and 0 n to correspond they ; cot and tan with 4 of on intersecti the from resulting states bound two are there 2 R When b) ( ). /( 1 . 1 yields which , 73909 . 0 ) 2 /( relation by the determined is energy ing correspond the Thus, . 73909 . 0 by y numericall given is cos of solution The . - 1 with tan of on intersecti by the given is energy ing correspond The 0. n to s correspond state bound This . since state bound one only is there , 1 ) 2 /( when Fig.5 (a)From 1 1 0 0 2 0 2 2 0 2 2 0 2 0 0 2 2 0 0 0 2 2 ma E E ma R V ma R o n o 122
- , 22 . 7 9 . 1 2 , 12 . 2 03 . 1 2 are energies ing correspond The ly. respective , 9 . 1 and 03 . 1 yield , sin 4 4 tan , cos 4 4 tan 2 2 1 2 1 2 1 2 2 0 2 0 2 0 1 0 2 1 1 2 2 1 1 1 2 0 0 2 2 0 0 0 ma E E ma ma E E ma 123
- The Harmonic Oscillator The harmonic oscillator is one of those few problems that are important to all branches of physics. It provides a useful model for a variety of vibrational phenomena that are encountered, for instance, in classical mechanics, electrodynamics, statistical mechanics, solid state, atomic, nuclear, and particle physics. In quantum mechanics, it serves as an invaluable tool to illustrate the basic concepts and the formalism. Classical treatment : and the solution is Where is the angular frequency of oscillation . 2 2 dt x d m kx F ) cos( ) sin( ) ( t B t A t x 124
- A particle oscillating in a harmonic potential 125
- its graph is a parabola. V x Turning point Kinetic energy zero; potential energy max. Turning point Turning point Classical particle can never be past turning point. Particle can be stationary at bottom of well, know position, x = 0; know momentum, p = 0. 126
- Which can be reduced to where x0 = ћ/m is a constant that has dimensions of length. 0 ) ( 2 ) ( 4 0 2 2 2 2 x x x mE dx x d The analytic method: this approach consists in using the power series method to solve the following differential Schrodinger equation: ) ( ) ( 2 1 ) ( 2 2 2 2 2 2 x E x x m dx x d m 127
- The solution of differential equation is expressed in terms of some special functions, the Hermit polynomials. The energy eigenvalues which are discrete or quantized: Wave functions that are physically acceptable and satisfy that Schrödinger equation are given: where Hn(y) are nth order polynomials called Hermite polynomials: ,...) 2 , 1 , 0 ( 2 1 n n En ) ( ! 2 1 ) ( 0 2 0 2 0 2 x x H e x n x n x x n n 128
- 129
- 130
- From this relation it is easy to calculate the first few polynomials: 2 2 2 2 ) 1 ( ) ( y n y n e dy d e y H En Hn(y) n ½ ħ 3/2 ħ 5/2 ħ 7/2 ħ 9/2 ħ 11/2 ħ 1 2y 4y2 – 2 8y3 – 12y 16y4 - 48y2 +12 32y5 -160y3 +120y 0 1 2 3 4 5 131
- We will deal with the physical interpretations of the harmonic oscillator results when we study the second method. Algebraic method: The Hamiltonian operator can be expressed in terms of the operators m P m m m P H 2 2 2 2 2 2 2 2 2 m q m P p q p H ˆ ˆ and ˆ ˆ ˆ ˆ 2 ˆ 2 2 and then introduce two-non-Hermitian, dimensionless operators: p i q a p i q a ˆ ˆ 2 1 ˆ , ˆ ˆ 2 1 ˆ † p q i p q q p i p q i p q p i q p i q a a ˆ , ˆ 2 ) 2 ˆ 2 ˆ ( 2 1 ) ˆ ˆ ˆ ˆ 2 ˆ 2 ˆ ( 2 1 ) ˆ ˆ )( ˆ ˆ ( 2 1 ˆ † ˆ i p p m m p q ˆ , ˆ 1 ˆ 1 , ˆ ˆ , ˆ 132
- Hence therefore where is known as the number operator or occupation number operator, which is clearly Hermitian. N̂ 133
- a H a a H a , , Can also show Now, we know that energy eigenstates of H must also be eigenstates of N, so we’ll denoted those (normalized) orthogonal eigenstates by ∣n〉, such that N∣n〉=n∣n〉 and H∣n〉 = En∣n〉 with En = ħ(n+1/2). They only constraint on n that we know right now is that it must be real number. Since we have 134
- It shall be useful to know a few mathematical relations: The Matrix Representation of Various Operators Her we look at the matrix representation of several operators in the N-space. In the matrix form, represented in the |n basis, 135
- 136
- In quantum mechanics the delta potential is a potential well mathematically described by the Dirac delta function - a generalized function. Qualitatively, it corresponds to a potential which is zero everywhere, except at a single point, where it takes an infinite value. This can be used to simulate situations where a particle is free to move in two regions of space with a barrier between the two regions. For example, an electron can move almost freely in a conducting material, but if two conducting surfaces are put close together, the interface between them acts as a barrier for the electron that can be approximated by a delta potential. The Delta-Function Potential 137
- The Dirac delta function, δ(x), is defined informally as follows: δ(x-a) would be a spike of area 1 at the point a. If you multiply δ(x-a) by an ordinary function f(x), it’s the same as multiplying by f(a): f(x) δ(x-a) = f(a)δ(x-a) δ(x) x because the product is zero anyway except at the point a. In particular, 138
- 139
- Let’s consider a potential of the form V (x) = -αδ (x) where α is some constant. This potential has a delta function well located at x = 0 (so was the infinite square well). Solving Schrödinger’s equation with delta function potential This potential yields both bound states (E < 0) and scattering states (E > 0). V (x) x x=0 E >0 E <0 140
- First consider bound states, away from x = 0, V (x) = 0 , so (E is negative, by assumption, so k is real and positive). energy) negative 0 ( ) ( ) ( 2 2 2 2 E x E dx x d m ) ( 2 ) ( 2 2 2 2 k x mE dx x d 2 mE k negative real + The general solution is (x) = A e -kx + B e kx x Potential going up at x=0 x>0 B=0 X<0 A=0 2(x) = B e -kx 1(x) = A e kx What happen in the boundary condition 141
- Boundary condition matching 1- is always continuous, and 2- d/dx is continuous except at points where the potential [V(x) ]is finite. ψ continuous 1(x=0) = 2(x=0) A = B d/dx is continuous d/dx at the boundary condition x 2 1 x=0 0 0 x kx x kx Be Ae 1 1 ) ( ) ( ) ( 2 2 2 2 x E x V x x m -αδ (x) 142
- ) ( ) ( ) ( 2 2 2 2 x E x V x x m dx x E dx x x dx x x m ) ( ) ( ) ( ) ( 2 2 2 2 0 ) 0 ( 2 2 dx d m 0 ) 0 ( 2 2 m dx d 0 0 1 0 2 dx d dx d ) 0 ( 2 2 m Bke Bke kx kx B A kx Be m Bk Bk 2 2 (0) 143
- kx Be m Bk Bk 2 2 B m Bk Bk 2 2 1 2 2 2 m k 2 2 m k mE k 2 2 2 2 m E 0 0 ) ( 2 2 x e x e B x x m x m 144
- Normalized bound state solution (x) x 2 ) ( 2 2 m E e m x x m 145
- In the region x > 0, V (x) is again zero, and the general solution is of the form, (x) = F e-kx + G ekx which reduces as G ekx as x (x) = F e-kx , (x > 0). 1- is always continuous, and 2- d/dx is continuous except at points where the potential [V(x) ]is finite. In this case the first boundary condition F = B, so The general solution is (x) = A e -kx + B e kx , but the first term blows up as x -, so A = 0: (x) = B e kx , (x < 0). 146
- The delta function must determine the discontinuity in the derivative of , at x = 0. The idea is to integrate the Schrodinger equation, from -ϵ to +ϵ , and then take the limit as ϵ 0: The right hand side becomes zero as ϵ 0 , then: 147
- k (x) k e- kx k ekx Bound state wave function for delta function potential. 0 x 148
- or you can write it in another way Ordinarily, the limit on the right is again zero, and hence d/dx is continuous. But when V(x) is infinite at the boundary, that argument fails. In particular, if V(x) = -αδ(x), then: 149
- and hence ∆(d/dx) = -2Bk. And (0) = B. So the equation says The corresponding energy is Finally, we normalize 150
- Evidently the delta-function well, regardless of its “strength” α, has exactly one bound state. The wave function can be written as: There is only one bound state for any given value of α. 151
- Scattering States: Scattering states have E 0 , for x < 0, away from x = 0, V(x)=0 so the Schrödinger equation becomes The general for x < 0 is : = A e ikx + B e –ikx . At this point we cannot eliminate either the A or B term. From the region away from the delta function 0 1 x Be Ae ikx ikx 0 2 x Ge Fe ikx ikx 152
- Boundary condition matching is continuous 1(0) = 2(0) A + B = F + G 0 1 x Be Ae ikx ikx 0 2 x Ge Fe ikx ikx d/dx at the boundary condition ) ( ) ( ) ( 2 2 2 2 x E x V x x m ) 0 ( 2 ) ( ) ( ) ( 2 2 2 2 2 m dx d x E x V x x m ) ( 0 1 1 B A ik dx d Be Ae ikx ikx 153
- ) ( 0 2 2 G F ik dx d Ge Fe ikx ikx ) ( 2 ) ( ) ( 2 B A m B A ik G F ik k m 2 2 ) 2 1 ( ) 2 1 ( i B i A G F 2 equations and 4 unknowns 0 1 x Be Ae ikx ikx 0 2 x Ge Fe ikx ikx B A G F ) 2 1 ( ) 2 1 ( i B i A G F 154
- x=0 A B F G Initial conditions ⇒Ics = scattering from the left ⇒G=0 A is incident wave Reflection and Transmission A i F A i i B 1 1 , 1 A i F A i i B 1 1 , 1 2 2 2 2 2 2 2 1 1 , 1 A F T A B R 1 T R 2 1 1 , 2 1 1 2 2 2 2 E m T m E R 155
- For higher energy the reflection coefficient decreases, that means more penetrated. In transmission coefficient the dominator is very small and the transmission coefficient approach to one Delta function barrier 2 1 1 2 1 1 2 2 2 2 E m T m E R V(x)= - aδ(x) a< 0 Classically: a > 0 (well) T = 1 and R = 0 a < 0 (barrier) T = 0 and R = 1 Q.M : is the same Ex; Tunneling 156
- For x 0 we have (x) = F e ikx + G e –ikx and at x = 0 (x) is continuous so we can write A + B = F + G . Differentiating (x) we find that: Going back to our previous integration of the Schrodinger equation, we have the result that For the scattering states (0) = A + B and we have 157
- Let , divide by ik and collect the A and B terms on one side and we get F – G = A (1 + 2iβ) – B(1- 2iβ) . What do the terms in the Schrodinger equation Represent? A e ikx represents a wave incident from the left and F e ikx is a wave travelling from the potential well to the right. 0 V(x) 0 x Feikx Ge- ikx Aeikx Be-ikx Scattering from a delta –function well 158
- Suppose G e –ikx is zero, then B e –ikx represents a wave reflected from the potential well, moving to the left. we put G = 0 and solve the two equations A + B = F + G and F – G = A (1+2iβ) – B (1-2iβ) . We can define reflection and transition coefficient, R and T which are the probabilities of reflection and transmission from the potential well. and 159
- and R + T = 1 In terms of energy we can write: The higher energy, the greater the probability of transmission (which seems reasonable). 160
- CHAPTER FOUR Angular Momentum Commutation Relations The Uncertainty Relations for Angular Momentum 161
- Angular momentum is as important in classical mechanics and quantum mechanics is the motion of a particle in a central force field where the potential function V depends only on the distance r between the particle and the origin of the coordinate system. It is particularly useful for studying the dynamics of systems that move under influence of spherically symmetric, or central, potentials. The orbital angular momenta of these systems are conserved. Additionally , angular momentum plays a critical role in the description of molecular rotations, the motion of electron in atoms, and the motion of nucleons in nuclei. 162
- In classical mechanics the angular momentum of a particle with momentum P and position r is defined as cross product of position vector and momentum vector In quantum mechanics the classical vectors become operators. More precisely, they give us triplets of operators: p r L L p r and , ) ˆ , ˆ , ˆ ( ) ˆ , ˆ , ˆ ( ) ˆ , ˆ , ˆ ( z y x z y x L L L L p p p p z y x r 163
- x y z x y z p p p L x y z x y z p p p i j k L The Cartesian component of are L̂ x z y y x z z y x L yp zp L zp xp L xp yp ˆ ˆ ˆ x y z L i y z z y L i z x x z L i x y y x x y z z x y y z x p y p x L p x p z L zp p y L ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ i x y z d d d dx dy dz i j k L 164
- Clearly, angular momentum does not exist in a one-dimensional space. We should mention that the components Commutation relations: , ˆ of square the and , ˆ , ˆ , ˆ L L L L z y x 2 2 2 2 ˆ ˆ ˆ ˆ z y x L L L L have we , ˆ , ˆ , ˆ , ˆ , ˆ , ˆ since and ˆ and , ˆ , ˆ do so and commute mutually ˆ and , ˆ , ˆ Since i p z i p y i p x p p p z y x z y x z y x y x z x z y z y x L i L L L i L L L i L L ˆ ˆ , ˆ , ˆ ˆ , ˆ , ˆ ˆ , ˆ 165
- We will use a " ladder operator " technique, introduce raising and lowering operators, L L and , The different components of L do not commute with each another, but they do commute with the squared magnitude of the angular momentum vector: 0 ˆ , ˆ ˆ , ˆ ˆ , ˆ 2 2 2 z y x L L L L L L 2 2 2 ˆ , , , 0 x y z L L L L L L y x L i L L ˆ ˆ ˆ ) ˆ ˆ ( 2 1 ˆ , ) ˆ ˆ ( 2 1 ˆ L L i L L L L y x 166
- In addition, satisfy L L L L L L L L z z ˆ ˆ , ˆ , ˆ 2 ˆ , ˆ , 0 ˆ , ˆ2 These relations leads to z z L L L L L ˆ ˆ ˆ ˆ ˆ 2 2 Which in turn yield 2 2 ˆ ) ˆ ˆ ˆ ˆ ( 2 1 ˆ z L L L L L L 167
- The different components of angular momentum do not commute • Lx, Ly and Lz are not compatible observables • They do not have simultaneous eigenfunctions (except when L = 0) • We can not have perfect knowledge of any pair at the same time But, the different components all commute with L2 • L2 and each component are compatible observables • We can find simultaneous eigenfunctions of L2 and one component 168
- The Uncertainty relations for angular momentum Recalling the generalized uncertainty relation for two operators 2 2 2 2 , i B A B A B A b a , 2 1 2 2 2 A A A 2 2 2 B B B i p x x ˆ , ˆ 169
- We can write down uncertainty relation for the components of angular momentum using the commutators. For example, we find z z y x y x L i L i i L L L L 2 2 2 , z y x L i L L , 170
- CHAPTER FIVE Quantum Mechanics In Three Dimensions •Schrodinger Equation in Spherical Coordinates •Separation of Variables • The Angular Equation • The radial Equation • The Box Potential • The Hydrogen Atom 171
- The generalization to three dimensions is straightforward. Schrodinger's equation says the Hamiltonian operator H is obtained from classical energy by the standard prescription (applied now to y and z, as well as x): or Schrodinger Equation in Spherical Coordinates 172
- for short. Thus where is the Laplacian, in Cartesian coordinates. The potential of finding the particle in the infinitesimal volume d3r = dx dy dx is |Ψ(r,t)|2 d3r, and the normalization condition reads 173
- With the integral taken, over all space. If the potential is independent of time. there will be complete set of stationary states, where the spatial wave function ψn satisfies the time- independent Schrodinger equation: The general solution to the (time-independent) Schrodinger equation is with the complex constants cn determined by initial wave function, Ψ(r,0), in the usual way. (If the potential admits continuum states, then the sum in above equation becomes an integral). n=1,2,3,…. r d r r c n n 3 0 , ) ( 2 ,.. 2 , 1 3 n n c r d 174
- Separation of Variables Typically, the potential is function only of the distance from the origin. In that case it is natural to adopt spherical coordinates, (r,θ,) (see Fig.1). In spherical coordinates the Laplacian takes the form, In spherical coordinates, then, the time-independent Schrodinger equation reads We begin by looking for solutions that are separable into products: ψ (r,θ,) = R(r) Y (θ,) 175
- Then Fig.1: Spherical coordinates: Radius r, polar angle θ, and azimuthal angle z x y r p ȓ θ 176
- x y z r r cos r sinsin Note that = 0 = 2 In math notation, and are swapped Switch from Cartesian to Spherical Coordinates •r is the distance from the origin to the point • is the angle compared to the z-axis • is the angle of the projected “shadow” compared to the x-axis sin cos sin sin cos x r y r z r 0 0 0 2 r 177
- dividing by Y R and multiplying by -2mr 2/ћ2: 178
- The Angular Equation Multiplying by Y sin2θ Separation variables Y (θ,) = (θ) Ф() 179
- Since Φ(ϕ) must be periodic with period 2π, m must be an integer m = 0, ±1, ± 2, … The significance of m is essentially the same as in two dimensions: If we fix r and θ and let ϕ vary, then we move around a circle about the z-axis 180
- z r 0 radius 181
- 182
- The solution is Where is the associated Legendre function 183
- The radial Equation The angular part of the wave function, Y(θ,), is the same for all spherically symmetric potentials; the actual shape of the potential, V(r), affects only the radial part of the wave function, R(r), which is determined by Let u (r ) r R(r ) , So that R= u/r , dR/dr = {r(du/dr) – u}/r2, (d/dr)[r2(dR/dr)] = rd2u/dr2 , and hence: 184
- This is called the redial equation; it is identical in form to the one-dimensional Schrodinger equation. Example: Consider the infinite spherical well, Outside the well the wave function is zero; inside the well the radial equation 185
- Where The boundary condition u (a) = 0. The case l = 0 is easy, The actual radial wave function is R(r) = u(r) / r, and [cos(kr)] / r blows up as r 0. So we must choose B = 0. The boundary condition then requires sin(ka) = 0, and hence ka = n, for some integer n. The allowed energies are evidently 186
- The Box Potential The Rectangular Box Potential: The potential energy of the particle is considered to be equal zero inside the box and it is infinity at the boundaries(surface) and in the remaining space. Consider first the case of a spinless particle of mass m confined in a rectangular box of sides (a, b, c) which can be written as V(x, y, z) = Vx(x) + Vy(y) + Vz(z), with 187
- the potentials Vy(y) and Vz(z) have similar forms. The wave function ψ(x, y, z) must vanish at the walls of the box, x = 0, x = a, y = 0, y = b, z = 0 and z = c . a b c y x z V(x,y,z) =∞ V(x,y,z)=0 188
- S.E in three dimension is E z y x m 2 2 2 2 2 2 2 2 E m 2 2 2 The boundary condition 0 ) , , ( ) 0 , , ( 0 ) , , ( ) , 0 , ( 0 ) , , ( ) , , 0 ( c y x y x z b x z x z y a z y Using the separation variable ) ( ) ( ) ( ) , , ( z Z y Y x X z y x The total Hamiltonian is break down in three cases l dimensiona one for case same the ) ( ) ( 2 2 2 2 x X E dx x X d m x x a n A x X x x sin ) ( is solution the 189
- The same for Y and Z y b n A y Y y y sin ) ( z c n A z Z z z sin ) ( b c a z y x z y x dz dy dx 0 0 0 1 ) , , ( ) , , ( abc A A A z y x 8 c z n b y n a x n abc z y x z y x n n n z y x sin sin sin 8 ) , , ( 2 2 2 2 2 2 2 8 c n b n a n m h E z y x n n n z y x 190
- Two particles problems The Hydrogen Atom The hydrogen atom consists of an electron and a proton. The H atom is a bound state of a proton and an electron. The approximation of the potential energy of the electron- proton system is electrostatic. From Coulomb’s law, the potential energy: r e r V 0 2 4 ) ( -e (electron) r +e (proton) The hydrogen atom 191
- Spherical polar coordinate system 2 2 2 2 2 2 2 sin 1 sin sin 1 1 1 r r r r r Angular momentum operator ) ( 2 2 2 2 z y x L L L L 2 2 2 2 2 sin 1 sin sin 1 L 2 2 L 2 2 2 2 2 2 1 r L r r r r 192
- The hydrogen atom potential energy is given by: e- P+ r V(r) e2 r V (potential) 0 r The time-independent Schrödinger equation in three dimensions is then: E r e z y x m 0 2 2 2 2 2 2 2 2 4 2 The radial equation: It describes a particle with angular momentum L = √l(l+1). which behaves like a particle in a one-dimensional effective potential of the form 193
- The problem is to solve this equation for u (r) and determine the allowed electron energies E. Let (For bound states, E < 0 , so is real.) The radial equation becomes We define a new variable ρ so that 194
- As , the constant term in the brackets dominates, so approximately The general solution is u () = A e- + B e but e blows up (as ), so B = 0. Evidently, u () A e- , as 0 the centrifugal term dominates; approximately, then The general solution is u () = C l+1 + D -l , but -l blows up (as 0 ), so D = 0. Thus u () C l+1 195
- For small . The next step is to peel off the asymptotic behavior, introducing the new function v (): u () = l+1 e - v () , And the radial equation 196
- Finally, we assume the solution, v (), ca be expressed as a power series in : Differentiating term by term, Differentiating again, 197
- This recursion formula determines the coefficients. Now let's see what the coefficient look like for large j (this corresponds to large , where the higher powers dominate). 198
- and hence u () = A l+1 e To be complete So the allowed energies are: This is the famous Bohr formula. 199
- Where is the so-called Bohr radius. Evidently the spatial wave functions for hydrogen are labeled by three quantum numbers (n , l , and m): nlm (r,θ,) = Rnl(r) Yl m (θ,) , 200
- CHAPTER SEX Approximation Methods for Stationary States • perturbation theory • The variational method • WKB method • Time-Independent Perturbation Theory •Nondegenerate Perturbation Theory 201
- Most problems encountered in quantum mechanics cannot be solved exactly. Exact solutions of the Schrodinger equation exist only for a few idealized system. To solve general problems, one must restore to approximation methods. In this chapter we consider approximation methods that deal with stationary states corresponding to time-independent Hamiltonians. To study problems of stationary states, we focus on three approximation methods: perturbation theory, the variational method, and WKB method. Perturbation theory is designed to predict the eigenvalues and eigenfunctions of a perturbed Hamiltonian H that differs only slightly from an unperturbed Hamiltonian H0 whose eigenvalues and eigenfunctions are known. The difference between H and H0 is called the perturbation. 202
- The variational method was initially derived from a mathematical theorem that formulates a relation between differential equations with boundary conditions, integral equations, and matrices of infinite order. In situations where exact solutions to the problem could not found, the matrix representations lead to convenient and fairly reliable approximate solutions. The WKB method is useful for finding the energy eigenvalues and wave functions of systems for which the classical limit is valid. Unlike perturbation theory, the variational and WKB methods do not require the existance of a closely related Hamiltonian that can be solved exactly. 203
- Perturbation Theory Time-Independent Perturbation Theory Suppose that we solved the time-independent Schrodinger equation for some potential and obtained a complete set of orthonormal eigenfunctions 0 n and corresponding eigenvalues E0 n. nm m n n n n E H 0 0 0 0 0 0 This is the problem that we completely understand and know solutions for. Now we slightly perturb the potential. For example, we raise a little bit the bottom of the infinite square well or put a little bump there: x a V( x) Infinite square well with small perturbation little bump 204
- The problem of the perturbation theory is to find eigenvalues and eigenfunctions of the perturbed potential, i.e. to solve approximately the following equation: H H H E H n n 0 , This method is most suitable when Ĥ is very close to a Hamiltonian Ĥ0 that can be solved exactly. In this case Ĥ can be split into two time-independent parts using the known solutions of the problem perturbation 0 0 0 0 n n n E H Suppose we change this potential only slightly; e.g., we could add a slight ‘bump’ in the bottom of the well. It is not likely that we can solve for the S.E of this new Hamiltonian H exactly, but let’s try to find an approximate solution. 0 0 0 0 n n n E H 205
- H = H0 + H΄ rewrite the Hamiltonian introducing explicitly a parameter ¸ that determines how small is a particular term Where 1 ≥ λ ≥ 0 Ĥ ΄ = λŴ The application of the approximation methods to the study of stationary states consists of finding the energy eigenvalues En and the eigenfunctions |n of a time- independent Hamiltonian Ĥ that does not have exact solutions: Ĥ |n = En |n . or The eigenvalue becomes: (Ĥ0 + λŴ ) |n We consider two separate cases depending on whether the exact solutions of Ĥ0 are nondegenerate or degenerate. Each of these two cases requires its own approximation scheme. 206
- Nondegenerate Perturbation Theory non-degeneracy occurs when each eigenstate corresponds to a unique energy. If Ĥ0 has no degenerate eigenvalues: that is, for every energy En (0) there corresponds only one eigenstate |n : n n n E H ) 0 ( 0 ˆ Where the exact eigenvalues En (0) and exact eigenfunction n are known. The main idea of perturbation theory consists in assuming that the perturbed eigenvalues and eigenstates can both be expanded in power series in the parameter λ. 207
- When λ = 0 the two above expression yield the unperturbed solutions: En = En (0) and |n = |n . or n = n Here En (1) is the first order correction to the nth eigenvalue, and n (1) is the first order correction to the nth eigenfunction; En (2) and n (2) are the second order corrections, and so on. The lowest order (λ0) this yields H0 n 0 = En 0 n 0 . ......] )[ ( 2 2 1 0 0 n n n H H ] .....)[ ( 2 2 1 0 2 2 1 0 n n n n n n E E E ...... ) ( ) ( 1 2 0 2 0 1 0 0 0 n n n n n H H H H H ... ] ( ) ( 0 2 1 1 2 0 2 0 1 1 0 0 0 n n n n n n n n n n n n E E E E E E , . )......... ( 0 0 0 0 0 n n n E H , )....... ( 0 1 1 0 0 1 0 1 n n n n n n E E H H , )... ( 0 2 1 1 2 0 1 2 0 2 n n n n n n n n E E E H H 208
- The first order (λ1) The second order (λ2) multiplying by (n 0) and integrating. But H 0 is Hermitian, so 209
- and this cancels the first term on the right. Moreover, , so This is the fundamental result of first order perturbation theory; 210
- So if m = n, we are in trouble (also 2nd order energy correction). Having m=n means we have two (or more) states with the same energy = Degenerate states 211
- Example: The unperturbed wave functions for the infinite square well are the equation Suppose first that we perturb the system by simply raising the "floor" of the well by a constant amount V0 as shown Fig. In that case H' = V0, and the first order correction to the energy of the nth state is: 0 : n Hamiltonia on Perturbati V H 212
- V(x) x a V0 Constant perturbation over the whole well. 213
- The corrected energy levels, then, are En En 0 + V0 ; they are simply lifted by the amount V0 . The only surprising thing is that in this case the first order theory yields the exact answer. Evidently, for a constant perturbation all the higher corrections vanish. If, on the other hand, the perturbation extends only halfway across the well (as in Fig.) V(x) x a V0 a/2 214
- then, In this case every energy level is lifted by V0 /2. That’s not the exact result, presumably, but it does seem reasonable as a first order approximation. Q)- Find the first order correction to the wave function. Second-Order Energies Now, cut the perturbation to only a half-way across the well 215
- 216
- Typically, the perturbation (H) will “break” the degeneracy: As we increase λ (from 0 to 1), The common unperturbed energy E 0 splits into two(as in Fig.). The essential problem is this: when we turn off the perturbation, the upper state reduces down to one linear combination of a 0 and b 0 , and the lower state reduce to some other linear combination, but we don’t know a priori what these good linear Combination will be. For this reason we can’t even Calculate the first-order energy λ 1 E0 E Lifting of a degeneracy by a perturbation. 217
- because we don’t know what unperturbed states to use. For the moment, therefore, let’s just write “good” unperturbed states in the general form Keeping α and β adjustable. We want to solve Schrodinger equation, H = E with H = H 0 + λ H and E = E0 + λE1 + λ2E2 + … , = 0 + λ 1 + λ2 2 + … 0 0 0 b a 218
- Degenerate Perturbation Theory If the unperturbed states are degenerate that is, if two (or more) distinct states (a 0 and b 0) share the same energy then ordinary perturbation theory fails. However the perturbation usually removes the degeneracy, so the levels split. Twofold Degeneracy We have two states a 0 and b 0 that are degenerate, i.e they have the same energy E0, Spilt states will have orthogonal. b b a a E H E H 0 0 0 0 ˆ ˆ ˆ ˆ 0 0 0 0 0 0 0 0 b b a a E H E H 1 , 0 0 0 0 0 0 0 b b a a b a 0 1 1 0 0 1 0 E E H H 219
- Linear combination of these states 0 0 0 b a is also an eigenstate of H0 with eigenvalue E0 . We want to solve H H H E H 0 , ......... ....., 1 0 1 0 E E E 0 1 1 0 0 0 E E H H This time we multiply this equation from the left by a 0 and integrate, i.e take inner product with a 0. 0 0 1 1 0 0 0 0 1 0 0 a a a a E E H H 1 0 0 1 0 0 n a n a E E H 1 0 0 0 n b a a E H 0 0 0 b a 220
- 0 0 1 0 0 1 0 0 0 0 b a a a b a a a E E H H 1 0 ab aa b a a a w w H H E 0 0 0 0 1 ab aa w w E 1 ) , , ( , 0 0 b a j i H w j i ij Wij are known since we know i 0 we can calculate them Similarly, the inner product with b 0 yields βE 1 = αWba + βWbb now solve this system of equations for E1 where ab aa w w E 1 bb ba w w E 1 Notice that the W’s are (in principle) known “matrix elements” of H, with respect to the unperturbed wave functions a 0 and b 0 221
- bb ab ba ab ab ab bb ba aa ab ab aa w w w w w E w w w E w E w w w E 1 1 1 1 2 2 1 1 2 1 1 1 1 1 1 4 ) ( 2 1 0 ) ( ) ( ) ( 0 if ) )( ( ) ( ) ( ab ab aa ab aa ab ba ba ab bb aa bb aa ba ab bb aa aa bb ba ab aa w w w w w E w w w w w w w w E E w w w E w E w E w w w w E E Fundamental result of degenerate perturbation theory: two roots correspond to two perturbed energies (degeneracy is lifted). 222
- aa bb w E w E 1 1 0 If 0 If 0 0 0 0 1 b b bb a a aa H w H w E if we could guess some good linear combinations a 0 and b 0, then we can just use nondegenerate perturbation theory. This is the fundamental result of degenerate perturbation theory; the two roots correspond to the two perturbed energies. 223
- But what if α is zero? In that case β = 1, Wab = 0, and E1 = Wbb. This is actually include in the general result with plus sign (the minus sign corresponds to α = 1 , β = 0). Then using E E0 1 E 1 E d=2 , a b 0 0 ' bb b b W H 0 0 ' aa a a W H 0 0 ' ab a b W H 0 ab ba W W If aa a W E 1 0 bb b W E 1 0 224
- Problem: Suppose we put a delta-function bump in the center of the infinite square well 2 a x H where is a constant (a) Find the first-order correction to the allowed energies. Explain why energies are not perturbed for even n. (b) Find the first three nonzero terms in the expansion below of the correction to the ground state, 0 0 0 0 0 m n m m n n m n E E H 225
- The Variational Principle There are many problems of wave mechanics which cannot be conveniently treated either by direct solution of the wave equation or by use of perturbation theory. The Helium atom is such a system. No direct method of solving the wave equation has been found for this atom, and the application of perturbation theory is unsatisfactory because the first approximation is not accurate enough. One of the approximation methods that is suitable for solving such problems is the virational method, which is also called the Rayleigh-Ritz method. The variational method is one way of finding approximations to the lowest energy eigenstate or ground state, and some excited states. But not very useful for the study of excited states. 226

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