Quantum Mechanics found in a nutshell. Work in progress, chapter 2 updated.

Transcription

1 Quantum Mechanics found in a nutshell. Work in progress, chapter updated. Content.Introduction Homework:... 4 Homework, One Dimension Square well Homework,, 3, 4, 5, 6, Linear potentials Homework Harmonic oscillator Homework, 3, Raising and Lowering Operators.... Homework 5, Exponential Potential Bound States... Homework Continuum States Homework, Delta-Function Potential Homework 7, 8, Number of Solutions Normalization Homework,, Wave Packets Homework Homework Approximate Methods WKBJ Homework,, 3, 4, Bound States by WKBJ Harmonic Oscillator Morse Potential Symmetric Ramp... 38

2 3..4 Discontinuous Potentials Homework 5, 6, Electron Tunneling Homework 5, Variational Theory Half-Space Potential Harmonic Oscillator in One Dimension Homework 7, 8, 9,,3, Homework Spin and Agular Momentum Operators, Eigenvalues, and Eigenfunctions Commutation Relations Raising and Lowering Operators Eigenfunctions and Eigenvalues Homework Representations... 5 Homework, 3, Rigid Rotations Homework 5, The Addition of Angular Momentum Homework 7, 8, Homework Two and Three Dimensions Plane Waves in Three Dimensions Homework:, Plane Waves in Two Dimensions Central potentials Central Potentials in 3D Homework: 3, 4, 5, 6, 7, 4, 5, Central Potential in D Homework Coulomb Potentials Bound States Confluent Hypergeometric Functions Homework 9, Hydrogen Eigenfunctions... 79

3 Homework,, Continuum States WKBJ Three Dimensions Homework 8, D Hydrogen Atom Homework Two Dimension Hydrogen -like Atoms Quantum Defect Homework WKBJ Derivation Expectation values Variational Theory Hydrogen Atom: n = Homework, Hydrogen Atom: l = Helium Atom Homework, 3, Free Particles in a Magnetic Field Gauges Eigenfunctions and Eigenvalues Homework 6, 7, Density of States Quantum Hall Effect Flux Quantization... Homework Matrix Methods and Perturbation Theory H and Ho Matrix Methods Coupled Spins... 4 Homework,, Tight-Binding Model... 4 Homework 4,5, The stark Effect

4 Homework Perturbation Theory General Formulas Homework 8,,, Harmonic Oscillator in Electric Field Homework 9, Continuum States Homework Green s Function The Polarizability Literature dr.l.noordzij@leennoordzij.nl Exercises, Remarks and Questions. Based on Quantum Mechanics in a Nutshell by Mahan. References will be made to Quantum Mechanics, The Theoretical Minimum by Susskind, The Feynman Lectures on Physics and The Principles of Quantum Mechanics by Dirac..Introduction. In this chapter Mahan summarizes some basics of Quantum Mechanics. Having consummated for example Susskind s book, this chapter is accessible. Remark: On page 4 in paragraph.3 reference is made to the Schrödinger equation. The equation number should read (.6) instead of (.8), a printing error. On top of page 5 Eq. (.5) is given: b n = d 3 rf(r)φ n (r). I think this Eq. should read b n = d 3 rf(r)φ n (r). Then you will find with Eq. (.), after substitution of Eq. (.4) in the latter expression for b n, b n = b n. In paragraph.5 on The Heisenberg Representation, Eq. (.59) should read: ψ ϯ (r, t) = ψ (r, t = )e ith/ħ. Homework: Homework,. Exercise. Prove that e L ae L = a + [L, a] + [L, [L, a]] + [L, [L, [L, a]]] +! 3! where (a, L) are any operators. Use will be made of the Taylor series expansion e x = r=. (C.) 4 x r r!

5 After expansion of e L ae L into Taylor series of the operators, the above sum of commutators can be found. Well, this is not exactly a proof. It is a demonstration that the expansion works. How to prove it in a more formal way? That is not so easily done. Taking a close look at the expansion in commutators for example the third expression of the commutators on the right hand side, you will find [L, [L, [L, a]]] = 3 (3 3! 3! k= k ) L3 k a( L) k. This indicates a general expression for the n th expression: [L,. [L, [L, a]]. ] = n (n n! n! k= k ) Ln k a( L) k, (C.) where the dots on the left-hand side of (C.) indicates: L, [ and the dots on the right-hand side: ]. Now something can be done with induction. So we assume (C.) to be true and then prove it for n +. First, let s illustrate induction for the fourth expression of commutators: [L, [L, [L, [L, a]]]] = 4 (4 4! 4! k= k ) L4 k a( L) k. (C.3) The left-hand side of (C.3) can be written as [L, [L, [L, [L, a]]]] = L [L, [L, [L, a]]] [L, [L, [L, a]]] L, (C.4) where we neglect, since this factor sows up at the left-hand and right-hand side. With 4! [L, [L, [L, a]]] = 3 ( 3 k= k ) L3 k a( L) k, the right-hand side of (C.4) can be written as: [L, [L, [L, [L, a]]]] = 3 ( 3 k= k ) L4 k a( L) k 3 ( 3 k= k ) L3 k a( L) k+. (C.5) After some algebra collecting the expressions for k =, k =, k = and k = 3 we obtain: L 4 a 4L 3 al + 6L al 4LaL 3 + al 4. Well, it comes as no surprise, this equals 4 ( 4 k= k ) L4 k a( L) k. Induction works here. Now for n + : [L, [L,. [L, [L, a]]. ]] = n+ (n + (n+)! (n+)! k= k ) Ln+ k a( L) k. (C.6) Again we neglect (n+)!. With (C.) the left-hand side of (C.6) becomes: n ( n k= k ) Ln+ k a( L) k + n ( n k= k ) Ln k a( L) k+, (C.7) where we have included +L into L n+ k and L into ( L) k+. The first term of this expression (k = ) gives L n+ a, the last one (k = n) gives a( L) n+. What about the k term? Well, after carefully inspecting (C.7) you need two terms: ( n k ) Ln+ k a( L) k and ( n k ) Ln (k ) a( L) k (= ( n k ) Ln+ k a( L) k ). So we need the k term, the left part of (C.7), and the (k )term, the right part of (C.7) in order to obtain equal powers of the operators. So collecting equal powers of the operators, the k term of (C.7) finally reads: {( n k ) + ( n k )}Ln+ k a( L) k. From a straightforward (n+)! procedure the term between {} becomes: = (n + k!(n+ k)! k ). To summarize: The first term is L n+ a, the k term is ( n + k ) Ln+ k a( L) k and the last 5

6 one is a( L) n+. We know after collecting the terms, this can be written as n+ ( n + k= k ) Ln+ k a( L) k. By induction we have proven (C.6) to be correct. Exercise. If F is any operator that does not explicitly depend on time, show that F = in an eigenstate of H with discrete eigenvalues. With Eq..5 we have F (t) = i [H, F]. For the time independent Schrödinger t ħ equation we have with eigenvectors E j and eigenvalues ε j : H E j = ε j E j. The eigenvalues of the Hermitian operator H are real. So Eq..5 can be written as: t F (t) = i ħ [H, F] = i ħ E j [HF FH] E j = i ħ E j ε j( F F)ε j E j = = iε j ħ E j (F F) E j. And iε j E ħ j (F F) E j =. Remark: In the above exercise we learned that for any operator not explicitly depending on time the time derivative is zero in an eigenstate of the Hamiltonian operator with discrete eigenvalues. In addition we can add this applies also for a more general state when the set of eigenvectors is complete. The general state vector A, say, can be expanded in the eigenvector base. Then again: F (t) = i A [HF FH] A =. Note: A can be written t ħ as: A = α j E j and A [HF FH] A = E j α j ε j (F F) ε j α j E j.. One Dimension. This chapter is about the one-dimensional Schrödinger equation, the square well, the harmonic oscillator and various types of potential all leading to exact solutions.. Square well. Remark: In section. Mahan introduced the expression: binding energy. It is clear that there is a direct relation between binding energy and de eigenvalues. Mahan started this section with the square well of infinite positive potential- walls and width a. Inside this well, < x < a, the potential V(x) is zero. d The Schrödinger equation Eq. (.3) is: ( ħ + E) ψ =. m dx The boundary conditions are: ψ(x) =, at x = and ψ(x) = at x = a. No solutions exist for E <. Mahan gives the physical meaning of this statement. The mathematics are: for E <, the solution of the above wave equation is similar to Eq. (.7): ψ(x) = C 3 e αx + C 4 e αx. So for x =, C 3 = C 4 and for x = a, C 3 (e ax e ax ) =. With a, C 3 = : no solution for E <. As mentioned above inside the well the potential V(x) is zero. This is not of vital importance. We could have set inside the well V(x) equal to a positive value, V say, or a negative one, V say. In the first case, V, to find solutions V + E > and in the second one V + E >. The second example is equivalent to the first. The third example is for a potential well/box V(x) given by Eq. (.3) and Figure.. V(x) = for x <, V(x) = V for < x < a and V(x) = for a < x; V is positive. 6 t

7 Mahan discussed bound states first. A bound state describes a system where a particle is subject to a potential such that the particle has a tendency to remain localized in one or more regions of space. Caveat: in Eq.(.4) it is less confusing to write p = p B for bound states with eigenvalues E = E B. Again for x <, ψ(x) =. In the region a < x with V(x) =, the Schrödinger equation is ( ħ m dx + E) ψ =. With E <, bound states, the eigen function decays exponentially. For < x < a the d Schrödinger equation is ( ħ V(x) + E) ψ =, where V(x) = V m dx, and V + E >. With V + E > the boundary conditions can be fulfilled and the general solutions are given by Eqs. (.5) and (.7). A bound state is found for < E + V < V. Mahan found a minimum eigenvalue E > V + π E 4 a with E a = ħ ma. In Eq. (.36) Mahan defined the strength of the potential: g = V and he introduced a dimensionless binding energy ε E a through E = εv. Plug ε and g into E > V + π E 4 a and we obtain ε < ( π g ). Then with ε > the expression for ε is < ε < ( π g ). Now we see that the smallest or minimum binding energy is found for ε, then g = π. This is also the minimum value for g or minimum critical value g c as represented by Eq. (.38). Mahan derived the minimum value of the coupling strength g c with help of Eq. (.37). In the middle of page Mahan writes: the two equations in (.4)-.6), except now α = k <. This is confusing. α is related with bound states E B and k is related with continuous states E k. So α k. At the bottom of page Mahan used the expression unrenormalized. I was not aware Mahan has discussed the subject of renormalization. The fourth example with constant potentials discussed by Mahan is, when a particle reflects at a barrier. The transmission is discussed for a particle entering from the right and from the left (Fig..4). The equations derived by Mahan are extended to the transmission and reflection of a barrier of any shape and thickness. In Fig..5, the fifth example, a barrier of any shape(height) and thickness is represented by a grey box. I think to find expressions for R L, T RL, etc, expressed in the amplitude of the incoming wave and the wave numbers k and p, the geometry of the barrier has to be known. In the text below Figure.5 the equations for the fifth example are the same as for a step function potential. Fig..5 consists of two parts: (a) and (b). The text below Eq.(.7) : In figure.5b, take the complex conjugate of the wave coming from the left. Looking at what Mahan did, the complex conjugate of the left incoming wave in Fig..5(a) is taken, given in Eq. (.73). Furthermore Mahan stated: Since Schrödinger s equation is invariant under complex conjugate operation,.. The Hamiltonian is invariant too? In addition Mahan mentioned: In fact, the complex conjugate is the time- reversal operation:.. Is there a time reversal operator? Time is not explicitly present. Mahan constructed the complex conjugate by multiplying the two eigen functions representing the particle entering from the right and the left by R L and T LR respectively. Why these two coefficients? I will come back to this question later on. d 7

8 From the resulting equations he derived for example: T RL = p k T LR, Eq.(.8). Now, look at Eq. (.7) example four, there we see: T = p T. This is exactly the same as the k result in Eq.(.8), example number five. Let s return to example four and use the same procedure as used in example five. That is to say, we don t use the matching at x =. We find: for x < : I = R + T T (C.) and R = R I. So I =, and I = ; for x > : T R + R T = (C.) and T = T I. So I =, and I =. Now we look at the conservation of currents. For a particle entering from the left, with the notation of Mahan: v p T + v k R = v k I. (C.3) For a particle entering from the right: v k T + v p R = v p I. (C.4) With I =, Eq. (C.) can be written as = R + T T equivalent to Eq..77. Eq. (C.) can be written as: R = R T being equivalent to Eq..8. With I =, (C.3) and the notation of Mahan: T = k p ( R ) equivalent to Eq..79 With I =, (C.4) and the notation of Mahan: T = p k ( R ) equivalent to Eq..8. T 8 (C.5) (C.6) The above equations give the solution T = p T, equivalent to Eq..77. k From Eq. (.53) we learn with the assumptions E > V and V >, k to be larger than p. What does that mean for R and T with I = given in Eqs. (.56) and (.57)? Well, it looks like T is larger than one. What is the physical meaning of that? Does the particle wave, the transmitted one get a kind of boost from the positive potential step? Above I raised the question why Mahan used the coefficients R L and T LR to construct the time-reversed state ψ L. With the amplitudes of the incoming waves from the right and from the left equal to unity and equating ψ L equal to Aψ L + Bψ R the general coefficients are found to be A = R L and B = T RL. Use has been made of the equations (.7), (.7) and (.73.). Furthermore for x >, the coefficient of e ipx in the expression for ψ L is forced to zero. Consequently R L T LR + T LR R R =. This leads directly to Eq. (.8). See also Eq.(.76). Homework,, 3, 4, 5, 6, 7. Exercise. Derive the numerical value in ev for the bound-state energy E of an electron in the onedimensional square-well potential: V(x) = { V for x < b for x > b where b =. Å and V =. ev. What is the critical value of coupling strength g c for this potential? To begin with: when the potential is symmetric, the eigenfunctions are either of even or odd parity(mahan, page 7). Even parity eigenfunctions are in this case cosine functions.

9 Odd parity eigenfunctions are sine functions. The Schrödinger equation: for x < b: d ψ(x) + β ψ(x) = and β = m (V dx ħ + E). This gives the solution for ψ(x) = C sin(βx) + C cos (βx) (C..9) for x > b: d ψ(x) α ψ(x) = and α = m E. dx ħ ψ(x) = C 3 e αx + C 4 e αx or ψ(x) = C 3 e α x. (C..) Now we start with even parity eigenfunctions: So x < b: ψ(x) = C cos (βx) and (C..) x > b: ψ(x) = C 3 e α x, ψ vanishes for x. (C..) We match both functions and their derivatives at x = b (matching at b creates no new information): with Eqs. (C..) and (C..) : C cos(βb) = C 3 e αb and C β sin(βb) = C 3 αe αb. Dividing this two expressions leads to the eigenvalue equation: tan (βb) = α or tan β ( V +E ) = E and E V +E b = E b ħ mb. Bound states exists for V < E <. So the tangent function is positive: < E+V < π. The left-hand E b side of this expression produces no new information. The right-hand side leads to E < V + π E 4 b. (C.3) The lower bound state can be found for E > V. The upper bound state is given by Eq. (C.3). What about the factor π E 4 b? Let us look at the numerical value. Using the mass of the electron we have: E b = 3,75 ev. Bound state can be found for: < E < 8.4 ev. The upper bound is positive, consequently, < E <. Well, this could have been said on beforehand. Are there bound states actually? Look again at Eq. (C..3). We also have V < E <. So V < E < V + π 4 E b. This is inconclusive. Using E = εv and g = V, the eigenvalue equation tan ( V +E ) = E E b E b V +E, can be written as : tan ( g ( ε)) = ε. The critical binding is found for ε. Then tan (g) = and ε the minimum value of g c =. g is defined as g = V. With numerical values: g = or g =.5 E b 3.75 Now we apply the odd parity eigenfunction: x < b: ψ(x) = C sin(βx). (C..4) Again we match both functions and their derivatives, Eqs. (C.4) and (C.), at x = b (matching at b gives no new information): C sin(βb) = C 3 e αb and C β cos(βb) = C 3 αe αb. Dividing these two expressions: tan (βb) = β α. Then the minimum eigenvalue is found from E+V E b > π or E > V + π 4 E b. Plug in the numerical values and E > 8.4 ev. Since E has to be < we conclude: no bound states. The potential well is too shallow. What about the critical value of the coupling strength g c? 9

10 g is defined as g = V. With numerical values: g = or g =.5. E b 3.75 For the odd parity eigenfunction we can use the minimum coupling strength or minimum potential strength g c, given by Eq. (.38), g c = π. Consequently g < g c and there are no bound states. What conclusion can be drawn? Only a bound state of the even parity type? Well, according to Fitzpatrick, no odd-parity bound states can be found for g < g c. Exercise. Make a graph similar to Figure.8 of the solutions of Eq. (.37) over the range of < g <. Show that new bound state start at g cn = π(n + ). Remark: I think Figure.8 should read Figure. after reading the text. As given by Mahan, the eigenvalue equation Eq. (.37) can be solved for various values of g using graph paper or with help of a computer. The lowest bound state started at g c = π. Eq. (.38) states that the eigenvalues start at the following critical values: g c = π (n + ) = π (n + ). I think that Eq.(.33) also gives the maximum eigenvalue. So Eq.(.34) becomes π(n + ) > V +E > π (n + ). So V E a + π E a > E > V + π E 4 a. By computing, numerical or graphical, the solutions of the eigenvalue equation tan g ( ε) = ε ε for the various values of g, you can check the critical values of g = g cn = π(n + ). An eigenvalue by inspection: ε =. This eigenvalue is found with all calculations. For g > π you will find additional values for ε. For example: g =, we have ε =. and ε =. For g = 5π, you will find for 3 ε:.58,.875 and (With help of WolframAlpha). Exercise 3. A particle in a one-dimensional square well of infinite sides is confined to the interval < x < L. The eigenfunctions are given in Eq. (.9). Prove the relation n= φ n (x)φ n (x ) = Cδ(x x ), by evaluating the summation, and determine the constant C. So prove sin (nπx L n= ) sin ( nπx ) = Cδ(x L L x ). (C..5) For the left-hand side of (C..5) we write: (e inπ(x x ) L L n= + e inπ(x x ) L e inπ(x+x ) L e inπ(x+x ) L ). 4 Change the sign of n in the first and fourth term in the above expression, then we have: n= e inπ(x x ) L ( L e inπ(x+x ) L ). (C..6)

11 We use the delta function representation for the expression in (C.6) and obtain, using δ[ π(x x ) L ] = L π δ(x x ) (Dirac), : π (δ(x x ) δ(x + x )). We know < x < L. Likewise < x < L, consequently δ(x + x ) = for < x < L. This finally leads to: sin (nπx L n= L ) sin ( nπx ) = δ(x L x ). So C =. Exercise 4. For the potential shown in Figure., prove Eq. (.5) by doing the integral. ψ b and ψ k are the bound and continuum eigenfunctions. Remark: I think Figure. is meant, since figure. illustrates bound sates only. The integral for orthogonality reads: = ψ B ψ k dx, Eq. (.5). This equation is correct for real wave functions. In general the orthogonality relation is given by Eq.(.). Then Eq.(.5) is = ψ B ψ k dx. The bound states are given by Eq.(.9):, x < ψ B = { C sin(p B x), < x < a C 3 e αx, a < x The continuum states are given by Eq.(.4): a ψ k = { sin(p k x), D sin(kx + δ), a < x Mahan used δ as a phase shift in Eq.(.4). To find out about the continuum- and bound wave functions to be orthogonal we calculate the constants C, etc. of the wave functions. For the bound states the procedure to find the constant by means of normalization is rather straightforward as shown by Mahan. C is given in Eq.(.44): C = (a sin(p Ba) + sin (p B a) ) /, (C..7) p α where p and α are given by Eqs.(.4) and (.5). With Eq.(.3): C 3 = C sin (p B a)e αa. (C..8) For the continuum states delta function normalization is used, discussed in section.8.3 by Mahan. The delta function is represented by: πδ(k k ) = e ix(k k ) dx. In section.8.3, Mahan derived the normalization constant D =. So with Eq.(.47) we find for a : a = sin (ka+δ). sin (p k a) Now with the constants for the wave functions we can try to prove : (C..9) = ψ B ψ k dx. All solutions of the Schrödinger s equation must be orthogonal to all others with the same potential energy function, Mahan page. We have the expressions for the wave functions. So let us find out about the orthogonality. Is the following equality true: δ(x a) = π n= eik(x a), Algebraic and Analytical Methods,

12 a = C sin(p B x) a sin(p k x)dx + C a 3 e αx D sin(kx + δ)dx. The first integral is reads: C a ( sin(p p B p B p k ) a sin (p k p B +p B + p k )a). k The second integral in (C..) is evaluated by means of integration by parts. This results into: ( + α k ) e αx sin(kx + δ) dx = a k e aα ( α k sin(ka + δ) + cos(ka + δ)). (C..) (C..) Caveat: with Eq.(.6) and Eq.(.4) we could find ( + α k) =. However, keep in mind for α bound states E = E B < and for k continuous states E = E k >. a Substitute the integral and (C..) into (C..) and find out whether the right-hand side of (C..) is. To this end we use Eq.(.3), (C..9) and D =. C sin (ka+δ) ( sin(p sin (p k a) p B p B p k ) a sin (p k p B +p B + p k )a) + k + kc sin(p B a) ( α sin(ka + δ) + cos(ka + δ)) =? (C..) k +α k We divide (C..) by C and sin (ka + δ), both, then we have ( sin(p sin (p k a) p B p B p k ) a sin (p k p B +p B + p k )a) + k + k sin(p Ba) ( α + cot(ka + δ)) =? (C..3) k +α k = p k tan (ka+δ) k Substitute into (C..3),, Eq.(.49), and multiply by sin (p tan (p k a) ka) ( sin((p sin (p k a) p B p B p k )a) sin((p k p B +p B + p k )a)) + k sin(p Ba) ( α + +cot(ka + δ)). k k +α k We are still trying to find out whether this expression is zero. Rewrite this expression by expending the two first sin functions: p k sin(p B a) cot(p k a) p B cos (p B a) p B p k + k sin(p Ba) ( α + cot(ka + δ)) =? (C..4) k +α k What about the factor (p B p k )? Back to some physics. With the Eqs. (.4), (.6) and (.4): (k +α ) (p B p k ) (k +α ) =. Substitute this result into (C.4) : (p k sin(p B a) cot(p k a) p B cos(p B a)) + k sin(p B a) (cot(ka + δ) + α k ) =? Now = p k tan (ka+δ) k tan (p k a), Eq.(.49). Substitute this into (C.5) : sin (p p B a) k p sin(p tan (p k a) B cos(p b a) p B a) k α sin(p tan(p k a) Ba) =? Well, this results into: tan(p B a) + p B =? Eq.(.3): it is zero indeed and = ψ B ψ k dx. α (C.5) Exercise 5. Consider in one dimension the solutions of the Schrödinger s equation for the half-space problem and V > :

13 V =, x V(x) = { V, < x < a V =, x > a Find an expression for the phase shift δ(k) for the two cases (a) < E < V and (b) V < E. a) < E < V and V >. Bound states. These are no real bound states. < x < a : Schrödinger s equation: d ψ(x) dx α ψ(x) = and α = m ħ (V E B ). The general solution is ψ(x) = C e αx + C e αx. For x =, V(x) =, so ψ = and we have C = C. The wave function is ψ(x) = C (e αx e αx )= C sinh (αx). x > a: d ψ(x) dx + E B ψ(x) =. With general solution expressed with help of phase shift δ: ψ(x) = a sin (k B x + δ) and k B = m E ħ B. At x = a the wave function (C..6) and its derivative are continuous: C sinh(αa) = a sin (k B a + δ), αc cosh(αa) = k B a cos(k B a + δ). tanh (αa) Divide these two equations and we have = tan (k Ba+δ). α k B The phase shift is δ = k B a + arctan [ k B tanh(αa)]. α b) Now V < E k. Continuous states. < x < a : Schrödinger s equation: d ψ(x) dx + p ψ(x) = and p = m ħ (E k V ). For x =, we have ψ = and ψ(x) = a sin(px). x > a : the wave function is ψ(x) = D sin (k k x + γ) and k k = m E ħ k. At x = a the wave function and its derivative are continuous: a sin(pa) = D sin (k k a + γ), a p sin(px) = Dk k cos (k k a + γ). Again, divide these two equations and we find tan (pa) The phase shift is γ = k k a + arctan [ k k p tan(pa)]. p = tan(k ka+γ) k k. (C..6) (C..7) (C..8) (C..9) Exercise 6. Consider a particle of energy E > V > approaching a potential step from the left. The Hamiltonians in the two regions are: p, x < m H { L p + V m, x > R where m L m R and V(x) = V Θ(x). Θ(x) is the Heaviside function. 3

14 Find the amplitude of the transmitted (T) and reflected wave (R), by matching at x = the amplitude of the wave function and the derivative dψ L = dψ R. Show that this choice of m L dx m R dx matching conserves the current of particles. For x < : d ψ L + Eψ dx L =, and k = m LE. ħ Then ψ L = Ie ikx + Re ikx. For x > : d ψ R dx + (E V )ψ R =, and p = m R ħ (E V ). Then ψ R = Te ipx. Now at x =, ψ L = ψ R and I + R = T. Furthermore: m L dψ L k = dψ R dx m R dx (I R) = p T. m R m R With (C..3) and (C..3) we find: R = I km R pm L km R +pm L. T = I km R. km R +pm L at x = and (C..3) (C..3) (C..3) (C..33) Now we check the matching of the derivatives by calculating the particle currency given by Eq.(.58): j = ħ mi [ψ dψ dx dψ ψ ], the current operator. dx The incoming particle: ψ i (x) = Ie ikx, then j i = ħk m L I. The reflected wave : ψ r = Re ikx, then j r = ħk m L R. (C..34) (C..34) The transmitted wave: ψ t = Te ipx, then j t = ħp T. (C..36) m R With j t j r = j i and the equations (C..3)-(C..36) we indeed find the proposed matching condition justified. Exercise 7. A particle of mass m moves in a one-dimensional square-well potential with walls of infinite height: the particle is constrained in the region L/ < x < L/. a. What are the eigenvalues and eigenfunctions of the two lowest states in energy? b. What is the expectation value of the energy for a state that is an equal mixture of these two states? c. For the state in (b), what is the probability, as a function of time, that the particle is in the right-hand side of the well? a) See Mahan page 7. The wave function: ψ n (x) = L sin [k n (x + L )], with k n = nπ, and E L n = ħ k n. m So ψ n = [sin (nπ x) cos L L (nπ ) + cos (nπ x) sin L (nπ)]. As we see, the boundary conditions are fulfilled. (C..37) ψ n (x) = L sin (nπx) n is even, (C..38) L 4

15 ψ n (x) = cos (nπx) n is odd. (C..39) L L The eigenfunctions are: ψ (x) = cos (πx ) and ψ L L (x) = L sin (πx), (C..4) L The eigenvalues are, Eq. (.) and (.3): E = ħ π, and E ml = 4E. (C..4) b) The expectation value of the energy for a state that is an equal mixture of the two states in (C..4). The mixed state: ψ m = a(ψ (x) + ψ (x)). The factor a is found by normalization. ψ m = [cos (πx) + sin (πx )] L L L L/ ψ m dx =. Substituting the two wave functions (C..4) and we find L/ a = L. So ψ m = [cos (πx) + sin (πx)]. (C..4) L L L The expectation value for the energy: L/ L/ d H = ψ m ( ħ m dx )ψ mdx. (C..43) So the expectation value of the Hamiltonian will produce the expectation value of the energy. Reminder: in Dirac notation of bra s, kets, eigenvectors and eigenvalues we have H = Ψ H Ψ = Ψ E Ψ = E Ψ Ψ = E. We substitute (C..4) into (C..43) and find: H = ħ 5π = E ml m. We see H = E m = (E + E ). (C..44) c) For the state in (b), ψ m, what is the probability, as a function of time, that the particle is in the right-hand side of the well, < x < L/? It is about time-dependency. So we have to return to (C..4) and include e ie nt/ħ, where n = and n =. With (C..4) : ψ m = [cos (πx ) L L e iet/ħ + sin ( πx ) L e iet/ħ ], (C..45) and E and E given in (C..4). The probability as a function of time to find that the particle is in the right-hand side of the well, < x < L/, is: L/ ψ m P(t) = ψ m dx. Substitute (C..45) into P(t), you will obtain after some goniometrical manipulation: P(t) = + 4 3π cos[(e E )t ]. ħ Hence with two states we will find time dependency for the probability. 5

16 . Linear potentials. In section. Mahan deals with linear potentials. The Schrödinger equation is converted into an Airy s equation. Below Eq. (.94) Mahan writes: The last line shows that the integrand is an exact differential, which vanishes at both end points. The value of the sine function as t is not obvious, since it oscillates. However, it is to be zero anyway. For any unbound state? In Fig..6 the potential (a) and the wave function (b) is shown. There is no numerical relation between (a) and (b). The energy level in (a) is not an eigenvalue. It illustrates the point where z =, the turning point. Homework 8. Exercise 8. Derive an equation for the eigenvalues of the potential V(x) = F x, where F > for values < x <. So we have a wedge, ramp, type symmetrical potential well with infinite walls. Schrödinger s equation is ħ d ψ(x) + ( F x + E)ψ(x) =. m dx Keep in mind: When the potential is symmetric, the eigen functions are either of even or odd parity (Mahan). We use the transformations of Eq. (.85) and (.86): x = ( ħ m )/3, z = x E/F x. Schrödinger s equation reads: ( d dz z) ψ(z) =. This is Airy s equation and the solution is ψ(z) = C Ai( x E F). x The eigenvalues of the eigenfunctions for ψ(x) with odd parity are found with: ψ() = = C Ai( E ). The allowed eigenvalues for odd parity are: x F z =.338, 4 = 4.879, z 6 = 5.56, z 8 = , etc. The eigenvalues of the eigenfunctions with even parity are found with: dψ E = = d dx dx Ai( x F), at x =. The allowed eigenvalues are: x z =.88, z 3 = 3.48, z 5 = 4.8, z 7 = 6.633, etc. The above values for zeros z n can be found from tables (Abramowitz and Stegun) or with help of WolframAlpha. Also a few values are given by Mahan page 9. As in the example given by Mahan for linear potentials z n <, consequently E n (= x Fz n) ) >..3 Harmonic oscillator. Remark: On top of page 3 Mahan mentioned the question about the polynomial as a solution to the Hamiltonian on page 9. Solution to the Hamiltonian or the solution to the Schrödinger equation? Mahan gave the eigen values and eigenvectors(functions) of the Schrödinger equation in the Eqs. (.7-.9). A differential equation is a mathematical problem that is incompletely 6

17 specified: the boundary conditions are needed to completely determine the eigenfunctions, Mahan. Which are the boundary conditions for the harmonic oscillator where the potential exists over all space? The potential is a symmetric function, consequently we find even parity eigenfunctions for dψ =, at x = ; odd-parity eigenfunctions are found for ψ() =. In dx addition for x, V(x), so ψ(x) =. Do these conditions produce the Hermite polynomials, the eigenvalues and the eigenvectors(functions) as solutions? Or are there just the Hermite polynomial s for n = (ε n )? Schrödinger s equation is:[ d dξ ξ + ε] ψ =. Now we substitute the Hermite polynomial for the bound state ψ n = C n H n (ξ)e ξ / into the Schrödinger s equation and obtain: H n ξ dh n + d H n + ε dξ dξ nh n =. The constants drop out of the differential equation. With help of Eq.(.) we rewrite the differential equation. dh n dξ = nh n. Consequently: d H n dξ H n ξ dh n dξ = n dh n dξ 7 = 4n H n. We substitute this result into + d H n dξ + ε nh n = and obtain (ε n )H n 4n(ξH n nh n ) =. With help of equation (.), ξh n = H n+ + nh n, the equation for the eigenvalues is ε n = n +. So the Hermite polynomials are solutions to Schrödinger s equation. The constant C n is found by normalization : = = dξψ n ψ l. This can be written as δ nl = dξψ n ψ n and orthogonality is found by dξψ n ψ l. With help of generating functions C is found and equals N n. This approach differs from what has been written on page 3 and 3. It differs indeed. Mahan writes on page 3: Generating functions are useful for evaluating integrals of harmonic oscillator functions. They will used to prove (.5): I nl = dξψ n (ξ)ψ l (ξ) = N n N l means to prove the first part I nl = dξh n (ξ)h l (ξ)e ξ, Eq.(.8). I assume Mahan dξψ n (ξ)ψ l (ξ), since the second part represent the integral of the wave functions in terms of Hermite polynomials. So the second equality sign is an identity. I further assume I nl to be the unity matrix. So what is Mahan basically doing? I think to prove I nl = δ nl. This is what he concluded on page 3. I think this is quantum mechanics upside down. δ nl = dξψ n ψ l is part of the physics: the probability to find a particle in the whole interval is. Furthermore the bound states are orthogonal. Even orthonormal after normalization. Mahan derived Eq. (.4): dξh n (ξ)h l (ξ)e ξ = δ nl N n. Well, plugging into this equation the unknown constant C n, we have ψ C n C n (ξ)ψ l (ξ)dξ = δ nl, and C l N l N n = N n. This is n concluded by Mahan:.., and also derives the value of the normalization constant N n. Let s have a look once more at Schrödinger s equation is:[ d dξ ξ + ε] ψ =.Playing with WolframAlpha you will find for n =, ε = : ψ = C πe ξ / erfi(ξ) + C e ξ /, where erfi(ξ) = ξ π et dt. Consequently C = and C = N. So ψ = N H e ξ /, even parity. For n =, ε = 3 a solution is produced which is not so easy to recognise: ψ = C D (i ξ) + C D ( ξ). The D n -functions can be found in Whittaker and Watson.

18 There the relation with the Whittaker function W k,j can be found. Abramowitz and Stegun list the Whittaker function too. Again in the above expression for ψ, C = and ψ = N ξe ξ /, odd parity. For n =, ε = 5, WolframAlpha produces again error functions and a Gaussian function, the wave function ψ is of even parity. Susskind found the entire energy spectrum, eigenvalues, without solving the differential equation. Based on the sometimes called ladder operator developed by Dirac. Susskind did not solve the differential equation. On the other hand he just proposed as ground state a Gaussian distribution and found in this way the lowest ground energy E. At the bottom of page 3, above Eq. (.9), in the line on the two generating functions e ξ, has to be included. See Eq. (.). Now we will pay some attention to deriving Eqs. (.8) and (.9). We write Eq. (.) as ξh n = H n+ + nh n, Eq.(.7). This equation is converted into eigenfunctions by multiplying Eq. (.7) with the appropriate factor given in Eq. (.8). Then on the left of Eq. (.7) appears ξψ n. On the right appears N n exp( ξ ) with the Hermite polynomials Mahan converted N n on the right of the resulting equation to N n+ and N n respectively with help of Eq. (.9). The result for N n+ is: N n = (n + )N n+, and for N n : N n = n N n. We use both expressions for N n. N n+ with H n+ and N n with H n. Then N n nh n = n N n H n and N n H n+ = n+ N n+h n+, where we left exp( ξ ) out of account on both sides of these expressions. In using Eq. (.8) for the right hand side of Eq. (.7) with the above expressions including exp( ξ ) you will obtain Eq. (.8) and consequently Eq. (.9). The latter equation is given in the Dirac notation. Nota bene: keep in mind, by converting N n to N n±, we used Eq. (.9) and N n+ = [ π(n + )! n+ ] = [(n + ) π(n!) n ] = [(n + )] N n, etc. With Eq. (.9) Mahan derived the matrix element of Eq. (.5) n x l. In Eq. (.3), e.g., δ n,l+ = for l + = n, or l = n. In order to derive the Hamiltonian, expressions such as ξ n are needed. The basis for that is Eq. (.9). For the Hamiltonian we need also the derivative d =. dx x dξ In Eq. (.36) we find d H dξ n(= nh n ) by differentiating H n given in Eq.(.3) and using Eq.(.). Like we did for the matrix element of x, Eq. (.39) for the matrix element for p is constructed. For the Hamiltonian we need the second derivative of the wave function. I consider the subscript of the Kronecker Delta in Eq. (.4) a bit confusing. Finally, Eq.(.44), the eigenvalues are found: E n = ħω (n + ). To me the relation between the boundary conditions for the differential equation of the quantum harmonic oscillator d 8

19 and the eigenvalues is not clear. By plugging ψ n of Eq. (.8) into the Schrödinger equation, using Eqs. (.)-(.) you can find the expression for ε n at ξ =. First derive d dξ ψ n : d dξ ψ n = ξψ n + n N n N n ψ`n where use has been made of Eq. (.36). To find the second derivative differentiate this expression once more and we arrive at: d ψ dξ n = ψ n ξ d ψ dξ n nξ N n ψ N n + 4n(n )N n e ξ H n. (C.46) n This expression can be rewritten with help of the recursion relation Eq. (.): H n = ξ n (ξ H n n H (n ) n+ H (n ) n. (C..47) With Eq. (.8), (C.46) and (C.47) we have for Schrödinger s equation: ψ n (n + ) ξ d ψ dξ n nξ N n ψ N n + 4ξ n ψ n n n n N n ψ N n+ ξ ψ n + εψ n =. n+ For even parity,ψ n, at ξ =, this this gives: ε = n +. Homework, 3, 4. Exercise. Find the exact eigenvalues for the potential x < V(x) = { Kx x > Hint: With a little thought, the answer may be found by doing no derivation. A little thought: For the above potential function we have ψ(x) = at x =. Mahan dealt with the Schrödinger s equation in section.3: the Harmonic oscillator. There the potential function is V(x) = Kx for < x <. A symmetrical potential function where we can find odd parity and even parity wave functions. The general solution is given in terms of Hermite H polynomials. In this exercise we do not have a symmetrical potential function and the boundary condition is: ψ() =. And this is the only boundary condition at x =. From the recursion relations we have H n+() = nh n and H =. Furthermore H () = and H () =. So we learn for n is odd and nonnegative ψ() =, and n is even ψ(). Use has been made of the general solution for ψ n (ξ) = N n H n (ξ)e ξ /, Eq.(.8), and N n. The eigenvalues are E n = ħω(n + ) for n is odd. I do not know whether or not I gave sufficient little thought. Well, look at this problem from a different perspective. We have the harmonic oscillator and the infinite potential wall at x =. So we know ψ() =. The boundary condition leading to odd parity solutions for the wave function. Consequently we can mirror the oscillator at x =. Picture that and we immediately have the exact eigenvalues for the mirrored situation: the eigenvalues are E n = ħω(n + ) for n is odd. Exercise 3. Evaluate the following integral for the harmonic oscillator using generating functions: 9

20 M nl = dxφ n (x)xφ l (x). Evaluate to prove something? To prove what? M nl is the matrix element of the observable represented by the position operator. To evaluate means: find x [ nδ n,l+ + (n + )δ n,l ]? As Mahan writes: Generating functions are useful in evaluating integrals of harmonic oscillator functions. In the section of the harmonic oscillator Mahan evaluated the matrix element n x l = M nl of the position operator. This is done with the Dirac approach with bra s and kets. a). We could follow the procedure of page 3 for the normalization. φ n (x) = ψ x n ( x ) = ψ x x n (ξ). Then we evaluate the integral M nl = x With the Hermite polynomials this becomes: M nl = x N l N n Using Eqs.(.9) and(.) the integral becomes: M nl = N l N n dξ[h n Now we make use of: N l+ N l dξψ n (ξ)ξψ l (ξ) using generator functions l+ H l+ + H n N l N l l H l ]e ξ. dξh n dξh n (ξ)ξh l (ξ)e ξ. (ξ)h l (ξ)e ξ = δ nl N l N n. Then we find for the matrix element: M nl = x [ l + δ n,l+ + lδ n,l ], or M nl = [ nδ n,l+ + n + δ n,l ]. Well, this is Eq.(.3). Did I use generating functions or not? See the discussion in the above section.3 on the harmonic oscillator. b).on the other hand we could proceed as follows: We have M nl = x N l N n dξh n (ξ)ξh l (ξ)e ξ. Multiply both generating functions, Eqs. (.9) and (.), and include ξe ξ. Then integrate over all ξ, we obtain with the expression for M nl : zn y l M n,l n,l n!l! N n N l = πx (z + y)e zy. Well, what do you prefer? Exercise 4. Evaluate the following integral for the harmonic oscillator using generating functions: M nl (q) = dxφ n (x)φ l (x)e iqx. With generating functions follow the procedure b) of exercise 3. a) zn y l M n,l(q) n,l = πe zy+i(z+y)qx q x /4. n!l! N n N l b) Another approach. We have: M nl (q) = N l N n dξh n (ξ)h l (ξ)e ξ e iqxξ. For convenience we substitute: a = iqx. Then: M nl (q) = N l N n e a 4 dξhn (ξ)h l (ξ)e (ξ a) = e a 4 δ nl = e q x /4 δ nl..4 Raising and Lowering Operators. Remark: I refer here again to the work of Dirac and the ladder operator. Eq. (.8): C is found with the selection of A and B. The operator for finding the observables, energy levels, is the Hamiltonian H which can be

21 written as Eq. (.73): H = ωħ(a ϯ a + ). So the raising and lowering operators are acting together. How to measure their effects separately? Homework 5, 6. Exercise 5. For the harmonic oscillator, evaluate the following (s is constant): a. [a, H] b. [a ϯ, H] c. e sh ae sh d. e sh a ϯ e sh ad a) For the Hamiltonian we use: H = ħω(a ϯ a + ). [a, H] = ah Ha = ħω(aa ϯ a a ϯ aa) = ħω(aa ϯ a ϯ a)a = ħωa = ħω (ξ + d dξ ) with ξ = x/x. ad b) Similar to ad a): [a ϯ, H]= ħωa ϯ (aa ϯ a ϯ a) = ħωa ϯ = ħω (ξ d dξ ). ad c) e sh ae sh = esh (ξ + d ) dξ e sh = esh (ξe sh sh dh se ) = dh (ξ s dξ d With H = ħ m dx Kx. ad d) e sh a ϯ e sh = esh (ξ d ) dξ e sh = esh (ξe sh sh dh + se ) = dh (ξ + s ). dξ dξ Exercise 6. The squeezed state is the operator (λ is a constant) S(λ) = N(λ)e λaϯ. This is also called a coherent state. a. Find the normalization constant N(λ) such that S ϯ S =. I assume N(λ) and λ to be complex and S(λ) Hermitian. Now S ϯ S = N e λ a e λaϯ. Since the exponentials are separated we need not to apply the Feynman-Glauber Theorem. With help of the Eqs. (.57), (.6) or (.8)- (.84) we have: M = N e λ a e λaϯ = N. Consequently N =. b. Evaluate the commutator [a, S]. For the commutator we write: as Sa = N exponents of k we finally obtain [a, S] = λn a (λaϯ ) k k= k! λ k k= k! N (λa ϯ ) k k= k! (a ϯ ) k = λne λaϯ = λs. dξ ). a. Collecting equal Use has been made of: aa ϯ a ϯ a = or aa ϯ = a ϯ a +. Furthermore use has been made of induction. In order to prove : k= a (λaϯ ) k k! (λa ϯ ) k k= k! λ k a = λ k= (C..48) (k )! we presume this expression to be correct for k = m and prove that the expression is true for k = m +. So is the following m + expression correct:

22 a(λa ϯ ) m+ (m+)! (λaϯ ) m+ a (m+)! = λ(λaϯ ) m m!? Let s find out. We can rewrite the latter expression by using aa ϯ = a ϯ a +. Then we obtain: λa ϯ m ) m+ {a(λaϯ (λaϯ ) m a } + λ(λaϯ ) m = λ(λaϯ m )? The expression between {} is just the m-th m! m! (m+)m! m! term of the commutator (as Sa), the left hand side of (C..48). This equals according to the presumption of induction λ(λaϯ ) m. So λa ϯ m ) m+ {λ(λaϯ (m )! } = λ(λaϯ ) m? m! (m )! Rearranging and dividing by λ m+ we arrive at m(a ϯ ) m + (a ϯ ) m = (m + )(a ϯ ) m. Et voilà (m + )(a ϯ ) m = (m + )(a ϯ ) m. c. Evaluate the commutator [a ϯ, S]. Since S is a power series in a ϯ, [a ϯ, S] commute and equals. Or a ϯ S Sa ϯ = a ϯ e λaϯ e λaϯ a ϯ = [ aϯ (λa ϯ ) k k= k! (λaϯ ) k a ϯ ] =. k! d. Show that S is an eigenstate of the lowering operator a and find its eigenvalue. So we have to prove as = S S, where the eigenvalue is indicated by S. Above(ad. b) we found as Sa = λs. Then as = (λs + Sa). We know a =. Consequently as = λs. We may conclude S to be an eigenstate and S = λ to be its eigenvalue..5 Exponential Potential. Both bound states and continuum states are discussed..5. Bound States. In the section on bound states the exponent β in the second term in Eq.(.) should have the minus sign. Homework 9. Exercise 9. Derive the equation for the bound states of the potential V(x) = V e x /a, for all < x < and V >. The derivation follows closely the procedure of exercise 8 with respect to a symmetrical potential. Due to this symmetry, the wave function is described by Eq.(.98). The solution to this differential equation of section.5.(mahan) can be used for the eigenvalues of odd parity wave functions. The equation for the bound states : J β (g) =, Eq.(.3). β, g and E a are given by Eqs. (.95)-(.97). The equation for even parity wave functions is found from the condition dψ = at x =. So d dx J β(ge x /a) )=. With this equation you can also find some additional critical values of the coupling strength g c. These values are: g c = 3.837, 7.56,.734, etc.(see Fig..). These values are found with WolframAlpha and are equal to the values found from J (g) =. For β, I could only find solutions(with WolframAlpha) for integer values of β. For dx

23 example: β =, we find with J (g) = : g = 5.356, 8.47,.796, etc..5. Continuum States. Eq. (.) describes continuum states. Mahan writes: The prefactor in front is inserted to make ψ(k. x) a real function.. The wave function in general is not a real function. Mahan wanted to make the wave function real. Eq.(.) represents a wave going to the left and a wave going to the right. The latter can be interpreted as a reflected wave due to the infinite exponential potential barrier. In order to find Eq. (.3) use has been made of: Γ(z ) = (Γ(z)) (Abramowitz and Stegun): the complex conjugate of the gamma function equals the gamma function with complex argument.. Then you will obtain: = Γ(+iK) Γ(+iK) [Γ(+iK) Γ( ik) ], and = Γ( ik) Γ(+iK) [Γ(+iK) Γ( ik) ]. The second term in Eq.(.) can also be rewritten using a phase factor: ( g ) ik = Γ( ik) Γ(+iK) eiγ, then Γ(+iK) eiγ = ( g ) ik Γ( ik) Γ( + ik). With help of the above expression for ln [Γ(+iK)]. May be it comes as no surprise: Γ( ik), we can find for γ = Kln Γ( ik) (g) i γ = δ (Eq.(.4) and Eq.(.) can be written as Eq.(.5). On page 44 Mahan writes below Eq.(.8): The current j(x) does not depend on x since all the cross terms cancel to zero. Well, this is a mathematical argument. A physical argument is: there is nothing at x that can create a dependency on x. It is confusing to use δ to express the phase shift; top page 44 and 53. Does ψ(k, x) in Eq. (.) represent a real wave function? Homework,. Exercise. Find the transmission coefficient of a continuum wave going from left to right for the potential V(x) = V e x /a. For a continuum wave we have E > and Schrödinger s equation is ħ d ψ(x) + (E V)ψ(x) =. (C..49) m dx E V = E + V e x /a. (C..5) We transform the wave equation with y = e x /a and use the definitions of Eq.(.6). We arrive at the following Schrödingers equation: (y d + y d + dy dy g y + K ) ψ(y) =. (C..5) Now what to do? First we know that we are considering a continuum wave going from the left to the right. We might expect the particle wave not to be trapped in the potential sink or well, E >. For x and V(x), we can approximate the wave function by ψ(x) = Ie ikx, with k = me and I is the amplitude. For x we can approximate the wave function, the ħ 3

24 transmitted wave, by ψ(x) = Te ikx. T is the amplitude. Since no particles are created or disappear I = T. There only can be a phase factor between the two waves. Define the transmission coefficient as T, then the coefficient equals. There is no reflection at the potential well. could we look at the problem in another way? Well, we have equation (C..5). Rewrite this equation as: (y d dy + y d dy (ig) y + K ) ψ(y) =. We can use Eq.(.)-Mahan-for a wave going to the write. Due to the term with ig instead of g the phase factor changes with a factor Kiπ/ for the incoming wave as well as for the transmitted wave. So the transmission coefficient is. Exercise. Consider the potential in one dimension: ħ V(x) = g mx, where g > is a dimensionless parameter and solve for x >. a. Find the solution to Schrödinger s equation. b. Derive the phase shift by considering the form of the eigenfunction at large positive x. Hint: Try a solution of the form xj ν (kx) and use the fact that lim J ν (z) = cos [z π (ν + )]. z πz a) Schrödinger s equation: ( d g + dx x k ) ψ(x) =, with k = me ħ. At x =, ψ() = since V(x). Now there is an infinite potential barrier at x =, we may expect at x with a particle wave going to the left a reflected particle wave similar to the example of section.5. (Mahan). Mahan proposes a solution of the form xj ν (kx). After substitution of this particular Bessel function you will find a relation between ν and g. Substitution of z for kx Schrödinger s equation reads: ( d g + ) ψ(z) =. Playing with dz z this equation using WolframAlpha for various values of g, for example g = /4, you will find as a solution: ψ(z) = C zj (z) + C zy (z). (C..5) After playing with various values of g you could guess a relation between ν and g. This is shown in the table below. I g ν ν 3 3/ 3/4..53? / 5/4 / 3/ 3/4 g = ν 4 g = ν 4 g = ν 4 4

25 /4 / / /8 3/ 3/8 g = ν 4 g = ν 4 For g =, I put a question mark into the table since with Wolfram Alpha I got the following expression: ψ(z) = C π [ sin(z) cos(z)] + C z π [ sin(z) cos(z) ]. (C..53) z Again by substituting g =. and g =.99 into Schrödinger s equation I found with WolframAlpha the values for ν as given in the table. Now you could guess ν to be.5. With a sort of reverse engineering you will obtain (Chisholm and Morris): zj 3/ = π [sin(z) cos(z)], so ν = 3/ indeed. z To have ψ = at z for any value of ν, C = in equation (C..5) for ψ(z). However g = is a special case as shown in (C..53). For this value of g no solution exists. The other way: with help of Abramowitz and Stegun in the section on Other differential equations in the chapter on Bessel functions, you will find the differential equation w (z) + (λ ν 4) w(z) =. The solution of this equation is, among other solutions: z w = z J ν (λz). Now for our Schrödinger s equation λ = and ν = g. Well, that is all 4 there is. Playing with WolframAlpha creates a lot of fun. b) The eigenfunction at large positive x is given by Mahan(see above) and can also be found in Abramowitz and Stegun. Then the phase shift is π (ν + ). Or written in the parameters of Schrödinger s equation: π ( + g + 4 )..6 Delta-Function Potential. mε In Eq. (.5) the expression containing Eψ(x ) vanishes due to: lim Eψ(x ε ħ ). In Eqs. (.34) and (.35) the subscript B indicates bound states, I presume. Homework 7, 8, 9. Exercise 7. Find the transmission and reflection coefficients, from left to right, of a particle scattering of the potential V(x) = V Θ(x)+λδ(x), λ > in one dimension, where E > V. Find the transmission and reflection coefficients from right to left and verify the relations the relations found from time reversal. The barrier described above I consider to be one of the barriers as described by Mahan on top of page 5. Schrödinger s equation is d ψ (E V(x))ψ =. There are no bound states. Nb.: a delta-function potential I consider to be a potential of infinite height and infinite thinness. My first thought was: there is no transmission. Well, then I supposed the infinite thinness sufficient for tunnelling). We start with a particle wave from the left. The wave function is: + m d x ħ 5