Solutions to chapter 3 problems

Transcription

1 Chapter 8 Solutions to chapter 3 problems Solution to Exercise 3.. a We calculate the right-han sie of Eq. 3.5 using the ecomposition 3.3 we call the integration variable x : x ψ ψx x x x 3. ψx δx x x ψx. 8. The last equality follows from the efinition of the elta function [see Eq. C.3]. b Let us act with the operator ˆ x x x upon an arbitrary state ψ. accoring to the properties of the outer prouct, We have, Î ψ x x ψ x 3.5 x ψxx 3.3 ψ. The operator Î acting on any state returns the same state, i.e. it is the ientity operator. c We insert the ientity operator 3.6 into ψ ψ : ψ ψ ψ x x x ψ ψ x x ψ x ψ xψ xx ψx x. Solution to Exercise 3.. All the relations from Ex. 3. rely upon Eq. 3.5, which, { in turn, is a consequence of Eq. 3.. If the latter relation were replace by x x ϵx x if xx 0 if x x, Eq. 8. woul take the form x ψ ψx x x x ψx ϵx x x 0. The last equality is vali because the function ψx ϵx x takes a nonzero value at only one point in R an its value its finite. The integral of such a function is zero. 49

2 50 CHAPTER 8. SOLUTIONS TO CHAPTER 3 PROBLEMS Solution to Exercise 3.3. Applying Eq. 3.7, we fin that ψ ψ ψx x. The left-han sie of this equation is because ψ is a physical state. Solution to Exercise 3.4. a Integrating the square absolute value of the wavefunction over the real axis we have an thus b ψ top hat x x A A a b a. x A b a, b using Eq. C.9 we fin so ψ Gauss x x A A e x x A π /4. e x x π, Solution to Exercise 3.5. Accoring to Eq. 3.5, the wavefunction of the state x 0 is x x 0 δx x 0. Solution to Exercise 3.6. In accorance with the efinition of the continuous-variable observable, ˆx x x x x x x x x x x x x x δx xx x x. Solution to Exercise 3.7. a We insert the ientity operator 3.6 at both sies of Â: b Using the same approach,  ˆÂˆ x x xâ ψ  ψ ψ x x x x x  x x xx Ax, x x x xx. x x x  ψ x x x x ψ x  x x ψ x x ψ xax, x ψx x x.

3 5 c Here we insert the ientity operator only at the right-han sie of Â: x  ψ x  x x x ψ x  x x ψ x Ax, x ψx x. Similarly, ψ  x ψ x x x  x ψ x ψ x x  x x Ax, x x. e Accoring to the properties of ajoint operators see Ex..74, A x, x x  x x  x A x, x. f Inserting the ientity operator between  an ˆB, we fin x  ˆB x x  x x x ˆB x x  x x ˆB x x Ax, x Bx, x x. Solution to Exercise 3.8. Using the efinition 3. of the continuous-variable observable, ψ ˆx ψ ψ x x x x ψ Solution to Exercise 3.0. x ψ x x ψ x x ψx x x prx x. a If a 000-kg car is moving at a velocity 0 m/s 7 km/h, its momentum is p 0 4 kg m/s. Using the tabulate value of π m kg / s, we fin the e Broglie wavelength equal to λ π /p m. b The average translational velocity of molecules of a gas is v 3k B T/m, an the momentum is p 3k B T m, where k B is the Boltzmann constant k J/K, T 300 K is the room temperature an m M air /N A kg is the average molecular mass where M air 0.08 kg/mol is the molar mass of air an N A is the Avogaro number. We fin p kg m/s an hence λ.7 0 m.

4 5 CHAPTER 8. SOLUTIONS TO CHAPTER 3 PROBLEMS c The kinetic energy of the electron is p /m eu, where m kg is the electron mass, e Col is the electron charge an U 0 5 V is the voltage. We fin p.9 0 kg m/s an λ m. Because the e Broglie wavelength of the electron is much smaller than the wavelength of light, the electron microscope achieves much higher resolution than optical. By analogy with part b, we fin the mass of rubiium atoms as m 0.087/6 0 3 kg kg an their momentum p 3k B T m kg m/s. The e Broglie wavelength is m0.78 µm. This wavelength is comparable to the istance between atoms in the conensate, which leas to quantum effects in interaction between atoms. Solution to Exercise 3.. Using the metho of inserting ientity, we write p Equation 3.8 is proven similarly. x x p x 3.4 Solution to Exercise 3.. Accoring to Eq. 3.7, + π e i px x x.. p p ψ p xψ p xx π e i p p x x C.9 π π δp p δp p Solution to Exercise 3.3. To convert between the position an momentum bases, we apply our usual trick of inserting the ientity operator: ψx x ψ x p p p ψ Solution to Exercise x p p ψ p + e i px ψp p; π ψp p ψ p 3.4 p x x ψ x + e i px ψx x. π x x x ψ a Again, we use the wavevector basis rather than the momentum basis: ψk C.5 C.4 C.5 p 0 x π F[ei /4 e x a ]k x a π F[e /4 ]k k 0 where k 0 p 0 π /4 e ik k 0a F[e x ]k k 0 π /4 e ik k 0a e k k 0 / π /4 e ik k0a e k k0 /.

5 53 Now we can rewrite this result in the momentum basis using Eq. 3.36: ψp π /4 e ip p 0a/ e p p 0 /. 8. b The probability ensity prx ψx e x a π is a Gaussian curve centere aroun x a. The expectation of the position i.e. the average value of the position observe must be at x a. Similarly, k k 0 an thus p p 0. Solution to Exercise 3.7. Writing the momentum observable as ˆp p p p p, we fin x ˆp x p x p p x p π p e i p x x p. We now notice that the integran can be expresse as p e i p x x i x ei p x x. The orer of ifferentiation an integration can be inverte: x ˆp x C.9 π i e i p x x p x π i x π δx x i x δx x. Solution to Exercise 3.8. Using the result of the Ex. 3.7, x ˆp ψ x ˆp i i x x x x ψ x ˆp x x ψ x [ ] x δx x ψx x [ i x ψx δx x ψx x ] Solution to Exercise 3.9. Using the result of Ex. 3.7 an 3.8, x ˆp ψ x ˆp x x x ˆp ψ x ˆp x x ˆp ψ x [ ] [ i x δx x i ] x ψx x i [ δx x ] x x ψx x [ ] x x ψx x ψx

6 54 CHAPTER 8. SOLUTIONS TO CHAPTER 3 PROBLEMS Solution to Exercise 3.0. Left for the reaer as an inepenent exercise. Solution to Exercise 3.. a e iˆpx0/ x e iˆpx0/ π π x + x 0. e iˆpx/ p p e iˆpx+x 0/ p p b since the operator isplaces each position eigenstate by x 0, it isplaces the whole wavefunction ψx as shown in Fig. 3.. The new wavefunction is then given by ψ x x e iˆpx 0/ ψ 3.44 ψx x 0. x e iˆpx 0/ x x ψ x x x + x 0 x ψ δx + x 0 xψx x c Since the operator isplaces the whole wavefunction by x 0, it must also as x 0 to the mean position value. Formally this can be expresse as follows. For the mean position value in the state ψ e iˆpx0/ ψ, we have x ψ x x x 0 x ψx x 0 x x ψx x + x 0 ψx x where x ψ is the mean position value in state ψ. x ψ + x 0, 8.3 To fin the transformation of the mean momentum value uner the isplacement operator, we notice that the action of this operator in the momentum basis is simply a multiplication Hence ψ p ψp an ψp e i ˆpx 0 / ψ p e ipx 0/ ψp. 8.4 p ψ p ψ p p p ψp p p ψ. 8.5 The fact that the uncertainties of the position an momentum of the isplace state are the same as those in the original state is, again, intuitive Fig. 3.. A rigorous proof can be ome

7 as follows. For the position: x ψ x x x 0 x x ψ ψ x x ψ x 0 ψx x0 x x x ψ ψx x x x ψ ψ 55 x ψ. 8.6 For the momentum: p ψ p p ψ ψ p p ψ ψ p p p p ψ ψp p p ψ 8.7 Solution to Exercise 3.. e iˆxp 0/ p + eiˆxp 0/ π π p + p 0. Other properties are proven similarly to Ex. 3.. e iˆpx/ x x e iˆxp+p 0/ x x Solution to Exercise 3.4. We recall that the probability to etect a certain value of momentum is prp p ψ ψp, where the wavefunction ψp in the momentum basis is the Fourier transform of the wavefunction ψx in the position basis. Because the latter is real, ψp ψ p Ex. C.6 an thus prp pr p. The expectation value of the momentum observable is given by because p prp is an o function. Solution to Exercise 3.5. p p prpp 0 a Because the position operator is Hermitian, x ˆx x x an thus x ˆxˆp ψ ˆx x ˆp ψ 3.8 i x x ψx.

8 56 CHAPTER 8. SOLUTIONS TO CHAPTER 3 PROBLEMS b Let us enote ˆx ψ ϕ ; then ϕx x ˆx ψ xψx. Therefore x ˆpˆx ψ x ˆp ϕ i ϕx i [xψx] i ψx i x x x x ψx. Note that the above relation can also be foun using the technique of inserting ientity. The reaer can try this inepenently. c Using the two results above, we fin x [ˆx, ˆp] ψ x ˆxˆp ψ x ˆpˆx ψ i ψx. Therefore, applying the operator [ˆx, ˆp] to any ψ is equivalent to multiplying this state by i. We conclue that [ˆx, ˆp] i ˆ. Accoring to the uncertainty principle, a nonzero commutator between the position an momentum means that these two quantities can never be etermine precisely simultaneously: ˆx ˆp Solution to Exercise 3.6. Substituting the uncertainty principle.6, we fin: ˆx ˆp 4 i ˆ Solution to Exercise 3.7. To fin x we use the efinition.53 of the mean square uncertainty, as well as the fact that x 0 in accorance with Ex. 3.6: x 3.50 x prxx + π + π 3 π. x e x x x e x x t e t t replace variable: t x/ Similarly, Eq. 3.39, k / an thus p /. The prouct of the uncertainties is x p 4, which is the minimum allowe by the uncertainty principle. Solution to Exercise 3.8. a The wavefucntion in the momentum representation for convenience, we use the physically ientical wavevector representation can be foun using the stanar conversion formula 3.35.

9 57 The Fourier transformation has to be applie to both x A an x B. Ψk A, k b C.8 π π π π π πδk A + k B δk A + k B. Ψx A, x B e ik Ax A e ik Bx B x A x B δx A x B e ik Ax A ik B x B x A x B e ik Ax A ik B x A x A e ik A+k B x A x A b The wavefunction Ψx A, x B δx A x B of the system in the position basis implies that the positions of Alice s an Bob s particles must be ientical. If Alice etects her particle at a position x 0, Bob s particle will project onto a state with the same position, i.e. x 0. c Similarly, because Ψk A, k b δk A + k B, Alice s etection of wavevector k 0 or momentum p 0 k 0 will project Bob s state onto k 0 or p 0. Solution to Exercise 3.9. We rewrite the Schröinger equation ψ i Ĥ ψ i ] [V ˆx + ˆp ψ m in the position basis using the result of Ex. 3.9: ψx, t x ψ i x V ˆx ψ i x ˆp m ψ i V xψx, t i ψ ψx, t. m x Solution to Exercise In the absence of potential, the Hamiltonian is a function of the momentum: Ĥ ˆp /m. An eigenstate p of the momentum is therefore automatically an energy eigenstate with the eigenvalue E p /m. Accoring to the general solution.77 of the Schröinger equation, this state evolves as follows: p e i Et p i p e m t p. Assuming that the wavefunction of the momentum eigenstate at the moment t 0 is given by the e Broglie wave 3.4, its evolution can be written in the position basis as ψ p x, t e i px p i m t. π To fin the phase velocity, we rewrite the above as ψ p x, t π e ikx iωt π e ikx ω k t, where k p/ is the wavevector an ω p /m is the oscillation frequency. Over the time t, the wave translates by a istance x ω/kt p/mt, so the phase velocity is v ph ω/k p/m. Solution to Exercise 3.3.

10 58 CHAPTER 8. SOLUTIONS TO CHAPTER 3 PROBLEMS a We foun the wavefunction in the wavevector representation in Ex. 3.6: ψk π /4 e k k 0 /. Since ψ ψk k k an each wavevector eigenstate k is also an eigenstate of the Hamiltonian with eigenvalue E k k /m, we have for the evolution of state ψ ψk, t b Defining k k 0 + δk, we rewrite Eq. 8.0 as ψk, t π /4 e k k 0 / e i kt/m. 8.0 e i π/4 k 0 t m e i k 0 δkt δk t m e i m. 8. Let us now inverse Fourier transform this result back into the position basis. The first exponential in the equation above is a k-inepenent phase factor, which is not affecte by the Fourier transform. The last exponential is a Gaussian function, whose inverse Fourier transform is also a Gaussian. The secon exponential is a linear phase factor, whose effect on the Fourier transform translates, accoring to Eq. C.5, into the shift of the position by k 0 t/m. Finally, since the whole expression Eq. 8. is given in terms of δk k k 0, we must apply C.4, obtaining the factor e ik 0x. The resulting wavefunction is as follows: ψx, t i t / e i k 0 t π /4 m e ik 0 x e x k 0 δkt m / i t m, 8. m c Expression 8. represents a Gaussian wavepacket centere at x p 0 /mt accoring to Ex The fining of the position uncertainty procees similarly to Ex. 3.7, but we have to take into account the complexity of the Gaussian exponent in Eq. 8.. We fin pr x ψ x, tψx, t 4 + t / π m e x k 0 δkt m / 4 + t m 8.3 Using the integral 3.50 we fin x x x pr x x + t m Solution to Exercise 3.3. a Accoring to Eq. 8.4, the with of the Gaussian wavepacket behaves for large t accoring to x t. 8.5 m The conition that p 0 greatly excees the momentum uncertainty of the initial wavepacket means, in accorance with Ex. 3.7, that p 0 /. This means that the travele istance, p 0 t/m, is much greater than t/m, i.e. it is much greater than x in accorance with Eq b We rewrite Eq. 8.5 as t x m/. Substituting x 0 3 m, 0 0 m an m 0 30 kg we fin t ns.

11 Solution to Exercise In accorance with Eq. 3.55, we have v gr ω k k m. For all e Broglie waves constituting a wavepacket with a mean momentum p 0 an a small momentum uncertainty, we have k p 0 / an hence v gr p 0 /m. This result is ientical to the classical one. Solution to Exercise The time erivative of the operator s expectation value can be foun using the rule for the erivative of a prouct: x A ψ x  ψ ψ  ψ + ψ A ψ. 8.6 The time erivative of the state ψ is etermine from the Schröinger equation: ψ i/ Ĥ ψ. Because the Hamiltonian is a Hermitian operator, we also have ψ i/ ψ Ĥ. Substituting these relations into Eq. 8.6, we fin Ȧ i ψ Ĥ ψ i ψ ÂĤ ψ i ψ [Ĥ, Â] ψ. Solution to Exercise a Accoring to the previous result, x i/ [Ĥ, ˆx], so we nee to fin the commutator between the Hamiltonian Ĥ ˆp /m + V ˆx an the position operator ˆx. The potential energy is a function of the position an thus commutes with it. To fin the commutator between the kinetic energy an the position, we use Eq..5: [ ] ˆp m, ˆx Ex ˆp[ˆp, ˆx] + [ˆp, ˆx]ˆp i ˆp m m an thus x p /m b Here we nee to fin the commutator between the Hamiltonian an the momentum operator; the only non-commuting term in the Hamiltonian is the potential energy V ˆx. We accomplish our task by analyzing the action of the commutator [V ˆx, ˆp] on an arbitrary state ψ in the position basis. We set ϕ V ˆx ψ an write x [V ˆx, ˆp] ψ x V ˆxˆp ψ x ˆpV ˆx ψ Consequently, V x x ˆp ψ x ˆp ϕ [because x is an eigenstate of V ˆx] V x i ψx i [V xψx] x x V x i ψx + i [V x]ψx + i V x x x x ψx i ψx x V x. ṗ i ψ [V ˆx, ˆp] ψ i ψ x x [V ˆx, ˆp] ψ ψ xψx x V xx. The latter expression can be formally written as ṗ ψ ˆV x ψ. 59

12 60 CHAPTER 8. SOLUTIONS TO CHAPTER 3 PROBLEMS Solution to Exercise 3.4. We can rewrite the time-inepenent Schröinger equation 3.64 as which can be simplifie to m x ψx V 0 Eψx, x ψx κ ψx, where κ mv 0 E/ oes not epen on x. This secon-orer ifferential equation has two linearly-inepenent solutions: ψx Ae ±κx. 8.7 Note that κ is real only if E < V 0, i.e. the total energy is below the potential energy level. Otherwise, κ becomes imaginary an the solution 8.7 takes the form of the e Broglie wave: where k iκ me V 0 / is a real wavevector. ψx Ae ±ikx, 8.8 Solution to Exercise Let us rewrite the time-inepenent Schröinger equation 3.64 as follows: ψx [E V x]ψx. m x If both V x an ψx are finite for all x, so is the right-han sie of the above equation. This means that ψx/x is finite for all x as well. This implies in turn that the first erivative of the wavefunction is continuous for all x, because otherwise ψx/x woul be singular at some points. Because ψx/x is continuous, it must be finite for all x. Therefore, ψx must be finite for all x as well. Solution to Exercise a Because the Hamiltonian Ĥ is a Hermitian operator, its eigenstates form an orthonormal basis. Our goal is to prove that such a basis can be compose of states with real wavefunctions. Suppose this is not the case, i.e. there exists a Hamiltonian eigenstate ψ with eigenvalue E which cannot be expresse as a linear combination of eigenstates with real wavefunctions. Consier the state ψ escribe by the wavefunction ψ x, where ψx is the wavefunction of the state ψ. If ψx is an eigenwavefunction of the Hamiltonian i.e. satisfies the timeinepenent Schröinger equation, so must ψ x; therefore, ψ must also be an eigenstate of the Hamiltonian with the same eigenvalue. Now consier the following states: ψ ψ + ψ ; ψ [ ψ ψ ]/i. Accoring to properties of complex numbers, the wavefunctions of these states are real. Because ψ an ψ are linear combinations of ψ an ψ, they are also eigenstates of Ĥ with eigenvalue E. Furthermore, the state ψ can be expresse as a linear combination ψ ψ + i ψ of energy eigenstates with real eigenvalues. We have arrive at a contraiction. b By the same logic, consier an energy eigenstate ψ with eigenvalue E an wavefunction ψx. If ψx satisfies the time-inepenent Schröinger equation, the wavefunction ψ x oes, too, so the state ψ with this wavefunction is also an eigenstate of the Hamiltonian. We now construct an energy eigenstate with an even wavefunction, ψ ψ + ψ,

13 6 an one with an o wavefunction, ψ ψ ψ. The state ψ can then be expresse as a linear combination of ψ an ψ : ψ ψ + ψ. Solution to Exercise As iscusse in Ex. 3.4, energies E below a constant potential level V 0 are associate with eigenwavefunctions ψx Ae ±κx, with κ mv 0 E/. Because of the normalization conition, the wavefunctions cannot have components that exponentially grow at infinity, an thus we must have { Ae ψx κx at x + A e κx at x In other wors, ψx 0 for x ±, so we have a boun state. Conversely, if the energy excees the potential at infinity, the eigenwavefunctions ten to ψx Ae ikx + A e ikx, with k me V 0 /. If at least one of the factors A or A oes not vanish, the state is not boun. Solution to Exercise Because the potential is an even function of x, it suffices to look for even an o solutions of the time-inepenent Schröinger equation. Let us consier these two cases separately. A general o solution is of the form Be κx, x < a/ ψx A sin kx, a/ x a/ 8.9 Be κx, x > a/ with k κ me, 8.0 mv0 E. 8. Because the potential is finite, both the wavefunction ψx an its erivative ψ x must be continuous. Writing these conitions for the bounary of the box x a/, we fin or A sin kx xa/ Be κx xa/ ; Ak cos kx xa/ κbe κx xa/ A sin ka Ak cos ka Be κa/ ; 8. κbe κa/. 8.3 These equations result in restrictions on energy values at which the time-inepenent Schröinger equation has a solution. To see this, let us ivie Eqs. 8. an 8.3 by each other. We obtain The continuity conition for x a/ yiels the same set of equations. cot ka κ k. 8.4

14 6 CHAPTER 8. SOLUTIONS TO CHAPTER 3 PROBLEMS This equation relates k an κ. Another relation between these quantities is ue to Eq. 8.0, which can be incorporate into our calculations as follows. Let us enote ka/ θ an κa/ θ. From Eq. 8.0 we then have θ + θ θ 0, where Equation 8.4 now takes the form mv0 a θ cot θ θ 8.6 θ or θ cot θ θ This equation contains only one unknown variable, θ, which is relate to the evergy eigenvalue. Unfortunately, this equation is transcenental an cannot be solve in elementary functions. A generic even solution is given by Be κx, x < a/ ψx A cos kx, a/ x a/ 8.8 Be κx, x > a/ Proceeing in a fashion similar to the o case, we fin the continuity conitions for the bounary of the well an the transcenental equation for θ A cos ka Ak sin ka tan θ Be κa/ ; 8.9 κbe κa/, 8.30 θ θ Figure 8.: Graphic solution to transcenental equations 8.7 an 8.3 for θ 0 5. Solution to Exercise When V 0 is infinite, so is the right-han sie of Eqs. 8.7 an 8.3. The tangent in the left-han sie of Eq. 8.3 takes on a positive infinite value when

15 θ j + π/, an the cotangent in 8.7 when θ πj, where j is an arbitrary natural number. The general solution in the limit V 0 can then be written as θ nπ/, with n being an arbitrary natural number: an even n j prouces an o solution, an an o n j + an even solution. Substituting θ ka/, we fin wavevector values k n nπ/a, which correspon to energy eigenvalues E n k m π n ma. We notice that the oscillating parts of the wavefunctions, insie the box, a sin nπx a, even n ψ n x a cos ; 8.3 nπx a, o n. vanish at x ±a/. This implies accoring to Eqs. 8. an 8.9 that B 0 for both o an even cases, an that the wavefunction vanishes outsie the box. We can now fin the normalization constant A. To this en, we integrate the square absolute value of the wavefunction over the real axis. We fin, for both even an o solutions, so the norm is A /a. ψx x a/ a/ ψx x A a, Solution to Exercise Since V 0 a W 0 we can rewrite Eq. 8.5 as mw0 θ 0 63 a Because a tens to zero an W 0 is a constant, θ 0 also tens to zero. The bol curve in Fig. 8. shrinks to a vertical line just next to the orinates axis. Therefore we have only one, even, energy eigenstate, an we rewrite Eq. 8.3 using the fact that tan θ θ for small θ: or Therefore θ θ θ θ 4 + θ θ θ ± + 4θ0. Because θ 0 is small, we approximate to the secon orer the reason why we nee this will become clear shortly + 4θ 0 + θ 0 θ 4 0. Then the two roots of Eq can be rewritten as θ [ θ 0 θ 4 0 θ 0 + θ Because we are looking for a boun solution, we expect θ to be real, so we choose the first root, θ θ 0. Since θ 0 mv 0 a/ an θ mea/, we see that E ma θ V 0 ma V0 V 0 mw It is now clear why we neee the secon orer Taylor expansion. The energy level of the boun state is almost at the top of the well. The coefficient κ etermining the behavior of the wavefunction

16 64 CHAPTER 8. SOLUTIONS TO CHAPTER 3 PROBLEMS outsie the well is given by Eq. 8.. Ha we kept only the first orer, we woul obtain E V 0, an no information about κ. Now we can however calculate mv0 E κ W 0m As we see, this coefficient is inepenent of a in the limit a 0 as long as V 0 a is kept constant. Let us now fin the normalization coefficient. In the limit a 0, we nee to only take into account the part of the wavefunction that is localize outsie the well when calculating the norm. Using Eq. 8.8, we have so B κ. Solution to Exercise ψx B 0 e κx x B, 8.39 a Because the potential is an even function of x, we can restrict to even an o wavefunctions. At x 0, the potential is zero. The energy of a boun state must then be negative, so a generic o solution must be of the form ψx { Be κx, x < 0 Be κx, x > with κ m E/. Unless B 0 i.e. ψx 0, the o wavefunction has a iscontinuity at x 0, i.e. unphysical. The even solution is given by ψx { Be κx, x < 0 Be κx, x > 0 It is continuous at all x, but its erivative has a iscontinuity: ψ x x0 Bκ. This is not a problem, because the potential has a iscontinuity at x 0. The solution 8.40 is vali for an arbitrary κ at all values of x except x 0. In orer to fin out for which values of κ the time-inepenent Schröinger equation 3.74 is satisfie at x 0, let us integrate this equation over an infinitesimal interval aroun this point: +0 0 Using the Newton-Leibniz axiom as well as Eq. C.5, we fin m +0 m x ψx [E V x]ψx. 8.4 ψx x Given that, accoring to Eq. 8.40, 0 x+0 ψx x x 0 W 0 ψ ψ0 B; 8.43 ψx x x+0 Bκ; 8.44 ψx x x 0 Bκ, 8.45

17 65 we fin from Eq. 8.4 that κ W 0 m/ an thus E W 0 m. The normalization factor B κ is foun from the conition ψx x. Solution to Exercise The particle is initially prepare in the boun state of the original potential see Ex. 3.53: ψ 0 x κ 0 { e κ 0 x, x < 0 e κ 0x, x > 0 with κ 0 V 0 m/. After the suen change of the potential, the boun state is given by another wavefunction, ψ x κ { e κ x, x < 0 e κ x, x > 0 with κ V 0 m/. The probability that the particle will remain in the boun state of the new potential is given, accoring to the Secon Postulate, by the inner prouct pr ψ 0 ψ ψ0xψ xx + κ 0 κ e κ0x e κx x 0 κ 0 κ κ 0 + κ 8 9. Solution to Exercise The general solution associate with energy eigenvalue E is { Ae ik 0 ψ bar E, x x + Be ik0x, x < 0 Ce ikx + De ikx, x 0, 8.46 where k 0 me/, k me V 0 /. The four amplitues A, B, C, D must be chosen so that the wavefunction an its erivative are continuous at the barrier, i.e. A + B C + D; ik 0 A B ik C D; One of the parameters provies normalization an can be set to an arbitrary value. This leaves us with three parameters an two equations; therefore, for each energy value, there are two linearly inepenent solution sets. They can be chosen to correspon to the following physical situations: an initial e Broglie wave approaching from the left B A k 0 k k 0 + k ; C A k 0 k 0 + k ; D 0, 8.47 an an initial e Broglie wave approaching from the right A 0 A 0; B D k k 0 + k ; C D k k 0 k 0 + k. 8.48

18 66 CHAPTER 8. SOLUTIONS TO CHAPTER 3 PROBLEMS Of course, any linear combination of these solutions is also a solution. Solution to Exercise 3.6. The initial wavepacket can be written in the wavevector basis as 3.39 /4 ψ0 e iκa e κ / k 0 + κ κ, 8.49 π where κ is small compare to k 0 an k ue o the secon an fourth assumptions above. To simplify the calculation, we replace the momentum eigenstate k in the above equation by an energy eigenstate ψ bar κ 8.46, which we write in the form x ψ bar κ Ae ik 0+κx θ x + Be ik 0+κx θ x + Ce i k 0 +κ mv 0 x θx In Eq A-wave, with A / π, is ientical to k 0 + κ, while the B- an C- waves are orthogonal to it. As we shall see below, B- an C-terms o not moify the initial state, but emerge, as separate wavepackets, only after the initial packet reaches the barrier. The amplitue factors B an C are given by Eq. 3.79a. We can neglect the variation of the amplitues B an C as a function of the small κ. Furthermore, using k k0 mv 0 / an neglecting terms that are quaratic with respect to κ, an we can replace in Eq k 0 + κ mv 0 k 0 + k 0κ mv 0 k + k 0 κ/k k + κ k 0 k. Knowing that the state ψ bar κ is an eigenstate of the Hamiltonian with energy E κ k 0 + κ /m /mk0 + k 0 κ, we write the evolution of the state ψ, again neglecting quaratic terms: /4 ψt e i k π 0 t/m e iκa+ k 0 m t e κ / ψ bar κ κ. 8.5 The overall phase factor e i k 0 t/m can be neglecte. We shall now calculate the integral 8.5 for each wave in Eq separately. A-wave. Applying stanar Fourier transform rules C., we obtain ψ A x, t Aθ xe ik 0x /4 π Aθ xe ik 0x π /4 π e iκa+ k 0 m t e κ / e iκx κ k x a 0 e m t. This is a Gaussian wavepaket centere at the point x a + k0 m t an propagating with the spee k 0 /m in the positive irection. When the wavepacket reaches the barrier i.e. at t bar am k 0, it isappears ue to the factor θ x. Before this happens, the total probability associate with this wavepacket is pr A ψ Ax, t x πa. B-wave is treate similarly, except that the integral is the inverse Fourier transform. We obtain ψ B x, t Bθ xe ik 0x π /4 π k x+a+ 0 e m t. This wavepaket is a mirror image of the previous one. At t 0, it is locate at x a but is invisible ue to the factor θ x. It propagates in the negative irection. Once it reaches the barrier simultaneously with the A-packet, it becomes visible. This wavepacket is associate with the reflection of the particle from the barrier. The total probability associate with this wavepacket is pr B πb. C-wave. ψ C x, t Cθxe ik x /4 π e iκa+ k 0 m t e κ / e iκ k 0 k x κ

19 Cθxe ik x /4 π π e k 0 x a k 0 k m t. This packet is narrower than the other two by a factor k 0 /k. It begins to exist at t t bar an propagates in the positive irection at a spee k /m. This wavepacket is associate with the particle transmitte through the barrier an has the probability pr C πc k /k 0. A irect calculation shows that pr B + pr C. Solution to Exercise Proceeing similarly to Ex we fin that the solution is a linear combination of six wavefunctions as shown in Fig. 3.7 an is thus a function of six parameters. For each of the two interfaces, there are two continuity conitions for the wavefunction an its erivative: A + B C + D; ik 0 A B k C D; Ce k L + De k L E + F ; k Ce k L De k L ik 0 E F, where k 0 me/, k mv 0 E/. Again, each energy value is twice egenerate: the linearly inepenent solutions correspon to the matter waves approaching from the left F 0 an from the right A 0. We are intereste in the first option an solve the above equations assuming an arbitrary E an working our way to the left. We then fin the relation between the incient, transmitte an reflecte amplitues: A E [ cosh k L + i 67 k k ] 0 sinhk L ; 8.5 k 0 k [ B E i k + k ] 0 sinhk L k 0 k The transmission an reflection coefficients are then given by Eqs Solution to Exercise We rescale the position an momentum operators accoring to ˆX Aˆx, ˆP B ˆp. We then have, for the counterpart classical observables, P/X B/Ap/x. Because p max /x max mω, we fin P max /X max B/Amω. The requirement that P max X max implies that A B mω. On the other han, the commutator of the rescale observables satisfies [ ˆX, ˆP ] AB[ˆx, ˆp] i AB. Since we nee this commutator to equal i, we obtain the secon equation: AB. Solving these two equations for A an B, we fin the esire result. Solution to Exercise Since has the same imension as the prouct of the position an momentum, i.e. kg m /s, the imension of mω/ is m i.e. the same as x an that of mω is kg m/s i.e. the same as p. Solution to Exercise As iscusse in Sec. 3., rescaling a continuous variable affects the normalization of its eigenstate: if ˆX ˆx mω/, then X /mω /4 x. Similarly, the renormalize eigenstatewe of the rescale momentum ˆP ˆp/ m ω is P m ω /4 p. Now for the e Broglie wave we have X P /4 m ω /4 x p e ixp/ e ixp. mω π π

20 68 CHAPTER 8. SOLUTIONS TO CHAPTER 3 PROBLEMS Using this result, we argue similarly to Ex. 3.3 to obtain the conversion expressions for wavefunctions in the rescale observable basis: ψx ψp + π ψp e ip X P ; + ψxe ip X X. π Solution to Exercise Using the relations foun in Ex. 3.69, we fin X ˆP ψ 3.4 mω mω i m ω x /4 m ω x ˆp ψ /4 i x ψ m ω x m ω i i X ψ. X mω X ψ X ψ X The expression for the position operator in the momentum basis is obtaine similarly. Solution to Exercise 3.7. Solution to Exercise 3.7. H ˆp m + mωˆx mω ˆP m + mω ˆX mω ω ˆX + ˆP a Because the position an momentum operators are Hermitian, ˆX ˆX an i ˆP i ˆP. Therefore, â ˆX + i ˆP / ˆX i ˆP /. b [â, â ] [ ˆX + i ˆP, ˆX i ˆP ] [ ˆX, ˆX] i[ ˆX, ˆP ] + i[ ˆP, ˆX] + [ ˆP, ˆP ]. c The position an momentum operators are expresse through â an â by solving Eqs an Ĥ 3.88 ω ω ˆX + ˆP [â 4 ω + â + i ] â â [ 4 ω â + â + ââ + â â + ] i â + â ââ â â 4 ω[ââ + â â] 4 ω[â â + + â â] [ â â + ].

21 69 e [â, â â].5 â [â, â] + [â, â ]â â; [â, â â].5 â [â, â] + [â, â ]â â. Solution to Exercise a In orer to verify if the state â n is an eigenstate of the photon number operator â â, let us subject this state to the action of this operator an employ the result 3.9, rewritten in the form â ââ ââ â â: as was require. â ââ n ââ â â n ân â n n â n, b Similarly, from Eq. 3.9 we fin â ââ â â â + â an thus â ââ n â â â + â n â n + â n n + â n. Solution to Exercise a Let ψ â n. From the previous exercise, we know that ψ is an eigenstate of â â with eigenvalue n, i.e. ψ A n, where A is some constant. We nee to fin A. To this en, we notice that ψ n â an calculate But on the other han, ψ ψ n â â n n. ψ ψ A n n A, where in the last equality we have use the fact that the eigenstates of the number operator are normalize. From the last two equations, we fin A n. The phase of A is arbitrary. By convention, it is chosen equal to zero, so A is real an positive: A n. b Similarly, if ϕ â n B n +, then, on one han, an on the other han ϕ ϕ n ââ n n â â + n n +, ϕ ϕ B n + n + B. Therefore invoking a similar convention, B n +. Solution to Exercise n 3.93b â n â â â n n n n n n! Solution to Exercise The vacuum state obeys the equation â 0 0, or ˆX + i ˆP In orer to fin the wavefunction in the position basis, we use Eq to write the momentum operator in this basis. Equation 8.55 then becomes X + ψx 0. X

22 70 CHAPTER 8. SOLUTIONS TO CHAPTER 3 PROBLEMS This is a first orer orinary ifferential equation whose solution is ψx Ae X /, where A is the normalization constant, calculate in the usual manner: ψ ψ ψx x A e X x A π. Requiring the norm of ψ to equal, we fin A π /4. The wavefunction in the momentum basis is calculate similarly. Solution to Exercise a The single-photon Fock state is obtaine from the vacuum state by applying a single creation operator. Using Eq. 3.84, we express the creation operator in the position basis as â ˆX i ˆP X X π /4 an thus the wavefunction of the state â 0 is ψ X X X e X / Xe X π/4 The two-photon Fock state is obtaine by applying the creation operator to the single-photon state: 3.93b â. In the position basis, X ψ X X ψ x X π /4 Xe X / X /. π /4 X e X /. b We now show by inuction that Eq escribes the wavefunction of the Fock state n. First, from the efinition of the Hermite polynomial, H n X n e X n X n e X, we fin that H 0 X an thus the wavefunction of the vacuum state obtaine from Eq is ψ 0 X π /4 e X /, which is consistent with Eq Secon, assuming that if Eq is vali for a specific Fock state n, we nee to prove it to be also vali for the next Fock state n + â n / n +. We apply the creation operator in the position basis: n + â n + n X /X H n X n + π /4 n n! e X / [ π /4 XH n Xe X / H nx n+ n +! X [ π /4 XH n X H nx n+ n +! X π /4 n+ n +! H n+xe X /, e X e X / ] / / H n X e X ] X

23 which, accoring to Eq. 3.96, is the wavefunction of the state n +. In the final step in the above transformation, we have applie the known recursion relation for the Hermite polynomials, H n+ X X H n X H nx X H nx. Solution to Exercise For an arbitrary Fock state n, we have n ˆX n n â + â n n n n + n + n + Similarly, n ˆP n 0. For the uncertainties, we have X n ˆX n n ââ + ââ + â â + â â n n [ nn n + n + n + n n + n + n + n + ] n The same answer hols for the momentum uncertainty: P n +. Solution to Exercise Both 0 an are energy eigenstates with the eigenvalues being, respectively, ω/ an 3 ω/. The evolution of the superposition of these states is then given by ψt e iωt/ 0 + e 3iωt/. The expectation value of the position observable is then given by X 3.89 ψt â + â ψt 0 + e iωt â + â 0 + e iωt. The only nonvanishing matrix elements in the expression above are 0 â â 0. We then conclue that X e iωt + e iωt cos ωt. Similarly, for the momentum observable we fin X i 0 + e iωt â â 0 + e iωt i e iωt + e iωt sin ωt. The trajectory in the phase space is a clockwise circle with the center in the origin an a raius of. Solution to Exercise We will perform the calculation in the position basis. Acting similarly to Ex. 3.76, we rewrite Eq as X + ψx Re α + iim αψx. X

24 7 CHAPTER 8. SOLUTIONS TO CHAPTER 3 PROBLEMS Substituting Eq. 3.00, we fin X + ψ α X X PαXα ei X + e ipαx e X Xα π /4 X PαXα ei [X + ip α X X α ] e ipαx e X Xα π /4 X α + ip α ψ α X, 8.58 so Eq hols provie that X α Re α an P α Im α. The wavefunction 3.0 in the momentum basis is obtaine from that in the position basis by means of Fourier transform, similarly to Ex Solution to Exercise 3.8. The right-han sie of Eq consists of sequential application of the position an momentum isplacement operators, as well as an overall phase factor, to the vacuum state Let us first just apply the position isplacement: ψ e ix α ˆP 0. Working in the position basis, we obtain the wavefunction ψ x ψ 0 X X α 3.96 Now if we apply the momentum isplacement, we have ψ e ip α ˆX ψ, e X Xα /. π/4 ψ x π /4 eipαx e X Xα /. Upon multiplying by the phase factor e ip αx α /, we obtain the wavefunction that is ientical to that of the coherent state The proof of the secon equality in Eq is left for the reaer as an inepenent exercise. Solution to Exercise 3.8. The right-han sie of Eq consists of sequential applications of the position isplacement by X α an momentum isplacement by P α. As we know from Ex. 3. an 3., each of these operators as the isplacement value to the mean value of the position or momentum, respectively, while leaving the expectation value of the other canonical observable as well as the uncertainties of both observables. Base on this fact, an using Eq. 3.0, we obtain the esire result. Solution to Exercise Let us apply the Baker-Hausorff-Campbell formula.70 to Eq We have  ip α ˆX an ˆB ix α ˆP. Then the commutator [Â, ˆB] P α X α [ ˆX, ˆP ] ip α X α is a number which we enote as c, so the Baker-Hausorff-Campbell formula is applicable. In our notation, the right-han sie of Eq is ientical to the right-han sie of Eq..70, so we can write α eâ+ ˆB 0 e ip α ˆX ix α ˆP Let us now transform this result as follows. α expip α ˆX ixα ˆP 0 ˆXα + i exp ˆP α ˆX i ˆP ˆX α i ˆP α ˆX + i ˆP 0 exp αâ α â

25 Now since the commutator [αâ, α â] α [â, â] α, is also a number, we can use the Baker-Hausorff-Campbell formula again an obtain Eq In orer to simplify this equation, let us ecompose e α â 0 α n â n 0 0. n! n0 The last equality above hols because, since â is the annihilation operator, all the terms in the sum vanish except for n 0. Substituting this result into Eq. 3.06, we obtain Eq Solution to Exercise Decomposing Eq into the Taylor series, we fin α e α / e αâ 0 e α / α n n! â n e α / n0 n0 73 α n n! n. 8.6 The same result can also be partially obtaine using a more intuitive argument, not involving isplacement operators. Let us assume some ecomposition of the coherent state into the number basis, α α n n, 8.6 n0 an apply the efinition 3.99 of the coherent state to this ecomposition. For the left-han sie of Eq. 3.99, we have in accorance with Eq. 3.93a, â α n n α n â n n0 α n n n n α n + n + n n 0 At the same time, the right-han sie of 3.99 can be written as α α αα n n n 0 Equalizing both sies, we fin a recursive relation so α n + αα n n +, 8.65 α αα 0 ; α αα α α 0 ; α 3 αα 3 α3 α 0 6 ;..., 8.66

26 74 CHAPTER 8. SOLUTIONS TO CHAPTER 3 PROBLEMS or in general α n αn α 0 n! It remains to fin such a value of α 0 that state 8.6 is normalize to one. We fin α α α n α 0 α n n! n0 If we look carefully at the sum in the above expression, we will fin it to be the Taylor ecomposition of e α, so we have α α α 0 e α. Setting α α, we fin or n0 α 0 e α 8.69 α 0 e iφα e α / The quantum phase factor e iφ α is a matter of convention, but it must be consistent with the convention chosen for the phase of the coherent state wavefunction This phase cannot be etermine base on these elementary consierations. However, a rigorous erivation shown in the beginning of this solution emonstrates that φ α 0 for all α. Combining Eqs an 8.70 we obtain α n e α / αn n!. 8.7 Solution to Exercise For the mean number of excitation quanta, we write: n npr n n0 e α n0 e α n n α n n! n n α e α α n n! n 0 α n n. 8.7! As we know from calculus, the sum equals e α. Accoringly, n α. For the mean square number of excitation quanta, the calculation is similar, but somewhat more complicate: n e α n0 e α n n n n α n n! n α n n! α e α n + α n n! n 0 α e α [ n 0 [ α n pr n + n 0 n α n n! + n 0 ] pr n n 0 ] α n n! 8.7 α [ α + ]. 8.73

27 75 Hence the variance of n is n n n α. Solution to Exercise a The energy of each pulse is E p P/f 0 8 J. b The energy of each photon is ω πc λ J. The mean number of photons per pulse is n E p / ω c The variance in the number of photons per pulse equals n n. Hence the uncertainty in this number equals n 0 5. The relative uncertainty is the inverse of this number: n / n If we reuce the laser power by a factor of 0 6, the mean number of photons per pulse will reuce by the same factor. On the other han, the uncertainty in that number will only reuce by a factor of 0 3. Therefore, the relative uncertainty will increase by a factor of 0 3 an become Solution to Exercise Given that the coherent state is ecompose into the Fock basis accoring to Eq an that each Fock state is an eigenstate of the Hamiltonian with eigenvalue ωn + /, we fin e iĥt/ α e α / n e α / n α n n! e iĥt/ n α n n! e iωn+/t n e iωt/ e α / n αe iωt n n! n e iωt/ e iωt α. 8.74