[The factor (2πℏ)−1/2 is arbitrary and may be absorbed into β and γ.] Though ψE(x) will satisfy the equation even if E<0, are these functions in the Hilbert space?
Solution.
In X basis, equation (5.1.3) is
⟨x∣H∣E⟩=E⟨x∣E⟩
which becomes
−2mℏ2dx2d2ψE(x)=EψE(x)
The most general solution is
ψE(x)=A+e+ℏi2mEx+A−e−ℏi2mEx
where
⟨x∣E⟩=β⟨x∣E,+⟩+γ⟨x∣E,−⟩
βγ=A+2πℏ=A−2πℏ
If E<0, these solutions are not in the Hilbert space, since then the two terms grow exponentially as x→±∞.
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Exercise 5.1.3 We have seen that there exists another formula for U(t), namely, U(t)=e−iHt/ℏ. For a free particle this becomes
It is clear that when this function is differentiated any number of times we get another function confined to the interval ∣x∣⩽L/2. Consequently the action of
U(t)=exp[ℏi(2mℏ2t)dx2d2]
on this function is to give a function confined to ∣x∣⩽L/2. What about the spreading of the wave packet?
5.2 The Particle in a Box
Exercise 5.2.1 A particle is in the ground state of a box of length L. Suddenly the box expands (symmetrically) to twice its size, leaving the wave function undisturbed. Show that the probability of finding the particle in the ground state of the new box is (8/3π)2.
Exercise 5.2.2 (a) Show that for any normalized ∣ψ⟩,⟨ψ∣H∣ψ⟩⩾E0, where E0 is the lowest-energy eigenvalue. (Hint : Expand ∣ψ⟩ in the eigenbasis of H.)
(b) Prove the following theorem: Every attractive potential in one dimension has at least one bound state. Hint: Since V is attractive, if we define V(∞)=0, it follows that V(x)=−∣V(x)∣ for all x. To show that there exists a bound state with E<0, consider
ψα(x)=(πα)1/4e−αx2/2
and calculate
E(α)=⟨ψα∣H∣ψα⟩,H=−2mℏ2dx2d2−∣V(x)∣
Show that E(α) can be made negative by a suitable choice of α. The desired result follows from the application of the theorem proved above.
Exercise 5.2.3 Consider V(x)=−aV0δ(x). Show that it admits a bound state of energy E=−ma2V02/2ℏ2. Are there any other bound states? Hint: Solve Schrödinger's equation outside the potential for E<0, and keep only the solution that has the right behavior at infinity and is continuous at x=0. Draw the wave function and see how there is a cusp, or a discontinuous change of slope at x=0. Calculate the change in slope and equate it to
∫−ε+ε(dx2d2ψ)dx
(where ε is infinitesimal) determined from Schrödinger's equation.
Exercise 5.2.4 Consider a particle of mass m in the state ∣n⟩ of a box of length L. Find the force F=−∂E/∂L encountered when the walls are slowly pushed in, assuming the particle remains in the nth state of the box as its size changes. Consider a classical particle of energy En in this box. Find its velocity, the frequency of collision on a given wall, the momentum transfer per collision, and hence the average force. Compare it to −∂E/∂L computed above.
Exercise 5.2.5 If the box extends from x=0 to L (instead of −L/2 to L/2) show that ψn(x)=(2/L)1/2sin(nπx/L),n=1,2,…,∞ and En=ℏ2π2n2/2mL2.
Exercise 5.2.6 Square Well Potential. Consider a particle in a square well potential:
V(x)={0,V0,∣x∣⩽a∣x∣⩾a
Since when V0→∞, we have a box, let us guess what the lowering of the walls does to the states. First of all, all the bound states (which alone we are interested in), will have E≤V0. Second, the wave functions of the low-lying levels will look like those of the particle in a box, with the obvious difference that ψ will not vanish at the walls but instead spill out with an exponential tail. The eigenfunctions will still be even, odd, even, etc.
(1) Show that the even solutions have energies that satisfy the transcendental equation
ktanka=κ(5.2.23)
while the odd ones will have energies that satisfy
kcotka=−κ(5.2.24)
where k and iκ are the real and complex wave numbers inside and outside the well, respectively. Note that k and κ are related by
k2+κ2=2mV0/ℏ2(5.2.25)
Verify that as V0 tends to ∞, we regain the levels in the box.
(2) Equations (5.2.23) and (5.2.24) must be solved graphically. In the (α=ka,β=κa) plane, imagine a circle that obeys Eq. (5.2.25). The bound states are then given by the intersection of the curve αtanα=β or αcotα=−β with the circle. (Remember α and β are positive.)
(3) Show that there is always one even solution and that there is no odd solution unless V0⩾ℏ2π2/8ma2. What is E when V0 just meets this requirement? Note that the general result from Exercise 5.2.2b holds.
5.3 The Continuity Equation for Probability
Exercise 5.3.1 Consider the case where V=Vr−iVi, where the imaginary part Vi is a constant. Is the Hamiltonian Hermitian? Go through the derivation of the continuity equation and show that the total probability for finding the particle decreases exponentially as e−2Vit/ℏ. Such complex potentials are used to describe processes in which particles are absorbed by a sink.
Exercise 5.3.2 Convince yourself that if ψ=cψ~, where c is constant (real or complex) and ψ~ is real, the corresponding j vanishes.
Exercise 5.3.3 Consider
ψp=(2πℏ1)3/2ei(p⋅r)/ℏ
Find j and P and compare the relation between them to the electromagnetic equation j=ρv, v being the velocity. Since ρ and j are constant, note that the continuity Eq. (5.3.7) is trivially satisfied.
Exercise 5.3.4 Consider ψ=Aeipx/ℏ+Be−ipx/ℏ in one dimension. Show that j=(∣A∣2−∣B∣2)p/m. The absence of cross terms between the right- and left-moving pieces in ψ allows us to associate the two parts of j with corresponding parts of ψ.
5.4 The Single-Step Potential: A Problem in Scattering
Exercise 5.4.1 Evaluate the third piece in Eq. (5.416) and compare the resulting T with Eq. (5.4.21). [Hint: Expand the factor (k12−2mV0/ℏ2)1/2 near k1=k0, keeping just the first derivative in the Taylor series.]
Exercise 5.4.2 (a) Calculate R and T for scattering of a potential V(x)=V0aδ(x). (b) Do the same for the case V=0 for ∣x∣>a and V=V0 for ∣x∣<a. Assume that the energy is positive but less than V0.
Exercise 5.4.3 Consider a particle subject to a constant force f in one dimension. Solve for the propagator in momentum space and get