1. The case of an infinite square well whose right wall expands at a
constant velocity (v) can be solved exactly2. A complete set of
solutions isEquation 1...where w(t) ≡ a + vt is the (instantaneous)
width of the well and E'n ≡ n2π2h2/2ma2 is the nth allowed energy of the
original well (width a). The general solution is a linear combination
of the Φ's:...the coefficients cn are independent of t.(a) Check that
Equation 1 satisfies the time-dependent Schròdinger equation, with the
appropriate boundary conditions.(b) Suppose a particle starts out (t =
0) in the ground state of the initial well:...Show that the expansion
coefficients can be written in the form...where α = mva/2 π 2h is a
dimensionless measure of the speed with which the well expands.
(Unfortunately, this integral cannot be evaluated in terms of elementary
functions.)(c) Suppose we allow the well to expand to twice its
original width, so the "external" time is given by w(Te) = 2a. The
"internal" time is the period of the time-dependent exponential factor
in the (initial) ground state. Determine Te and Ti, and show that the
adiabatic regime corresponds to a « 1, so that exp(—iαz2) ≡ 1 over the
domain of integration. Use this to determine the expansion coefficients,
cn. Construct Ψ(x, t), and confirm that it is consistent with the
adiabatic theorem.(d) Show that the phase factor in Ψ (x, t) can be
written in the form...where En(t) = n2π2h2/2mw2 is the instantaneous
eigenvalue, at time t. Comment on this result. Get solution
2. Check that Equation 10.31 satisfies the time-dependent Schrodinger equation for the Hamiltonian in Equation 10.25. Also confirm Equation 10.33 and show that the sum-of the squares of the coefficients is 1, as required for normalization. ... Get solution
3. (a) Use Equation 10.42 to calculate the geometric phase change when the infinite square well expands adiabatically from width W1 to width W2. Comment on this result.(b) If the expansion occurs at a constant rate (dw/dt = v), what is the dynamic phase change for this process?(c) If the well now contracts back to its original size, what is Berry's phase for the cycle? Get solution
4. The delta function well (Equation 2.114) supports a single bound state (Equation 2.129). Calculate the geometric phase change when α gradually increases from ..., what is the dynamic phase change for this process? V (x) = -αδ(x) Get solution
5. Show that if ... is real, the geometric phase vanishes. (Problems 10.3and 10.4 are examples of this.) You might try to beat the rap by tracking an unnecessary (but perfectly legal) phase factor onto the eigenfunctions: ... where ... is an arbitrary (real) function. Try it. You'll get a nonzero geometric phase, all right, but note what happens when you put it back into Equation 10.23. And for a closed loop it gives zero. Moral: For nonzero Berry's phase, you need (i) more than one time-dependent parameter in the Hamiltonian, and (ii) a Hamiltonian that yields nontrivially complex eigenfunctions. V(x) = -αδ(x) ... Get solution
6. Work out the analog to Equation 1 for a particle of spin 1.Equation 1... Get solution
7. (a) Derive Equation 10.67 from Equation 10.65. (b) Derive Equation 10.79, starting with Equation 10.78. ... ... Get solution
8. A particle starts out in the ground state of the infinite square well (on the interval 0 ≤ x ≤ a). Now a wall is slowly erected, slightly off-center:...Where f(t) rises gradually from 0 to ∞. Accoroding to the adiabatic theorem, the particle will remain in the ground state of the evolving Hamiltonian.(a) Find (and sketch) the ground state at t → ∞. Hint: This should be the ground state of the infinite square well with an impenetrable barrier at a/2 + ∑. Note that the particle is confined to the (slightly) larger left “half” of the well.(b) find the (transcendental) equation for the ground state of the Hamiltonian at time t. Answer:...Where z ≡ ka, T ≡ maf(t)/h2, δ ≡ 2∑/a, and k ≡ ....(c) Setting δ = 0, solve grahically for z, and show that the smallest z goes from π to 2 π as T goes from 0 to ∞. Explain this result.(d) Now set δ = 0.01, solve numerically foe z, using T = 0,1,5,20,100. And 1000.(e) Find the prbalility pr that the particle is in the right “half” of the well, as a function of z and δ. Answer : Pr = 1/[1+(I+/I-)], where I± ≡ [1±δ-(1/z) sin (z(1±δ))] sin2[z(1 ... δ)/2]. Evalute this expression numerically for the T’s in part (d). Comment on your results.(f) Plot the ground state wave function for those same values of T and δ. Note how it gets squeezed into the left half of the well, as the barrier grows. Get solution
9. Suppose the one-dimensional harmonic oscillator (mass m, frequency ω) is subjected to a driving force of the form ..., where f(t) is some specified function (I have factored out ...for notational convenience; f (t) has the dimensions of length). The Hamiltonian is ... Assume that the force was first turned on at time .... This system can be solved exactly both in classical mechanics and in quantum mechanics.2t (a) Determine the classical position of the oscillator, assuming it started from rest at the origin ... Answer: ... (b) Show that the solution to the (time-dependent) Schrodinger equation for this oscillator, assuming it started out in the nth state of the undriven oscillator ... is given by Equation 2.61), can be written as ... ... Get solution
10. The adiabatic approximation can be regarded as the first term in an adiabatic series for the coefficients cm(t) in Equation 10.12. Suppose the system starts out in the nth state; in the adiabatic approximation, it remains in the nth state, picking up only a time-dependent geometric phase factor (Equation 10.21):...(a) Substitute this into the right side of Equation 10.16 to obtain the "first correction" to adiabaticity:... (1)This enables us to calculate transition probabilities in the nearly adiabatic regime. To develop the "second correction," we would insert Equation 1 on the right side of Equation 10.16, and so on.(b) As an example, apply Equation 1 to the driven oscillator (Problem 10.9). Show that (in the near-adiabatic approximation) transitions are possible only to the two immediately adjacent levels, for which...(The transition probabilities are the absolute squares of these, of course.) Get solution
2. Check that Equation 10.31 satisfies the time-dependent Schrodinger equation for the Hamiltonian in Equation 10.25. Also confirm Equation 10.33 and show that the sum-of the squares of the coefficients is 1, as required for normalization. ... Get solution
3. (a) Use Equation 10.42 to calculate the geometric phase change when the infinite square well expands adiabatically from width W1 to width W2. Comment on this result.(b) If the expansion occurs at a constant rate (dw/dt = v), what is the dynamic phase change for this process?(c) If the well now contracts back to its original size, what is Berry's phase for the cycle? Get solution
4. The delta function well (Equation 2.114) supports a single bound state (Equation 2.129). Calculate the geometric phase change when α gradually increases from ..., what is the dynamic phase change for this process? V (x) = -αδ(x) Get solution
5. Show that if ... is real, the geometric phase vanishes. (Problems 10.3and 10.4 are examples of this.) You might try to beat the rap by tracking an unnecessary (but perfectly legal) phase factor onto the eigenfunctions: ... where ... is an arbitrary (real) function. Try it. You'll get a nonzero geometric phase, all right, but note what happens when you put it back into Equation 10.23. And for a closed loop it gives zero. Moral: For nonzero Berry's phase, you need (i) more than one time-dependent parameter in the Hamiltonian, and (ii) a Hamiltonian that yields nontrivially complex eigenfunctions. V(x) = -αδ(x) ... Get solution
6. Work out the analog to Equation 1 for a particle of spin 1.Equation 1... Get solution
7. (a) Derive Equation 10.67 from Equation 10.65. (b) Derive Equation 10.79, starting with Equation 10.78. ... ... Get solution
8. A particle starts out in the ground state of the infinite square well (on the interval 0 ≤ x ≤ a). Now a wall is slowly erected, slightly off-center:...Where f(t) rises gradually from 0 to ∞. Accoroding to the adiabatic theorem, the particle will remain in the ground state of the evolving Hamiltonian.(a) Find (and sketch) the ground state at t → ∞. Hint: This should be the ground state of the infinite square well with an impenetrable barrier at a/2 + ∑. Note that the particle is confined to the (slightly) larger left “half” of the well.(b) find the (transcendental) equation for the ground state of the Hamiltonian at time t. Answer:...Where z ≡ ka, T ≡ maf(t)/h2, δ ≡ 2∑/a, and k ≡ ....(c) Setting δ = 0, solve grahically for z, and show that the smallest z goes from π to 2 π as T goes from 0 to ∞. Explain this result.(d) Now set δ = 0.01, solve numerically foe z, using T = 0,1,5,20,100. And 1000.(e) Find the prbalility pr that the particle is in the right “half” of the well, as a function of z and δ. Answer : Pr = 1/[1+(I+/I-)], where I± ≡ [1±δ-(1/z) sin (z(1±δ))] sin2[z(1 ... δ)/2]. Evalute this expression numerically for the T’s in part (d). Comment on your results.(f) Plot the ground state wave function for those same values of T and δ. Note how it gets squeezed into the left half of the well, as the barrier grows. Get solution
9. Suppose the one-dimensional harmonic oscillator (mass m, frequency ω) is subjected to a driving force of the form ..., where f(t) is some specified function (I have factored out ...for notational convenience; f (t) has the dimensions of length). The Hamiltonian is ... Assume that the force was first turned on at time .... This system can be solved exactly both in classical mechanics and in quantum mechanics.2t (a) Determine the classical position of the oscillator, assuming it started from rest at the origin ... Answer: ... (b) Show that the solution to the (time-dependent) Schrodinger equation for this oscillator, assuming it started out in the nth state of the undriven oscillator ... is given by Equation 2.61), can be written as ... ... Get solution
10. The adiabatic approximation can be regarded as the first term in an adiabatic series for the coefficients cm(t) in Equation 10.12. Suppose the system starts out in the nth state; in the adiabatic approximation, it remains in the nth state, picking up only a time-dependent geometric phase factor (Equation 10.21):...(a) Substitute this into the right side of Equation 10.16 to obtain the "first correction" to adiabaticity:... (1)This enables us to calculate transition probabilities in the nearly adiabatic regime. To develop the "second correction," we would insert Equation 1 on the right side of Equation 10.16, and so on.(b) As an example, apply Equation 1 to the driven oscillator (Problem 10.9). Show that (in the near-adiabatic approximation) transitions are possible only to the two immediately adjacent levels, for which...(The transition probabilities are the absolute squares of these, of course.) Get solution
