Physics 486 Discussion 5 Piecewise Potentials

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1 Physics 486 Discussion 5 Piecewise Potentials Problem 1 : Infinite Potential Well Checkpoints 1 Consider the infinite well potential V(x) = 0 for 0 < x < 1 elsewhere. (a) First, think classically. Potential energy maps V(! r ) are ways of representing forces :! F =! V. What sort of physical situation does this infinite well represent? One thing to think about: is it possible for a particle of finite energy E = T + V to be outside the region 0 < x < 1? Why or why not? (b) Write down the general solution for the position-dependent wavefunction ψe(x) for energy E > 0. (If you re not sure what to do, follow the steps in the STRATEGY SUMMARY appendix.) Since our potential V(x) is a piecewise function, ψe(x) will have different forms in the left-hand region x < 0, in the well 0 < x < 1, and in the right-hand region x > 1. GUIDANCE : This is a case where it is easier to work with real sines and/or cosines instead of complex e ikx forms. Also remember that the general solution to a diff.eq. like the SE must have free parameters. (c) As we presented in class, when working with piecewise potentials, it is crucial to apply the following boundary conditions at the edges between regions where V has different form : a. Wavefunctions are always continuous. b. Wavefunctions have continuous derivatives, except at points where V is infinite (see appendix) c. Wavefunctions are zero in any region where V =. Apply boundary conditions #a and #c to the three-piece general solution ψe(x) you found in the previous part. You should find that this restricts the wavenumbers of your sines and/or cosines, and therefore the allowed energies, to a discrete set of allowed values kn & En for n = 1, 2, 3, FYI : At last we have found a system that has a quantized energy spectrum! The free particle does not because its motion is not periodic; a particle trapped in this potential well will certainly behave in a period way, hence quantization! (d) You now have time-independent solutions ψn(x). Turn these into time-dependent solutions Ψn(x, t). (e) What do these solutions mean? Plot the probability density P(x, t) = Ψn*Ψn for n = 1, 2, 3. Does it dependent on time? (f) Your plots make it obvious what the expectation values < x > and widths σx are in position but momentum is much harder to see from a wavefunction. The expectation value < p > is zero because each state represents the superposition of a left- and right-going wave but < p 2 > is more interesting since the sign of momentum doesn t enter. < p 2 > is the kinetic energy times 2m. How does it depend on the energy quantum number n? IMPORTANT : We have not normalized our solutions Ψn(x, t) do we need to? e.g. for these expectation value calculations? If you are unsure, ask!!! Also, before you integrate Ψ* ( ˆp) 2 Ψ over x to get the expectation value, you might find it interesting to plot Ψ* ( ˆp) 2 Ψ vs x it has an x-dependent pattern, just like Ψ*Ψ. 1 Q1 (a) Inward impulse forces at x = 0 & 1; region outside well is classically forbidden particle would have T < 0 (impossible). (b) ψe(x) = 0 outside well; ψe(x) = A sin kx + B cos kx inside well with k² = 2mE/ ² (c) ψn(x) = An sin knx where kn = nπ En = ( n π) ² / 2m (d) Ψn(x,t) = An sin nπx exp( i n² π² / 2m) (e) no t-dependence (f) normalized: A n = 2 / a <p 2 > = 2mEn

2 Problem 2 : Step Potential Consider the step potential V(x) = 0 if x 0 V 0 if x > 0 where V0 is a positive constant. Checkpoints on last page A particle with total energy E = T + V starts in the left-hand region x < 0 and travels to the right (+x direction). Eventually it will encounter the step at x = 0. Our task is to discover what happens. (a) Think classically for a minute. First, potential energy maps V(! r ) are ways of representing forces :! F =! V. Second, the particle s total energy E = T + V is conserved. Why? A brief recap: The Lagrangian T V has no explicit dependence on time ( L/ t = 0), and dh/dt = L/ t, so the Hamiltonian H of this system is conserved (dh/dt = 0). No time-dependent constraints are part of this problem (i.e. natural coordinates are in use) so the Hamiltonian H of this system is equal to T + V. (T + V) of this system is conserved. (Just refreshing your memory to help it gel; you won t be asked about these classical details in this course.) SO: A classical particle of mass m starts at x = with momentum p0 in the +x direction. At some time t = 0 it reaches the origin x = 0. Describe the motion of the particle by specifying x(t) and p(t) in these three cases: (a1) before it arrives at the origin, i.e. for times t < 0 (a2) after it arrives at the origin (t > 0) in the case E > V0 (a3) after it arrives at the origin (t > 0) in the case E < V0 Switch to quantum! We now want to describe the state of a quantum particle that is not localized in position but has definite momentum and energy (back to plane waves!). Our initial condition is simply that the particle is incident from the left (x = ) with energy E and momentum pointing to the right (+x direction). Our solution will be a single wavefunction Ψ(x,t) for the particle, though it will have different functional forms in the regions x < 0 and x > 0 due to the piecewise nature of our potential. The quantum description does not at all resemble the classical this happened, then this happened description; rather it describes a steady beam of particles coming in from the left and responding to the forces they encounter. Bound-State Case : Let the particle have definite energy E in the range 0 < E < V0 (b) Write down the general solution for the position-dependent wavefunction ψ(x). Since our step potential V(x) is a piecewise function, ψ(x) will have a different form in the left-hand region x < 0, and in the right-hand region x > 0. Also think : we are trying to describe a particle that comes in from the left (x = ) with energy E but when it hits x = 0 it encounters an impulse force (classically, Fx = dv/dx) in the region x < 0 there will thus be both an incident wave propagating to the right and a reflected wave propagating to the left. (c) Now apply the boundary conditions at the edges between regions (i.e. at x = 0). (d) Since we have plane waves in our wavefunction, it is not normalizable, so we will not be able to calculate any meaningful expectation values. However we can compare the amplitudes-squared of the incident and reflected waves to determine the ratio R Aincoming ² / Areflected ² of their probability densities. R is called the reflection coefficient; if the incoming wavefunction represents a beam of particles, R tells you the fraction of beam particles that are reflected from the step at x = 0. Calculate its value! Scattering-State Case : Now let the particle have definite energy E in the range E > V0 (e) Repeat steps (b), (c), and (d) for this case, where the regions x > 0 and x < 0 are both classically-allowed. You will find that the reflection coefficient R is now less than 1 because some particles do continue onward despite the leftward impulse force at x = 0.

3 STRATEGY SUMMARY Finding the GENERAL SOLUTION Ψ(x, t) for a TIME-INDEPENDENT POTENTIAL V(x) To find the general solution Ψ(x, t) of the Schrödinger Equation (SE) for a given potential energy V(x), our strategy is to find a class of solutions that form the basis elements of a complete set of solutions; the general solution is then just the set of { all possible linear combination of the basis solutions }. To find a complete set of basis solutions we do this: (1) Apply separation of variables to the SE gives solution form Ψ E (x,t) =ψ E (x) f E (t) where i E/! the time-dependent part f E (t) = e ( )t the space-dependent part ψe(x) solves the time-independent SE :!2 2m ψ + V(x)ψ = Eψ E E E E is the separation constant and can a priori take on any value Note / recall that these separated solutions ΨE(x, t) are energy eigenstates : Ê Ψ i! Ψ / t = E Ψ E E E so they have definite energy E stationary states : the probability densities P E (x,t) Ψ E * Ψ E =ψ E * (x)ψ E (x) are time-independent (2) Solve the time-independent SE with the given potential V(x) & energy eigenvalue E get all ψe(x) Special Case : PIECEWISE POTENTIALS V(x) A common special case is when the potential V(x) is a piecewise function of position, i.e. when V(x) has different forms in different regions of x. The step potential in problem 2 is a perfect example: it has one form, V(x) = 0, in the region x < 0 and another form, V(x) = V0, in the region x > 0. Solving such cases requires a new tool: the boundary conditions that all wavefunctions must obey. We ll prove them in class and use them here. Here s how you obtain ψe(x) for a piecewise potential V(x) that has form V1(x) for x = to x1, form V2(x) for x = x1 to x2,, form Vn(x) for x = xn 1 to. (2.1-piecewise) Solve the time-independent SE separately for each piece Vj (x) of the potential get all ψe, j (x) in each region j = xj 1 to xj. (2.2-piecewise) Patch together the solutions ψe, j (x) from the different regions j by applying these crucial boundary conditions on the wavefunction at the boundaries xj between regions : a. Wavefunctions are always continuous ψe, j (xj) = ψe, j+1(xj) b. Wavefunctions have continuous derivatives ψ E, j (xj) = ψ E, j+1(xj) except at points where V is infinite; there, the discontinuity in the derivative is lim ψ E, j+1 (x j + ε) ψ E, j (x j ε) = 2m ε 0! lim x j +ε V(x)ψ (x)dx 2 ε 0 x j ε c. Wavefunctions are zero in any region where V =

4 SPECIFIC SOLUTION for a given INITIAL WAVEFUNCTION Ψ(x, 0) If you are additionally given some initial conditions usually a starting wavefunction Ψ(x,0) then you must find the specific solution = the specific linear combination of basis solutions { ΨE(x, t) = ψe(x) e iet/ } that satisfies the initial conditions. Since our basis solutions so far have all been sinusoidal (in real or complex form, it doesn t matter), Fourier analysis is the perfect tool to determine the linear combination that works. (3.0) First, make sure the given starting shape is normalized ensure dx Ψ(x,0) 2 = 1 (3.1-sinusoidal-basis) If your basis solutions ψe(x) are sinusoidal, they have a wavelength, and so can be characterized by their wavenumber k /λ as follows: ψe(x) = e ikx. This k is related to E via the dispersion relation E(k) = ω(k) that results from solving the time-independent SE. Obtain the relation between k and E. (3.2-sinusoidal-basis) We will now label the solutions ψk(x) instead of ψe(x), because for each energy eigenvalue E there are usually two possible solutions: one with positive k and one with negative k. Apply a Fourier transform to the starting wavefunction Ψ(x, 0) to determine the amplitudes A(k) that represent how much of each solution ψk(x) is needed to build the starting wavefunction: Calculate the amplitude function A(k) = dx Ψ(x,0)e ikx that builds the starting shape Ψ(x,0) = dk A(k)e ikx from the solutions ψe(x) = e ikx. (3.3-sinusoidal-basis) Finally, build the final time-dependent wavefunction Ψ(x, t) by superposing the sinusoidal basis solutions as shown above, this time with the time-dependent part fe(t) = e i (E/ ) t included along with each position-dependent part ψe(x) = e ikx. Note: this is where you need that dispersion relation E(k). Build the final solution Ψ(x, t) = dk A(k)e ikx e ie(k)t/!

5 Comparison with CLASSICAL MECHANICS Here s how you solve for the motion of a system in Classical Mechanics (e.g. 3 spheres connected by springs): Describe the system with n generalized coordinates q1, qn, where n is the number of degrees of freedom (#dof) of the system. Write down the equations of motion (EOMs) of the system by e.g. obtaining the kinetic and potential energies T and U of the system in terms of your coordinates qi and their derivatives, forming the Lagrangian L = T U, then evaluating the Lagrange equations L/ qi=d( L/ (dqi/dt))/dt for i = 1,, n. These EOMs are n 2 nd -order differential equations. Solve the EOMs to obtain their general solution { q1(t),, qn(t) }. The expressions { qi(t) } in the general solution will have 2n free parameters. They are needed because the EOMs, being differential equations, only specify derivatives, not values. We are always free to specify the state of the system at some time t0 and this requires 2n variables e.g. n positions & n speeds. If we are given a set of initial conditions, e.g. n positions q1(t0),, qn(t0) and n momenta p1(t0),, pn(t0) at some starting time t0, we must also find the specific solution q1(t),, qn(t) that matches the given conditions. The specific solution has no free parameters as the 2n free parameters of the general solution are fixed by the initial conditions. These familiar steps have equivalents in Quantum Mechanics; let s try to build a translation table. In QM, we are currently considering only 1-particle systems in 1 spatial dimension, so that s one degree of freedom (n = 1) and one generalized coordinate q which we will choose to be x. CM EOMs to solve : Lagrange Equation for Schrödinger Equation for system-specific PE U(x) system-specific PE V(x) General solution looks like : x(t) Ψ(x, t) = A(E) ΨE(x, t) = solution function = all possible linear combinations with 2 free parameters of a complete set of basis solutions { ΨE(x, t) = ψe(x) e i (E/ ) t } Initial conditions sufficient to specify x(t 0 ) = given value x 0 Ψ(x, t0) = given function ψ0(x) a unique specific solution :!x(t 0 ) = given value v 0 How to find the specific solution : find the 2 free parameters find coefficients A(E) via Fourier by solving the 2 equations transform of starting shape ψ0(x), x(t 0 )= x0 &!x(t 0 ) = v 0 then put back in t-dependence : Ψ(x, t) = A(E) ψe(x) e i (E/ ) t QM Checkpoints for Problem Q2 (a1) p(t) = 2mE p0, x(t) = p0t / m (a2) p(t) = 2m(E V 0 ) p1, x(t) = p1t / m (a3) p(t) = 2mE p0, x(t) = p0t / m (b) for x < 0, ψ(x) = Ae ikx + Be ikx with k² = 2mE/ ² for x > 0, ψ(x) = Ce Kx with K² = 2m(V0 E)/ ² (c) B = A(K+ik)/(K ik), C = A+B = A 2ik / (K ik) (d) R = B ²/ A ² = 1 (e) R = ( E E V 0 ) 4 / V 0 2

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