Physics 505 Homework No. 3 Solutions S3-1

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1 Physics 55 Homework No. 3 s S More on Bloch Functions. We showed in lecture that the wave function for the time independent Schroedinger equation with a periodic potential could e written as a Bloch function e iqx u q x, where u q is periodic with the same period as the potential. a Show that u q satisfies the differential equation where E is the energy. h2 d 2 u q i h2 q du q 2 m + h2 q 2 u q + V u q = Eu q, We just plug the Bloch function into the Schroedinger equation: Ee iqx u q = h2 = h2 = h2 = h2 d 2 2 e iqx u q + V xe iqx u q e iqx du q + iqeiqx u q + V xe iqx u q d e iqx d2 u q + iqeiqx du q 2 + iqeiqx du q q2 e iqx u q + V xe iqx u q e iqx d2 u q 2 + 2iqeiqx du q q2 e iqx u q + V xe iqx u q. After dividing out the traveling wave and some rearrangement, the desired result appears. In fact, any wave 1D wave function can e written in the aove form with u q not necessarily periodic and will satisfy the aove equation. The key point is that u q is periodic with the same period l as the potential. This means we only need to solve for u q in an interval of length l, [, l] or [ a, a], say. 2a = l. What are the oundary conditions on u q at the ends of the interval? You should consider the case when V is well ehaved at the ends of the interval. u q must e periodic with period l. Therefore it must have the same value and the same derivative at oth ends of the interval. Note that if we consider the periodic lattice of δ-functions, we have a prolem if the δ-functions occur at the ends of the intervals. The prolem can e avoided y shifting the interval to place the δ-functions somewhere else. Alternatively, if δ-functions occur at the ends of the intervals, then the oundary conditions are that u q must have the same value at oth ends of the interval and the difference in slopes u q must e given y the δ-function.

2 Physics 55 Homework No. 3 s S3-2 c Solve u q for the case of a periodic potential containing δ-function potential wells every 2a the same case we did in lecture. V x = λ + n= δx a nl. Since the u q must e periodic, it doesn t matter where you place the interval of length l. You might put it at [, l] so the potential well is in the middle of the interval. The equation for u q is a linear, homogeneous equation with constant coefficients. So, we should try a function like expipx. Plugging this into the equation from part a gives everywhere except at the δ-function in the potential, where k = h 2 p 2 + h2 pq m + h2 q 2 u q = h2 k 2 u q, E/ h 2. Simplifying, p + q 2 = k 2, or p = q ± k. If we put the δ-function at the middle of the interval, so the interval runs from a to +a and the delta function is at, then the solution in the left half is and the solution in the right half is Ae i q + kx + Be i q kx, Ce i q + kx + De i q kx. The continuity of u q and its derivative at the ends of the interval gives Ae i+q ka + Be i+q + ka = Ce i q + ka + De i q ka i q + kae i+q ka + i q kbe i+q + ka = i q + kce i q + ka + i q kde i q ka. A little algera gives the solution for C and D in terms of A and B, C = Ae i+q kl, D = Be i+q + kl,

3 Physics 55 Homework No. 3 s S3-3 with l = 2a. Next we need to satisfy the conditions at the location of the δ-function potential well at x =. Continuity of u q gives A + B = Ae i+q kl + Be i+q + kl. Setting the discontinuity of the slope equal to the integral of the delta function times / h 2 gives i q + kae i+q kl + i q kbe i+q + kl i q + ka i q kb = 2A + B, where = mλ/ h 2. These two equations can e rearranged A 1 e iq kl + B 1 e iq + kl =, A 2 + iq k 1 e iq kl + B 2 + iq + k 1 e iq + kl =. We now have two homogeneous equations in two unknowns. These have non-trivial solutions only when the determinant of the matrix of coefficients vanishes. After a it of algera, one arrives at cosql = coskl k sinkl, exactly the same expression for q we had when we solved the eigenvalue prolem in lecture. There are solutions for real q as long as the right side is etween 1 and +1. When it is outside this range, q will e imaginary and the Bloch function is not an acceptale solution. This will lead to energy gaps exactly as in lecture. Once q is known, the ratio B/A can e determined. B A eiq kl = 1. 1 eiq + kl 2. The square well and the δ-function well. In lecture we discussed the energy levels in a square well V = V for x < a and V = for x > a. We also discussed the ound state of the δ-function potential well, V = λδx. Show that in the limit V, a, such that 2V a λ, the ground state energy and the ground state wave function of the square well ecome the same as those for the δ-function well. We found that the energies of the even states of the squarewell are given y solutions of the transcendental equation q tanqa = κ with q = V + E/ h 2 and κ = E/ h 2. Outside the well, the wave function is exp κ x. Considering the argument of the tangent, we note that as a, qa since q contains the square root of

4 Physics 55 Homework No. 3 s S3-4 V. In the limit, then, κ q 2 a V a/ h 2. E inside the square root can e ignored since V. Thus κ mλ/ h 2. This is exactly what we found for the δ-function potential well with ψ = exp κ x and ground state energy E = h 2 κ 2 / = mλ 2 /2 h The variational method. This is a suject I touched on in lecture ut decided to save all the fun for the homework! Suppose you have a potential for which a ground state exists the minimum energy of a stationary state is finite. You want to know the ground state energy ut the potential is sufficiently complicated that you are unale to get an exact solution for the ground state wave function or the eigenvalue prolem. The variational method consists in picking a random function which has at least one adjustale parameter, calculating the expectation value of the energy assuming the function you picked is the wave function of the system and then varying the parameters to find the minimum energy. Of course, the function you pick must e square integrale normalizale. Show that the minimum energy estimated this way has a lower ound which is the ground state energy. You may assume the spectrum is discrete. With a discrete spectrum, we have energies E n, n =, 1, 2,... where E is the ground state energy and E E n for all n. The trial wave function assumed normalized can e expanded in orthonormal stationary states. ψ trial = c n ψ n, n= Then the expectation value of the energy in the state ψ trial is E = ψ trial H ψ trial = E n c n 2 n= E c n 2 = E c n 2 = E. n= 4. Variational method example. Use the variational method to estimate the ground state energy of the harmonic oscillator with Hamiltonian H = p 2 /+mω 2 x 2 /2. We already know that the ground state wave function is a Gaussian, so don t use a Gaussian trial wave function! That would e cheating. Pick some other simple function and see how close to hω/2 you get. Note that if you pick a function with discontinuities in the function or its derivative, you need to treat the discontinuities properly. My trial function is a triangle. ψx = 1 x /a for x < a and ψx = for x > a. The parameter is a. Note that it s even the potential is even, so that s a good thing and n=

5 Physics 55 Homework No. 3 s S3-5 it s maximum where the potential well is deepest which you would guess is the place the particle is most likely to e found. A ad thing aout this trial function is that it has discontinuities in slope which as we ve seen correspond to δ-functions in the potential and our potential is smooth. To start, we calculate the normalization constant. ψ ψ = 2 a 1 x/a 2 = 2a a 3 1 x/a3 = 2a 3. Next we calculate p 2 / using the unnormalized ψ. Here we need to e a little careful ecause the p 2 involves a second derivative and the trial function has discontinuities in slope. We get around this y calculating p ψ pψ rather than ψ p 2 ψ. p 2 = 2 1 = 2 h2 = h2 ma. a a h i a 1 2 dψ h i dψ Now we calculate mω 2 x 2 /2 again with the unnormalized function. 1 a 2 mω2 x 2 = mω 2 x 2 1 x a 2 a = mω a3 4 + a3 5 = 1 3 mω2 a 3. So, our the energy is rememering to put in the normalization constant E = 3 h2 a + mω2 a We differentiate this with respect to a 2 and find that at the minimum energy, a 2 = 3 h/mω and the minimum energy for this function is 3 3 E min = hω = hω = 5 Even with this crude function, the result is not ad!. 1 1 = hω 2 hω

6 Physics 55 Homework No. 3 s S3-6 For additional fun not to e turned in. What wold you do if you wanted to estimate E 1 as well as E using the variational method? 5. Bound states in a periodic potential. In lecture we discussed the one-dimensional periodic potential made from delta functions, V x = λ + n= δx a nl, where l = 2a, so there are δ-functions at 2n + 1a where n is any integer. In lecture we looked for energy eigenfunctions with energy E >. In this prolem you are asked to find the energy eigenvalues with energy E <. These correspond to the ound state of the single delta function potential. If E <, then define κ y Then in the n th interval, the solution is h 2 κ 2 = E. ψ n = A n e κx + B n e κx, where A n and B n are constants to e determined later. We must join the solutions for each interval at the oundaries set y the δ-functions. ψ must e continuous: A n+1 e κa + B n+1 e κa = A n e κa + B n e κa. The derivative of the wave function must have a step given y the integral of the wave function times the potential: κa n+1 e κa κb n+1 e κa κa n e κa κb n e κa = λ 1 h 2 A n+1 e κa + B n+1 e κa + A n e κa + B n e κa. 2 Define = mλ/ h 2. Then the two equations aove can e summarized in matrix form as e κa e κa An+1 κ + e κa κ + e κa e = κa e κa An κ e κa κ e κa. B n+1 B n The inverse of the left hand matrix aove is 1 2κ κ + e κa e κa κ e κa e κa.

7 Physics 55 Homework No. 3 s S3-7 Multiplying oth sides of the aove equation y this inverse yields An+1 1 = k e κl κ An An B n+1 κ 1 + κ e κl = D B n B n Note that the matrix D has unit determinant so it has an inverse. Let the eigenvalues of D e d 1 and d 2. By the same argument we used in lecture, we only have a solution when d 1 = d 2 = 1 and d 1 = d 2. We find the eignevalues from the characteristic equation:. or det 1 k e κl d κ κ 1 + κ e κl d =, d 2 2d coshκl κ sinhκl + 1 =. s for the eigenvalues are: d = ξ ± ξ 2 1, ξ = coshκl l κl sinhκl. In order that d = 1, we must have 1 < ξ < +1. So the next step is to decide for what values of l and κl we can meet this condition. A good way to find out is to use excel or matla to generate various plots. However, the following things can e said. First of all, cosh = 1 and sinh = and sinhx < coshx for all x. For large x, coshx expx/2, sinhx expx/2 and the difference etween the two is exp x for all x. As x, sinhx/x 1. So, if l = that is, no potential, there is no solution. This is to e expected. If l > there are always solutions and a single and of energies for which solutions exist. If l < 2, the and extends up to E =. If l > 2, a zero energy solution does not exist and the and is elow zero. For very large l, the and will e a narrow range with κl l since coshκl sinhκl for large κl. Note that for a single delta function, the solution was κ =, so when the potential is large, the ound state from a single δ-function does not reach the adjacent δ-functions and there is little change to the energy of the ound state leading to a relatively narrow and of energies. On the other hand, when l < 1, the ound state wave functions do overlap the adjacent δ-functions, so there are significant changes to the energies leading to a relatively road and of energies. You weren t asked to find the energy eigenfunctions. However, it is interesting to note that since d = 1 for a solution, we can write d ± = exp±iql and generate the eigenfunctions just as we did for the positive energy solutions. So, we will find traveling wave solutions even in this case!

Physics 505 Homework No. 4 Solutions S4-1

Physics 505 Homework No. 4 Solutions S4-1 Physics 505 Homework No 4 s S4- From Prelims, January 2, 2007 Electron with effective mass An electron is moving in one dimension in a potential V (x) = 0 for x > 0 and V (x) = V 0 > 0 for x < 0 The region