Section 4: Harmonic Oscillator and Free Particles Solutions

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1 Physics 143a: Quantum Mechanics I Section 4: Harmonic Oscillator and Free Particles Solutions Spring 015, Harvard Here is a summary of the most important points from the recent lectures, relevant for either solving homework problems, or for your general education. This material is covered in the middle part of Chapter of [1]. An important problem in quantum mechanics is the simple harmonic oscillator, with Ĥ = ˆp m + 1 mωˆx. 1) We can also quantize this by studying the solutions to the time-independent Schrödinger equation directly. After non-dimensionalizing x = /mωξ, and ɛ = E/ω, we find An asymptotic analysis as ξ ±, shows that we must take d ψ dξ = ξ ɛ ) ψ. ) ψ = hξ)e ξ /, 3) and furthermore that the function h leads to a ψ which, for generic ɛ, is not normalizable ψ e ξ / ). When ɛ is an odd integer, h reduces to a Hermite polynomial, and ψ is normalizable. From the differential equation perspective, the requirement of normalizability is what leads to a discrete energy spectrum. The free particle has Hamiltonian Ĥ = ˆp m, 4) and is defined on the entire real line. The solutions to the time independent Schrödinger equation are not normalizable: they are plane waves, of energy ψ k = e ikx, 5) Ek) = k m. 6) Normalizable solutions can be constructed to the time-dependent Schrödinger equation: Ψx, t) = The normalizability condition becomes 1 = dk π φk)e ikx iek)t/. 7) dk φk). 8) 1

2 If we have a wave packet with φk) very narrow and localized around k = k 0, then we can use the stationary phase approximation to show that x x 0 1 de dk t t 0) v g t t 0 ). 9) In general v g is defined as the group velocity of the wave packet it corresponds to the velocity at which the wave packet appears to move. For a quantum mechanical free particle, 1 de dk = k 0 m, 10) and noting that k 0 = p 0 is the average momentum of the particle at least for simple wave packets), we see that we recover on average) the classical trajectory. If we have a Hamiltonian of the form Ĥ = ˆp + V ˆx), 11) m and look for stationary states ψ of energy E, then in any region where E > V x), ψ will oscillate and propagate forever. Beyond a classical turning point, in regions where E < V, ψ will in general exponentially decay so that the solution is normalizable!). If there are two regions: I: x 1 < x < x, and II: x 3 < x < x 4 x < x 3 ) where V < E, then a particle starting in region I can quantum mechanically tunnel into region II, by propagating through the classically forbidden region where E > V. Consider scattering off of the potential V x) = { 0 x < 0 V 0 x > 0. 1) We begin by looking for stationary states with 0 < E < V 0. To do this, we express ψx < 0) in terms of plane waves, and ψx > 0) as an exponentially decaying function; in terms of me mv0 E) k =, κ =, 13) { e ψ E x) = ikx + re ikx x < 0 Ae κx x > 0, r = ik + κ ik κ. 14) To get a good solution to the Schrödinger equation we require continuity of ψ and dψ/ at x = 0. All higher derivatives will not be continuous due to the jump in V x)). We find the r above when we demand these continuities. Note that the reflection probability r = 1 every incident particle will get reflected. For E > V 0, we instead find { e ikx + re ikx x < 0 me ψ E x) = te ik x x > 0, V0 ) k =. 15) This time we have r = k k k + k, t = k k + k. 16) Note that r + t 1. This is because particles that are transmitted are moving slower the flux of probability incident, reflected and transmitted is: J probability probability amplitude velocity of particles 17)

3 and now J inc + J ref = J trans, as J inc = k m 1, J ref = k m r, J trans = k m t. 18) Problem 1 Hermite Polynomials): In this problem we re going to properly understand the Hermite polynomials by exploiting the beautiful harmonic oscillator algebra. For simplicity we work in dimensionless units: we set = m = ω = 1. So a = x + ip, Note that [a, a] = [a, a ] = 0 and [a, a ] = 1, and a = x ip, p = i d. aψ n = nψ n 1, a ψ n = n + 1ψ n+1. In these units, the harmonic oscillator wave functions are ψ n x) = e x / π 1/4 n/ n! H nx), where H n x) is the n th Hermite polynomial n = 0, 1,, 3,...). a) Begin with the fact that H 0 x) = 1. By constructing ψ n from the creation operators, derive the Rodrigues formula: H n x) = e x d ) n e x. Solution: We start with the identity ψ n = e x / π 1/4 n/ n! H nx) = 1 a n ψ 0 = 1 1 x d )) n e x / n! n! π 1/4. Evidently what we need to show is that H n x)e x / = x d ) n e x /. To do this, we note that, for any smooth function fx): x d ) fx) = xfx) df = / d ) ex fe x /. Evidently: H n x) = e x / x d ) n [ e x / = e x / e x / d ) n e /] x e x /. We can multiply through and cancel off all the internal factors of e ±x / in the long string of operators above. This leaves us with the Rodrigues formula. 3

4 b) It is not obvious from the Rodrigues formula that H n x) is a polynomial. Derive the following recursion relation: H n+1 x) = xh n x) dh nx). and show how H n is always a polynomial. Solution: Analogously to in the previous part, we stick back in some exponential factors into the derivative product: H n+1 x) = e x d ) e x e x d ) n e x = e x d ) e x H n x) = e x xe x) H n x) e x x dh nx) = xh n x) dh nx). It is now much easier to see how H n x) is a polynomial. We do this recursively we know that H 0 = 1 is a polynomial. Suppose H n x) is a polynomial. Then the derivative of H n is also a polynomial, and so is xh n x). The sum of two polynomials is a polynomial, so H n+1 is a polynomial. c) Use the relations between x, p, a and a, acting on the wave function ψ n, to derive the following pair of identities: xh n x) = 1 H dh n x) n+1x) + nh n 1 x), = nh n 1 x). Solution: We know that, rearranging the definitions of a, a to find x and p: Now let s study Plug in for ψ n in terms of H n : x = a + a. xψ n = a + a n n + 1 ψ n = ψ n 1 + ψ n+1. e x / x π 1/4 n/ n! H nx) = e x / π 1/4 n 1)/ n 1)! n H n 1x) + e x / n + 1 π 1/4 n+1)/ H n+1 x) n + 1)! Cancel out a few factors, and find the identity: xh n x) = 1 H n+1x) + nh n 1 x). For the second identity, we don t have to do anymore work. Just use the recursive formula from part b) to get rid of H n+1 in our new identity: xh n x) = 1 [ xh n x) dh ] nx) + nh n 1 x). A simple rearranging of this gives us our second identity. d) A generating function for Hermite polynomials is defined as Gx; z) 4 z n n! H nx),

5 as n G z = H n x). z=0 Show that the generating function for Hermite polynomials is Gx; z) = e xz z. Solution: This is tricky, but the key place to start is to notice that π 1/4 Ge x / G = π 1/4 z n n! H nx)e x / n = n! zn ψ n x). Next, note that a 1 G = x ) G = x Now combine these equations: x ) G = x n n! zn a ψ n = n + 1) n n + 1)! zn ψ n+1 n + 1) n+1 n + 1)! zn ψ n+1 = z G a G = 1 x + ) n G = x n! zn aψ n = x + ) G = z x G Multiply this equation by π 1/4 e x / and find: The solution to this PDE is x G = z G + G z. x z)g = G z. Gx; z) = G 0 x)e xz z ; n n 1)! zn ψ n 1 G 0 x) is the initial condition for the PDE at z = 0. But G 0 x) = Gx; 0) = H 0 x) = 1. We conclude G = e xz z as claimed. Problem Scanning Tunneling Microscope): A scanning tunneling microscope STM) is used to determine where electrons are on the surface of a material. The basic idea is as follows: a thin metal tip is placed a distance a from the surface of a metal. In reality, the surface of the metal is not entirely smooth electronic wave functions stick out farther in some places than in others. As we ll compute in this problem, it is possible, though unlikely, for an electron from the metal to tunnel into the STM tip. Not surprisingly, the closer the tip is to the electronic wave function, the larger the rate of tunneling events will be. So we scan the STM tip over the surface of the metal and, by measuring the rate of these tunneling we can tell how much the electronic wave functions stick out from the surface of the metal. 5

6 a δ We end up with the following cartoon. For simplicity, we only model the dynamics of the electrons perpendicular to the surface, and assume that electrons are approximately bound in place along the surface, due to the presence of attractive Coulomb interactions with a nearby ion. An electron of mass m moves in the following potential: x < 0 V x) = 0 0 < x < a, ϕ x > a with 0 < < ϕ. The former constraint ensures that it takes some energy to pull an electron out of a metal, and the latter ensures that tunneling events from the STM tip into the metal are essentially negligble we won t worry about this point, as a lab set-up is more complicated and would avoid this issue entirely). The electron starts at x < 0, in the metal of interest. a) Approximately find the scattering states where the electron is incident on the barrier from the left, in the limit where a, if the electron has incident wave number k, assuming that 1 k m. Solution: If our scattering state has energy E, then we have to solve the Schrödinger equation d ψ = E V )ψ, m with V the piece-wise constant function above. The energy E can be found by noting that we must begin constructing our solution by assuming an incident plane wave from the left, so ψ = e ikx +Ae ikx for x < 0. This requires that so we conclude that k ψ = E ))ψ = E + )ψ, m E = k m. 1 Remember that physically, the probability that the particle can tunnel through the barrier becomes extremely small as a. What does this imply about the dominant contribution to the wave function in the region 0 < x < a? 6

7 The assumption at the start of the problem is then equivalent to E < 0. This means that in the region 0 < x < a, we will have E < V, and thus exponentially growing/decaying wave functions: ψ = Be κx + Ce κx, where me κ. And finally, in the right hand region, we have outgoing waves only since by assumption no electrons come from the STM tip into the metal), so the wave function here is ψ = F e ik x, k = mϕ + E). Now, let s discuss how to solve this kind of equation. Because we only have one outgoing plane wave on the right, it s easiest to start at the boundary x = a. We require that F e ik a = Be κa + Ce κa ik F e ik a = κce κa Be κa ) continuity of ψ), continuity of dψ/). Now in general, k and κ are comparable numbers, but we re interested in the limit a. The set of equations above implies that Be κa and Ce κa are comparable in size, since we ve got two equations relating them to F, up to the factor of κ/k in the latter. But a. This means that C/B e κa everywhere except for the x = a interface, we can entirely neglect the growing mode! This makes sense, because if a then the wave function should not be able to leak through the barrier at all. We conclude that the intermediate wave function is approximately given by Be κx. Now this makes life simple at x = 0: We obtain that 1 + A B continuity of ψ), ik1 A) κb B continuity of dψ/). 1 + iκ/k, A 1 iκ/k 1 + iκ/k. Note that A 1, so at this order, every electron gets reflected! Not quite of course there are really tiny exponential factors that are subleading, associated with the Cs, that will reduce the reflection probability ever so slightly. Now we can calculate those since we know what B is. In particular: ) F e ik a 1 ik = Be κa + Ce κa Ce κa + Be κa = Be κa. κ F = Be κa ik a 1 + iκ/k = 4e κa ik a 1 + iκ/k)1 ik /κ). Since we have approximately found A, B and F, we have approximately found our scattering state. b) Approximate that the incident energy of the electrons is comparable to, and estimate the width of the vacuum gap a c necessary before the likelihood of a tunneling event occuring is high. Obtain both an analytic estimate, as well as a numerical estimate, using that for a typical metal, J 3 ev, and the mass of the electron is m kg. Solution: The probability that any given electron will tunnel across is given by F = 16e κa 1 + κ /k )1 + k /κ ). 7

8 The overwhelmingly important factor here is e κa. So approximately when a a c = 1 κ the probability of tunneling becomes high. Let s plug in for numbers. Using κ m /: a c m m. c) Typical STM measurements work at a 10a c, so that the probability of tunneling events is very low. Approximate that whenever an electron gets reflected, it must travel about an atomic distance before another electron is incident on the barrier. Using dimensional analysis, and noting that the interatomic distance scale in a metal is about δ 0.3 nm, estimate the time required before we can see a single electron tunnel into our STM. Solution: Let s estimate the time in between electrons attempting to tunnel through the barrier. This is semiclassical logic, but it is reasonable and will be made rigorous later in the course. The typical velocity of the electron is E v m m 7 m 105 s. We have to travel a distance of δ m before another electron has a chance to pass through. Thus the waiting time is t δ v s. Actually this estimate is probably a bit small due to complications about electrons in solids, but it s fine for our purposes. The probability is dominated by the e κa factor. Our estimate of the time it takes for an electron to tunnel through is then [ t tunnel e κ10ac)] 1 m = e s. Although the probability for tunneling is extremely low, tunneling events still occur very rapidly on human time scales. d) Now suppose that we move the STM tip to another location on the surface where the thickness of the distance between the STM tip and the surface electrons has grown to a + δ. By how much will the electric current flowing through the STM tip reduce? Solution: The electric current will reduce by a factor proportional to the rate at which electrons can tunnel across the barrier, which will be a factor e κa+δ) e κa e κδ e This is an extremely dramatic effect that is easily detectable in a measurement of the current. e) To what extent does the answer in this problem depend on ϕ? What if ϕ was a function of x? Solution: Although ϕ does enter into F via the denominator, the dependence on ϕ is extremely weak compared to the exponential suppression. So it s unlikely that altering the potential in the STM tip will do very much to alter our qualitative answer. And there s a lot we haven t yet taken into account, so you shouldn t take anything derived here more seriously than an order of magnitude estimate. 8

9 In reality, the net electric current flowing is sensitive to ϕ. But calculating the electric current requires a bit more solid-state physics, and the most dramatic effect is the low probability for tunneling, which we re already able to calculate fairly reliably. The power of the STM is that as shown in part d) it is an extraordinarily sensitive measurement. We are thus able to image the electronic wave functions on the surface of a metal. This technique was pioneered in [], which later got some of the authors a Nobel Prize. [1] D. J. Grififths. Introduction to Quantum Mechanics Prentice Hall, nd ed., 004) [] G. Binnig, H. Rohrer, C. Gerber and E. Wiebel. Surface studies by scanning tunneling microscopy, Physical Review Letters ). 9

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