QM1 - Tutorial 5 Scattering Yaakov Yudkin 3 November 017 Contents 1 Potential Barrier 1 1.1 Set Up of the Problem and Solution...................................... 1 1. How to Solve: Split Up Space.......................................... 1..1 Case 1: E > V 0.............................................. 1.. Case : E < V 0.............................................. 3 1..3 Example.................................................. 4 Example: Step Potential Barrier 5 3 Example: Delta Barrier 6 1 Potential Barrier 1.1 Set Up of the Problem and Solution Potential Barrier Consider the potential V (x) = { V 0 0 < x < L 0 else A particle of mass m is released from x = with a denite momentum k in the +ˆx direction. Since the potential at x = is that of a free particle the energy of the particle is E = k /m. After some time the particle reaches x = 0. The question is: Will the particle cross into the region where V = V 0 and eventually reach x = + or will it be reected back and go to x =? Classical Solution In classical mechanics a particle with E > V 0 will pass through the potential while a particle with E < V 0 will bounce back. We can say this with absolute certainty. Quantum Solution In quantum mechanics this is not necessarily the case. No matter what the energy is, there is always a probability to be transmitted and there is always a probability to be reected. As we will see these probabilities depend on the energy E. But we should stress that a particle with E > V 0 has a probability of bouncing back while a particle with E < V 0 has a probability of passing through (aka tunneling). The objective of quantum scattering theory is to determine these probabilities. Probability Current The quantum solution is not to surprising. We know that quantum theory is probabilistic. In fact, it is wrong to talk about a particle coming from x = with a denite momentum k, because, if we know the momentum how can we know its position. The correct way to think about this is in terms of a probability current J (x). The probability of nding the particle at a certain location ows like a current. Given a particle with a wave function ψ (x) the probability current J (x) is J (x) = m I [ ψ (x) ] dψ (x) dx For all scattering problems, as we will see, J (x) = J is independent of location. It is just a number (with the correct units). 1
Transmission and Reection Coecients The probability current of the incoming particle is. At x = 0 it splits up into two currents: a reected current J R and a transmitted current J T. The probabilities to be transmitted of reected, also called the coecients of transmission and reection, are given by Since the total probability is 1 we must have J T, R = R + 1 This can also be expressed in terms of probability currents. The incoming currents are equal to the outgoing currents. = J R + J T The relation R + 1 is a trivial consequence of this observation. 1. How to Solve: Split Up Space In order to nd the transmission and reection coecients we need to nd the probability currents for which, in turn, we need to know the wave function. This is simple. All we have to do is solve the time independent Schrodinger equation. H op ψ (x) = Eψ (x), H op = p op m + V (x op) Since V (x) changes abruptly in space lets solve the equation in parts d [ m dx ψ 1 (x) ] = Eψ 1 (x) x < 0 d m dx + V 0 ψ (x) = Eψ (x) 0 < x < L d m dx ψ 3 (x) = Eψ 3 (x) x > L By rearranging the three equations (only two are dierent) we get ψ 1 (x) = me ψ 1 (x) x < 0 ψ (x) = m(e V0) ψ (x) 0 < x < L ψ 3 (x) = me ψ 3 (x) x > L Next dene k and q as follows k = me, q = m (E V 0) Remember that E is given in the question. It is the energy with which the particle is released form x =. We can then write the three equations as ψ 1 (x) = k ψ 1 (x) x < 0 ψ (x) = q ψ (x) 0 < x < L ψ 3 (x) = k ψ 3 (x) x > L J R 1..1 Case 1: E > V 0 Solution of Dierential Equations Both k and q are real. The solution of the three dierential equations are ψ 1 (x) = Ae ikx + Be ikx x < 0 ψ (x) = ψ (x) = Ce iqx + De iqx 0 < x < L ψ 3 (x) = F e ikx + Ge ikx x > L Interpretation The six parts of ψ (x) can be understood in the following way Ae ikx : wave describing particle released at x = and traveling towards x = 0 Be ikx : Ce iqx : reected wave traveling back to x = transmitted wave
The transmitted part of the wave now continues until it reaches x = L. The same thing happens there. De iqx : reected wave traveling back to x = 0 F e ikx : transmitted wave traveling to x = + Since the wave is not reected again after x = L the part Ge ikx cannot exist. It does not suit the initial condition (particle starting at x = ). We thus set G = 0. Probability Currents The incoming probability current is associated with the rst part of ψ (x). By plugging A exp (ikx) into the equation for the current we get = [ m I A e ikx d ] dx Aeikx = k m A For the reected probability current we use the second part of ψ (x) and obtain J R = [ m I B e ikx d ] dx Be ikx = k m B Now for the transmitted current. We are interested in the probability of being fully transmitted, so reaching the region x > L. A particle reaching this region will necessarily continue to x = +. Hence we use the part F exp (ikx) for the transmitted probability current J [ m I F e ikx d ] dx F eikx = k F m Transmission and Reection Coecients 1.. Case : E < V 0 It is now straight forward to get F A, R = B A The Dierence Regions 1 and 3 are unchanged. The dierence is in the region of the barrier (0 < x < L). Since E < V 0 the value of q is negative. Hence we dene g = q = m (V 0 E) which is positive. The dierential equation in region is now ψ (x) = g ψ (x) to which the solution is (no i) In total we thus have ψ (x) = Ce gx + De gx 0 < x < L ψ 1 (x) = Ae ikx + Be ikx x < 0 ψ (x) = ψ (x) = Ce gx + De gx 0 < x < L ψ 3 (x) = F e ikx x > L Transmission and Reection Coecients Notice that, since regions 1 and 3 are unchanged, the probability currents are the same as in the previous case. Thus also the transmission coecients are the same. The values of the constants though are dierent. F A, R = B A 3
1..3 Example Statement of the Problem following potential Find the wave function of a particle of mass m and energy E being scattered by the 0 x < x 1 V 1 x 1 < x < x V (x) = V x < x < x 3 V 3 x 3 < x < x 4 V 4 x > x 4 All V i are positive and we have V > E > V 1 and E > V 4. Also nd an expression for T and R in terms of the integration constants A, B,.... Figure 1: Potential V (x) and energy E for the example. Wave Function The Schrodinger equations, after some simple rearranging, for the dierent regions are ψ 1 (x) = me ψ 1 (x) x < x 1 ψ (x) = m(e V1) ψ (x) x 1 < x < x ψ 3 (x) = m(v E) ψ 3 (x) x < x < x 3 ψ 4 (x) = m(e+v3) ψ 4 (x) x 3 < x < x 4 ψ 5 (x) = m(e V4) ψ 5 (x) x > x 4 Notice that for ψ 3 (x) we have taken out an overall minus sign so that the coecient will be positive. Also note the plus sign in the equation for ψ 4 (x) due to the negative potential in that region. Lets now replace the factors of ψ i on the rhs by ki so that the equations become ψ 1 (x) = k1ψ 1 (x) x < x 1 ψ (x) = kψ (x) x 1 < x < x ψ 3 (x) = k3ψ 3 (x) x < x < x 3 ψ ψ 4 (x) = k 4ψ 4 (x) x 3 < x < x 4 5 (x) = k 5ψ 5 (x) x > x 4 The solution of the dierential equations are ψ 1 (x) = A 1 e ik1x + B 1 e ik1x x < x 1 ψ (x) = A e ikx + B e ikx x 1 < x < x ψ (x) = ψ 3 (x) = A 3 e k3x + B 3 e k3x x < x < x 3 ψ 4 (x) = A 4 e ik4x + B 4 e ik4x x 3 < x < x 4 ψ 5 (x) = A 5 e ik5x x > x 4 4
The incoming an reected current are deduced from the part containing A 1 and B 1 respec- Probability Currents tively = k 1 m A 1, J R = k 1 m B 1 The transmitted current of interest is the one reaching x = +, hence it is deduced from ψ 5. We get Transmission and Reection Coecients J k 5 m A 5 The reection coecient is the ratio of J R and. We get R = B 1 The transmission coecient is a little more subtle. We have to divide J T by. Notice that the k i 's do not cancel. The transmission coecient thus contains more than just the ratio of the integration constants! We have k 5 A 5 k 1 A 1 = E V 4 A 5 E A 1 Example: Step Potential Barrier Statement of the Problem incident on the potential A 1 Consider a particle with energy E (and momentum k) released from x = that is V (x) = { V 0 x > 0 0 x < 0 Find the transmission and reection coecients T and R for the case E > V 0. Classical Solution Since E > V 0 a classical particle will cross x = 0 without doubt. It will denitely reach x = +. Wave Function As the particle (or better - the probability current) hits the wall some of it is reected while some of it is transmitted. The wave function is thus ψ(x) = { Ae ikx + Be ikx x < 0 Ce iqx x > 0 where k = me, q = m (E V0 ) Transmission and Reection Coecients From the discussion above it is easy to nd R = J R = B A, J T = q C k A Hence, all we have to do is nd the integration constants. Finding the Integration Constants A, B and C In order to nd the constants we impose boundary conditions. Here they are simple. Both the wave function and its derivative must be continuous at x = 0. Hence we get ψ(0 ) = ψ(0 + ) A + B = C ψ = ψ ik (A B) = iqc x x 0 0 + We now have two equations and three unknowns. Therefore there is no way to nd all three constants. But we are not interested in all three. All we need is the two ratios B/A and C/A. Lets therefore divide both equations by A. This step is simple but crucial for succeeding. We obtain two equations with two unknowns. 1 + B A = C A ( ik 1 B ) = iq C A A 5
The Ratios B/A and C/A We plug C/A from the rs equation into the second and get ( k 1 B ) ( = q 1 + B ) B A A A (k + q) = k q B A = k q k + q For C/A we plug this result back into the rst equation. C A = 1 + k q k + q = k k + q Transmission and Reection Coecients Lets start with the reection coecient R. It is the ratio B/A squared. R = k q k + q Plugging in k and q we get The transmission coecient T is Obviously we got R + 1. R = ( ) E E V0 = V0 E 1 V0 E E + E V0 V0 E + 1 V0 E q k k k + q qk = 4 (k + q) = 4 1 V0 E V0 E + 1 V0 E Limiting Cases After nishing such computation it is always a good idea to check dierent limits. That way we can get a feeling for the physics going on and check our intuition. For E V 0 (so V 0 /E 0) we get 1 and R = 0. The particle will denitely cross the barrier. We see that in this limit our classical intuition is correct. For E = V 0 we get 0 and R = 1. In fact one gets 0 for all energies E < V 0. We did not show this here but you will do so at home. The two regimes E > V 0 and E < V 0 must match at E = V 0. Thus we can explain this phenomena. 3 Example: Delta Barrier Statement of the Problem For an incoming particle of mass m and energy E > 0 nd the transmission and reection coecient for the barrier V (x) = βδ (x) where β > 0. Schrodinger Equation We plug in the potential and write the down the Schrodinger eqaution d ψ (x) + βδ (x) ψ (x) = Eψ (x) m dx Wave Function Lets split up space and solve the time independent Schrodinger equation in each area. Here, there are two dierent regions: x < 0 and x > 0. Since δ (x 0) = 0 the Schrodinger equation in both regions is The wave function is thus ψ(x) = d ψ (x) = Eψ (x) m dx { Ae ikx + Be ikx x < 0 Ce ikx x > 0 6
Transmission and Reection Coecients The coecients are given by R = J R = B A, J T = C A Hence, all we have to do is nd the integration constants. But due to the delta function this is a little more involved than usual. Boundary Conditions The two ratios B/A and C/A are found by imposing two boundary conditions. The rst one is continuity of the wave function ψ(0 ) = ψ(0 + ). From this condition we get A + B = C 1 + B A = C A The second boundary condition is the discontinuity (i.e. not continuous) of the derivative of the wave function. Discontinuity of the Derivative Lets see the discontinuity. The Schrodinger equation is We integrate both sides over x from ɛ to ɛ. m ɛ ψ(x) + βδ(x)ψ(x) = Eψ(x) m x ɛ x ψ(x)dx }{{} ψ (ɛ) ψ ( ɛ) ɛ +β ɛ δ(x)ψ(x)dx } {{ } ψ(0) ɛ = E ψ(x)dx ɛ }{{} 0 In the limit ɛ 0 the rhs vanishes. This is because the integral of a continuous function is continuous and thus takes the same value for ±ɛ in the limit ɛ 0. We thus get ψ x ψ ɛ x = mβ ɛ ψ(0) Solve for the Ratios Imposing the boundary conditions we get A + B = C 1 + B A = C A ikc ika + ikb = mβ C ik C A ik + ik B A = mβ C A By plugging in C/A from the rst equation into the second we get ( ik 1 + B ) ik + ik B A A = mβ ( 1 + B ) B A A = mβ i k mβ Plugging the result back into the rst equation gives 1 + mv 0 i = C k mv 0 A C A = i k i k mβ Transmission and Reection Coecients It is now easy to compute C 4 k A = 4 k + m β R = B m β A = 4 k + m β We see that R + 1. A particle is either transmitted or reected. It cannot disappear. Limiting Cases for V 0 In the limit V 0 0 we get 1 and R = 0. When there is no potential there is no reection. In the limit V 0 we get 0 and R = 1. When the potential is innite nothing passes. 7
Limiting Cases for E we obtain We can write the solution in terms of the energy E. Using E = k m 4 k = me E mβ and E + mβ R = E + mβ In the limit E 0 we get 0 and R = 1. The particle does not have enough energy to pass though the potential. In the limit E we get 1 and R = 0. The particle ies across the potential without even noticing it exists. 8