Quantum Mechanical Tunneling

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1 Chemistry 460 all 07 Dr Jean M Standard September 8, 07 Quantum Mechanical Tunneling Definition of Tunneling Tunneling is defined to be penetration of the wavefunction into a classically forbidden region Since the wavefunction is non-zero in the classically forbidden region, the probability density ψ *ψ also is non-zero; thus, there is a finite probability that the particle will exist in a region that it could never be in if it were obeying classical mechanics In this situation, we say that the particle has tunneled into the classically forbidden region [ region is said to be classically forbidden if the particle's energy is less than the potential energy in the region, E < V ] Example : Half-Infinite Well We have previously considered a particle in a half-infinite well, shown in igure or the case E < V 0, the wavefunction of the system extends into region II Since the energy is less than the potential in this region, the system exhibits tunneling III I II V=V 0 x=0 x=l igure Potential energy for a particle in a half-infinite well in one dimension or the specific case found in Project, with L=50 bohr, and V 0 = 45 hartrees, the first two wavefunctions are shown in igure The dashed line at x=50 bohr or x=l) indicates the position of the finite potential well, and thus any wavefunction extent to the right of the dashed line is an illustration of tunneling ψx) 03 ψx) x bohr) x bohr) a) igure a) Ground state E =07 hartrees) and b) first excited state E =070 hartrees) wavefunctions for the particle in a half-infinite well b)

2 or energies closer to the top of the well, the tunneling proportion increases Shown in igure 3 are the wavefunctions corresponding to the fourth and fifth energy levels for this system ψx) ψx) x bohr) -08 x bohr) a) igure 3 a) Third excited state E 4 =75 hartrees) and b) fourth excited state E 5 =46 hartrees) wavefunctions for the particle in a half-infinite well Notice that for the fourth excited state, with an energy very close to the top of the barrier E 5 =46 hartrees compared to a barrier of 45 hartrees), the tunneling is substantial or this system, we can determine the tunneling probability by integrating the probability density from x=l to x= or the st energy level the tunneling probability is 0005, while for the 5 th energy level the tunneling probability is 08 b) Example : Tunneling Through a Barrier of inite Width nother simple example of tunneling is the case of a barrier of finite width, illustrated in igure 4 V=V 0 I II III x=0 x=l igure 4 Potential energy for a particle interacting with a finite barrier in one dimension The potential energy of the system may be described by the following equation, V x) = # 0, % $ V 0, % & 0, x < 0 0 x L x > L ' % % ) The method of solution is similar to what we did for the particle in a half-infinite well However, there are two degenerate solutions, one in which the particle is initially traveling to the right and one in which the particle is initially traveling to the left Only one of these degenerate cases needs to be considered Here, we will assume that the particle is traveling to the right, so it is initially incident on the barrier from region I

3 3 or the case E < V 0, the general solutions may be given as ψ I x) = e ik x + Be ik x ψ II x) = C e kx + De k x ψ III x) = e ikx The wave vectors k and k are defined as k = me! and k = m V 0 E)! In region I, and in general for unbounded regions where the potential energy is zero, we use exponentials with imaginary exponents as the preferred form of the wavefunction The two parts of the wavefunction written in this form can be related to waves traveling in the positive and negative x directions, respectively Notice that in region II, we have used exponentials with real rather than imaginary exponents This is the preferred form in regions where the total energy E is less than the potential energy V 0 lso notice that in region III, there is no wave traveling to the left This is because we started with a wave traveling to the right, and in region III, no wave traveling to the left can form because there is nothing for the wave to reflect from Matching the wavefunctions and their first derivatives at the boundaries x=0 and x=l yields conditions among the arbitrary constants, B, C, D, and Of particular interest is a quantity called the transmission coefficient T The transmission coefficient T is related to the ratio of the probability density current that is transmitted through the barrier to the incident probability density current The probability density current S is defined as S = v ψ *ψ, where v is the particle velocity Thus, the transmission coefficient T is T = S tr S in, where S tr is the transmitted probability density current and S in is the incident probability density current or the specific case here, the transmitted wave in region III has the form ei k x Since this form has a momentum eigenvalue given by! k, the velocity is v tr = p tr m =! k m Similarly, the incident wave in region I has the form ei k x Since this form has a momentum eigenvalue given by! k, the velocity is v in = p in m =! k m

4 4 Substituting, we can calculate the transmission coefficient T, T = S tr S in = v trψ * tr ψ tr v in ψ * in ψ in = T =!k m * e ik x e ik x!k m * e ik x e ik x Using the relations obtained from matching the wavefunctions and first derivatives at the boundaries, the ratio of / may be determined or this system, the transmission coefficient is T =, or T = + sinh k L) 4E # E & % V 0 $ V 0 ' In this equation, the function sinh is the hyperbolic sine function It is defined sinhbx = ebx e bx or values of the energy less than V 0, the transmission coefficient only gets large for energies close to the top of the barrier, as illustrated in igure 5 T E/V 0 igure 5 Transmission coefficient for a particle tunneling through a finite barrier in one dimension

5 5 Example 3: Double Well Potential common model that is used to represent inversion in molecules like ammonia is the double well potential shown in igure 6 V=V 0 L x=0 x=a x=c x=d igure 6 Double well potential This potential mimics the inversion of ammonia through its planar form, illustrated in igure 7 a) b) igure 7 a) Inversion motion of ammonia through the planar transition state b) Potential energy of inversion of ammonia as a function of the out-of-plane angle Classically, the ammonia molecule in its ground state does not have enough energy to go over the barrier Therefore, a classical ammonia molecule would not experience inversion In quantum mechanics, though, the ammonia molecule experiences tunneling through the barrier and so undergoes inversion

6 6 ppendix : ull Solution of inite Barrier Problem for E < V 0 or the case E < V 0, the general solutions may be given as The wave vectors k and k are defined as ψ I x) = e ik x + Be ik x ψ II x) = C e kx + De k x ψ III x) = e ikx k = me! and k = m V 0 E)! The boundary conditions can be applied by matching the wavefunctions and their first derivatives at x=0 and x=l This leads to a set of four equations, ) = ψ II 0) ) = ψ # II 0) ψ I 0 ψ # I 0 ψ II L ψ # II L ) = ψ III L) ) = ψ # III L Upon substitution of the forms of the wavefunctions in each region into the matching equations, we get ) + B = C + D ) ) ik B) = k C D) 3) C e k L + D e k L 4) k C e k L D e k L = e ik L ) = ik e ik L Note that there are four matching equations, but there are five arbitrary constants,, B, C, D, and In order to get the transmission coefficient T, we will need the ratio / Recall that the transmission coefficient T for this system is defined T = So we need to solve the matching equations to get, B, C, and D in terms of The easiest way to do this is to take linear combinations of pairs of the equations Taking the sum and difference of equations ) and ) yields 5) = )+ ) = 6) = ) ) B = i k C + + i k D k k + i k C + i k D k k

7 7 or convenience in manipulating the equations, define α = k k Then, equations 5) and 6) simplify to 5) = 6) B = iα ) C + + iα ) D + iα ) C + iα ) D Taking the sum and difference of equations 3) and 4) yields 7) = 3)+ 4) C = 8) = 3) 4) D = + i k e k L e ik L k i k e kl e ikl k or convenience in manipulating the equations, define β = k k Then, equations 7) and 8) simplify to 7) C = 8) D = + iβ ) e kl e ik L iβ ) e k L e ik L Equations 7) and 8) express the coefficients C and D in terms of Since equations 5) and 6) give and B in terms of C and D, substitution of equations 7) and 8) into them will give and B in terms of as well Since we are primarily interested in the relation between and for the transmission coefficient, we will focus on substitution of equations 7) and 8) into equation 5) 5) = = = iα ) C + +iα 4 iα ) D ) + iβ) e kl e ik L 4 eik L iα + iβ + α β iα ) iβ) e kl e ik L ) e k L + + iα iβ + α β) e k L rom the definitions of α and β, it can be shown that α β = Substituting, = 4 eik L iα + iβ) e k L + + iα iβ) e k L Rearranging, = ) iα e k L e k L ) + iβ e k L e k L ) 4 eik L e k L + e k L = e ik L cosh k L ) + i α β ) sinh k L)

8 8 Here, the definitions of the hyperbolic functions sinh and cosh have been used to rewrite the exponentials, cosh bx = sinh bx = ebx + e bx ) ebx e bx ) Though we want the ratio /, it is actually easier to first solve for the ratio /, = eik L cosh k L) + i α β ) sinh k L) To get /, we use the relation = * " % " % $ ' $ ' # & # & The complex conjugate is * = e ik L cosh k L) i α β ) sinh k L) Combining, = e ik L cosh k L = cosh k L ) ) i α β ) sinh k L) i α β ) sinh k L) eik L cosh k L cosh k L) + i α β ) sinh k L) ) + i α β ) sinh k L) Note that a + i b) a i b) = a + b Using this, the ratio can be simplified = cosh k L ) = cosh k L) + i α β ) sinh k L ) 4 α β ) sinh k L ) cosh k L ) + i α β ) sinh k L) The ratio may be written in terms of either sinh or cosh by using the identity cosh a sinh a = Substituting and combining, = + sinh k L) + = + = + 4 α β ) sinh k L ) α β) sinh k L ) 4 α + β + ) sinh k L)

9 9 Here, the relation α β = has again been employed to simplify the expression Using the definition of α and β, we can show that α + β + = V 0 E V 0 E ) = E E V 0 V 0 Substituting into the expression for / yields = + 4 α + β + ) sinh k L) ) = + sinh k L 4E E V 0 V 0 inally, inverting the equation to get T = / gives T = = + sinh k L 4E E V 0 V 0

10 0 ppendix : Hyperbolic unctions Definitions Hyperbolic sine sinh), cosine cosh), tangent tanh) and other related functions are defined in terms of exponential functions The hyperbolic sine function sinh) is defined sinhbx = ebx e bx The hyperbolic cosine function cosh) is defined in a similar fashion, coshbx = ebx + e bx inally, the hyperbolic tangent function tanh) is defined as the ratio of sinh and cosh, tanhbx = sinhbx coshbx = ebx e bx e bx + e bx Useful Relations The hyperbolic functions have many properties that are analogous to the ordinary trigonometric functions or example, they satisfy the relation cosh bx sinh bx = The hyperbolic sine function sinh) is an odd function, while the hyperbolic cosine function is even, sinh bx) = sinh bx) cosh bx) = cosh bx) Derivatives The derivatives of the hyperbolic functions are similar to those of the ordinary trigonometric functions, d sinh x = cosh x dx d cosh x = sinh x dx d dx tanh x = tanh x

11 Graphical Behavior Graphs of the hyperbolic functions sinh x ), cosh x ), and tanh x ) are shown in the figures below

Sample Quantum Chemistry Exam 1 Solutions

Chemistry 46 Fall 217 Dr Jean M Standard September 27, 217 Name SAMPE EXAM Sample Quantum Chemistry Exam 1 Solutions 1 (24 points Answer the following questions by selecting the correct answer from the