1. Thegroundstatewavefunctionforahydrogenatomisψ 0 (r) =


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1 CHEM 5: Examples for chapter Thegroundstatewavefunctionforahydrogenatomisψ (r = 1 e r/a. πa (a What is the probability for finding the electron within radius of a from the nucleus? (b Two excited states of hydrogen atom are given by the following wavefunctions: Solution: ψ 1 (r = A(+λre r a and ψ (r = Brsin(θcos(φe r a Proceed in the following order: 1 obtain λ from the orthogonality requirement between ψ and ψ 1, use the normalization requirement sepratately for ψ 1 and ψ to get constants A and B, respectively. (a The probability P for finding the electron within a (a ball with radius a ; denoted by V below is: P = = 4 a V ψ ψ dτ = 1 πa a 5a e 4 a e r/a 4πr dr }{{} =dτ = 1 5e. Note that the 4π in the volume element above comes from the fact that the function in the integral does not depend on the angles θ and φ and therefore the angular part can be integrated independently to yield 4π (i.e. the original volume element dτ = r sin(θdθdφdr is then effectively just 4πr dr. Above we have used the following result from a tablebook: x e ax = aax a (x xa + a 1
2 This was applied in the following form: a e r/a r dr = e r/a /a = a e 5a e 4 ( r r + /a ( /a ( a +a + a + a 4 + a 4 = a 5a e 4 (b First recall the wavefunctions: ψ (r,θ,φ = 1 πa e r/a 1 /a ( /a ψ 1 (r,θ,φ = A(+λre r/(a,ψ (r,θ,φ = Bsin(θcos(φre r/(a First we calculate λ from the orthogonality: ψ 1ψ dτ = 4π ψ 1(rψ (rr dr = 4πA πa (+λre r/(a r dr = (+λre r/(a r dr = e r/(a r dr+λ 16a λ 81 a4 = e r/(a r dr = a +λa 4 = λ = 1 a Then determine A from the normalization condition:
3 π π ψ 1 (r,θ,φ r sin(θdrdθdφ r=θ= φ= = 4πA ( r a e r/a r dr r= = 4πA 4 r= r e r/a dr 4 a r= r e r/a dr+ 1 a r= r 4 e r/a dr = 4πA (4! (1/a 4! a (1/a + 1 4! 4 a (1/a 5 = 4πA ( 8a 4a +4a = πa a = A = = πa 4 πa B can also be determined from normalization: π π ψ (r,θ,φ r sin(θdrdθdφ r= θ=φ= π = B r 4 e r/a dr sin (θdθ π cos (φdφ r= ( 4 B (1/a 5 = 1 4 πa 5 θ= φ= 4 π = B πa 5 = 1 B = Above we have used the following integrals (tablebook: 1 πa 5
4 x n e ax = n! a n+1 sin (ax = cos(ax 1a cos (ax = x + sin(ax 4a cos(ax 4a. (a Which of the following functions are eigenfunctions of d/ and d / : exp(ikx, cos(kx, k, and exp(ax. (b Show that the function f(x,y,z = cos(axcos(bycos(cz is an eigenfunction of the aplacian operator ( and calculate the corresponding eigenvalue. (c Calculate the standard deviation r for the ground state of hydrogen atom ψ (r. (d Calculatetheexpectationvalueforpotentialenergy(V(r = e 4πǫ r in hydrogen atom ground state ψ (r. Express the result finally in the units of ev. Solution: (a d(eikx = ike ikx = constant original function. Thus this is an eigenfunction of the given operator. = d (ikeikx = k e ikx = constant original function. Thus this is an eigenfunction of the given operator. d (e ikx d(k = d (k =. This could be considered as en eigenfunction with zero eigenvalue. d(eax = axe ax constant original function, thus not an eigenfunction. d (e ax = d(axeax = a(ax +1e ax constant original function, thus not an eigenfunction. = k sin(kx. Not an eigenfunction. d(cos(kx 4
5 d (cos(kx = d( ksin(kx = k cos(kx. This is an eigenfunction (with eigenvalue k. (b f(x, y, z = cos(ax cos(by cos(cz. The partial derivatives are: and f(x,y,z x f(x,y,z y f(x,y,z z = a sin(ax cos(by cos(cz = b cos(ax sin(by cos(cz = c cos(ax cos(by sin(cz f(x,y,z = a cos(axcos(bycos(cz x f(x,y,z = b cos(axcos(bycos(cz y f(x,y,z = c cos(axcos(bycos(cz z Therefore f(x,y,z = (a + b + c f(x,y,z = constant original function and this is an eigenfunction with the corresponding eigenvalue (a +b +c. (c The standard deviation can be calculated as: ψ (r = 1 πa e r/a ˆr = 1 πa ˆr = 1 πa ˆr = 9a 4 ˆr ˆr = a 9a ˆr ˆr = e r/a r } 4πr {{ dr} = 4 a5 a dτ 4 = a e r/a r4πr }{{ dr} = a dτ 4 = 4a 4 a.87a 5
6 (d The potential energy expectation can be calculated as: ψ (r = 1 e r/a and V(r = e πa 4πǫ r ˆV = e 4π ǫ a r= e r/a 1 r } 4πr {{ dr} = e πǫ a dτ = e a πǫ a 4 = e 7. ev 4πǫ a r= e r/a rdr. Consider function ψ(x = ( π 1 4 e x /. Using (only this function, show that: (a Operators ˆx and ˆp x do not commute. (b Operators ˆx and the inversion operator Î commute (Îx = x. Note that consideration of just one function does not prove a given property in general. Solution: (a ( π ψ(x = 1/4e x / ˆp x = i h d and [ˆx,ˆp x]ψ(x = (ˆxˆp x ˆp xˆxψ(x To obtain the commutator, we need to operate with ˆx and ˆp x : ( π 1/4x d π 1/4x (ˆxˆp x ψ(x = i h / e x = i h( e x / ( π 1/4 d ( (ˆp xˆxψ(x = i h xe x / ( π 1/4x ( π 1/4e = i h e x / x i h / When these are subtracted, we get: 6
7 ( π 1/4e x (ˆxˆp x ψ(x (ˆp xˆxψ(x = i h / When the wavefunction ψ(x is removed, we have: [ˆx,ˆp x ] = i h Because the commutator between these operators is nonzero, it means that they are complementary. (b The commutator for the given function is: [ ] ˆx,Î ψ(x = ( ( π 1/4e ˆx Î Îˆx x / ( π 1/4e ( = x ( x π / ( x ( π 1/4e ( = x x π 1/4e / x x / = ˆx and Î commute for the given function. 1/4e ( x / 4. Aparticleisdescribedbythefollowingwavefunction: ψ(x = cos(χφ k (x + sin(χφ k (x where χ is a parameter (constant and φ k and φ k are orthonormalized eigenfunctions of the momentum operator with the eigenvalues + hk and hk, respectively. (a What is the probability that a measurement gives + hk as the momentum of the particle? (b What is the probability that a measurement gives hk as the momentum of the particle? (c What wavefunction would correspond to.9 probability for a momentum of + hk? (d Consider another system, for which ψ =.9ψ 1 +.4ψ + c ψ. Calculate c when ψ 1, ψ and ψ are orthonormal. Use the normalization condition for ψ. Solution: 7
8 (a The probability for measuring + hk is cos (χ. We have a superposition of the eigenfunctions of momentum and therefore the squares of the coefficients for each eigenfunction give the corresponding probability. (b The probability for measuring hk is sin (χ. (c Given cos (χ =.9 (hence cos(χ = ±.95 we use the normalization condition cos (χ + sin (χ = 1 where we solve for sin(χ = ±.. The overall sign for the wavefunction does not matter and therefore we have two possibilities for our wavefunction: ψ =.95e ikx ±.e ikx. (d Normalizationcondition: 1 = (.9 +(.4 +c. Thusc = ± Consider a particle in the following onedimensional infinitely deep box potential: {, when x a V(x =, when x > a Note that the position of the potential was chosen differently than in the lectures. The following two wavefunctions are eigenfunctions of the Hamiltonian corresponding to this potential: ψ 1 (x = 1 a cos ( πx a and ψ (x = 1 a sin ( πx a with the associated eigenvalues are E 1 = 1 ev and E = 4 ev. Define a superposition state ψ as ψ = 1 (ψ 1 +ψ. (a What is the average energy of the above superposition state (e.g. < Ĥ >? (b Plot ψ 1 and ψ and determine the most probable positions for a particle in these states. (c What are the most probable positions for the particle given by wavefunction ψ (x = sin( πx where the box potential is now located between [, ]. Solution: 8
9 (a We calculate the expectation value (Ĥψ 1 = E 1 ψ 1 and Ĥψ = E ψ : Ĥ = ψ (xĥψ(x = 1 (ψ 1(x+ψ (xĥ(ψ 1(x+ψ (x = 1 (ψ 1(x+ψ(x(E 1 ψ 1 (x+e ψ (x = 1 (E 1 +E =.5 ev (b For a = 1 both ψ 1 (x (one maximum and ψ (x (maximum and minimum are shown below: The most probable values for position can be obtained from the squared wavefunctions: ψ 1 hasthemaximumvalueatr = whereasψ hastwomaximaat ±.5. Note that on average both will give an outcome of ˆr =. (c The most probable positions are given by the square of the wavefunction ψ (x = (/ 1/ sin(πx/. The probability function 9
10 is then ψ (x sin (πx/. The extremum points for this function can be obtained by: d ( sin (πx/ = sin(πx/cos(πx/ = πx x = n ( πx = nπ or = n+ 1 π n+1/ or x = Second derivatives can be used to identify the extrema: d ( sin (πx/ cos ( πx ( πx sin At x = n the values are positive which means that these correspond to (local minima. For x = n+1/ the values are nega tive and these points correspond to (local maxima. For example, when = the probability function looks like: 6. Calculate the uncertainty product p x for the following wavefunctions: (a ψ n (x = ( 1/sin a a with x (particle in a box (b ψ n (x = N v H v ( ( xexp Solution: x (harmonic oscillator 1
11 (a For momentum: ( 1/ ψ n (x = sin, where x ( 1/ ( ˆp x = sin i h d ( 1/ sin }{{}}{{}}{{} ψn (x ˆp x ψ n(x = i h sin d ( sin = πni h ˆp x = = h = n π h sin cos } {{ } = ( 1/ ( sin i h d } {{ } ψn sin Now we can calculate p = = }{{} ˆp x d sin sin sin } {{ } = ( 1/ sin }{{} ψ n = n π h ˆp x ˆp x = nπ h. For position 11
12 we have: ˆx = = ˆx = ( 1/ sin x xsin = / ( 1/ sin = (4n π 6 4π n ( 1/ sin ( 1/ x sin = x sin = (π n 6π n }{{} = ( 1 π n Now x = ˆx ˆx ( = 1 π n = Since we have both x and p, we can evaluate the uncertainty product: n π p x = h π n. Thesmallestvalueisobtainedwithn = 1: p x.568 h > h. (b First recall that: ( ψ n (x = N v H v x e x /, where N v = 1 ( 1/4 v v! π Also the following relations for Hermite polynomials are used (lecture notes & tablebook: H v+1 = yh v vh v 1 H v (yh v (ye y dy = {, if v v π v v!, if v = v 1
13 Denote y = x and hence dy =. Now ˆx is given by: ˆx = N v = N v = N v H v ( x e x / xh v ( x e x / H v (ye y / yh v (ye y / dy H v (yy H }{{} v (ye y dy = = 1 H v+1(y+vh v 1 (y The last two steps involved using the recursion relation for Hermite polynomials as well as their orthogonality property. Next we calculate ˆx : ˆx = N v / = N v / = π N v / (yh v (y e y dy ( 1 H v+1(y+vh v 1 (y e y dy ( v+1 (v +1! +v v 1 (v 1! 4 = 1 ((v +1+v = 1 ( v + 1 = h ( v + 1 µk (v Combiningtheabovecalculationsgives x = + 1 h µk. Next we calculate p: ˆp = i h d and dy =. ˆp = = i hn v N v H v (ye y /ˆp ( N v H v (ye y / dy H v (ye y / d ( H v (ye y / dy 1
14 Above differentiation of the eigenfunction changes parity and therefore the overall parity of the integrand is odd. Integral of odd function is zero and thus ˆp =. For ˆp we have: ˆp = N v H v (ye y /ˆp N v H v (ye y / dy Theoperatormustalsobetransformedfromxtoy: ˆp = ( i hd/ = ( i h d/. The above becomes now: ( N v H v (ye y / ( h d N dy v H v (ye y / dy = h N v = h N v = h N v ( H v (ye y / d H dy v (ye y / dy = h Nv ( v 1 H v (y (y 1H v (y yh v(y+h v(y e y dy }{{} = vh v(y H v (y [ (y 1H v (y vh v (y ] e y dy Hv(ye y dy+ }{{} = π v v! y Hv(y e }{{} =( 1 H v+1(y+vh v 1 (y y dy = h [ Nv ( v 1 π v v!+ 4 1 π v+1 (v +1!+v ] π v 1 (v 1! = h [ Nv π v v! v +1 v +1 v ] [ = h v + 1 ] = h [ mk v + 1 ] 14
15 Therefore we have p = x p = (v + 1 h mk [ v + 1 ]. Overall we then have h mk h mk [ v + ] 1 = h(v + 1 h. 7. Scanning tunneling microscopy (STM is a technique for visualizing samples at atomic resolution. It is based on tunneling of electron through the vacuum space between the STM tip and the sample. The tunneling current (I ψ where ψ is the electron wavefunction is very sensitive to the distance between the tip and the sample. Assume that the wavefunction for electron tunneling through the vacuum is given by ψ(x = Bexp Kx with K = m e (V E/ h and V E =. ev. What would be the relative change in the tunneling current I when the STM tip is moved from x 1 =.5 nm to x =.6 nm from the sample (e.g. I 1 /I =?. Solution: The current is proportional to the square the wavefunction ( ψ(x : I ψ(x = B e Kx I 1 = e K(x 1 x I K = m e (V E / h }{{} = ev ( ( kg(. ev( J/eV = ( Js K(x 1 x = K ( m 1.45 I 1 I 4. 1/ 1/ 8. Show that the sperhical harmonic functions a Y,, b Y, 1 and c Y, are eigenfunctions of the (threedimensional rigid rotor Hamiltonian. What are the rotation energies and angular momenta in each case? 15
16 Solution: Ĥψ(r,θ,φ = Eψ(r,θ,φ where Ĥ = h Λ = 1 sin (θ φ + 1 sin(θ θ sin(θ θ I Λ Since h is constant, it is sufficient to show that spherical harmonics I are eigenfunctions of Λ. We operate on the given spherical harmonics by Λ. (a Λ Y (θ,φ = Λ 1 = (no angular dependency. The rotation π energy is E = h =. Also I = h = (since ˆ = h Λ. (b Λ Y 1 (θ,φ =...differentiation... = 6Y 1 energy is E = h I (c Expression for Y (θ, φ. The rotation h ( 6 =. Also I = 6 h = 6 h. was not given in the lecture notes. We use Maxima to obtain the expression: Y (θ,φ = 5 7e φi sin (θ 8. Then we 5π needtoevaluate: Λ Y (θ,φ =...differentiation... = 1Y (θ,φ. From this we can get the rotational energy as E = 6 h. Also I = 1 h = 1 h. 9. Calculate the zcomponent of angular momentum and the rotational kinetic energy in planar (e.g. dimensional; rotation in xyplane rotation for the following wavefunctions (φ is the rotation angle with values between [,π[: (a ψ = e iφ (b ψ = e iφ (c ψ = cos(φ (d ψ = cos(χe iφ +sin(χe iφ Solution: 16
17 The Hamiltonian for D rotation is Ĥ = ˆ z = h d where the I I dφ rotation axis is denoted by z and φ is the angle of rotation. (a Operatefirstby ˆ z on e iφ : ˆ z e iφ = i h d dφ eiφ = he iφ. ThusĤeiφ = h I eiφ. (b The same reasoning gives: ˆz e iφ = i h d dφ e iφ = he iφ and Ĥe iφ = 4 h I e iφ. (c The given wavefunction is not an eigenfunction of ˆ z : ˆ z cos(φ = i h( sin(φ (expectation value is zero. However, it is an eigenfunction of Ĥ: Ĥcos(φ = h cos(φ = h cos(φ. I dφ d I (d Operation by ˆ z would change the plus sign in the middle of the wavefunction into a minus, hence this is not an eigenfunction of ˆ z. It is an eigenfunction of Ĥ: Ĥ(cos(χe iφ + sin(χe iφ = h I ( (i cos(χe iφ +( i sin(χe iφ = h I ( cos(χe iφ +sin(χe iφ. 17
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