= ( Prove the nonexistence of electron in the nucleus on the basis of uncertainty principle.

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1 Worked out examples (Quantum mechanics). A microscope, using photons, is employed to locate an electron in an atom within a distance of. Å. What is the uncertainty in the momentum of the electron located in this way? Also find the uncertainty in energy of electron in electron volt for this situation. Solution: Here the uncertainty in position( x) =. Å =. m, Uncertainty in momentum( p) =? We have Heisenberg s uncertainty relation x p ћ. We can take x p ћ Or, p = h x = = kg ms Now uncertainty in energy of the electron= p = (5.7 4 ).5 7 m 9. 3 =.6 9 = 95 ev. Prove the nonexistence of electron in the nucleus on the basis of uncertainty principle. Solution: We have uncertainty relation: x p ћ Or, p ћ, which implies that p must at least equal to x p = ћ = If electron lies inside the nucleus, then uncertainty in position must not greater x x than r. Hence x = r = 4, where r is the radius of the nucleus. p = = 5.7 kg. Kinectic energy of the elctron = p = (5.7 ).5 m 9. 3 =.6 3 = 95 MeV. Thus kinetic energy of the electron must at least equal to 95 MeV. But experiment shows that electrons emitted from β-decay is of the order 3 MeV. Hence it is concluded that electrons do not remain in the nucleus. 3. Find the normalization constant of the energy eigen function of the particle in a box of length L. Solution: See particle in a box. Answer: L 4. Find the commutation relation between position and momentum.

2 Solution: Since x x op and p x p op = iħ x. Let ψ be a well defined wave function, then [xp x p x x]ψ = x( iħ) ψ x ( iħ) x (xψ) = iħx ψ ψ x + iħx + iħψ x x x = iħψ xp x p x x = iħ Or, [x, p x ] = iħ. Similarly [y, p y ] = [z, p z ] = iħ. But [x, p y ] = [z, p y ] = =. It is noted that since [x, p y ] =, it results that x and p y have common eigen functions and can be measured simultaneously with accuracy. Since [x, p x ], it follows that both x and p x can not be measured both simultaneously with accuracy. 5. What are the eigen values of parity operator? Solution: Let ψ be any well behaved wave function and be the parity perator. Let α be the eigen value of the operator. We have eigen value equation: ψ(x) = α ψ(x) (i) Or, ψ(x) = {α ψ(x) } = α ψ(x) = α ψ(x) (ii) We have definition of parity operator : ψ(x) = ψ( x) Or, ψ(x) = { ψ(x)} = {ψ( x)} = ψ(x). (iii) From equation (ii) and (iii) we get:

3 α = α = ±. Hence eigen value of parity operator is + and. If the eigen function ψ(x) has parity +, then it is said to have even parity while for parity -, the eigen function is said to have odd parity. 6. If the state function is given by ψ(x, t) = e i(p x xħ iωt). Find the eigen value of momentum operator. Solution: Given ψ(x, t) = e i(p x xħ iωt). Operating both side by momentum operator p x(= iħ ) we get: x p xψ = iħ x x ei(p xħ iωt) = iħ ip x ħ = p x ψ ψ Comparing with the operator eigen value equation of a operator F op ψ = fψ, we get eigen value of the momentum operator p x is equal to p x. 7. Explain whether energy and momentum can be measured simultaneously. Solution: Consider ψ be the well behaved wavefunction, E x and P x are the energy and momentum operators of the particle, along x axis. Then we have E x = P x is the momentum itself. Now, [P x, E x ]ψ = [P x, P x m ] ψ = m [P x, P x ]ψ = m [P x, P x. P x ]ψ = m [P x (P x, P x ) + P x (P x, P x ) ]ψ = m, since eigen value of momentum operator Since the momentum commutes with energy, so they can be measured simultaneously. 8. If L = r p is the angular momentum operator, find the commutator [L x, L y ]. Solution: See operator for the solution. 9. Write the momentum and time independent energy operators in quantum mechanics. Explain with proper theory, whether these physical quantities can be measured simultaneously. Solution: we have momentum operator is, 3

4 p x = iħ x (in one dimension) and time independent energy operator is H = ħ m d dx + V(x). In the case of free particle, i.e. V(x) =. Hence H = ħ Now, we can prove that [H, p x] = [See operator] Hence these observables can be measured simultaneously with high accuracy.. Calculate the probability density corresponding to ψ(x) = eikx x. Solution: The probability current density J x = ħ im (ψ ψ x ψ ψ ) (i) x d. m dx where, ψ = e ikx x, ψ = x x eikx + x eikx (ik) and ψ = x x e ikx + x e ikx ( ik). Putting these values in equation (i) we get: J x = ħ {e ikx im x ( x eikx + x eikx (ik)) eikx x ( x e ikx + x e ikx ( ik))} = ħ im [( x 3 + ik x ) + ( x 3 + ik x )] = ħ ik im x Or, J x = ħk mx This is the expression of required probability current density for the given wave function.. The wave function ψ(r, θ, φ) of the hydrogen atom for s state is ψ = ( is normalized. Solution: We have given the wave function : ψ = ( The normalized wave function satisfy the condition ψ dv =. (i) In spherical polar coordinates, the volume element: a 3) e r a a 3) e r a. Show that the wave function dv = r sin θ dr dθ dϕ 4

5 Hence equation (i) becomes a3 e r a r dr dφ sin θ dθ = () To solve this equation, put r a = x. Then dx = dr a. e r ( a Then a r dr = e x )3 x dx = ( a )3!, where we have used e x x n dx = n!. Hence left hand side of equation () becomes: a 3 ( a )3! [φ] [ cos θ] = 8 =!. Hence normalizing condition i.e. equation (), is satisfied by the given wave function ψ. Hence ψ is the normalized wave function.. Calculate the average value of /r for an electron in the s state of the hydrogen atom. Solution: The expectation value of /r is given by = r () r ψ dv () The wave function for a s state electron is ψ,, = (a 3 ) Or, ψ = e r a a3 e r a Hence equation () becomes : r = a3 = r and also volume element in spherical coordinates is given by : dv = r sin θ dr dθ dϕ ( ) r e a r dr sin θ dθ dϕ r a3 e a r dr sin θ dθ dϕ = a 3 ( a )! ()() Or, r = a 5

6 3. A particle on a straight line is described by ψ(x) = +ix +ix. Normalize it and explain where this particle is most likely found. Solution: We have given wave function ψ(x) = +ix and +ix ψ (x) = ix ix. Let M be the normalization constant. Then using normalization condition M ψ (x)ψ(x)dx = Or, M Or, M Or, M +x + x Take x x = t +x 4 dx = x + x (x x ) + +x dx = dx = ( + ) dx = dt. Hence x dt M dx =. The integration gives : (t) +( ) M [tan (t/ )] = Or, M ( + ) This gives M = ( ) Then the normalized wave function will be ) +ix. ψ(x) = ( +ix The maximum probability of finding particle at distance x for which dp =, where P is the probability density equal to ψ ψ. Hence d dx ( ψ ψ) = dx Or, d dx (+x +x 4) =. Or, (+x4 )x (+x )4x 3 (+x 4 ) = Using d dx (u v ) = u v uv Or, x =. Hence the particle will have maximum probability of finding at x =. v 4. Find the probability current density of the wave function ψ = e ikx + Ae ikx in the region x < where the potential V=. Hint : Use J x = ħ ψ ψ (ψ ψ ) where ψ = im x x eikx + Ae ikx and ψ = e ikx + Ae +ikx. The final result is obtained as J x = ħk m [ A ] 5. Compare the reflection coefficient with that of transmission coefficient in the case of a free electron moving towards a potential step of height Kev if the energy of the particle is.5 kev. Solution: Energy of the electron E=.5 Kev and that of potential height V = Kev. Hence E > V. In this case, 6

7 Reflection coefficient R = (n n ) m(e V ). Now R T = (n n ) (n +n ) and transmission coefficient T = 4n n (n +n ) where n = me and n = = ( E E V ) 4n n 4 E E V = (.5.5 ) = Normalize the harmonic oscillator wave function given by ψ n = N n H n e μ Solution: ψ n (μ) = N n H n (μ)e μ..() where N n is called a normalization constant. To obtain the value of the constant, we start from the generating function e S +Sμ = H n (μ)s n n= n! Similarly for the index t e t +tμ = H m (μ)t n m= m! Multiplying both the equations (e S +Sμ ). (e t +tμ ) = H n (μ)s n n= H m (μ)t n n! m= m! Multiply both side by e μ and then integrate with respect to μ from to + we get: () Considering only left side (e S +Sμ ). (e t +tμ ) e μ dμ = exp( s t + sμ + tμ μ )dμ + = exp { (s + t sμ tμ + μ )} dμ + = exp [ {(μ s) + t tμ + st st}] dμ = e st exp{ (μ s t) } d(μ s t) = e st. 7

8 Or, (e S +Sμ ). (e t +tμ ) e μ dμ = { + st + (st) + + (st)n + }!! n! = (st) n n= n! Now equation () becomes (3) If equal powers of st are equated in this series: n n! = n! H n! n (μ)e μ dμ, If n = m. H n (μ)e μ dμ = n! n...(#) But, H n (μ)h m (μ)e μ dμ =, if n m. By normalizing condition, ψ n (μ) dμ = N n H n (μ)e μ dμ = N n n! n = N n = ( ) n n! 7. Find the energy expectation value of particle described by the wave function Aexp (iax bt) moving between L and L. Solution: We have wave function ψ = Aexp (iax bt) and ψ = Aexp ( iax bt). The normalization condition is L ψ ψ dx = L L Or, Aexp (iax bt)aexp ( iax bt) dx = L L Or, A e bt dx = L Or, A e bt (L L ) = Or, A = ebt (L L ) Hence wave function ψ = e bt (L L ) exp (iax bt) The energy expectation value of the particle is given by L E x = ψ (iħ ) ψ dx t L 8

9 L = Aexp ( iax bt) (iħ ) t L L L L L Aexp (iax bt) dx = Aexp ( iax bt) Aexp (iax bt)( b) dx = A exp ( bt) ( iħb)dx exp ( bt)( iħb) L dx [ Putting the value of A ] (L L ) L = ebt = iħb (L L ) = iħb (L L ) 8. What is the significance of commutation relation in quantum mechanics? Hint: see operator. 9. Explain the concept of probability density, normalization and expectation value in quantum mechanics.. Show that parity operator commutes with Hamiltonian of a system if the potential V(x) = V( x) Solution: Let H be the Hamiltonian operator and be the parity operator. Then we can write H = ћ m d dx + V(x) Then H ψ(x) = ћ [H ψ(x)] = ћ Similarly, d m dx d m dx d = ћ = ћ H [ ψ(x)] = [ ћ = ћ m m dx d m dx d ψ(x) + V(x)ψ(x) ψ(x) + {V(x)ψ(x)} m dx d ψ( x) ψ( x) + V( x)ψ( x) ψ( x) + V(x)ψ( x).. (i) [ Given V(x) = V( x)] + V(x)] ψ( x) dx From equation (i) and (ii), we get: [H ψ(x)] H [ ψ(x)] = + V(x)ψ( x) (ii) Or, [ H H ] = [H, ] =. Hence the parity operator commutes with Hamiltonian of a system in the given condition V(x) = V( x). An electron of mass 9. 3 kg is constrained to move in a -D potential well of width.5 Angstrom. Compare the energy values of electron in the ground state to second excited states if the walls of the well are impenetrable. Solution: Given mass of an electron m e =9. 3 kg, width of the well L=.5 Å =.5 m. Here energy value of the electron inside the well is E n = h n For the ground state, n =, for the second excited, n = 3. Then the ratio of energy is now E E =. Estimate the de-broglie wavelength of an electron of kinetic energy of kilo electron volt. 8mL h n 8mL h n 3 8mL = n n = 3 9 Hint: Use λ = h p, where h is Planck s constant = joule; momentum p = me with energy E = KeV = 3 ev = joule. The answer is.3867 Å. 3. Under what circumstances is an atomic electron s probability density distribution spherically symmetric? Why? Solution: According to quantum mechanics, the wave function in spherical coordinate system is given by ψ(r, θ, φ) = ψ(r, Θ, φ) = R(r)Θ(Θ)Φ(φ). Then the probability density P(r, θ, φ) = ψψ = R(r)Θ(Θ)Φ(φ)R (r)θ (θ)φ (φ) 9

10 = R(r) Θ(θ) Φ(φ). (i) We have solution of φ equation in hydrogen atom is Φ = e±imφ ( see hydrogen atom) Hence Φ Φ = = constant. Hence equation (i) becomes P(r, θ, φ) = R(r) Θ(θ) Hence the probability density P depends only on the radial probability ( R(r) ) and polar probability ( Θ(θ) ) but not on the azimuthal coordinate ϕ. This results electron distribution is symmetric about z-axis. These numericals are only for practice you should also practice other numricals problems [ ] from Murugeshan and other books of quantum mechanis