Tutorial Questions on Special Relativity

TUTORIAL 5 Introductory Quantum Mechanics

1. Which of the following wave functions cannot be solutions of Schrodinger's equation for all values of x? 2 2 Why not? (a)  = A sec x; (b)  = A tan x; (c)  = A ex ; (d)  = A e-x (Beiser, Ex. 3, pg. 197)

Solution

Only (d) is could be a solution as it contains no divergence. It also 0 when x .

2. The wave function of a certain particle is  =A cos2x for /2

Solution 3. The expectation value 狁x of a particle trapped in a box a wide is a/2 (0xa), which means that 2 its average position is the middle of the box. Find the expectation value 狁x n in the stationary 2 state n. What is the behaviour of 狁x n as n becomes infinity. Is this consistent with classical physics? (Beiser, Ex. 17, pg. 198) Solution 2 np x The solution of the ground state wave function for a particle in an infinite box isy (x )= sin . n a a 2anp x 2 a2 np x2=y x 2 y dx = x 2 sin 2 dx = y 2 sin 2 ydy , where we have used the substitution n 蝌n n 3 3 - a0 a n p 0 np x y = . Using integration by parts, a np np y3 ycos 2 y (1- 2 y 2 )sin 2 y n 3p 3 n p n 3 p 3 n p y2sin 2 ydy= - + = - cos( 2p n) = - 0 6 4 80 6 4 6 4

2 2 2 骣1 1 \x =y x y dx = a 琪 - 2 2 . - 桫3 2p n 1 As ninfinity, \ x2 a 2 , which is the expectation value of \ x2 in the classical limit, for which 3 the probability distribution is independent of position in the box, and n-independent.

4. Find the probability that a particle in a box L wide can be found between x = 0 and x = L/n when it is in the nth state. (Beiser, Ex. 19, pg. 198)

Solution * The probability density is given by pn( x )=y n ( x ) y n ( x ) , where the wave function f a particle in a box is 2 np x y (x )= sin ; The probability to find the particle between x and x within the box is n L L 2 1

x2 x 2 x2 *2 2 np x骣 x x 2 p nx Pn( x2 , x 1 )=y n ( x ) y n ( x ) dx = sin dx =琪 - sin . Here, set x = L/n, x = 0, 蝌 L L桫 L2 np L 2 1 x1 x 1 x1 L/ n 骣x x2p x 骣 1 L 1 P( L / n ,0)=琪 - sin = 琪 - sin 2p n -( 0 - 0) = 桫L2p L0 桫 n 2 p n n

5. What is the physical meaning of |y |2 dx = 1  dx = 1? (Krane, Q3, pg/ 168) - Solution Due to the probabilistic interpretation of the wave function, the particle must be found within the region in which it exists. Statistically speaking, this means that the probability to find the particle in the region where it exists must be 1. Hence, the square of the wave function, which is interpreted as the probably density to find the particle at an intervals in space, integrated over all space must be one in accordance with this interpretation. Should the wave function is not normalised, that would lead to the consequence that the probability to find the particle associated with the wave function in the integrated region not being unity, which violates the probabilistic interpretation of the wave function.

6. What are the dimensions ofy (x ) ? (Krane, Q4, pg. 168) Solution From |y |2 dx = 1 , [y (x ) ]2[x] = [y (x ) ]2 L = 1, hence [y (x ) ]2 = 1/L, so that [y (x ) ]=1/ L -

7. What happens to the probability density in the infinite well when n  ? Is this consistent with classical physics? (Krane, Q6, pg. 168)

Solution Consider a particle moving with constant speed, bouncing back and forth between the walls in a box of

length L. In classical mechanics the probability to find the particle between the interval x1 and x2 is simply 1 the ratio of the interval to the length L, P( x , x ) =( x - x ) . The quantum probability of a particle in classical 2 1L 2 1 a box should reduce to this classical limit when n.

The probability to locate the particle in a box between the intervals x1 and x2 in state n in general is given by x x2 2 * 骣x x2p nx 骣x2 x 22p nx 2 骣 x 1 x 1 2 p nx 1 Pn( x2 , x 1 )=y n ( x ) y n ( x ) dx =琪 - sin =琪 - sin - 琪 - sin = 桫L2 np L桫 L 2 n p L 桫 L 2 n p L x1 x1

骣x2 x 1 骣 x 12p nx 1 x 2 2 p nx 21 1 骣 2 p nx 1 2 p nx 2 琪- + 琪sin - sin =( x2 - x 1) + 琪 x 1 sin - x 2 sin 桫L L 桫2 np L 2 n p L L 2 n p 桫 L L sin nx 1 which is n-dependent in general. In the limit of n, the terms vanish and P ( x , x ) =( x - x ) , n n 2 1L 2 1

which is just the probability to locate a particle between the interval intervals x1 and x2 in classical mechanics.

8. How would the solution to the one-dimensional infinite potential energy well be different if the potential

energy were not zero for 0xL but instead had a constant value U0. What would be the energies of the excited states? What would be the wavelengths of the standing de Broglie waves? Sketch the behavior of the lowest two wave functions. (Krane, Q6, pg. 168)

Solution Be reminded that the energies of the excited states depend on possible the stationary modes constrained

by the condition L = nn/2, which in tern quantize the possible momentum of the particle in the box via 2 pn = h/n and this in terns quantize the energy states of the particle via En= p n / 2 m .

In the case of zero potential, the energy of the particle in the box is equal the kinetic energy of the

particle. However, if the potential U0 is a constant, has a positive value, and assumed to be less than E1, the kinetic energy of the particle will be reduced from En  En’ = En - U0. The reduction in kinetic energy

causes a reduction in the linear momentum from pn = 2mEn  pn’= 2mEn '= 2 m( E n - U0 ) < p n . A reduction in momentum means an increase on the particle’s wavelength in the box from 2 ln=h/ 2 mE n� l ' n = h / 2 mE n ' - h / 2 m( E n U0 ) such that(ln/ l ' n) =(E n - U0 ) / E n < 1. Hence, the energy 2 2 2 骣h1 骣 h 1 骣 h 1 骣En - U0 states in the box is modified from En=琪2� E' n = 琪 2 = 琪 2 琪 - E n U0 , 桫2ml 桫 2 m ln ' 桫 2 m l n桫 E n

which means that the energy states is shifted from En� E n U0 . 9. A particle in an infinite well is in the ground state with an energy of 1.26 eV. How much energy must be added to the particle to reach the second excited state (n = 3)? The third excited state (n = 4)? (Krane, P4, pg. 170)

10. An electron is trapped in a one-dimensional well of width 0.132 nm. The electron is in the n = 10 state. (a) What is the energy of the electron? (b) What is the uncertainty in its momentum? (c) What is the uncertainty in its position? How do these results change as n  ? Is this consistent with classical behavior? (Krane, P9, pg. 170)

11. Consider a particle moving in a one-dimensional box with walls at x = -L/2 and x = +L/2. (a) Write the wave functions and probability densities for the states n = 1, n = 2, and n = 3. (b) Sketch the wave function and probability densities. (Hint: Make an analogy to the case of a particle in a box with walls at x = 0 and x = L) (Serway, M & M, P11, pg. 228)

Solution By applying the boundary conditions that the solution must vanish at both ends, i.e.  (x  L / 2)  (x  L / 2)  0 , the solution takes the form  2 nx  cos (odd n)  L L L L  n (x)   for   x 2 nx 2 2  sin (even n)  L L This question is tantamount to re-analyse the same physical system in a shifted coordinates, x  x – L/2. The normalisation and energies shall remain unchanged under the shift of coordinate system x  x – L/2. Both of these quantities depends only on the width of the well but not on the coordinate system used. We L L get for - ＃x 2 2 1 2 2 骣p x 2 2 骣p x y (x) = cos琪 ; P1 ( x) = cos 琪 1 ( L) 桫 L ( L) 桫 L 1 2 2 骣2p x 2 2 骣2p x y (x) = sin琪 ; P2 ( x) = sin 琪 2 ( L) 桫 L ( L) 桫 L 1 2 2 骣3p x 2 2 骣3p x y (x) = cos琪 ; P3 ( x) = cos 琪 3 ( L) 桫 L ( L) 桫 L

12. A particle of mass m is placed in a one-dimensional box of length L. The box is so small that the particle’s motion is relativistic, so that E = p2/2m is not valid. (a) Derive an expression for the energy levels of the particle using the relativistic energy- momentum relation and the quantization of momentum that derives from confinement. (b) If the particle is an electron in a box of length L = 1.0010-12m, find its lowest possible kinetic energy. By what percent is the nonrelativistic formula for the energy in error? (Serway, M & M, P14, pg. 228)

Solution nl h nh (a) Still, = L so p = = 2 l 2L 2 1 2 K=轾 c2 p 2 + mc 2 - mc 2 = E - mc 2 臌 ( ) ( ) 2 1 2 轾骣nhc 2 2 En =犏琪 + ( mc ) , 臌桫2L 2 1 2 轾骣nhc 22 2 Kn =犏琪 +( mc) - mc 臌桫2L -12 -31 -14 (b) Taking L = 10 m , m = 9.11 10 kg , and n = 1 we find K1 = 4.69 10 J=0.29MeV . The nonrelativistic result is 2 2 醋 -34 h (6.63 10 J s) -14 E1 = = = 6.03 10 J=0.38 MeV 8mL2 8( 9.11 10- 31 kg)( 10 - 24 m 2 )

Comparing this with K1 , we see that this value is too big by 29%.