Concept explainers
(a)
Prove that for integers
(a)
Answer to Problem 5.38P
It has been proved that for integers
Explanation of Solution
Consider
Since,
The denominator vanishes only when
For
For
Conclusion:
It has been proved that for integers
(b)
The commutation relations for the ladder operators,
(b)
Answer to Problem 5.38P
The commutation relations for the ladder operators,
Explanation of Solution
Given the ladder operator
Solving for
Dropping the two terms involving commutators of two coordinates or two derivatives. The remaining commutators are
Hence it is proved that
Similarly solving for
Dropping the two terms involving commutators of two coordinates or two derivatives. The remaining commutators are
Similarly solving for
Dropping the two terms involving commutators of two coordinates or two derivatives. The remaining commutators are
Equating the two commutator,
Conclusion:
Hence it is proved that the commutation relations for the ladder operators,
(c)
Show that
(c)
Answer to Problem 5.38P
It is proved that
Explanation of Solution
Solving for
Since
Using the above equation to solve for
Further solving,
Since,
Hence the first relation is proved,
Solving for
Since
Using the above equation to solve for
Further solving,
Since,
Hence it is proved that
Conclusion:
It is proved that
(d)
Show that
(d)
Answer to Problem 5.38P
It has been proved that
Explanation of Solution
Solving for
Squaring on both sides,
Solving for
Substitute equation (IV) and (V) in
Solving for
The middle term in the above equation vanish when it is summed over
Adding the results
Conclusion:
It has been proved that
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Chapter 5 Solutions
Introduction To Quantum Mechanics
- Which of the following statements is not true regarding a massspring system that moves with simple harmonic motion in the absence of friction? (a) The total energy of the system remains constant. (b) The energy of the system is continually transformed between kinetic and potential energy. (c) The total energy of the system is proportional to the square of the amplitude. (d) The potential energy stored in the system is greatest when the mass passes through the equilibrium position. (e) The velocity of the oscillating mass has its maximum value when the mass passes through the equilibrium position.arrow_forwardA spherical bob of mass m and radius R is suspended from a fixed point by a rigid rod of negligible mass whose length from the point of support to the center of the bob is L (Fig. P16.75). Find the period of small oscillation. N The frequency of a physical pendulum comprising a nonuniform rod of mass 1.25 kg pivoted at one end is observed to be 0.667 Hz. The center of mass of the rod is 40.0 cm below the pivot point. What is the rotational inertia of the pendulum around its pivot point?arrow_forwardFor each expression, identify the angular frequency , period T, initial phase and amplitude ymax of the oscillation. All values are in SI units. a. y(t) = 0.75 cos (14.5t) b. vy (t) = 0.75 sin (14.5t + /2) c. ay (t) = 14.5 cos (0.75t + /2) 16.3arrow_forward
- A lightweight spring with spring constant k = 225 N/m is attached to a block of mass m1 = 4.50 kg on a frictionless, horizontal table. The blockspring system is initially in the equilibrium configuration. A second block of mass m2 = 3.00 kg is then pushed against the first block, compressing the spring by x = 15.0 cm as in Figure P16.77A. When the force on the second block is removed, the spring pushes both blocks to the right. The block m2 loses contact with the springblock 1 system when the blocks reach the equilibrium configuration of the spring (Fig. P16.77B). a. What is the subsequent speed of block 2? b. Compare the speed of block 1 when it again passes through the equilibrium position with the speed of block 2 found in part (a). 77. (a) The energy of the system initially is entirely potential energy. E0=U0=12kymax2=12(225N/m)(0.150m)2=2.53J At the equilibrium position, the total energy is the total kinetic energy of both blocks: 12(m1+m2)v2=12(4.50kg+3.00kg)v2=(3.75kg)v2=2.53J Therefore, the speed of each block is v=2.53J3.75kg=0.822m/s (b) Once the second block loses contact, the first block is moving at the speed found in part (a) at the equilibrium position. The energy 01 this spring-block 1 system is conserved, so when it returns to the equilibrium position, it will be traveling at the same speed in the opposite direction, or v=0.822m/s. FIGURE P16.77arrow_forwardA blockspring system oscillates with an amplitude of 3.50 cm. The spring constant is 250 N/m and the mass of the block is 0.500 kg. Determine (a) the mechanical energy of the system, (b) the maximum speed of the block, and (c) the maximum acceleration.arrow_forwardA 50.0-g object connected to a spring with a force constant of 35.0 N/m oscillates with an amplitude of 4.00 cm on a frictionless, horizontal surface. Find (a) the total energy of the system and (b) the speed of the object when its position is 1.00 cm. Find (c) the kinetic energy and (d) the potential energy when its position is 3.00 cm.arrow_forward
- A block of unknown mass is attached to a spring with a spring constant of 6.50 N/m and undergoes simple harmonic motion with an amplitude of 10.0 cm. When the block is halfway between its equilibrium position and the end point, its speed is measured to be 30.0 cm/s. Calculate (a) the mass of the block, (b) the period of the motion, and (c) the maximum acceleration of the block.arrow_forwardA 500-kg object attached to a spring with a force constant of 8.00 N/m vibrates in simple harmonic motion with an amplitude of 10.0 cm. Calculate the maximum value of its (a) speed and (b) acceleration, (c) the speed and (d) the acceleration when the object is 6.00 cm from the equilibrium position, and (e) the time interval required for the object to move from x = 0 to x = 8.00 cm.arrow_forwardAn object of mass m1 = 9.00 kg is in equilibrium when connected to a light spring of constant k = 100 N/m that is fastened to a wall as shown in Figure P12.67a. A second object, m2 = 7.00 kg, is slowly pushed up against m1, compressing the spring by the amount A = 0.200 m (see Fig. P12.67b). The system is then released, and both objects start moving to the right on the frictionless surface. (a) When m1 reaches the equilibrium point, m2 loses contact with m1 (see Fig. P12.67c) and moves to the right with speed v. Determine the value of v. (b) How far apart are the objects when the spring is fully stretched for the first time (the distance D in Fig. P12.67d)? Figure P12.67arrow_forward
- A block connected to a horizontal spring is in simple harmonic motion on a level, frictionless surface, oscillating with amplitude A around x = 0. Identify whether each of the following statements is true or false: (a) If x = ±A then |v|=|vmax| and |a|=|amax|. (b) If x = 0 then |v|=|vmax| and |a|= 0. (c) If v > 0 then a < 0. (d) If x > 0 then a < 0. (e) If x > 0 then v > 0.arrow_forwardA homogeneous rod with a length of L = 1 m and a mass of m = 0.5 kg is fixed in the middle so that it can rotate. Two springs with a constant k = 12 N / m are connected at both ends of the rod. The other ends of the bow are fixed and the springs are unstretched. The rod is rotated from its equilibrium position to Ɵ << and released. What is the angular frequency of the rod (rad / s)? g = 10 m / s2arrow_forwardA mass is 200 g and attached to a spring that has a force constant k = 2000 N/m. The amplitude is 20 cm. The mass is moving back and forth horizontally and assumes no friction. Calculate the speed of the mass when x=20cm. a. 0.01 m/s b. 15.65 m/s c. 0.1 m/s d. 0 m/s e. 17.32 m/sarrow_forward
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