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Show that the quite general wave group given in equation (6-21) is a solution of the free-particle Schr枚dinger equation, provided that each plane wave's w does satisfy the matter-wave dispersion relation given in (6-23).

Short Answer

Expert verified

Given general wave equation x,t=-+Akeikx-蝇迟dkis a solution of the free-particle Schr枚dinger equation -h22m2x,tx2=ihx,tt, provided, each plane wave鈥檚 satisfiesk=hk22m.

Step by step solution

01

Formula used:

wave function of a quantum-mechanical system.

Schr枚dinger equation is given by,

-h22m2(x,t)x2=ih(x,t)t 鈥.. (1)

Where, h is Plank鈥檚 constant, m is the mass of the object, x is the position of the object, t is the time, andis the Wave function

Wave Group (6-21) is,

(x,t)=-+A(k)ei(kx-蝇迟)dk 鈥.. (2)

Where, k is the wave number, A is the amplitude of the wave, and is theangular frequency.

Matter Wave dispersion relation (6-23) is,

(k)=hk22m 鈥.. (3)

02

Substituting wave group equation into Schrödinger equation:

-h22m2x2-+Ake-ikx-蝇迟dk=iht-+Ake-ikx-蝇迟dk-h22m-+-k2Ake-ikx-蝇迟dk=ih-+-i蝇Ake-ikx-蝇迟dk

Substituting hk22mfor (k) in above Equation, and you have

-h22m-+-k2Ake-ikx-蝇迟dk=ih-+-ihk22mAke-ikx-蝇迟dk-h22m-+-k2Ake-ikx-蝇迟dk=-h22m-+-k2Ake-ikx-蝇迟dk

03

Conclusion:

You can see in the above equation, LHS = RHS.

Hence, the given general wave equation (6-21) is a solution of the free-particle Schr枚dinger equation.

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Most popular questions from this chapter

Exercise 39 gives a condition for resonant tunneling through two barriers separated by a space width of2s, expressed I terms of factorgiven in exercise 30. Show that in the limit in which barrier widthL, this condition becomes exactly energy quantization condition (5.22) for finite well. Thus, resonant tunneling occurs at the quantized energies of intervening well.

In the wide-barrier transmission probability of equation (6-18), the coefficient multiplying the exponential is often omitted. When is this justified, and why?

To obtain a rough estimate of the mean time required for uranium-238 to alpha-decay, let us approximate the combined electrostatic and strong nuclear potential energies by rectangular potential barrier half as high as the actual 35 Mev maximum potential energy. Alpha particles (mass 4 u) of 4.3 Mev kinetic energy are incident. Let us also assume that the barrier extends from the radius of nucleus, 7.4 fm to the point where the electrostatic potential drops to 4.3 Mev (i.e., the classically forbidden region). Because U(1/r), this point is 35/4.3 times the radius of the nucleus, the point at which U(r) is 35 Mev. (a) Use these crude approximations, the method suggested in Section 6.3, and the wide-barrier approximation to obtain a value for the time it takes to decay. (b) To gain some appreciation of the difficulties in a theoretical prediction, work the exercise 鈥渂ackward鈥 Rather than assuming a value for U0, use the known value of the mean time to decay for uranium-238 and infer the corresponding value of U0, Retain all other assumptions. (c) Comment on the sensitivity of the decay time to the height of the potential barrier.

Given the same particle energy and barrier height and width, which would tunnel more readily: a proton or an electron? Is this consistent with the usual rule of thumb governing whether classical or non-classical behavior should prevail?

A method for finding tunneling probability for a barrier that is "wide" but whose height varies in an arbitrary way is the so-called WKB approximation.

T=exp[2122m(U(x)E)dx]

Here U(x) is the height of the arbitrary potential energy barrier.Whicha particle first penetrates at x=0 and finally exits at x=L. Although not entirely rigorous, show that this can be obtained by treating the barrier as a series of rectangular slices, each of width dx (though each is still a "wide" barrier), and by assuming that the probability of tunneling through the total is the product of the probabilities for each slice.

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