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For a total energy of 0, the potential energy is given in Exercise 96. (a) Given these, to what region of the x-axis would a classical particle be restricted? Is the quantum-mechanical particle similarly restricted? (b) Write an expression for the probability that the (quantum-mechanical) particle would be found in the classically forbidden region, leaving it in the form of an integral. (The integral cannot be evaluated in closed form.)

Short Answer

Expert verified

(a) Classically allowed region is -3b,3b.

(b) Forbidden region is 23∫0bA2X2e-x2ib2dx.

Step by step solution

01

Relationship between the total energy and the potential energy

The half of the potential energy is known as the total energy.

02

Step 2(a): The classically allowed region

As we know the potential energy is

Ux=h22mb4x2-3h22mb2Ux=22mb4x2-3×22mb2=22m-3b2+1b4x2

The potential energy is smaller than 0 when the region is classically allowed.

-3b2+1b4x2=0x2=3b2x=-3b,3b

So, the classically allowed region is -3b,3b.

The wave function extends infinitely far in both directions, so the quantum entity is not restricted to this region.

03

The probability of the particle

The probability for the particle is

∫-3b3bΨxΨ0xdx=∫-3b3bA2x2e-x2ib2dx=2∫03bA2x2e-x2ib2dx

Therefore, the probability that the particle would be found in the classical forbidden region is 2∫03bA2x2e-x2ib2dx.

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Most popular questions from this chapter

Air is mostly N2, diatomic nitrogen, with an effective spring constant of 2.3 x 103N/m, and an effective oscillating mass of half the atomic mass. For roughly what temperatures should vibration contribute to its heat capacity?

A finite well always has at least one bound state. Why does the argument of Exercises 38 fail in the case of a finite well?

It is possible to take the finite well wave functions further than (21) without approximation, eliminating all but one normalization constant C . First, use the continuity/smoothness conditions to eliminate A, B , andG in favor of Cin (21). Then make the change of variables z=x-L/2 and use the trigonometric relations

sin(a+b)=sinacosb+cosasinband

cos(a+b)=cosacosb-sinasinbon the

functions in region I, -L/2<z<L/2. The change of variables shifts the problem so that it is symmetric about z=0, which requires that the probability density be symmetric and thus that ψ(z)be either an odd or even function of z. By comparing the region II and region III functions, argue that this in turn demands that (α/k)sinkL+coskL must be either +1 (even) or -1 (odd). Next, show that these conditions can be expressed, respectively, as αk=tankL2 and αk=-cotkL2. Finally, plug these separately back into the region I solutions and show that

ψ(z)=C×{eα(z+L/2) â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰z<L/2coskzcoskL2 â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰-L/2<z<L/2e-α(z-L/2) â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰z>L/2


or

ψ(z)=C×{eα(z+L/2) â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰z<L/2-sinkzsinkL2 â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰-L/2<z<L/2e-α(z-L/2) â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰z>L/2

Note that Cis now a standard multiplicative normalization constant. Setting the integral of |ψ(z)|2 over all space to 1 would give it in terms of kand α , but because we can’t solve (22) exactly for k(or E), neither can we obtain an exact value for C.

Determine the expectation value of the momentum of the particle. Explain.

Obtain expression (5-23) from equation (5-22). Using ³¦´Ç²õθ=cos2(12θ)−sin2(12θ)and²õ¾±²Ôθ=2sin(12θ)cos(12θ), first convert the argument of the cotangent fromkLto12kL. Next, put the resulting equation in quadratic form, and then factor. Note thatαis positive by definition.

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