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Chapter 5: Q57E (page 191)

The uncertainty in a particle's momentum in an infinite well in the general case of arbitrary nis given bynπhL .

Short Answer

Expert verified

For n=0the uncertainty vanishes but is perfectly finite for all other values. Thus, so long asΔx2 is large enough (it is), the uncertainty principle is perfectly satisfied for all n>0.

Step by step solution

01

The concept and the formula used.

Heisenberg's uncertainty principle states that it is impossible to measure or calculate exactly, both the position and the momentum of an object.

Consider, energy E=0. Then, the momentum of the state must satisfyE=p22m. Now, there are two solutions forP corresponding to positive and negative momentum. The uncertainty can thus be calculated as follows:

ΔP2=⟨P2⟩−⟨P⟩2

=2mE

=n2π2ℏ2L2

ΔP=nπℏL

02

Conclusion

Clearly, forn=0 the uncertainty vanishes but is perfectly finite for all other values. Thus, so long asΔx2 is large enough (it is), the uncertainty principle is perfectly satisfied for alln>0 .

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Most popular questions from this chapter

A 2kg block oscillates with an amplitude of 10cm on a spring of force constant 120 N/m .

(a) In which quantum state is the block?

(b) The block has a slight electric charge and drops to a lower energy level by generating a photon. What is the minimum energy decrease possible, and what would be the corresponding fractional change in energy?

In the harmonic oscillators eave functions of figure there is variation in wavelength from the middle of the extremes of the classically allowed region, most noticeable in the higher-n functions. Why does it vary as it does?

If a particle in a stationary state is bound, the expectation value of its momentum must be 0.

(a). In words, why?

(b) Prove it.

Starting from the general expression(5-31) with p^in the place of Q⏞, integrate by parts, then argue that the result is identically 0. Be careful that your argument is somehow based on the particle being bound: a free particle certainly may have a non zero momentum. (Note: Without loss of generality,ψ(x) may be chosen to be real.)

What is the product ofΔxandΔp(obtained in Exercise 83 and 85)? How does it compare with the minimum theoretically possible? Explain.

Consider the delta well potential energy:

U(x)={0x≠0-∞x=0

Although not completely realistic, this potential energy is often a convenient approximation to a verystrong, verynarrow attractive potential energy well. It has only one allowed bound-state wave function, and because the top of the well is defined as U = 0, the corresponding bound-state energy is negative. Call its value -E0.

(a) Applying the usual arguments and required continuity conditions (need it be smooth?), show that the wave function is given by

ψ(x)=(2mE0h2)1/4e-(2mE0/ℏ)|x|

(b) Sketch ψ(x)and U(x) on the same diagram. Does this wave function exhibit the expected behavior in the classically forbidden region?
See all solutions

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