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Chapter 7: Q9CQ (page 278)

Knowing precisely all components of a nonzero L→would violate the uncertainty principle, but knowingthat L→is precisely zerodoes not. Why not?

(Hint:For l=0 states, the momentum vector p→ is radial.)

Short Answer

Expert verified

Knowingthat L→is precisely zerodoes not violate uncertainty principle because the components commute in this case making them measurable simultaneously.

Step by step solution

01

Angular momentum commutation relation

The angular momentum components follow the commutation relation

Li,Lj=iℏLkLj,Lk=iℏLiLk,Li=iℏLj ..... (I)

Here, ℏ is the reduced Planck's constant.

02

Step 2:Determining the commutation for zero angular momentum

Observables which do not commute cannot be measured simultaneously. Thus it is evident from equation (I) that components of angular momentum cannot be measured simultaneously. But if all the components are measured to be zero, the right hand sides of equations (I) become zero and the components commute. Thus simultaneous measurement of angular momentum components if the values are zero does not violate uncertainty principle.

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Most popular questions from this chapter

Question: Consider a cubic 3D infinite well of side length of L. There are 15 identical particles of mass m in the well, but for whatever reason, no more than two particles can have the same wave function. (a) What is the lowest possible total energy? (b) In this minimum total energy state, at what point(s) would the highest energy particle most likely be found? (Knowing no more than its energy, the highest energy particle might be in any of multiple wave functions open to it and with equal probability.)

In general, we might say that the wavelengths allowed a bound particle are those of a typical standing wave,λ=2L/n , where is the length of its home. Given that λ=h/p, we would have p=nh/2L, and the kinetic energy, p2/2m, would thus be n2h2/8mL2. These are actually the correct infinite well energies, for the argumentis perfectly valid when the potential energy is 0 (inside the well) and is strictly constant. But it is a pretty good guide to how the energies should go in other cases. The length allowed the wave should be roughly the region classically allowed to the particle, which depends on the “height” of the total energy E relative to the potential energy (cf. Figure 4). The “wall” is the classical turning point, where there is nokinetic energy left: E=U. Treating it as essentially a one-dimensional (radial) problem, apply these arguments to the hydrogen atom potential energy (10). Find the location of the classical turning point in terms of E , use twice this distance for (the electron can be on both on sides of the origin), and from this obtain an expression for the expected average kinetic energies in terms of E . For the average potential, use its value at half the distance from the origin to the turning point, again in terms of . Then write out the expected average total energy and solve for E . What do you obtain

for the quantized energies?

For a hydrogen atom in the ground state. determine (a) the most probable location at which to find the electron and (b) the most probable radius at which to find the electron, (c) Comment on the relationship between your answers in parts (a) and (b).

Consider a vibrating molecule that behaves as a simple harmonic oscillator of mass 10-27kg, spring constant 103N/m and charge is +e , (a) Estimate the transition time from the first excited state to the ground state, assuming that it decays by electric dipole radiation. (b) What is the wavelength of the photon emitted?

Question: An electron is trapped in a cubic 3D well. In the states (nx,ny,nz)= (a) (2,1,1) (b) (1,2,1)(c) (1,1,2), what is the probability of finding the electron in the region 0⩽x⩽L,L/3⩽y⩽2L/3,0⩽z⩽L. Discus any difference in these results.
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