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Question: Section 7.5 argues that knowing all three components of would violate the uncertainty principle. Knowing its magnitude and one component does not. What about knowing its magnitude and two components? Would be left any freedom at all and if so, do you think it would be enough to satisfy the uncertainly principle?

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Answer

Only the magnitude of L and the Z-component of L can be found accurately.

Step by step solution

01

Definition of the Uncertainty principle

The Uncertainty principle states that the momentum and position of a particular particle cannot be measured with great accuracy. In general, it explains that uncertainty will be there in measuring a particular variable for a specific particle.

02

Determination of the violation of uncertainty principal and magnitude if two components are known 

Write the expression for the sum of the square of the components.

Lx2+Ly2+Lz2=L2

Here, L is the magnitude of angular momentum vector, LX is the x-component of angular momentum vector, LY is the y-component of angular momentum vector, and LZis the z-component of angular momentum vector.

It is known that LZ can be known with certainty, by the equation:LZ=ml , and is also quantized and can be calculated by the following equation.

L=l(l+1)

Here, is the Plank’s Constant, and is the azimuthal quantum number

So, does not commute withLX or LY . Hence, if relation isLx2+Ly2+Lz2=L2 being used to calculate the value of LX and , LY then it will be uncertain.

Thus, only the magnitude of L and the z-component of L can be found accurately.

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Most popular questions from this chapter

The space between two parallel metal plates is filled with an element in a gaseous stale. Electrons leave one plate at negligible speed and are accelerated toward the other by a potential differenceVapplied between the plates. As Vis increased from 0, the electron current increases more or less linearly, but when Vreaches 4.9 V , the current drops precipitously. From nearly 0 , it builds again roughly linearly as Vis increased beyond 4.9 V .

(a) How can the presence of the gas explain these observations?

(b) The Gas emits pure “light” when Vexceeds 4.9 V . What is its wavelength?

We have noted that for a given energy, as lincreases, the motion is more like a circle at a constant radius, with the rotational energy increasing as the radial energy correspondingly decreases. But is the radial kinetic energy 0 for the largest lvalues? Calculate the ratio of expectation values, radial energy to rotational energy, for the(n,l,mt)=(2.1,+1)state. Use the operators

KErad=-h22m1r2r(rr)KErad=h2l(l+1)2mr2

Which we deduce from equation (7-30).

A simplified approach to the question of how lis related to angular momentum – due to P. W. Milonni and Richard Feynman – can be stated as follows: If can take on only those values mlh, whereml=0,±1,±l , then its square is allowed only valuesml2h2, and the average of localid="1659178449093" l2should be the sum of its allowed values divided by the number of values,2l+1 , because there really is no preferred direction in space, the averages of Lx2andLy2should be the same, and sum of all three should give the average of role="math" localid="1659178641655" L2. Given the sumrole="math" localid="1659178770040" 1Sn2=N(N+1)(2N+1)/6, show that these arguments, the average of L2 should be l(l+1)h2.

To conserve momentum, an atom emitting a photon must recoil, meaning that not all of the energy made available in the downward jump goes to the photon. (a) Find a hydrogen atom's recoil energy when it emits a photon in a n = 2 to n = 1 transition. (Note: The calculation is easiest to carry out if it is assumed that the photon carries essentially all the transition energy, which thus determines its momentum. The result justifies the assumption.) (b) What fraction of the transition energy is the recoil energy?

A mathematical solution of the azimuthal equation (7-22) is Φ(φ)=Ae-+Be- , which applies when D is negative, (a) Show that this simply cannot meet itself smoothly when it finishes a round trip about the z-axis. The simplest approach is to consider φ=0 and φ=2π. (b) If D were 0, equation (7-22) would say simply that the second derivative Φ(φ)of is 0. Argue than this too leads to physically unacceptable solution, except in the special case of Φ(φ) being constant, which is covered by the ml=0 , case of solutions (7-24).

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