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Knowing precisely all components of a nonzero \(L\) would violate the unceltainty principle, but knowing that \(\mathbf{L}\) is precisely zens does not. Why not? (Hint: For \(\ell=0\) states, the momentum vector \(\mathrm{P}\) is radial.)

Short Answer

Expert verified
Knowing that \( L \) is zero doesn't violate the Heisenberg's Uncertainty Principle because for these states of zero angular momentum, the momentum vector \( P \) is radial, meaning there's no 'sideways' momentum component to contribute to an angular momentum about the origin.

Step by step solution

01

Understand the Uncertainty Principle

The Heisenberg's Uncertainty Principle states that the more precisely the position of a particle is known, the less precisely its momentum can be known, and vice versa. In terms of angular momentum, it implies that we can't measure both components of angular momentum (say \( L_x \) and \( L_y \)) simultaneously with perfect accuracy. Thus, if we know precisely all components of a nonzero \( L \), it would violate this principle.
02

Apply the Uncertainty Principle for \( L_z \)

Consider now when \( L \) is zero. This means that we're in a state of zero angular momentum (\( \ell=0 \) state). For these cases, as the hint suggests, the momentum vector \( P \) is radial. This means that there's no 'sideways' momentum component to contribute to an angular momentum about the origin, so it's perfectly permissible to know that this angular momentum component is zero, because there's no associated momentum in the perpendicular direction that we'd be uncertain about.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Angular Momentum
Angular momentum in quantum mechanics refers to the quantity of rotation of a particle, which is a fundamental concept that describes how particles spin around a given point or axis.
It is a vector quantity, meaning it has both a direction and a magnitude.
Unlike in classical mechanics, where angular momentum is represented as a continuous quantity, in quantum mechanics, it is quantized, meaning it can only take on certain discrete values.

When dealing with nonzero angular momentum values, the Heisenberg Uncertainty Principle comes into play. This principle tells us that we cannot know both the position and the momentum of a particle to arbitrary precision, as there's a trade-off between the two. Similarly, it is impossible to measure all components of angular momentum with perfect precision.
For example, knowing both horizontal (\(L_x\)) and vertical (\(L_y\)) components of angular momentum would be subject to uncertainty, leading to imprecision, if we attempted to know both exactly.
The exception occurs when the total angular momentum is zero, which does not affect any measurement precision due to the absence of momentum components besides the radial one.
Quantum Mechanics
Quantum mechanics is the branch of physics that deals with the behavior, properties, and interactions of microscopic particles at the atomic and subatomic levels.
This field is fundamentally different from classical mechanics and introduces concepts like quantization, where physical quantities like energy, angular momentum, and charge appear in discrete values.
When particles are in a quantum state of zero angular momentum (\(\ell=0\)), it reveals interesting properties about their momentum vectors.

In such states, the principle of radial momentum is essential to understand. Here, the particle's momentum vector is aligned radially with respect to a point of origin. This radial alignment implies there's no transverse, or side, momenta and hence no angular momentum about that point.
This is why, when the total angular momentum is zero, it does not violate the Uncertainty Principle to know precisely this fact: all momentum is directed radially, rendering sideways momentum non-existent.
Momentum Vector
In physics and quantum mechanics, the momentum vector represents the momentum of a particle, which describes the quantity of its motion.
The vector nature of momentum means it has both a magnitude and direction, often used to describe linear and angular motion.

For particles in a state of zero angular momentum, it’s crucial to recognize that their momentum vectors are radial.
  • This means momentum is directed straight outwards or inwards, rather than having components that contribute to angular movement around a point.
  • The absence of sideways components is why it is plausible to know precisely the angular momentum is zero, as the Heisenberg Uncertainty Principle pertains to components leading to angular motion.
This sheds light on why in zero angular state scenarios, knowledge of exact angular momentum does not retake uncertainty into measurements.

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Most popular questions from this chapter

Classically, an orbiting charged particle radiates elecIromagnetic energy, and for an electron in atomic dimensions, if would lead to collupse in considerably less than the wink of an eye. (a) By equating the centripetal and Coulomb forces, show that for a classical charge \(-e\) of mass \(m\) held in circular orbit by its attraction to a fixed charge \(+e\), the following relationship holds: \(\omega=e r^{-1 / 2} / \sqrt{4 \pi \varepsilon_{0} m}\). (b) Electromagnetism tells us that a charge whose acceleration is a radiates power \(P=e^{2} a^{2} / 6 e_{f} f^{\prime}\). Shuw that thiv can also be expressed in terms of the orbit radius, as \(P=e^{6} /\left(96 m^{2} \varepsilon_{0}^{3} m c^{3} r^{\lambda}\right)\). Then calculate the energy lowt per orbit in terms of \(r\) by multiplying this power by the period \(T=2 \pi / \omega\) and using the formula from part (a) to eliminate \(\omega(\mathrm{c})\) in such a classical orbit. the total mechanical energy is half the potential energy (see Section 4.4), or \(E_{\text {orbu }}=-e^{2 / 8} \pi \varepsilon_{0} r .\) Calculate the change in energy per change in \(r, d E_{\text {obi }} l d r\). From this and the energy lost per orbit from part (b). determine the change in \(r\) per orbit and evaluate it for a typical orbit radius of \(10^{-10} \mathrm{~m}\). Would the electron's radius change much in a single orbit? (d) Argue that dividing \(d E_{\text {arbiu }} / d r\) by \(P\) and multiplying by dr gives the time required for \(r\) to change by \(d r\). Then. sum these times for all radii from \(r_{\text {initial }}\) to a final radius of \(0 .\) Evaluate your result for \(r_{\text {hiniul }}=10^{-10} \mathrm{~m}\). (One limitation of this extimate is that the electron would eventually be moving relativistically.)

Taking the \(n=3\) states as representative, explain the relationship between the complexity-numbers of nodes and antinodes- - - of hydrogen's standing waves in the radial direction and their complexity in the angular direction at a given value of \(n\). Is it a direct or inverse relationship. and why?

Consider an electron in the ground state of a hydrogen atom. (a) Calculate the expectation value of its potential energy. (b) What is the expectation value of its kinetic energy? (Hint: What is the expectation value of the total energy?

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?

Doubly ionized lithium. \(\mathrm{Li}^{2+}\), abvorbs a photon and juinps from the ground state to its \(n=2\) level. What was the wavelength of the photon?

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