/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 70 (a) At time \(t=0\) a large ense... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

(a) At time \(t=0\) a large ensemble of spin-1/2 particles is prepared, all of them in the spin-up state (with respect to the \(z\) axis). 74 They are not subject to any forces or torques. At time \(t_{1}>0\) each spin is measured some along the \(z\) direction and others along the \(x\) direction (but we aren't some along the \(z\) direction and others along the \(x\) direction (but we aren't told the results). At time \(t_{2}>t_{1}\) their spin is measured again, this time along the \(x\) direction, and those with spin up (along \(x\) ) are saved as a subensemble (those with spin down are discarded). Question: Of those remaining (the subensemble), what fraction had spin up (along \(z\) or \(x\), depending on which was measured) in the first measurement? (b) Part (a) was easy-trivial, really, once you see it. Here's a more pithy generalization: At time \(t=0\) an ensemble of spin-1/2 particles is prepared, all in the spin-up state along direction a. At time \(t_{1}>0\) their spins are measured along direction b (but we are not told the results), and at time \(t_{2}>t_{1}\) their spins are measured along direction c. Those with spin up (along \(\mathrm{c}\) ) are saved as a subensemble. Of the particles in this subensemble, what fraction had spin up (along b) in the first measurement? Hint: Use Equation 4.155 to show that the probability of getting spin up (along b) in the first measurement is \(P_{+}=\cos ^{2}\left(\theta_{a b} / 2\right)\), and (by extension) the probability of getting spin up in both measurements is \(P_{++}=\cos ^{2}\left(\theta_{a b} / 2\right) \cos ^{2}\left(\theta_{b c} / 2\right)\). Find the other three probabilities \(\left(P_{+\rightarrow}, P_{-+}, \text {and } P_{--}\right) .\) Beware: If the outcome of the first measurement was spin down, the relevant angle is now the supplement of \(\theta_{b c}\).

Short Answer

Expert verified
In the specific scenario of part (a), about 1/2 of the particles in the subensemble had spin up initially.

Step by step solution

01

Examine Initial Conditions

Initially, at time \( t = 0 \), all particles are spin up along the \( z \)-axis. Hence, we can represent the state of each particle as \( |+z\rangle \). Spin measurements are made along \( z \) at \( t_{1} \) and then along \( x \) at \( t_{2} \). We need to calculate the fraction of particles that have spin up along \( z \) or \( x \) in the first measurement and are spin up along \( x \) in the second measurement.
02

Spin State Evolution

After the first measurement along the \( z \) or \( x \) direction, the outcome doesn't affect the spin evolution until the second measurement. Between measurements, since no forces act on the particles, their state doesn't change. Consequently, the initial state transitions according to measurement probability, but not in physical state development.
03

Measurement Probabilities

When measuring spin along \( z \), initially prepared as \( |+z\rangle \), the probability of measuring spin up along \( z \) is 1. If measured along \( x \), use \( |+z\rangle = \frac{1}{\sqrt{2}}(|+x\rangle + |-x\rangle) \). Hence, the probability for spin up along \( x \) is \( \left|\frac{1}{\sqrt{2}}\right|^2 = \frac{1}{2} \).
04

Subensemble Preparation at t2

At \( t_{2} \), spin is measured along \( x \) and those with spin up are isolated. From Step 3, particles measured from \( |+z\rangle \) at \( t_1 \) have a 50% chance of spin up along \( x \) at \( t_2 \), and those initially \( |+x\rangle \) have a 100% chance to remain \( |+x\rangle \) at \( t_2 \). Thus, the subensemble community is a 50-50 mix from prior \( z \) and potential \( x \) measurements.
05

Fraction Calculation

The particles initially measured spin down on first measurement are excluded from subensemble as they do not meet second measurement \(|+x\rangle\) criteria. Fraction of those initially measuring spin up include those measurements at time \( t_1 \). If all particles measured on \( x \) at \( t_1 \), subensemble retains \( 1/2 \) from \( |+x\rangle\). Hence, description holds: the final fraction who were spin-up in the first MAJORITY case is usually \(1/2\).
06

General Case Analysis for Part B

In general, starting along direction \( a \), the spin up probability along \( b \) direction is \( P_+ = \cos^2(\theta_{ab}/2) \). Given it's spin up along \( c \) direction finally, probability of spinning up twice is \( P_{++} = \cos^2(\theta_{ab}/2) \cos^2(\theta_{bc}/2) \) by cascading rotational consistency. Since particle keeps same trajectory, trajectory config. supports precise bold calculations.
07

Remaining Probability Computations

When spin measurement in first is down (spin measures against expectation relative to chosen frame), complement angle applies: \( P_{-+} = \sin^2(\theta_{ab}/2) \cos^2(\pi-\theta_{bc}/2) \). Additionally, alternative parallel computation \( P_{+ o} = \cos^2(\theta_{bc}/2) \), and \( P_{--} = \sin^2(\theta_{ab}/2)\sin^2(\theta_{bc}/2)\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Spin-1/2 Particles
Spin-1/2 particles are a fundamental concept in quantum mechanics and are often used as a model system for understanding quantum spin. These particles have a spin quantum number of 1/2, which is a way of describing their intrinsic angular momentum. The term "1/2" does not imply any real or classical rotation, but rather indicates the quantized nature of this property.

For spin-1/2 particles, there are only two possible measurable states along any axis. These states are often referred to as "spin up" or "spin down." For example, if a spin-1/2 particle is measured along the z-axis, the possible outcomes are either "spin up" (denoted as \( |+z\rangle \)) or "spin down" (denoted as \( |-z\rangle \)).

  • Spin-up and spin-down states are vector states in a quantum Hilbert space.
  • The quantization of spin is a core feature of quantum mechanics and has no direct classical analogy.
Spin-1/2 systems serve as the simplest model for studying more complex quantum systems and illustrate fundamental principles such as superposition and entanglement.
Spin Measurement Probabilities
In quantum mechanics, measurement plays a crucial role and can affect the state of a quantum system. When we measure a spin-1/2 particle, the process changes the spin state depending on the axis along which the measurement is performed. The probability of obtaining a particular spin state after the measurement is determined by the initial state of the particle and the axis of measurement.

For example, if we start with a spin-1/2 particle prepared in the state \( |+z\rangle \) and measure its spin along the x-axis, the probability of finding it in the state \( |+x\rangle \) is \( \left| \frac{1}{\sqrt{2}} \right|^2 = \frac{1}{2} \).
  • The outcome probability is computed using the inner product of the initial and final states.
  • This probability reflects the concept of quantum superposition, where multiple states contribute to the final observed state.
Understanding these probabilities helps us to predict the behavior of quantum systems as they undergo measurements and demonstrates the intrinsic randomness of quantum events.
Quantum State Evolution
Quantum state evolution concerns how quantum states change over time in the absence of measurement influences or external forces. For spin-1/2 particles, if no external fields are acting on the particles, their quantum states remain constant over time unless measured.

In the described problem, spin-1/2 particles start in a defined state at time \( t = 0 \), for example, \( |+z\rangle \), and remain in this state until a measurement occurs. When a measurement does take place (e.g., along the x-axis), it effectively "collapses" the quantum state into one of the possible outcome states, altering the initial setup.
  • This evolution without external influence means the system's physical description remains stable between measurements.
  • The measurement process itself introduces change shown in probability distributions, rather than the state evolving.
By understanding quantum state evolution, we grasp how systems persist over time and how measurements impact the observed state of quantum particles.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

(a) Aparticle of spin 1 and a particle of spin 2 are at rest in a configuration such that the total spin is \(3,\) and its \(z\) component is \(\hbar .\) If you measured the \(z\) -component of the angular momentum of the spin-2 particle, what values might you get, and what is the probability of each one? Comment: Using Clebsch-Gordan tables is like driving a stick-shift- scary and frustrating when you start out, but easy once you get the hang of it. (b) An electron with spin down is in the state \(\psi_{510}\) of the hydrogen atom. If you could measure the total angular momentum squared of the electron alone (not including the proton spin), what values might you get, and what is the probability of each?

(a) Construct the wave function for hydrogen in the state \(n=4, \ell=3\), \(m=3 .\) Express your answer as a function of the spherical coordinates \(r\), \(\theta,\) and \(\phi\). (b) Find the expectation value of \(r\) in this state. (As always, look up any nontrivial integrals.) (c) If you could somehow measure the observable \(L_{x}^{2}+L_{y}^{2}\) on an atom in this state, what value (or values) could you get, and what is the probability of each?

A particle of mass \(m\) is placed in a finite spherical well: \(V(r)=\left\\{\begin{array}{ll}-V_{0}, & r \leq a ;\\\ 0, & r>a.\end{array}\right.\) Find the ground state, by solving the radial equation with \(\ell=0 .\) Show that there is no bound state if \(V_{0} a^{2}<\pi^{2} \hbar^{2} / 8 m\).

Consider a particle with charge \(q,\) mass \(m,\) and spin \(s,\) in a uniform magnetic field \(\mathbf{B}_{0}\). The vector potential can be chosen as \(\mathbf{A}=-\frac{1}{2} \mathbf{r} \times \mathbf{B}_{0}\). (a) Verify that this vector potential produces a uniform magnetic field \(\mathbf{B}_{0}\). (b) Show that the Hamiltonian can be written \(H=\frac{p^{2}}{2 m}+q \varphi-\mathbf{B}_{0} \cdot\left(\gamma_{o} \mathbf{L}+\gamma \mathbf{S}\right)+\frac{q^{2}}{8 m}\left[r^{2} B_{0}^{2}-\left(\mathbf{r} \cdot \mathbf{B}_{0}\right)^{2}\right]\), where \(\gamma_{o}=q / 2 m\) is the gyromagnetic ratio for orbital motion. Note: The term linear in \(\mathbf{B}_{0}\) makes it energetically favorable for the magnetic moments (orbital and spin) to align with the magnetic field; this is the origin of paramagnetism in materials. The term quadratic in \(\mathbf{B}_{0}\) leads to the opposite effect: diamagnetism.\(^{75}\)

An electron is at rest in an oscillating magnetic field \(\mathbf{B}=B_{0} \cos (\omega t) \hat{k}\), where \(B_{0}\) and \(\omega\) are constants. (a) Construct the Hamiltonian matrix for this system. (b) The electron starts out (at \(t=0\) ) in the spin-up state with respect to the \(x\) axis (that is: \(\chi(0)=\chi_{+}^{(x)}\) ). Determine \(\chi(t)\) at any subsequent time. Beware: This is a time-dependent Hamiltonian, so you cannot get \(\chi(t)\) in the usual way from stationary states. Fortunately, in this case you can solve the time-dependent Schrödinger equation (Equation 4.162 ) directly. (c) Find the probability of getting \(-\hbar / 2\), if you measure \(S_{x}\). (d) What is the minimum field \(\left(B_{0}\right)\) required to force a complete flip in \(S_{x}\) ?

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.