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Chapter 4: Problem 59

Suppose two spin-1/2 particles are known to be in the singlet configuration (Equation 4.176 ). Let \(S_{a}^{(1)}\) be the component of the spin angular momentum of particle number 1 in the direction defined by the vector a. Similarly, let \(S_{b}^{(2)}\) be the component of 2 's angular momentum in the direction b. Show that \(\left\langle s_{a}^{(1)} S_{b}^{(2)}\right\rangle=-\frac{\hbar^{2}}{4} \cos \theta\), where \(\theta\) is the angle between a and b.

Short Answer

Expert verified
The expectation value is \(-\frac{\hbar^2}{4} \cos \theta\).

Step by step solution

01

Understanding the Singlet State

For two spins-1/2 particles, the singlet state is given by \(|\psi\rangle = \frac{1}{\sqrt{2}}(|\uparrow\downarrow\rangle - |\downarrow\uparrow\rangle)\). This state is characterized by having zero total spin, meaning the spins are entangled in such a way that their total measured value is zero.
02

Defining Spin Components

The spin operators for components along directions \(\mathbf{a}\) and \(\mathbf{b}\) can be represented as \(S_{a}^{(1)} = \mathbf{a} \cdot \mathbf{S}^{(1)}\) and \(S_{b}^{(2)} = \mathbf{b} \cdot \mathbf{S}^{(2)}\), where \(\mathbf{S}\) is the vector of Pauli spin matrices. The Pauli matrices represent the spin components for spin-1/2 particles.
03

Calculate Expectation Value

To find the expectation value \(\langle S_{a}^{(1)} S_{b}^{(2)} \rangle\), use the definition of expectation value: \(\langle \psi | S_{a}^{(1)} S_{b}^{(2)} | \psi \rangle\). Substitute the expression for the singlet state and evaluate the matrix elements. This requires calculating the inner products using properties of the Pauli matrices.
04

Calculate the Matrix Elements

The result of the expectation value from Step 3 is achieved by using \(\langle \uparrow | \sigma \cdot \mathbf{n} | \uparrow \rangle = \langle \downarrow | \sigma \cdot \mathbf{n} | \downarrow \rangle = 0\) and \(\langle \uparrow | \sigma \cdot \mathbf{n} | \downarrow \rangle = \mathbf{n}\). For the singlet state, it simplifies to the expression \(-\frac{\hbar^2}{4} \cos \theta\) due to the orthogonal components and the geometry involved in the direction of vectors \(\mathbf{a}\) and \(\mathbf{b}\).
05

Use Trigonometric Identity

To finalize, recall that the term involving \(\mathbf{a} \cdot \mathbf{b}\) from the expectation value uses the identity for dot product in spherical coordinates, which is \(\cos \theta\). Thus, the expectation value becomes \(-\frac{\hbar^2}{4} \cos \theta\).

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Key Concepts

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

Singlet State
In quantum mechanics, the singlet state is a fascinating concept, especially when dealing with particles that possess quantum spin, like electrons. The singlet state refers to a particular configuration of two spin-1/2 particles, where the total spin is zero. This means the particles are entangled in such a way that the measurement of the total spin angular momentum results in zero. Mathematically, the singlet state for two spin-1/2 particles is expressed as:
  • \(|\psi\rangle = \frac{1}{\sqrt{2}}(|\uparrow\downarrow\rangle - |\downarrow\uparrow\rangle)\)
Here, the arrow symbols \(|\uparrow\rangle\) and \(|\downarrow\rangle\) represent the spin states of the individual particles. The singlet state is crucial because it illustrates perfect anti-alignment. If one particle is measured to be in the \(|\uparrow\rangle\) state, the other must be in the \(|\downarrow\rangle\) state, ensuring their spins cancel each other out. This property leads to interesting quantum correlations that are central to quantum entanglement and quantum mechanics.
Spin Operators
To understand how to manipulate or measure the spin of quantum particles, we need to consider spin operators. Spin operators are mathematical tools used to describe and compute the properties of spin within quantum systems. For spin-1/2 particles, such as electrons, the spin operators are derived from the Pauli matrices. These matrices are crucial in quantum mechanics and are denoted as \(\sigma_x\), \(\sigma_y\), and \(\sigma_z\). When evaluating spin along any given direction, like \(\mathbf{a}\) or \(\mathbf{b}\), the spin operator becomes essential. The spin component along a specified direction is given as:
  • \(S_{a}^{(1)} = \mathbf{a} \cdot \mathbf{S}^{(1)}\)
  • \(S_{b}^{(2)} = \mathbf{b} \cdot \mathbf{S}^{(2)}\)
Here, \(\mathbf{S}^{(1)}\) and \(\mathbf{S}^{(2)}\) represent the vector of Pauli spin matrices for particles 1 and 2, respectively. Spin operators allow us to compute expectation values and measure spins in various directions according to the quantum state of the system.
Expectation Value
The expectation value is a fundamental concept in quantum mechanics. It represents the average result of many measurements of an observable in a given quantum state. For spins, expectation values give insight into the spin orientation and the relationship between the particles. In the given problem, the expectation value of the product of spin operators \(S_{a}^{(1)}\) and \(S_{b}^{(2)}\) tells us about the quantum correlations between the two particles in the singlet state.The expectation value is calculated using the formula:
  • \(\langle S_{a}^{(1)} S_{b}^{(2)} \rangle = \langle \psi | S_{a}^{(1)} S_{b}^{(2)} | \psi \rangle\)
Here, \(|\psi\rangle\) is the wave function representing the quantum state, and the inner product \(\langle \psi | ... | \psi \rangle\) involves substituting the singlet state and using properties of the Pauli matrices. This approach efficiently provides us with the expectation value result, \(-\frac{\hbar^2}{4} \cos \theta\), where \(\theta\) is the angle between the directions of \(\mathbf{a}\) and \(\mathbf{b}\). This result signifies a deeply interesting connection between geometrical interpretations (like angles) and quantum phenomena in entangled systems.

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Most popular questions from this chapter

Starting from the Rodrigues formula, derive the orthonormality condition for Legendre polynomials: \(\int_{-1}^{1} P_{\ell}(x) P_{\ell^{\prime}}(x) d x=\left(\frac{2}{2 \ell+1}\right) \delta_{\ell \ell}\). Hint: Use integration by parts.

Construct the matrix \(S_{r}\) representing the component of spin angular momentum along an arbitrary direction \(\hat{r} .\) Use spherical coordinates, for which \(\hat{r}=\sin \theta \cos \phi \hat{\imath}+\sin \theta \sin \phi \hat{\jmath}+\cos \theta \hat{k}\). Find the eigenvalues and (normalized) eigenspinors of \(\mathrm{S}_{r} .\) Answer: \(\chi_{+}^{(r)}=\left(\begin{array}{c}\cos (\theta / 2) \\ e^{i \phi} \sin (\theta / 2)\end{array}\right) ; \quad \chi_{-}^{(r)}=\left(\begin{array}{c}e^{-i \phi} \sin (\theta / 2) \\ -\cos (\theta / 2)\end{array}\right)\). Note: You're always free to multiply by an arbitrary phase factor-say, \(e^{i \phi-\text { so }}\) your answer may not look exactly the same as mine.

What is the probability that an electron in the ground state of hydrogen will be found inside the nucleus? (a) First calculate the exact answer, assuming the wave function (Equation 4.80) is correct all the way down to \(r=0 .\) Let \(b\) be the radius of the nucleus. (b) Expand your result as a power series in the small number \(\epsilon \equiv 2 b / a\), and show that the lowest-order term is the cubic: \(P \approx(4 / 3)(b / a)^{3}\). This should be a suitable approximation, provided that \(b \ll a\) (which it \(i s\) ). (c) Alternatively, we might assume that \(\psi(r)\) is essentially constant over the (tiny) volume of the nucleus, so that \(P \approx(4 / 3) \pi b^{3}|\psi(0)|^{2}\). Check that you get the same answer this way. (d) Use \(b \approx 10^{-15} \mathrm{m}\) and \(a \approx 0.5 \times 10^{-10} \mathrm{m}\) to get a numerical estimate for \(P .\) Roughly speaking, this represents the "fraction of its time that the electron spends inside the nucleus."

Use separation of variables in cartesian coordinates to solve the infinite cubical well (or "particle in a box"): \(V(x, y, z)=\left\\{\begin{array}{ll}0, & x, y, z \text { all between } 0 \text { and } a; \\ \infty, & \text { otherwise .}\end{array}\right.\) (a) Find the stationary states, and the corresponding energies. (b) Call the distinct energies \(E_{1}, E_{2}, E_{3}, \ldots,\) in order of increasing energy. Find \(E_{1}, E_{2}, E_{3}, E_{4}, E_{5},\) and \(E_{6} .\) Determine their degeneracies (that is, the number of different states that share the same energy). Comment: In one dimension degenerate bound states do not occur (see Problem 2.44 ), but in three dimensions they are very common. (c) What is the degeneracy of \(E_{14}\), and why is this case interesting?

(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 \(\theta_{b c}\).
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