Problem 16 - A Modern Approach to Quantum Mechanics

Question

A spin- \(\frac{1}{2}\) particle is in the pure state \(|\psi\rangle=a|+\mathbf{z}\rangle+b|-\mathbf{z}\rangle\). (a) Construct the density matrix in the \(S_{z}\) basis for this state. (b) Starting with your result in (a), determine the density matrix in the \(S_{x}\) basis, where $$ |+\mathbf{x}\rangle=\frac{1}{\sqrt{2}}|+\mathbf{z}\rangle+\frac{1}{\sqrt{2}}|-\mathbf{z}\rangle \quad|-\mathbf{x}\rangle=\frac{1}{\sqrt{2}}|+\mathbf{z}\rangle-\frac{1}{\sqrt{2}}|-\mathbf{z}\rangle $$ (c) Use your result for the density matrix in (b) to determine the probability that a measurement of \(S_{x}\) yields \(\hbar / 2\) for the state \(|\psi\rangle\).

Step-by-Step Solution

Answer

The probability is \(\frac{1}{2} (|a|^2 + |b|^2 + 2\text{Re}(a^*b))\).

Step 1: Define the state in the Sz basis

The given state \(|\psi\rangle\) is \(|\psi\rangle = a|+\mathbf{z}\rangle + b|-\mathbf{z}\rangle\). In the \(S_z\) basis, this is represented as the column vector: \[|\psi\rangle = \begin{pmatrix} a \ b \end{pmatrix}\]

Step 2: Construct the density matrix in the Sz basis

The density matrix \(\rho\) is defined as \(\rho = |\psi\rangle\langle\psi|\). Thus:\[\rho = \begin{pmatrix} a \ b \end{pmatrix} \begin{pmatrix} a^* & b^* \end{pmatrix} = \begin{pmatrix} |a|^2 & ab^* \ a^*b & |b|^2 \end{pmatrix}\]

Step 3: Express the basis transformation for Sx

The transformation to the \(S_x\) basis involves the new states:\[|+\mathbf{x}\rangle=\frac{1}{\sqrt{2}}|+\mathbf{z}\rangle+\frac{1}{\sqrt{2}}|-\mathbf{z}\rangle, \quad| -\mathbf{x}\rangle = \frac{1}{\sqrt{2}}|+\mathbf{z}\rangle-\frac{1}{\sqrt{2}}|-\mathbf{z}\rangle\]The transformation matrix \(U\) is:\[U = \begin{pmatrix}\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}}\end{pmatrix}\]

Step 4: Transform the density matrix to the Sx basis

The density matrix in the \(S_x\) basis is given by \(\rho'_x = U\rho U^\dagger\). Calculate:\[U^\dagger = \begin{pmatrix}\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}}\end{pmatrix}\text{(since it's a symmetric matrix)}\]Compute \(\rho'_x\):\[\rho'_x = \begin{pmatrix}\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}}\end{pmatrix}\begin{pmatrix} |a|^2 & ab^* \ a^*b & |b|^2 \end{pmatrix}\begin{pmatrix}\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}}\end{pmatrix}\]This results in:\[\rho'_x = \begin{pmatrix}(\frac{|a|^2 + |b|^2}{2} + \text{Re}(a^*b)) &* \*&(\frac{|a|^2 + |b|^2}{2} - \text{Re}(a^*b))\end{pmatrix}\]The probabilities will involve the diagonal elements.

Step 5: Calculate the probability for Sx

The probability \(P(S_x = \frac{\hbar}{2})\) is given by the top-left entry of \(\rho'_x\):\[P(S_x = \frac{\hbar}{2}) = \frac{1}{2} (|a|^2 + |b|^2 + 2\text{Re}(a^*b))\]

Key Concepts

Spin-1/2 Particle

A spin-1/2 particle is a fundamental concept in quantum mechanics. It represents a simple quantum system with two possible spin states. These states are typically represented as \( |+z\rangle \) and \( |-z\rangle \), which correspond to the spin pointing "up" and "down" along the z-axis, respectively.
To describe these states, we use a vector in a two-dimensional complex space called the state vector or wave function. In the case of our exercise, the spin state of the particle is a linear combination \( |\psi\rangle = a|+z\rangle + b|-z\rangle \). Here, \( a \) and \( b \) are complex numbers that define the probabilities of measuring the particle in each state.

• The probabilities of measuring each state are given by \( |a|^2 \) and \( |b|^2 \).
• The total probability must be 1, leading to the normalization condition: \( |a|^2 + |b|^2 = 1 \).

Density Matrix

The density matrix is a powerful tool in quantum mechanics. It provides a complete description of a quantum system's statistical state. Unlike state vectors, which can only describe pure states, density matrices can also represent mixed states.
The density matrix \( \rho \) for a pure state \( |\psi\rangle \) is constructed as \( \rho = |\psi\rangle\langle \psi| \). For our given state, this becomes:
\[\rho = \begin{pmatrix} a \ b \end{pmatrix} \begin{pmatrix} a^* & b^* \end{pmatrix} = \begin{pmatrix} |a|^2 & ab^* \ a^*b & |b|^2 \end{pmatrix}\]

• The diagonal elements represent the probabilities of finding the system in each of the basis states.
• The off-diagonal elements, containing products like \( ab^* \), capture the coherence between states, which are crucial for quantum superpositions and interference phenomena.

Basis Transformation

In quantum mechanics, changing the basis can make certain calculations easier or reveal different insights about a system. A basis transformation involves switching from one set of basis vectors to another. Here, we change from the \( S_z \) to the \( S_x \) basis.
For this exercise, the transformation uses new states:
\[|+x\rangle = \frac{1}{\sqrt{2}}|+z\rangle + \frac{1}{\sqrt{2}}|-z\rangle, \quad |-x\rangle = \frac{1}{\sqrt{2}}|+z\rangle - \frac{1}{\sqrt{2}}|-z\rangle\]
The matrix that performs this transformation is:
\[U = \begin{pmatrix}\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \ \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}}\end{pmatrix}\]

• To find the density matrix in the new basis, compute \( \rho'_x = U\rho U^\dagger \).
• This involves matrix multiplication and offers a transformed view of the same physical system.

Probability Measurement

Probability measurement in quantum mechanics provides the likelihood of observing certain outcomes when measuring a quantum system's observable, like spin.
For the density matrix in the \( S_x \) basis, the probability that a measurement of \( S_x \) yields \( \hbar / 2 \) is given by the top-left entry of the transformed density matrix \( \rho'_x \). This is calculated as:
\[P(S_x = \frac{\hbar}{2}) = \frac{1}{2} (|a|^2 + |b|^2 + 2\text{Re}(a^*b))\]

• The diagonal elements of the density matrix represent the probabilities of each state in the chosen basis.
• This method ensures consistent predictions with the probabilistic nature of quantum mechanics.
• The result encapsulates intrinsic quantum interference effects through terms like \( \text{Re}(a^*b) \).