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A one-dimensional simple harmonic oscillator with natural frequency \(\omega\) is in initial state $$ |\alpha\rangle=\frac{1}{\sqrt{2}}|0\rangle+\frac{e^{i \delta}}{\sqrt{2}}|1\rangle $$ where \(\delta\) is a real number. a. Find the time-dependent wave function \(\left\langle x^{\prime} \mid \alpha ; t\right\rangle\) and evaluate the (time-dependent) expectation values \(\langle x\rangle\) and \(\langle p\rangle\) in the state \(|\alpha ; t\rangle\), i.e. in the Schrödinger picture. b. Now calculate \(\langle x\rangle\) and \(\langle p\rangle\) in the Heisenberg picture and compare the results.

Short Answer

Expert verified
The expectation values \(\langle x \rangle\) and \(\langle p \rangle\) are time-independent in both pictures. They are zero due to the symmetry of the states involved.

Step by step solution

01

Time Evolution of the Quantum State

To find the time-dependent wave function \(|\alpha; t\rangle\), we apply the time evolution operator \(U(t) = e^{-i\hat{H}t/\hbar}\) to the initial state \(|\alpha\rangle\). For a simple harmonic oscillator, the Hamiltonian is \(\hat{H} = \hbar\omega(\hat{a}^\dagger\hat{a} + \frac{1}{2})\). So, \(|\alpha; t\rangle = e^{-i\hat{H}t/\hbar}|\alpha\rangle\). Thus, \(|n; t\rangle = e^{-i(n+1/2)\omega t}|n\rangle\). Therefore, \(|\alpha; t\rangle = \frac{1}{\sqrt{2}}e^{-i\omega t/2}|0\rangle + \frac{e^{i\delta}}{\sqrt{2}}e^{-i3\omega t/2}|1\rangle\).
02

Position Representation of the Time-Dependent State

The wave function in the position representation is \(\langle x' | \alpha; t \rangle = \frac{1}{\sqrt{2}}e^{-i\omega t/2}\langle x' | 0 \rangle + \frac{e^{i\delta}}{\sqrt{2}}e^{-i3\omega t/2}\langle x' | 1 \rangle\). The wave functions \(\langle x' | 0 \rangle\) and \(\langle x' | 1 \rangle\) are the ground and first excited state wave functions of the harmonic oscillator. Substitute these normalized functions to find \(\langle x' | \alpha; t \rangle\).
03

Expectation Value of \(\langle x \rangle\) in the Schrödinger Picture

The expectation value of \(x\) is given by \(\langle x \rangle = \langle \alpha; t | \hat{x} | \alpha; t \rangle\). This involves calculating \(\langle n | \hat{x} | m \rangle\) for \(|n\rangle, |m\rangle\), using \(\hat{x} = \sqrt{\frac{\hbar}{2m\omega}}(\hat{a}+\hat{a}^\dagger)\). Compute this by substituting the expressions for \(\hat{a}, \hat{a}^\dagger\), and evaluating the inner products.
04

Expectation Value of \(\langle p \rangle\) in the Schrödinger Picture

Similar to \(\langle x \rangle\), use \(\langle p \rangle = \langle \alpha; t | \hat{p} | \alpha; t \rangle\), where \(\hat{p} = i\sqrt{\frac{\hbar m \omega}{2}}(\hat{a}^\dagger - \hat{a})\). Using the expressions for \(|\alpha; t \rangle\) and the operators \(\hat{p},\hat{a},\hat{a}^\dagger\), compute \(\langle p \rangle\).
05

Heisenberg Picture Calculation

In the Heisenberg picture, operators evolve with time, but states remain constant. The position and momentum operators evolve as \(\hat{x}_H(t) = e^{i\hat{H}t/\hbar}\hat{x}e^{-i\hat{H}t/\hbar}\) and \(\hat{p}_H(t) = e^{i\hat{H}t/\hbar}\hat{p}e^{-i\hat{H}t/\hbar}\). Calculate these using the commutation relations and find \(\langle x \rangle, \langle p \rangle\) similar to the Schrödinger calculations.
06

Compare Results

Compare the results obtained from the Schrödinger picture and the Heisenberg picture. Since they describe the same physics, \(\langle x \rangle\) and \(\langle p \rangle\) should be identical in both pictures.

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Key Concepts

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

Schrödinger Picture
In the Schrödinger picture of quantum mechanics, the focus is placed on how the state of a quantum system evolves over time. This is crucial for understanding time-dependent problems such as those involving a quantum harmonic oscillator.

In the Schrödinger picture, states evolve while operators remain constant. For the given initial state \( |\alpha\rangle = \frac{1}{\sqrt{2}}|0\rangle + \frac{e^{i\delta}}{\sqrt{2}}|1\rangle \), time evolution is governed by the time evolution operator \( U(t) = e^{-i\hat{H}t/\hbar} \). This operator utilizes the Hamiltonian, \( \hat{H} = \hbar\omega(\hat{a}^\dagger\hat{a} + \frac{1}{2}) \), specific to simple harmonic oscillators.

To find the time-dependent state \( |\alpha; t\rangle \), apply the evolution operator to the initial state. The exact forms of the ground and first excited states \( |0\rangle \) and \( |1\rangle \) are multiplied by exponential factors \( e^{-i\omega t/2} \) and \( e^{-i3\omega t/2} \), reflecting their energy levels.

This approach leads to a time-dependent wave function which is essential in finding properties like the position expectation value \( \langle x \rangle \) and the momentum expectation value \( \langle p \rangle \).
Heisenberg Picture
The Heisenberg picture offers a different perspective on quantum mechanics. Unlike the Schrödinger picture, in the Heisenberg picture, states are fixed and operators carry time dependence. This shifts the analysis from time-evolving states to operators that change with time.

For the quantum harmonic oscillator, the time evolution of operators such as position \( \hat{x}_H(t) \) and momentum \( \hat{p}_H(t) \) is described as \( e^{i\hat{H}t/\hbar}\hat{x}e^{-i\hat{H}t/\hbar} \) and \( e^{i\hat{H}t/\hbar}\hat{p}e^{-i\hat{H}t/\hbar} \), respectively.

This formulation uses commutation relations to handle the operators' time evolution, which ultimately affects calculations of expectation values. Whether in the Schrödinger or Heisenberg picture, operators like position and momentum derive their behavior from the Hamiltonian's impact. Such calculations typically involve mathematical operations through Heisenberg's equations of motion.

When comparing results from both pictures, expectation values like \( \langle x \rangle \) and \( \langle p \rangle \) are consistent—demonstrating the equivalence of these quantum mechanical views.
Time Evolution in Quantum Mechanics
Time evolution is a fundamental concept in quantum mechanics that provides insights into how quantum states change. In the context of the Schrödinger and Heisenberg pictures, time evolution is illustrated differently but leads to the same observable predictions.

In the Schrödinger picture, time evolution impacts the state vector. The operator \( U(t) = e^{-i\hat{H}t/\hbar} \) dynamically modifies the state's coefficients in a basis of energy eigenstates. For example, the quantum harmonic oscillator state is altered by energy-specific exponential terms. Pay attention to how factors of time appear in the expansion of the wave function and adjust the properties of the quantum state.

On the other hand, in the Heisenberg picture, time dependence shifts to operators. By employing the commutation relations and Heisenberg’s equation, operators such as \( \hat{x} \) and \( \hat{p} \) embody the time evolution. Here, the original quantum state remains unchanged but is subjected to time-evolving operators that reflect changes in the system.

Both frameworks demonstrate how time evolution dictates the shape and dynamics of a wave function and operators. This duality underscores the richness and versatility of quantum mechanical analysis and highlights the deep-seated equivalence in its predictions across different pictures.

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Most popular questions from this chapter

A particle of mass \(m\) is confined to a one-dimensional square well with finite walls. That is, a potential \(V(x)=0\) for \(-a \leq x \leq+a\), and \(V(x)=V_{0}=\eta\left(\hbar^{2} / 2 m a^{2}\right)\) otherwise. You are to find the bound-state energy eigenvalues as \(E=\varepsilon V_{0}\) along with their wave functions. a. Set the problem up with a wave function \(A e^{\alpha x}\) for \(x \leq-a, D e^{-\alpha x}\) for \(x \geq+a\), and \(B e^{i k x}+C e^{-i k x}\) inside the well. Match the boundary conditions at \(x=\pm a\) and show that \(k\) and \(\alpha\) must satisfy \(z=\pm z^{*}\) where \(z \equiv e^{i a k}(k-i \alpha)\). Proceed to find a purely real or purely imaginary expression for \(z\) in terms of \(k\) and \(\alpha\). b. Find the wave functions for the two choices of \(z\) and show that the purely real (imaginary) choice leads to a wave function that is even (odd) under the exchange \(x \rightarrow-x\). c. Find a transcendental equation for each of the two wave functions relating \(\eta\) and \(\varepsilon\). Show that even a very shallow well \((\eta \rightarrow 0)\) has at least one solution for the even wave function, but you are not guaranteed any solution for an odd wave function. d. For \(\eta=10\), find all the energy eigenvalues and plot their normalized wave functions.

Consider the neutron interferometer. Prove that the difference in the magnetic fields that produce two successive maxima in the counting rates is given by $$ B=\frac{4 \pi \hbar c}{|e| g_{n} \hbar l} $$ where \(g_{n}(=-1.91)\) is the neutron magnetic moment in units of \(-e \hbar / 2 m_{n} c\), and \(\lambda \equiv\) \(\lambda / 2 \pi\). This problem was in fact analyzed in the paper by Bernstein, Phys. Rev. Lett., 18 (1967) 1102.

A particle of mass \(m\) in one dimension is bound to a fixed center by an attractive \(\delta\)-function potential: $$ V(x)=-\lambda \delta(x) \quad(\lambda>0) . $$ At \(t=0\), the potential is suddenly switched off (that is, \(V=0\) for \(t>0\) ). Find the wave function for \(t>0\). (Be quantitative! But you need not attempt to evaluate an integral that may appear.)

Consider the spin-precession problem discussed in the text. It can also be solved in the Heisenberg picture. Using the Hamiltonian $$ H=-\left(\frac{e B}{m c}\right) S_{z}=\omega S_{z} $$ write the Heisenberg equations of motion for the time-dependent operators \(S_{x}(t)\), \(S_{y}(t)\), and \(S_{z}(t)\). Solve them to obtain \(S_{x, y, z}\) as functions of time.

A box containing a particle is divided into a right and a left compartment by a thin partition. If the particle is known to be on the right (left) side with certainty, the state is represented by the position eigenket \(|R\rangle(|L\rangle)\), where we have neglected spatial variations within each half of the box. The most general state vector can then be written as $$ |\alpha\rangle=|R\rangle\langle R \mid \alpha\rangle+|L\rangle\langle L \mid \alpha\rangle $$ where \(\langle R \mid \alpha\rangle\) and \(\langle L \mid \alpha\rangle\) can be regarded as "wave functions." The particle can tunnel through the partition; this tunneling effect is characterized by the Hamiltonian $$ H=\Delta(|L\rangle\langle R|+| R\rangle\langle L|) $$ where \(\Delta\) is a real number with the dimension of energy. a. Find the normalized energy eigenkets. What are the corresponding energy eigenvalues? b. In the Schrödinger picture the base kets \(|R\rangle\) and \(|L\rangle\) are fixed, and the state vector moves with time. Suppose the system is represented by \(|\alpha\rangle\) as given above at \(t=0 .\) Find the state vector \(\left|\alpha, t_{0}=0 ; t\right\rangle\) for \(t>0\) by applying the appropriate time-evolution operator to \(|\alpha\rangle\). c. Suppose at \(t=0\) the particle is on the right side with certainty. What is the probability for observing the particle on the left side as a function of time? d. Write down the coupled Schrödinger equations for the wave functions \(\langle R| \alpha, t_{0}=\) \(0 ; t\rangle\) and \(\left\langle L \mid \alpha, t_{0}=0 ; t\right\rangle\). Show that the solutions to the coupled Schrödinger equations are just what you expect from (b). e. Suppose the printer made an error and wrote \(H\) as $$ H=\Delta|L\rangle\langle R| $$ By explicitly solving the most general time-evolution problem with this Hamiltonian, show that probability conservation is violated.

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