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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

Use the WKB method to find the (approximate) energy eigenvalues for the onedimensional simple harmonic oscillator potential \(V(x)=m \omega^{2} x^{2} / 2\).

An electron is subject to a uniform, time-independent magnetic field of strength \(B\) in the positive \(z\)-direction. At \(t=0\) the electron is known to be in an eigenstate of \(\mathbf{S} \cdot \hat{\mathbf{n}}\) with eigenvalue \(\hbar / 2\), where \(\hat{\mathbf{n}}\) is a unit vector, lying in the \(x z\)-plane, that makes an angle \(\beta\) with the \(z\)-axis. a. Obtain the probability for finding the electron in the \(S_{x}=\hbar / 2\) state as a function of time. b. Find the expectation value of \(S_{x}\) as a function of time. c. For your own peace of mind show that your answers make good sense in the extreme cases (i) \(\beta \rightarrow 0\) and (ii) \(\beta \rightarrow \pi / 2\).

A particle with mass \(m\) moves in one dimension and is acted on by a constant force \(F\). Find the operators \(x(t)\) and \(p(t)\) in the Heisenberg picture, and find their expectation values for an arbitrary state \(|\alpha\rangle\). Use \(\langle x(0)\rangle=x_{0}\) and \(\langle p(0)\rangle=p_{0}\). The result should be obvious. Comment on how to do this problem in the Schrödinger picture, but do not try to work it through.

Make the definitions $$ J_{\pm} \equiv \hbar a_{\pm}^{\dagger} a_{\mp}, \quad J_{z} \equiv \frac{\hbar}{2}\left(a_{+}^{\dagger} a_{+}-a_{-}^{\dagger} a_{-}\right), \quad N \equiv a_{+}^{\dagger} a_{+}+a_{-}^{\dagger} a_{-} $$ where \(a_{\pm}\)and \(a_{\pm}^{\dagger}\) are the annihilation and creation operators of two independent simple harmonic oscillators satisfying the usual simple harmonic oscillator commutation relations. Also make the definition $$ \mathbf{J}^{2} \equiv J_{z}^{2}+\frac{1}{2}\left(J_{+} J_{-}+J_{-} J_{+}\right) $$ Prove $$ \left[J_{z}, J_{\pm}\right]=\pm \hbar J_{\pm}, \quad\left[\mathbf{J}^{2}, J_{z}\right]=0, \quad \mathbf{J}^{2}=\left(\frac{\hbar^{2}}{2}\right) N\left[\left(\frac{N}{2}\right)+1\right] $$

A particle of mass \(m\) moves in one dimension \(x\) under a potential energy \(V(x)\). a. For \(V(x)=-V_{0} b \delta(x), V_{0}>0, b>0\), find the bound-state energy eigenvalue \(E\). b. Generalize this to the "double delta function" potential $$ V(x)=-V_{0} \frac{b}{2}\left[\delta\left(x+\frac{a}{2}\right)+\delta\left(x-\frac{a}{2}\right)\right] $$ and find the bound-state energy eigenvalues and plot the corresponding eigenfunctions. Also show that you get the expected results as \(a \rightarrow 0\).

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