/*! 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 16 Consider a one-dimensional simpl... [FREE SOLUTION] | 91Ó°ÊÓ

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Consider a one-dimensional simple harmonic oscillator. a. Using $$ \left.\left.\begin{array}{l} a \\ a^{\dagger} \end{array}\right\\}=\sqrt{\frac{m \omega}{2 \hbar}}\left(x \pm \frac{i p}{m \omega}\right), \quad \begin{array}{l} a|n\rangle \\ a^{\dagger}|n\rangle \end{array}\right\\}=\left\\{\begin{array}{l} \sqrt{n}|n-1\rangle \\ \sqrt{n+1}|n+1\rangle \end{array}\right. $$ evaluate \(\langle m|x| n\rangle,\langle m|p| n\rangle,\langle m|\\{x, p\\}| n\rangle,\left\langle m\left|x^{2}\right| n\right\rangle\), and \(\left\langle m\left|p^{2}\right| n\right\rangle .\) b. Translated from classical physics, the virial theorem states that $$ \left\langle\frac{\mathbf{p}^{2}}{m}\right\rangle=\langle\mathbf{x} \cdot \nabla V\rangle \text { (3D) } \quad \text { or } \quad\left\langle\frac{p^{2}}{m}\right\rangle=\left\langle x \frac{d V}{d x}\right\rangle $$ Check that the virial theorem holds for the expectation values of the kinetic and the potential energy taken with respect to an energy eigenstate.

Short Answer

Expert verified
Steps verify matrix elements and confirm the virial theorem holds for SHM.

Step by step solution

01

Express Position and Momentum Operators

The operators for position \( x \) and momentum \( p \) in terms of the creation (\( a^\dagger \)) and annihilation operators (\( a \)) are given by:\[x = \sqrt{\frac{\hbar}{2m\omega}}(a + a^\dagger)\]\[p = i\sqrt{\frac{m\hbar\omega}{2}}(a^\dagger - a)\]These expressions will be used to calculate the required matrix elements.
02

Calculate \( \langle m | x | n \rangle \)

Substitute the expression for \( x \) from Step 1. Using the properties of the ladder operators:\[\langle m | x | n \rangle = \sqrt{\frac{\hbar}{2 m \omega}} \left( \sqrt{n} \delta_{m, n-1} + \sqrt{n + 1} \delta_{m, n+1} \right)\]Thus, this element is non-zero only if \( m = n \pm 1 \).
03

Calculate \( \langle m | p | n \rangle \)

Using the expression for \( p \) from Step 1, we obtain:\[\langle m | p | n \rangle = i\sqrt{\frac{m \hbar \omega}{2}} \left( \sqrt{n+1} \delta_{m, n+1} - \sqrt{n} \delta_{m, n-1} \right)\]This means the result is non-zero only when \( m = n \pm 1 \).
04

Evaluate \( \langle m | \{x, p\} | n \rangle \)

The anticommutator \( \{x, p\} = xp + px \) is obtained using the product of operators:\[\langle m | \{ x, p \} | n \rangle = \langle m | (xp + px) | n \rangle\]Apply expressions from previous steps while keeping the non-commutativity and linearity of expectation values in mind to find the elements when \( m = n\pm 1, n\pm 2 \).
05

Calculate \( \langle m | x^2 | n \rangle \)

The operator \( x^2 \) is computed using:\[\langle m | x^2 | n \rangle = \frac{\hbar}{2m\omega} (\langle m | a^2 | n \rangle + \langle m | (a^\dagger)^2 | n \rangle + 2 \langle m | a^\dagger a | n \rangle + \langle m | 1 | n \rangle)\]With the rules for ladder operator actions, non-zero terms depend on \( m \) being \( n, n \pm 2 \) or adjacent states.
06

Calculate \( \langle m | p^2 | n \rangle \)

The operator \( p^2 \) is similar:\[\langle m | p^2 | n \rangle = -\frac{m\hbar\omega}{2} (\langle m | a^2 | n \rangle + \langle m | (a^\dagger)^2 | n \rangle - 2 \langle m | a^\dagger a | n \rangle + \langle m | 1 | n \rangle)\]Again, use ladder properties to evaluate this based on \( m \) being \( n, n \pm 2 \).
07

Verify Virial Theorem for SHM

For a simple harmonic oscillator, the potential is \( V=\frac{1}{2}m\omega^2x^2 \). Check\[\left\langle \frac{p^2}{m} \right\rangle = \left\langle x \frac{dV}{dx} \right\rangle\]In this system, both terms equal \( \frac{1}{2} \hbar \omega\left(n + \frac{1}{2}\right) \), consistent with equality predicted by the theorem.

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Key Concepts

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

Ladder Operators
Ladder operators are a fundamental part of quantum mechanics, especially in the study of the quantum harmonic oscillator. They consist of two operators: the creation operator, denoted as \( a^\dagger \), and the annihilation operator, \( a \). These operators are used to move between different quantum states or energy levels of a system.

- The annihilation operator \( a \) lowers the quantum state, removing an energy quantum from it. It reduces the quantum number \( n \) by 1.
- Conversely, the creation operator \( a^\dagger \) raises the quantum state by adding an energy quantum, increasing the quantum number \( n \) by 1.

The application of these operators in calculating matrix elements in quantum harmonic oscillators reveals essential properties like \( \langle m | x | n \rangle \) or \( \langle m | p | n \rangle \). It showcases how these operators play a pivotal role in determining when results are non-zero, i.e., when \( m = n \pm 1 \). This behavior highlights how these operators streamline calculations within quantum systems.
Virial Theorem
The virial theorem provides deep insights into the balance of kinetic and potential energy within a stable quantum system. For quantum systems, it translates the behavior of these energies into an average sense, applicable to the harmonic oscillator as well.

In a one-dimensional harmonic oscillator, the theorem is expressed as \( \left\langle \frac{p^2}{m} \right\rangle = \left\langle x \frac{dV}{dx} \right\rangle \). This showcases how average kinetic energy and the derivative of the potential energy interact.

- In our case of the harmonic oscillator, this equality simplifies to show that both sides equal \( \frac{1}{2} \hbar \omega(n + \frac{1}{2}) \).
- This result confirms that in quantum harmonic oscillators, energy distribution between motion and the potential energy is balanced according to this theorem.

The virial theorem thus assures us that our computations of expectation values using ladder operators are consistent with the quantum harmonic oscillator's physical principles.
Expectation Values
Expectation values are critical for understanding measurable quantities in quantum mechanics. They provide average outcomes for measurements, essential for predicting quantum system behavior. Calculating them involves using operators like position \( x \) and momentum \( p \).

- For instance, \( \langle m | x | n \rangle \) and \( \langle m | p | n \rangle \) are used to determine the expected position and momentum respectively.
- These values are integral for describing the system's state in terms of observable quantities. They're calculated using the wave function's probability distribution in combination with respective operators.

Expectation values aid us in connecting the abstract mathematical framework of quantum mechanics with tangible, measurable aspects of a quantum system. In the case of a harmonic oscillator, these calculations reveal much about the state's transitions through the use of ladder operators.
Position and Momentum Operators
Position and momentum operators are central elements in quantum mechanics, representing the observable quantities of a system. In the context of the quantum harmonic oscillator, expressing these operators using ladder operators helps simplify calculations.

- The position operator \( x \) is expressed as \( x = \sqrt{\frac{\hbar}{2m\omega}}(a + a^\dagger) \).
- The momentum operator \( p \) is given by \( p = i\sqrt{\frac{m\hbar\omega}{2}}(a^\dagger - a) \).

Using these forms, the operators enable us to efficiently compute matrix elements like \( \langle m | x | n \rangle \) and \( \langle m | p | n \rangle \), which are foundational in determining expectation values.

These expressions underscore the utility of ladder operators in simplifying complex quantum operator manipulations, illustrating their effectiveness in revealing the dynamics of quantum systems like the harmonic oscillator.

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Most popular questions from this chapter

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.

Consider again a one-dimensional simple harmonic oscillator. Do the following algebraically, that is, without using wave functions. a. Construct a linear combination of \(|0\rangle\) and \(|1\rangle\) such that \(\langle x\rangle\) is as large as possible. b. Suppose the oscillator is in the state constructed in (a) at \(t=0\). What is the state vector for \(t>0\) in the Schrödinger picture? Evaluate the expectation value \(\langle x\rangle\) as a function of time for \(t>0\) using (i) the Schrödinger picture and (ii) the Heisenberg picture. c. Evaluate \(\left\langle(\Delta x)^{2}\right\rangle\) as a function of time using either picture.

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.

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.)

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) \(\beta \rightarrow \pi / 2\).

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