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In nondegenerate time-independent perturbation theory, what is the probability of finding in a perturbed energy eigenstate \((|k\rangle)\) the corresponding unperturbed eigenstate \(\left(\left|k^{(0)}\right\rangle\right)\) ? Solve this up to terms of order \(\lambda^{2}\).

Short Answer

Expert verified
The probability is approximately \(1 + 2\lambda \text{Re}(\langle k^{(0)} | k^{(1)}\rangle)\).

Step by step solution

01

Understand Perturbation Theory Basics

In nondegenerate time-independent perturbation theory, we start with the Hamiltonian of the system split into an unperturbed part \(H^{(0)}\) and a perturbation \(\lambda V\), where \(\lambda\) is a small parameter. The energy eigenstates \(|k^{(0)}\rangle\) are the solutions to the unperturbed Hamiltonian.
02

Express Perturbed Eigenstate

The perturbed eigenstate \(|k\rangle\) can be expressed as \(|k\rangle = |k^{(0)}\rangle + \lambda |k^{(1)}\rangle + \lambda^2 |k^{(2)}\rangle + \cdots\). Here, \(|k^{(1)}\rangle\) and \(|k^{(2)}\rangle\) represent the first and second-order corrections to the state.
03

Calculate Probability Amplitude

The probability amplitude of finding the unperturbed state \(|k^{(0)}\rangle\) in the perturbed state \(|k\rangle\) is given by the inner product \(\langle k^{(0)} | k \rangle\). Using the expression from Step 2, this yields \(\langle k^{(0)} | k \rangle = \langle k^{(0)} | (|k^{(0)}\rangle + \lambda |k^{(1)}\rangle + \lambda^2 |k^{(2)}\rangle)\).
04

Simplify Inner Product

Simplifying \(\langle k^{(0)} | k \rangle\) gives \(\langle k^{(0)} | k^{(0)}\rangle + \lambda \langle k^{(0)} | k^{(1)}\rangle + \lambda^2 \langle k^{(0)} | k^{(2)}\rangle\). Since \(|k^{(0)}\rangle\) is normalized, \(\langle k^{(0)} | k^{(0)}\rangle = 1\). Hence, \(\langle k^{(0)} | k \rangle = 1 + \lambda \langle k^{(0)} | k^{(1)}\rangle + \lambda^2 \langle k^{(0)} | k^{(2)}\rangle\).
05

Approximate Probability

The probability is the absolute square of the amplitude: \(|\langle k^{(0)} | k \rangle|^2\). Up to second order, this is approximately \(1 + 2\lambda \text{Re}(\langle k^{(0)} | k^{(1)}\rangle) + 2\lambda^2\text{Re}(\langle k^{(0)} | k^{(2)}\rangle) \), ignoring higher-order \(\lambda^2\) terms.

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Key Concepts

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

Nondegenerate Perturbation Theory
Nondegenerate perturbation theory is a vital concept in quantum mechanics that helps us understand how systems respond to small disturbances or changes in their environment. The theory deals specifically with systems where energy levels are distinct, meaning they are nondegenerate.

In simpler terms, when a quantum system is subjected to a minor external influence, its Hamiltonian can be broken down into two parts: the unperturbed Hamiltonian (\(H^{(0)}\)) that describes the system in its original state, and a perturbing Hamiltonian (\(\lambda V\)) representing the small change. Here, \(\lambda\) is a scaling parameter that defines the strength of the perturbation, assumed to be quite small.

This approach allows us to study how the original energy eigenstates, the states with definite energy of the unperturbed Hamiltonian, are modified by this perturbation. We then express these new states, called perturbed energy eigenstates, as a series expansion that accounts for different orders of perturbation.
Energy Eigenstates
Energy eigenstates are fundamental to understanding quantum systems. They are solutions to the Schrödinger equation associated with a particular energy value, called an eigenvalue. These states represent configurations in which the system can exist with a definite, unchanging energy.

When dealing with perturbation theory, the unperturbed energy eigenstates (\(|k^{(0)}\rangle\)) are what we initially focus on. These represent the state of the system before any external influence is applied.

Once a perturbation is introduced, however, these states shift to become perturbed energy eigenstates (\(|k\rangle\)). The change occurs as the system adjusts to the new Hamiltonian, which includes both the original and perturbing parts.
Probability Amplitude
Probability amplitude is an essential concept that quantifies the likelihood of transitioning between quantum states. It is expressed as the inner product between two states, indicating the overlap between them.

In the context of perturbation theory, the probability amplitude is calculated to assess how much the original, unperturbed state resembles the perturbed state after accounting for the perturbation.

Mathematically, the probability amplitude \(\langle k^{(0)} | k \rangle\) involves taking the inner product of the unperturbed state \(|k^{(0)}\rangle\) with the perturbed state \(|k\rangle = |k^{(0)}\rangle + \lambda |k^{(1)}\rangle + \lambda^2 |k^{(2)}\rangle + \cdots\). Simplifying this inner product gives the terms that show how the system transitions or remains similar under the influence of perturbation.

Taking the absolute square of this amplitude gives us the probability, revealing how likely the perturbed state is to still mirror the original.
Hamiltonian
The Hamiltonian is a central concept in quantum mechanics, analogous to the role of total energy in classical mechanics. It is an operator that dictates how a quantum system evolves over time.

In the realm of perturbation theory, the Hamiltonian of a system before any external changes are applied is known as the unperturbed Hamiltonian (\(H^{(0)}\)). This describes the intrinsic energies and interactions within the system.

When a perturbation occurs, we introduce a perturbing Hamiltonian (\(\lambda V\)), where \(\lambda\) is a small parameter indicating the extent of the change. The total Hamiltonian of the system then becomes a combination of these components: \(H = H^{(0)} + \lambda V\).

This decomposition allows us to tackle complex systems analytically by breaking them into more manageable subproblems, thereby understanding how slight alterations affect the system's overall behavior.

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Most popular questions from this chapter

A one-electron atom whose ground state is nondegenerate is placed in a uniform electric field in the \(z\)-direction. Obtain an approximate expression for the induced electric dipole moment of the ground state by considering the expectation value of ez with respect to the perturbed state vector computed to first order. Show that the same expression can also be obtained from the energy shift \(\Delta=-\alpha|\mathbf{E}|^{2} / 2\) of the ground state computed to second order. (Note: \(\alpha\) stands for the polarizability.) Ignore spin.

A simple harmonic oscillator (in one dimension) is subjected to a perturbation $$ H_{1}=b x $$ where \(b\) is a real constant. a. Calculate the energy shift of the ground state to lowest nonvanishing order. b. Solve this problem exactly and compare with your result obtained in (a).

Consider an isotropic harmonic oscillator in two dimensions. The Hamiltonian is $$ H_{0}=\frac{p_{x}^{2}}{2 m}+\frac{p_{y}^{2}}{2 m}+\frac{m \omega^{2}}{2}\left(x^{2}+y^{2}\right) $$ a. What are the energies of the three lowest-lying states? Is there any degeneracy? b. We now apply a perturbation $$ V=\delta m \omega^{2} x y $$ where \(\delta\) is a dimensionless real number much smaller than unity. Find the zerothorder energy eigenket and the corresponding energy to first order [that is, the unperturbed energy obtained in (a) plus the first-order energy shift] for each of the three lowest-lying states. c. Solve the \(H_{0}+V\) problem exactly. Compare with the perturbation results obtained in (b).

This chapter derived two of the three relativistic corrections to the one- electron atom, namely \(\Delta_{K}^{(1)}\) from "relativistic kinetic energy," and \(\Delta_{L S}^{(1)}\) from the spin-orbit interaction. A third term comes from the spread of the electron wave function in the region of changing electric field. The perturbation for this "Darwin term" is $$ V_{D}=-\frac{1}{8 m^{2} c^{2}} \sum_{i=1}^{3}\left[p_{i},\left[p_{i}, e \phi(r)\right]\right] $$ where \(\phi(r)\) is the Coulomb potential. Find \(\Delta_{D}^{(1)}\) and show that $$ \Delta_{n j}^{(1)} \equiv \Delta_{K}^{(1)}+\Delta_{L S}^{(1)}+\Delta_{D}^{(1)}=\frac{m c^{2}(Z \alpha)^{4}}{2 n^{3}}\left[\frac{3}{4 n}-\frac{1}{j+1 / 2}\right] . $$ In Section \(8.4\) we will compare this expression to the result of solving the Dirac equation in the presence of the Coulomb potential.

Compute the Stark effect for the \(2 s_{1 / 2}\) and \(2 p_{1 / 2}\) levels of hydrogen for a field \(\mathscr{E}\) sufficiently weak so that \(e \mathscr{E} a_{0}\) is small compared to the fine structure, but take the Lamb \(\operatorname{shift} \delta(\delta=1057 \mathrm{MHz})\) into account (that is, ignore \(2 p_{3 / 2}\) in this calculation). Show that for \(e \mathscr{E} a_{0} \ll \delta\), the energy shifts are quadratic in \(\mathscr{E}\), whereas for \(e \mathscr{E} a_{0} \gg \delta\) they are linear in \(\mathscr{E}\). Briefly discuss the consequences (if any) of time reversal for this problem. This problem is from Gottfried (1966), Problem 7-3.

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