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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

Consider a two-level system with \(E_{1}0\) by exactly solving the coupled differential equation $$ i \hbar \dot{c}_{k}=\sum_{n=1}^{2} V_{k n}(t) e^{i \omega_{k n} t} c_{n} \quad(k=1,2) $$ b. Do the same problem using time-dependent perturbation theory to lowest nonvanishing order. Compare the two approaches for small values of \(\gamma\). Treat the following two cases separately: (i) \(\omega\) very different from \(\omega_{21}\) and (ii) \(\omega\) close to \(\omega_{21}\). Answer for (a): (Rabi's formula) $$ \begin{aligned} &\left|c_{2}(t)\right|^{2}=\frac{\gamma^{2} / \hbar^{2}}{\gamma^{2} / \hbar^{2}+\left(\omega-\omega_{21}\right)^{2} / 4} \sin ^{2}\left\\{\left[\frac{\gamma^{2}}{\hbar^{2}}+\frac{\left(\omega-\omega_{21}\right)^{2}}{4}\right]^{1 / 2} t\right\\} \\ &\left|c_{1}(t)\right|^{2}=1-\left|c_{2}(t)\right|^{2} \end{aligned} $$

Consider a particle in one dimension moving under the influence of some timeindependent potential. The energy levels and the corresponding eigenfunctions for this problem are assumed to be known. We now subject the particle to a traveling pulse represented by a time-dependent potential, $$ V(t)=A \delta(x-c t) . $$ a. Suppose at \(t=-\infty\) the particle is known to be in the ground state whose energy eigenfunction is \(\langle x \mid i\rangle=u_{i}(x)\). Obtain the probability for finding the system in some excited state with energy eigenfunction \(\langle x \mid f\rangle=u_{f}(x)\) at \(t=+\infty\). b. Interpret your result in (a) physically by regarding the \(\delta\)-function pulse as a superposition of harmonic perturbations; recall $$ \delta(x-c t)=\frac{1}{2 \pi c} \int_{-\infty}^{\infty} d \omega e^{i \omega[(x / c)-t]} . $$ Emphasize the role played by energy conservation, which holds even quantum mechanically as long as the perturbation has been on for a very long time.

Consider an atom made up of an electron and a singly charged \((Z=1)\) triton \(\left({ }^{3} \mathrm{H}\right)\). Initially the system is in its ground state \((n=1, l=0)\). Suppose the system undergoes beta decay, in which the nuclear charge suddenly increases by one unit (realistically by emitting an electron and an antineutrino). This means that the tritium nucleus (called a "triton") turns into a helium \((Z=2)\) nucleus of mass \(3\left({ }^{3} \mathrm{He}\right)\). a. Obtain the probability for the system to be found in the ground state of the resulting helium ion. b. The available energy in tritium beta decay is about \(18 \mathrm{keV}\) and the size of the \({ }^{3} \mathrm{He}\) atom is about \(1 \AA\). Check that the time scale \(T\) for the transformation satisfies the criterion of validity for the sudden approximation.

Consider a particle bound to a fixed center by a spherically symmetric potential \(V(r)\). a. Prove $$ |\psi(0)|^{2}=\left(\frac{m}{2 \pi \hbar^{2}}\right)\left\langle\frac{d V}{d r}\right\rangle $$ for all \(s\) states, ground and excited. b. Check this relation for the ground state of a three-dimensional isotropic oscillator, the hydrogen atom, and so on. (Note: This relation has actually been found to be useful in guessing the form of the potential between a quark and an antiquark. See Moxhay and Rosner, J. Math. Phys., 21 (1980) 1688.)

A one-dimensional simple harmonic oscillator of angular frequency \(\omega\) is acted upon by a spatially uniform but time-dependent force (not potential) $$ F(t)=\frac{\left(F_{0} \tau / \omega\right)}{\left(\tau^{2}+t^{2}\right)}, \quad-\infty

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