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Chapter 7: Problem 16

Consider a system subject to the potential $$ V(x)=\frac{k}{2} x^{2}+\frac{\gamma}{6} x^{3}+\frac{\delta}{24} x^{4} $$ Calculate the ground-state energy of this system using a trial function of the form $$ \phi=c_{1} \psi_{0}(x)+c_{2} \psi_{2}(x) $$ where \(\psi_{0}(x)\) and \(\psi_{2}(x)\) are the first two even wave functions of a harmonic oscillator. Why did we not include \(\psi_{1}(x) ?\)

Short Answer

Expert verified
The ground-state energy comes mainly from \( \psi_0(x) \) and is corrected with perturbation contributions.

Step by step solution

01

Understand the Potential

The given potential is expressed as \( V(x)=\frac{k}{2} x^{2}+\frac{\gamma}{6} x^{3}+\frac{\delta}{24} x^{4} \). This potential includes quadratic, cubic, and quartic terms. The quadratic term corresponds to a harmonic oscillator potential, while the cubic and quartic terms are perturbations.
02

Trial Function Formulation

The trial function is \( \phi = c_{1} \psi_{0}(x) + c_{2} \psi_{2}(x) \), where \( \psi_{0}(x) \) is the ground state of the harmonic oscillator, and \( \psi_{2}(x) \) is the second excited state. Odd functions like \( \psi_{1}(x) \) are not included because they are not even functions and do not match the symmetry of the potential.
03

Write the Perturbation Hamiltonian

In this system, the Hamiltonian is \( H = H_{0} + V(x) \), where \( H_{0} \) is the harmonic oscillator Hamiltonian. We will use perturbation theory to evaluate the contribution of the cubic and quartic terms.
04

Use Perturbation Theory

We need to apply perturbation theory to estimate the ground-state energy. The first-order correction is calculated as the expectation value of the perturbation, \( \langle \psi_{0} | V(x) | \psi_{0} \rangle \), plus a correction from \( \psi_{2}(x) \). Higher-order terms can be ignored in a basic approximation.
05

Calculate Expectation Values

For a more precise correction, compute the expectation values such as \( \langle \psi_0 | x^3 | \psi_0 \rangle \) and \( \langle \psi_0 | x^4 | \psi_0 \rangle \). Keep symmetry in mind: odd-powered terms will give zero contributions when integrated over symmetric limits.
06

Evaluate Contributions from \(\psi_{2}(x)\)

Since \( \psi_{2}(x) \) is involved, calculate \( \langle \psi_2 | V(x) | \psi_2 \rangle \) similarly. Check interaction terms like \( \langle \psi_0 | V(x) | \psi_2 \rangle \), which contribute to second-order corrections.
07

Combine Results

Sum up the calculated contributions from \( \langle \psi_0 | V(x) | \psi_0 \rangle \), \( \langle \psi_2 | V(x) | \psi_2 \rangle \), and any necessary cross terms to find the correction to the ground-state energy. The lowest energy, primarily from \( \psi_0 \), is the approximate ground-state energy of the perturbed system.

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Key Concepts

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

Perturbation Theory
Perturbation theory is used to find an approximate solution to a problem that cannot be solved exactly. It is a powerful technique often applied in quantum mechanics. When dealing with complex systems, it allows us to start with a simple system we can solve, and then gradually incorporate the complexities as small adjustments or 'perturbations'.

In the context of calculating ground-state energy, perturbation theory helps in evaluating the impact of additional terms in the potential energy function, such as the cubic and quartic terms in this exercise. The problem is approached by first solving the basic harmonic oscillator, which we know and can easily manage. Then, those extra terms are considered as small adjustments, and their effects are calculated using this theory. By finding corrections, often first-order or second-order, the ground-state energy can be finely adjusted.

Important steps involve computing expectation values, like
  • The first-order correction, which is the average change in energy due to the perturbation.
  • Second-order corrections, needed when interaction terms like cross terms between wave functions are significant.
Use of perturbation theory prevents us from needing to solve the full problem analytically, which is usually difficult or impossible.
Harmonic Oscillator
The harmonic oscillator is a fundamental concept used in many areas of physics, particularly in quantum mechanics. It's a model where a particle is subject to forces that restore it to an equilibrium position, typically represented by a quadratic potential energy function \[ V(x) = \frac{k}{2} x^2 \].

In this exercise, the harmonic oscillator acts as the base system to which additional complexities (perturbations) are added. Its solutions are well-known, featuring even and odd wave functions that define the system's energy levels. The lowest energy level is associated with the ground state wave function, \( \psi_0(x) \), while \( \psi_2(x) \) represents the second excited state.

Harmonic oscillators are widely applicable because they approximate situations where systems oscillate or vibrate, such as molecules in chemistry or phonons in solid-state physics. By using the known solutions of a harmonic oscillator, it becomes easier to understand how extra potential terms affect the system's behavior.
Potential Energy Function
A potential energy function describes the potential energy of a system as a function of position or configuration. It essentially tells us how energy changes as a particle moves through space.

In this context, the potential energy function is given as \[ V(x) = \frac{k}{2} x^2 + \frac{\gamma}{6} x^3 + \frac{\delta}{24} x^4 \].

  • The quadratic term, \( \frac{k}{2} x^2 \), represents a harmonic oscillator potential and is symmetric.
  • The cubic term, \( \frac{\gamma}{6} x^3 \), breaks the symmetry and can introduce complexities like anharmonicity.
  • The quartic term, \( \frac{\delta}{24} x^4 \), further adds to the potential, possibly stabilizing or destabilizing certain states.
The symmetry of the potential largely influences which wave functions will contribute to the system's energy. Since odd functions do not match the symmetry of even potentials, \( \psi_1(x) \), which is an odd function, does not play a role here. Instead, even functions like \( \psi_0(x) \) and \( \psi_2(x) \) are used because they align symmetrically with the potential's even terms, allowing us to accurately calculate the related ground-state energy.

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Most popular questions from this chapter

We will derive the equations for first-order perturbation theory in this problem. The problem we want to solve is \\[ \hat{H} \psi=E \psi \\] where \\[ \hat{H}=\hat{H}^{(0)}+\hat{H}^{(1)} \\] and where the problem \\[ \hat{H}^{(0)} \psi^{(0)}=E^{(0)} \psi^{(0)} \\] has been solved exactly previously, so that \(\psi^{(0)}\) and \(E^{(0)}\) are known. Assuming now that the effect of \(\hat{H}^{(1)}\) is small, write \\[ \begin{array}{l} \psi=\psi^{(0)}+\Delta \psi \\ E=E^{(0)}+\Delta E \end{array} \\] where we assume that \(\Delta \psi\) and \(\Delta E\) are small. Substitute Equations 3 into Equation 1 to obtain \\[ \begin{array}{l} \hat{H}^{(0)} \psi^{(0)}+\hat{H}^{(1)} \psi^{(0)}+\hat{H}^{(0)} \Delta \psi+\hat{H}^{(1)} \Delta \psi \\ =E^{(0)} \psi^{(0)}+\Delta E \psi^{(0)}+E^{(0)} \Delta \psi+\Delta E \Delta \psi \end{array} \\] The first terms on each side of Equation 4 cancel because of Equation \(2 .\) In addition, we will neglect the last terms on each side because they represent the product of two small terms. Thus, Equation 4 becomes \\[ \hat{H}^{(0)} \Delta \psi+\hat{H}^{(1)} \psi^{(0)}=E^{(0)} \Delta \psi+\Delta E \psi^{(0)} \\] Realize that \(\Delta \psi\) and \(\Delta E\) are the unknown quantities in this equation. Note that all the terms in Equation (5) are of the same order, in the sense that each is the product of an unperturbed term and a small term. We say that this equation is first order in the perturbation and that we are using first-order perturbation theory. The two terms we have neglected in Equation 4 are second-order terms and lead to second-order (and higher) corrections. Equation 5 can be simplified considerably. Multiply both sides from the left by \(\psi^{(0) *}\) and integrate over all space to get \\[ \int \psi^{(0) *}\left[\hat{H}^{(0)}-E^{(0)}\right] \Delta \psi d \tau+\int \psi^{(0) *} \hat{H}^{(1)} \psi^{(0)} d \tau=\Delta E \int \psi^{(0) *} \psi^{(0)} d \tau \\] The integral in the last term in Equation 6 is unity because \(\psi^{(0)}\) is taken to be normalized. More important, however, is that the first term on the left side of Equation 6 is zero. Use the fact that \(\hat{H}^{(0)}-E^{(0)}\) is Hermitian to show that \\[ \int \psi^{(0) *}\left[\hat{H}^{(0)}-E^{(0)}\right] \Delta \psi d \tau=\int\left\\{\left[\hat{H}^{(0)}-E^{(0)}\right] \psi^{(0)}\right\\}^{*} \Delta \psi d \tau \\] But according to Equation \(2,\) the integrand here vanishes. Thus, Equation 6 becomes \\[ \Delta E=\int \psi^{(0) *} \hat{H}^{(1)} \psi^{(0)} d \tau \\] Equation 7 is called the first-order correction to \(E^{(0)}\). To first order, the energy is \\[ E=E^{(0)}+\int \psi^{(0) *} \hat{H}^{(1)} \psi^{(0)} d \tau+\text { higher order terms } \\]

Use the variational method to calculate the ground-state energy of a particle constrained to move within the region \(0 \leq x \leq a\) in a potential given by \\[ \begin{aligned} V(x) &=V_{0} x & & 0 \leq x \leq \frac{a}{2} \\ &=V_{0}(a-x) & & \frac{a}{2} \leq x \leq a \end{aligned} \\]

Use a trial function of the form \(e^{-\beta x^{2}}\) with \(\beta\) as a variational parameter to calculate the ground-state energy of a harmonic oscillator. Compare your result with the exact energy \(h v / 2 .\) Why is the agreement so good?

This problem involves the proof of the variational principle, Equation 7.4. Let \(\hat{H} \psi_{n}=\) \(E_{n} \psi_{n}\) be the problem of interest, and let \(\phi\) be our approximation to \(\psi_{e^{-}}\)Even though we do not know the \(\psi_{n}\), we can express \(\phi\) formally as $$ \phi=\sum_{n} c_{n} \psi_{n} $$ where the \(c_{n}\) are constants. Using the fact that the \(\psi_{n}\) are orthonormal, show that $$ c_{n}=\int \psi_{n}^{*} \phi d \tau $$ We do not know the \(\psi_{n}\), however, so Equation 1 is what we call a formal expansion. Now substitute Equation 1 into $$ E_{\phi}=\frac{\int \phi^{*} \hat{H} \phi d \tau}{\int \phi^{*} \phi d \tau} $$ to obtain $$ E_{\phi}=\frac{\sum_{n} c_{n}^{*} c_{n} E_{n}}{\sum_{n} c_{n}^{*} c_{\kappa}} $$ Subtract \(E_{0}\) from the left side of the above equation and \(E_{0} \sum_{n} c_{n}^{*} c_{n} / \sum_{n} c_{n}^{*} c_{n}\) from the right side to obtain, $$ E_{\theta}-E_{0}=\frac{\sum_{n} c_{n}^{*} c_{n}\left(E_{n}-E_{0}\right)}{\sum_{n} c_{n}^{*} c_{n}} $$ Now explain why every term on the right side is positive, proving that \(E_{\phi} \geq E_{0^{\prime}}\)

This problem shows that terms in a trial function that correspond to progressively higher energies contribute progressively less to the ground- state energy. For algebraic simplicity, assume that the Hamiltonian operator can be written in the form \\[ \hat{H}=\hat{H}^{(0)}+\hat{H}^{(1)} \\] and choose a trial function \\[ \phi=c_{1} \psi_{1}+c_{2} \psi_{2} \\] where \\[ \hat{H}^{(0)} \psi_{j}=E_{j}^{(0)} \psi_{j} \quad j=1,2 \\] Show that the secular equation associated with the trial function is \\[ \left|\begin{array}{cc} H_{11}-E & H_{12} \\ H_{12} & H_{22}-E \end{array}\right|=\left|\begin{array}{cc} E_{1}^{(0}+E_{1}^{(1)}-E & H_{12} \\ H_{12} & E_{2}^{(0)}+E_{2}^{(1)}-E \end{array}\right|=0 \\] where \\[ E_{j}^{(1)}=\int \psi_{j}^{*} \hat{H}^{(1)} \psi_{j} d \tau \quad \text { and } \quad H_{12}=\int \psi_{1}^{*} \hat{H}^{(1)} \psi_{2} d \tau \\] Solve Equation 1 for \(E\) to obtain \\[ \begin{aligned} E=\frac{E_{1}^{(0)}+E_{1}^{(1)}+E_{2}^{(0)}+E_{2}^{(1)}}{2} & \\ & \pm \frac{1}{2}\left\\{\left[E_{1}^{(0)}+E_{1}^{(1)}-E_{2}^{(0)}-E_{2}^{(1)}\right]^{2}+4 H_{12}^{2}\right\\}^{1 / 2} \end{aligned} \\] If we arbitrarily assume that \(E_{1}^{(0)}+E_{1}^{(1)} < E_{2}^{(0)}+E_{2}^{(1)},\) then we take the positive sign in Equation 2 and write \\[ \begin{array}{c} E=\frac{E_{1}^{(0)}+E_{1}^{(1)}+E_{2}^{(0)}+E_{2}^{(1)}}{2}+\frac{E_{1}^{(0)}+E_{1}^{(1)}-E_{2}^{(0)}-E_{2}^{(1)}}{2} \\\ \times\left\\{1+\frac{4 H_{12}^{2}}{\left[E_{1}^{(0)}+E_{1}^{(1)}-E_{2}^{(0)}-E_{2}^{(1)}\right]^{2}}\right\\}^{1 / 2} \end{array} \\] Use the expansion \((1+x)^{1 / 2}=1+x / 2+\cdots, x<1\) to get \\[ E=E_{1}^{(0)}+E_{1}^{(1)}+\frac{H_{12}^{2}}{E_{1}^{(0)}+E_{1}^{(1)}-E_{2}^{(0)}-E_{2}^{(1)}}+\cdots \\] Note that if \(E_{1}^{(0)}+E_{1}^{(1)}\) and \(E_{2}^{(0)}+E_{2}^{(1)}\) are widely separated, the term involving \(H_{12}^{2}\) in Equation 3 is small. Therefore, the energy is simply that calculated using \(\psi_{1}\) alone; the \(\psi_{2}\) part of the trial function contributes little to the overall energy. The general result is that terms in a trial function that correspond to higher and higher energies contribute less and less to the total ground- state energy.
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