Question

By taking the Fourier transform of both sides of the integral form of the Schrödinger equation, show that for bound-state problems \((E<0)\), the equation in "momentum space" can be written as $$ \tilde{\psi}(\mathbf{p})=-\frac{2 \mu}{(2 \pi)^{3 / 2} \hbar^{2}}\left(\frac{1}{\kappa^{2}+p^{2}}\right) \int \tilde{V}(\mathbf{p}-\mathbf{q}) \tilde{\psi}(\mathbf{q}) d^{3} q, $$ where \(\kappa^{2}=-2 \mu E / \hbar^{2}\).

Short answer

A Fourier transform is applied to the integral form of the Schrödinger equation, which then allows for recasting the problem in momentum space. The modified equation, which is used for bound-state problems, assumes a new form that includes a convolution with the Fourier transform of the potential \( \hat{V}(p) \).

Step-by-step solution

Analyzing the Schrödinger equation

The integral form of the Schrödinger equation is given by \[ \hat{H} \psi = E \psi \] where \( \hat{H} \) is the Hamiltonian operator. In 3D, \( \hat{H} \) is given by \[ \hat{H} = -\frac{\hbar^2}{2 \mu} \nabla^2 + V(x) \] where \( \mu \) is the reduced mass of the particle, \( \hbar \) is the reduced Planck’s constant, \( \nabla^2 \) is the Laplacian (which represents the second spatial derivative), and \( V(x) \) is the potential.

Fourier Transformation

The Fourier Transformation is defined as \[ \tilde{f}(p) = \frac{1}{(2\pi)^{3/2}} \int f(r) e^{-ip \cdot r} dr \] Apply this transformation to the Schrödinger equation, taking into account that the Fourier transform of a derivative is given by \[ FT{f'}(p) = ip FT{f}(p) \] This is important because the Laplacian is a second derivative, so the Fourier transform will bring down a factor of \( -p^2 \).

Apply the Fourier Transformation

Applying the Fourier transform to the Schrödinger equation gives \[ (-\frac{\hbar^2}{2 \mu} (ip)^2 + V(x)) \hat{\psi}(p) = E \hat{\psi}(p) \] This simplifies to \[ (-\frac{\hbar^2}{2 \mu} p^2 + V(x)) \hat{\psi}(p) = E \hat{\psi}(p) \]

Bound-state problems

For bound-state problems where \(E<0\), the Schrödinger equation will be further transformed, leading to the final equation.

Key concepts

Schrödinger Equation

The Schrödinger equation is a cornerstone of quantum physics, describing how quantum systems evolve over time. It's basically a kinda fancy-looking equation where we see

• The Hamiltonian operator \( \hat{H} \)
• An energy value \( E \)
• The wave function \( \psi \)

The Hamiltonian tells you the total energy of the system. It's got two main parts: kinetic energy and potential energy. In more detail, for a simple system it's given by \[ \hat{H} = -\frac{\hbar^2}{2 \mu} abla^2 + V(x)\]where:

• \( \hbar \) is the reduced Planck's constant, a tiny number that shows up everywhere in quantum mechanics.
• \( \mu \) is the reduced mass of interacting particles, and you can think of it like the 'weight' for the waving particles.
• \( abla^2 \) is the Laplacian, which tells you about how the wave function curves in space.
• \( V(x) \) represents the potential energy.

This equation basically helps you find the possible energy levels \( E \) a system can have.

Bound-state Problems

Bound-state problems in quantum mechanics focus on situations where particles are trapped inside a potential well because their energy is less than the potential around them, denoted by \( E < 0 \). This is kinda like how a marble can be stuck in a bowl and doesn't have enough energy to roll out.

In such problems, the solutions to the Schrödinger equation are special because they have to satisfy certain conditions:

• These solutions lead to discrete (quantized) energy levels, meaning not just any energy is possible.
• The wave functions of these solutions are normalizable, meaning they don’t go on forever but exist in a confined space.

Understanding bound states helps us make sense of how electrons orbit around a nucleus and why they have specific energy levels.

Momentum Space

Momentum space is a different way to look at wave functions, rather than in real space. Here, instead of describing a particle's position, we're describing its momentum. It's quite like having a different lens on reality that focuses on momentum rather than position.

Mathematically, getting to momentum space involves applying a transformation called the Fourier Transform. This move is powerful because it converts complicated differential equations (like the Schrödinger equation) into algebraic equations, which are often easier to handle.

In momentum space, the wave function \( \tilde{\psi}(\mathbf{p}) \) is expressed in terms of momentum \( \mathbf{p} \). This comes in super handy, especially for certain systems and problems like scattering or bound-state energies as it simplifies the math involved.

• Processes like differentiation in real space become multiplication in momentum space—which can be way simpler for solving equations.
• Studying interactions in momentum space can give insights into how particles scatter or interact without the hassle of working with spatial coordinates.

Overall, momentum space is a mighty tool in the quantum physicist's toolkit.