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Calculate the singular part of the GF for the three-dimensional free Schrödinger operator $$ i \hbar \frac{\partial}{\partial t}-\frac{\hbar^{2}}{2 \mu} \nabla^{2} $$.

Short Answer

Expert verified
The singular part of Green's Function in three dimensions for the Schrödinger operator would be obtained by Fourier solution to the wave equation.

Step by step solution

01

Identify the operator

The operator in question is the Schrödinger operator. In the given equation: \(i \hbar \frac{\partial}{\partial t}-\frac{\hbar^{2}}{2 \mu} \nabla^{2}\), we can recognize two components: the first term is \(i \hbar \frac{\partial}{\partial t}\), which acts on the time-component of the function, and the second term is \(-\frac{\hbar^{2}}{2 \mu} \nabla^{2}\), which is the spatial part, as \(\nabla^{2}\) denotes the Laplacian operator.
02

Find Green's function for the operator

Once we've identified the operator and its components, the next step is to find its Green's function (GF). Here we are interested in the three-dimensional GF, which satisfies \( (i \hbar \frac{\partial}{\partial t}-\frac{\hbar^{2}}{2 \mu} \nabla^{2}) G = \delta(x - x')\), where \(\delta\) denotes the Dirac delta function and x' represents some point in space.
03

Obtain singular part of the Green's function

The singular part of the Green's function refers to the part of the solution that becomes infinite as \(x \to x'\). It stems from the singularity of the Dirac delta function in the equation for the Green's function. In 3D the singular part of Green's function for the Schrödinger operator is obtained by `[fn \text{Fourier solution to the wave equation}]`

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Key Concepts

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

Green's Function
Green's function is a powerful tool in mathematical physics, especially when dealing with differential equations. It essentially acts as a bridge between the input of a system and its response. In our context, the Green's function of the Schrödinger operator serves to link a point-like impulse to the resulting wave function evolution in space and time.
For the three-dimensional Schrödinger equation, the Green's function must satisfy a specific form:
  • The expression for the Green's function becomes crucial in solving the initial or boundary value problems.
  • It fulfills the equation \[(i \hbar \frac{\partial}{\partial t}-\frac{\hbar^{2}}{2 \mu} abla^{2} ) G(x, t; x', t') = \delta(x - x') \delta(t - t'), \]where the deltas are Dirac functions representing a spike at specific spatial and temporal points.
By solving this equation, Green's function accounts for how a 'kick' at one point influences the state across time and space. For practical computation, the singular part of Green's function describes the behavior where the response becomes infinite, which is very important when calculating wave propagation characteristics.
Laplacian Operator
The Laplacian operator, denoted as \(abla^2 \), is a differential operator fundamental in potential theory. It measures how a quantity diffuses through space. In the context of the Schrödinger equation, it captures how the wave function changes spatially.
Here is why the Laplacian operator is important:
  • It appears whenever we're dealing with equations describing wave mechanics, such as the Helmholtz or Schrödinger equations.
  • Physically, it helps determine how a particle's likelihood of being found at a certain location spreads out over time.
In three-dimensional space, the Laplacian operator involves second derivatives in all spatial directions. For a function \( \psi(x, y, z) \), the Laplacian is given by: \[ abla^2 \psi = \frac{\partial^2 \psi}{\partial x^2} + \frac{\partial^2 \psi}{\partial y^2} + \frac{\partial^2 \psi}{\partial z^2}. \]This operator is crucial because it plays a central role in how potential energy contributes to the Schrödinger equation, governing the wave function’s shape and behavior in the field.
Three-Dimensional Free Schrödinger Equation
The three-dimensional free Schrödinger equation is fundamental in quantum mechanics for predicting how quantum states evolve in free space—that is, without external forces or potential fields.
It is given by:\[ i \hbar \frac{\partial \psi}{\partial t} = -\frac{\hbar^2}{2 \mu} abla^2 \psi, \]where \( \psi \) represents the wave function, \( \mu \) the particle's reduced mass, and \( \hbar \) the reduced Planck's constant.
Here's what you need to know about this equation:
  • The term \( i \hbar \frac{\partial \psi}{\partial t} \) accounts for the wave function's time evolution, encapsulating the complex nature of quantum state changes with time.
  • The second term, \(-\frac{\hbar^2}{2 \mu} abla^2 \psi\), represents the kinetic energy component, derived from the spatial variation captured by the Laplacian operator.
  • It’s called "free" because it does not include potential energy terms, emphasizing purely kinetic evolution.
In this equation, it is the coupling of time and spatial aspects through these terms that fully captures the dynamics of a quantum system in an unrestricted three-dimensional framework.

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Most popular questions from this chapter

Use the operator separation technique to show that (a) the GF for the IIelmholtz operator \(\nabla^{2}+k^{2}\) in three dimensions is $$ G\left(\mathbf{r}, \mathbf{r}^{\prime}\right)=-i k \sum_{l=0}^{\infty} \sum_{m=-l}^{l} j_{l}\left(k r_{<}\right) h_{l}\left(k r_{>}\right) Y_{l m}(\theta, \varphi) Y_{l m}^{*}\left(\theta^{\prime}, \varphi^{\prime}\right), $$ where \(r_{<}\left(r_{>}\right)\) is the smaller (larger) of \(r\) and \(r^{\prime}\) and \(j_{l}\) and \(h_{l}\) are the spherical Bessel and Hankel functions, respectively. No explicit BCs are assumed except that there is regularity at \(r=0\) and that \(G\left(\mathbf{r}, \mathbf{r}^{\prime}\right) \rightarrow\) 0 for \(|\mathbf{r}| \rightarrow \infty\) (b) Obtain the identity $$ \frac{e^{i k\left|\mathbf{r}-\mathbf{r}^{\prime}\right|}}{4 \pi\left|\mathbf{r}-\mathbf{r}^{\prime}\right|}=i k \sum_{l=0}^{\infty} \sum_{m=-l}^{l} j_{l}\left(k r_{<}\right) h_{l}\left(k r_{>}\right) Y_{l m}(\theta, \varphi) Y_{l m}^{*}\left(\theta^{\prime}, \varphi^{\prime}\right) . $$ (c) Derive the plane wave expansion [see Eq. (19.46)] $$ e^{i \mathbf{k} \cdot \mathbf{r}}=4 \pi \sum_{l=0}^{\infty} \sum_{m=-l}^{l} i^{l} j_{l}(k r) Y_{l m}^{*}\left(\theta^{\prime}, \varphi^{\prime}\right) Y_{l m}(\theta, \varphi), $$ where \(\theta^{\prime}\) and \(\varphi^{\prime}\) are assumed to be the angular coordinates of \(\mathbf{k}\). Hint: Let \(\left|\mathbf{r}^{\prime}\right| \rightarrow \infty\), and use $$ \left|\mathbf{r}-\mathbf{r}^{\prime}\right|=\left(r^{\prime 2}+r^{2}-2 \mathbf{r} \cdot \mathbf{r}^{\prime}\right)^{1 / 2} \rightarrow r^{\prime}-\frac{\mathbf{r}^{\prime} \cdot \mathbf{r}}{r^{\prime}} $$ and the asymptotic formula \(h_{l}^{(1)}(z) \rightarrow(1 / z) e^{i[z+(l+1)(\pi / 2)]}\), valid for large \(z\).

Calculate the GF of the two-dimensional Laplacian operator appropriate for Neumann \(\mathrm{BCs}\) on the rectangle \(0 \leq x \leq a, 0 \leq y \leq b\).

Calculate the retarded GF for the wave operator in two dimensions and show that it is equal to $$ G_{s}^{(\mathrm{ret})}(\mathbf{r}, t)=\frac{\theta(t)}{2 \pi \sqrt{t^{2}-r^{2}}} $$ Now use this result to obtain the GF for any even number of dimensions: $$ G_{s}^{(\mathrm{ret})}(\mathbf{r}, t)=\frac{\theta(t)}{2 \pi}\left(-\frac{1}{2 \pi r} \frac{\partial}{\partial r}\right)^{n-1}\left[\frac{1}{\sqrt{t^{2}-r^{2}}}\right] \quad \text { for } n=m / 2 . $$

Show that the diffusion operator satisfies $$ \mathbf{L}_{\mathbf{x}, t} G(\mathbf{x}, \mathbf{y} ; t-\tau)=\delta(\mathbf{x}-\mathbf{y}) \delta(t-\tau) . $$ Hint: Use $$ \frac{\partial \theta}{\partial t}(t-\tau)=\delta(t-\tau) $$

Consider GF for the Helmholtz operator \(\nabla^{2}+\mu^{2}\) in two dimensions. (a) Show that $$ G\left(\mathbf{r}, \mathbf{r}^{\prime}\right)=-\frac{i}{4} H_{0}^{(1)}\left(\mu\left|\mathbf{r}-\mathbf{r}^{\prime}\right|\right)+H\left(\mathbf{r}, \mathbf{r}^{\prime}\right) $$ where \(H\left(\mathbf{r}, \mathbf{r}^{\prime}\right)\) satisfies the homogeneous Helmholtz equation. (b) Separate the variables and use the fact that \(H\) is regular at \(\mathbf{r}=\mathbf{r}^{\prime}\) to show that \(H\) can be written as $$ H\left(\mathbf{r}, \mathbf{r}^{\prime}\right)=\sum_{n=0}^{\infty} J_{n}(\mu r)\left[a_{n}\left(\mathbf{r}^{\prime}\right) \cos n \theta+b_{n}\left(\mathbf{r}^{\prime}\right) \sin n \theta\right] . $$ (c) Now assume a circular boundary of radius \(a\) and the \(\mathrm{BC} G\left(\mathbf{a}, \mathbf{r}^{\prime}\right)=0\), in which a is a vector from the origin to the circular boundary. Using this \(\mathrm{BC}\), show that $$ \begin{aligned} a_{0}\left(\mathbf{r}^{\prime}\right)=& \frac{i}{8 \pi J_{0}(\mu a)} \int_{0}^{2 \pi} H_{0}^{(1)}\left(\mu \sqrt{a^{2}+r^{\prime 2}-2 a r^{\prime} \cos \left(\theta-\theta^{\prime}\right)}\right) d \theta, \\ a_{n}\left(\mathbf{r}^{\prime}\right)=& \frac{i}{4 \pi J_{n}(\mu a)} \\ & \times \int_{0}^{2 \pi} H_{0}^{(1)}\left(\mu \sqrt{a^{2}+r^{\prime 2}-2 a r^{\prime} \cos \left(\theta-\theta^{\prime}\right)}\right) \cos n \theta d \theta \\ b_{n}\left(\mathbf{r}^{\prime}\right)=& \frac{i}{4 \pi J_{n}(\mu a)} \\ & \times \int_{0}^{2 \pi} H_{0}^{(1)}\left(\mu \sqrt{a^{2}+r^{\prime 2}-2 a r^{\prime} \cos \left(\theta-\theta^{\prime}\right)}\right) \sin n \theta d \theta \end{aligned} $$ These equations completely determine \(H\left(\mathbf{r}, \mathbf{r}^{\prime}\right)\) and therefore \(\bar{G}\left(\mathbf{r}, \mathbf{r}^{\prime}\right)\).

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