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Chapter 2: Problem 20

Confirm that the completeness relation, eqn 1.25, may be expressed in terms of wavefunctions as \\[ \sum_{n} \psi_{n}(r) \psi_{n}^{*}\left(r^{\prime}\right)=\delta\left(r-r^{\prime}\right) \\] where \(\delta\left(r-r^{\prime}\right)\) is the Dirac \(\delta\) -function described in Section 2.1

Short Answer

Expert verified
The provided expression indeed represents the completeness of wavefunctions. The sum of the multiplication of a wavefunction and its complex conjugate over all states equals to the Dirac delta function, and this confirms the completeness relation.

Step by step solution

01

Understand the meaning of the equation

The given equation signifies the completeness relation in terms of wavefunctions, where \(\psi_{n}(r)\) represents the wavefunction and \(\delta\left(r-r^{\prime}\right)\) is the impulse function or Dirac delta function. This equation essentially says that the addition of all possible states (represented by n) multiplied by their complex conjugate is equal to the Dirac delta function.
02

Understand the properties of the Dirac delta function

The Dirac \(\delta\) function, \(\delta\left(r-r^{\prime}\right)\), is a function that is zero everywhere except at \(r= r'\) and its integral over the whole space is equal to 1. It's often used to represent a 'point source' or 'sink' in equations.
03

Check the Completeness

The expression is an example of the 'completeness relation', a central principle in quantum physics. A set of functions is 'complete' if any function can be written as a (possibly infinite) sum of functions from the set. For the given expression, it's assuming that the states \(\psi_{n}\) form a complete set, because when you sum over all of them and multiply by their complex conjugate, you get a Dirac delta function. It means any other function can be written as a sum of these functions. This is equivalent to the completeness of the wave functions.

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Most popular questions from this chapter

Locate the nodes of the harmonic oscillator wavefunction for the state with \(v=6 .\) Hint. Use mathematical software.

Determine the probability of finding the ground-state harmonic oscillator stretched to a displacement beyond the classical turning point. Hint. Relate the expression for the probability to the error function, erf \(z,\) defined as \\[ \operatorname{erf} z=1-\frac{2}{\pi^{1 / 2}} \int_{z}^{\infty} \mathrm{e}^{-y^{2}} \mathrm{d} y \\] and evaluate it using erf \(1=0.8427 .\) The error function is incorporated into most mathematical software packages.

Calculate the energies and wavefunctions for a particle in a one-dimensional square well in which the potential energy rises to a finite value \(V\) at each end, and is zero inside the well; that is \\[ \begin{array}{ll} V(x)=V & x \leq 0 \text { and } x \geq L \\ V(x)=0 & 0

Consider a harmonic oscillator of mass \(m\) undergoing harmonic motion in two dimensions \(x\) and \(y .\) The potential energy is given by \(V(x, y)=\frac{1}{2} k_{x} x^{2}+\frac{1}{2} k_{y} y^{2}\). (a) Write down the expression for the hamiltonian operator for such a system. (b) What is the general expression for the allowable energy levels of the two- dimensional harmonic oscillator? (c) What is the energy of the ground state (the lowest energy state)? Hint. The hamiltonian operator can be written as a sum of operators.

A particle of mass \(m\) is confined to a one-dimensional box of length \(L\). Calculate the probability of finding it in the following regions: (a) \(0 \leq x \leq \frac{1}{2} L,\) (b) \(0 \leq x \leq \frac{1}{4} L\) (c) \(\frac{1}{2} L-\delta x \leq x \leq \frac{1}{2} L+\delta x .\) Derive expressions for a general value of \(n\). Then evaluate the probabilities (i) for \(n=1\) (ii) in the limit \(n \rightarrow \infty\). Compare the latter to the classical expectations.
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