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Chapter 6: Problem 8

(a) Show that the parity operator \(\hat{\Pi}\) is Hermitian. (b) Show that the eigenvalues of the parity operator are 卤1

Short Answer

Expert verified
(a) The parity operator is Hermitian. (b) The eigenvalues of the parity operator are 卤1.

Step by step solution

01

Understand the Parity Operator

The parity operator, denoted as \(\hat{\Pi}\), is a linear operator that reflects the spatial coordinates of a function. For example, if \(\Psi(x)\) is a wavefunction, applying \(\hat{\Pi}\) gives \(\Psi(-x)\). Mathematically, \(\hat{\Pi}\Psi(x) = \Psi(-x)\).
02

Define a Hermitian Operator

An operator \(\hat{A}\) is Hermitian if it satisfies \(\langle \phi | \hat{A} \psi \rangle = \langle \hat{A} \phi | \psi \rangle \) for all states \(|\phi\rangle\) and \(|\psi\rangle\). This property ensures that the operator has real eigenvalues and orthogonal eigenvectors.
03

Show that \(\hat{\Pi}\) is Hermitian

Consider two arbitrary wavefunctions \(\phi(x)\) and \(\psi(x)\). The Hermitian condition requires \(\langle \phi | \hat{\Pi} \psi \rangle = \langle \hat{\Pi} \phi | \psi \rangle \).Calculate the left side: \[ \langle \phi | \hat{\Pi} \psi \rangle = \int \phi^*(x) \psi(-x) \, dx \]Make the substitution \(u = -x\), \(du = -dx\):\[ = \int \phi^*(-u) \psi(u) \, du \]Change \(u\) back to \(x\):\[ = \int \phi^*(-x) \psi(x) \, (-dx) = \int \phi^*(-x) \psi(x) \, dx \]Calculate the right side: \[ \langle \hat{\Pi} \phi | \psi \rangle = \int \phi^*(-x) \psi(x) \, dx \]Both integrals match, hence \(\langle \phi | \hat{\Pi} \psi \rangle = \langle \hat{\Pi} \phi | \psi \rangle \), showing that \(\hat{\Pi}\) is Hermitian.
04

Assume an Eigenfunction of \(\hat{\Pi}\)

Let \(\phi(x)\) be an eigenfunction of \(\hat{\Pi}\) with eigenvalue \(\lambda\). Then: \(\hat{\Pi} \phi(x) = \lambda \phi(x)\), which translates to \(\phi(-x) = \lambda \phi(x)\).
05

Apply the Parity Operator Twice

If \(\hat{\Pi}^2 \phi(x)= \phi(x)\), because the parity of \(\phi(-x)\) is \(\phi(x)\), then \(\hat{\Pi}^2 = \hat{I}\), where \(\hat{I}\) is the identity operator. Therefore, \(\hat{\Pi}^2 \phi(x) = \phi(x)\).
06

Solve for Eigenvalues

From \(\phi(-x) = \lambda \phi(x)\) and \(\phi(x) = \lambda^2 \phi(x)\), it follows \(\lambda^2 = 1\). Solutions to this equation are \(\lambda = \pm 1\). Thus, the eigenvalues of \(\hat{\Pi}\) are \(\pm 1\).

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Key Concepts

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

Hermitian Operator
A Hermitian operator plays a crucial role in quantum mechanics by ensuring real-valued eigenvalues, which link directly to observable physical quantities. This characteristic arises from the mathematical condition that the inner product \( \langle \phi | \hat{A} \psi \rangle = \langle \hat{A} \phi | \psi \rangle \) must hold true for any arbitrary states \( |\phi\rangle \) and \( |\psi\rangle \). Let's unpack what this means;

  • Real Eigenvalues: For an operator to be Hermitian, its eigenvalues must be real. This is significant because in quantum mechanics, measurements must yield real numbers. Hermitian operators ensure this outcome.
  • Orthogonal Eigenvectors: The eigenvectors of a Hermitian operator associated with distinct eigenvalues are orthogonal. This property is essential for constructing orthonormal bases in quantum mechanics.

In our context, the parity operator \( \hat{\Pi} \) is shown to be Hermitian. This establishes that operations like the parity transformation adhere to principles of real, measurable quantities in quantum systems.
Eigenvalues
Eigenvalues are a fundamental concept in quantum mechanics and linear algebra. When an operator acts on a function, producing a scaled version of the function, the scaling factor is the eigenvalue. Let's delve into their significance:

  • Definition: Mathematically, if \( \hat{A}\phi = \lambda\phi \), then \( \lambda \) is the eigenvalue corresponding to the eigenvector or eigenfunction \( \phi \).
  • Solution of Equations: Solving for eigenvalues involves finding solutions to \( \lambda \phi(x) \) equations. For the parity operator, this is seen as \( \phi(-x) = \lambda \phi(x) \).

In our exercise, we've shown that the parity operator \( \hat{\Pi} \) has eigenvalues of \( \pm 1 \). This outcome results from applying the operator twice, expressing how it returns to the original state with eigenvalues satisfying \( \lambda^2 = 1 \). The parity eigenvalues reflect the inherent duality in switching sign across dimensions.
Wavefunction Reflectivity
Wavefunction reflectivity refers to how a wavefunction changes when its spatial coordinates are inverted. This concept is crucial when understanding operators like the parity operator. Here are some insights on this topic:

  • Reflective Property: The parity operator \( \hat{\Pi} \) transforms a wavefunction \( \Psi(x) \) into \( \Psi(-x) \), essentially flipping it over the origin.
  • Implications in Symmetry: Reflectivity is connected to spatial symmetry considerations in quantum mechanics. Systems possessing symmetric properties tend to be more straightforward to analyze.

In the given problem, after operating \( \hat{\Pi} \) twice, the wavefunction reflects back to its original form. This is why the identity \( \hat{\Pi}^2 = \hat{I} \) applies, emphasizing that reflection happens twice, thus returning the wavefunction to its initial state. This reflectivity underlines the symmetry inherent in quantum systems.

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Most popular questions from this chapter

Consider two particles of mass \(m_{1}\) and \(m_{2}\) (in one dimension) that interact via a potential that depends only on the distance between the particles \(V\left(\left|x_{1}-x_{2}\right|\right),\) so that the Hamiltonian is \(\hat{H}=-\frac{\hbar^{2}}{2 m_{1}} \frac{\partial^{2}}{\partial x_{1}^{2}}-\frac{\hbar^{2}}{2 m_{2}} \frac{\partial^{2}}{\partial x_{2}^{2}}+V\left(\left|x_{1}-x_{2}\right|\right)\) Acting on a two-particle wave function the translation operator would be \(\hat{T}(a) \psi\left(x_{1}, x_{2}\right)=\psi\left(x_{1}-a, x_{2}-a\right)\) (a) Show that the translation operator can be written \(\hat{T}(a)=e^{-\frac{i a}{\hbar} \hat{P}}\) where \(\hat{P}=\hat{p}_{1}+\hat{p}_{2}\) is the total momentum. (b) Show that the total momentum is conserved for this system.

Consider the parity operator in three dimensions. (a) Show that \(\hat{\Pi} \psi(\mathbf{r})=\psi^{\prime}(\mathbf{r})=\psi(-\mathbf{r})\) is equivalent to a mirror reflection followed by a rotation. (b) Show that, for \(\psi\) expressed in polar coordinates, the action of the parity operator is \(\hat{\Pi} \psi(r, \theta, \phi)=\psi(r, \pi-\theta, \phi+\pi).\) (c) Show that for the hydrogenic orbitals, \(\hat{\Pi} \psi_{n \ell m}(r, \theta, \phi)=(-1)^{\ell} \psi_{n \ell m}(r, \theta, \phi).\) That is, \(\psi_{n \ell m}\) is an eigenstate of the parity operator, with eigenvalue \((-1)^{\ell} .\) Note: This result actually applies to the stationary states of any central potential \(V(\mathbf{r})=V(r) .\) For a central potential, the eigenstates may be written in the separable form \(R_{n \ell}(r) Y_{\ell}^{m}(\theta, \phi)\) where only the radial function \(R_{n \ell}-\) which plays no role in determining the parity of the state-depends on the specific functional form of \(V(r).\)

Show that the operator \(\hat{p}^{\prime}\) obtained by applying a translation to the operator \(\hat{p}\) is \(\hat{p}^{\prime}=\hat{T}^{\dagger} \hat{p} \hat{T}=\hat{p}\)

In this problem you will establish the correspondence between Equations 6.30 and 6.31. (a) Diagonalize the matrix \(^{16}\) \(M=\left(\begin{array}{cc}1 & -\varphi / N \\ \varphi / N & 1\end{array}\right)\) to obtain the matrix \(\mathrm{M}^{\prime}=\mathrm{SMS}^{-1}\) where \(S^{-1}\) is the unitary matrix whose columns are the (normalized) eigenvectors of M. (b) Use the binomial expansion to show that \(\lim _{N \rightarrow \infty}\left(\mathrm{M}^{\prime}\right)^{N}\) is a diagonal matrix with entries \(e^{-i \varphi \text { and } e^{i \varphi} \text { on the diagonal. }}\) (c) Transform back to the original basis to show that \(\lim _{N \rightarrow \infty} \mathrm{M}^{N}=\mathrm{S}^{-1}\left[\lim _{N \rightarrow \infty}\left(\mathrm{M}^{\prime}\right)^{N}\right] \mathrm{S}\) agrees with the matrix in Equation 6.31.

Consider a particle of mass \(m\) in a two-dimensional infinite square well with sides of length \(L\). With the origin placed at the center of the well, the stationary states can be written as $$ \psi_{n_{x} n_{y}}(x, y)=\frac{2}{L} \sin \left[\frac{n_{x} \pi}{L}\left(x-\frac{L}{2}\right)\right] \sin \left[\frac{n_{y} \pi}{L}\left(y-\frac{L}{2}\right)\right] $$ with energies \(E_{n_{x} n_{y}}=\frac{\pi^{2} \hbar^{2}}{2 m L^{2}}\left(n_{x}^{2}+n_{y}^{2}\right)\) for positive integers \(n_{x}\) and \(n_{y}\) (a) The two states \(\psi_{a b}\) and \(\psi_{b a}\) for \(a \neq b\) are clearly degenerate. Show that a rotation by 90 counterclockwise about the center of the square carries one into the other, \(\hat{R} \psi_{a b} \propto \psi_{b a}\) and determine the constant of proportionality. Hint: write \(\psi_{a b}\) in polar coordinates. (b) Suppose that instead of \(\psi_{a b}\) and \(\psi_{b a}\) we choose the basis \(\psi_{+}\) and \(\psi_{-}\) for our two degenerate states: \(\psi_{\pm}=\frac{\psi_{a b} \pm \psi_{b a}}{\sqrt{2}}\) Show that if \(a\) and \(b\) are both even or both odd, then \(\psi_{+}\) and \(\psi_{-}\) are eigenstates of the rotation operator. (c) Make a contour plot of the state \(\psi_{-}\) for \(a=5\) and \(b=7\) and verify (visually) that it is an eigenstate of every symmetry operation of the square (rotation by an integer multiple of \(\pi / 2\), reflection across a diagonal, or reflection along a line bisecting two sides). The fact that \(\psi_{+}\) and \(\psi_{-}\) are not connected to each other by any symmetry of the square means that there must be additional symmetry explaining the degeneracy of these two states. 42.
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