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Chapter 6: Problem 2

Show that, for a Hermitian operator \(\hat{Q},\) the operator \(\hat{U}=\exp [i \hat{Q}]\) is unitary. Hint: First you need to prove that the adjoint is given by \(\hat{U}^{\dagger}=\exp [-i \hat{Q}] ;\) then prove that \(\hat{U}^{\dagger} \hat{U}=1 .\) Problem \(\underline{3.5}\) may help.

Short Answer

Expert verified
The operator \(\hat{U} = \exp(i\hat{Q})\) is unitary because \(\hat{U}^\dagger = \exp(-i\hat{Q})\) and \(\hat{U}^\dagger \hat{U} = 1\).

Step by step solution

01

Define the concept of Hermitian operator

A Hermitian operator, \(\hat{Q}\), satisfies the property \(\hat{Q}^\dagger = \hat{Q}\). This means the operator is equal to its own adjoint (or Hermitian conjugate). Hermitian operators have real eigenvalues and are an important concept in quantum mechanics.
02

Determine the adjoint of the exponential operator

Given \(\hat{U} = \exp(i \hat{Q})\), we aim to find \(\hat{U}^\dagger\). Using the property of exponential operators, \(\exp( i \hat{Q})^\dagger = \exp(i \hat{Q})^* = \exp(-i \hat{Q}^\dagger)\). Since \(\hat{Q}\) is Hermitian (\(\hat{Q}^\dagger = \hat{Q}\)), we have \(\exp(-i \hat{Q})\). Thus, \(\hat{U}^\dagger = \exp(-i \hat{Q})\).
03

Verify the unitarity condition

To show the operator is unitary, we confirm that \(\hat{U}^\dagger \hat{U} = 1\). Compute \(\hat{U}^\dagger \hat{U} = \exp(-i \hat{Q}) \exp(i \hat{Q})\). Using the fact that the product of exponents with opposite signs is the identity, \(\exp(-i \hat{Q}) \exp(i \hat{Q}) = \exp(0) = 1\). Therefore, \(\hat{U}^\dagger \hat{U} = 1\), proving that \(\hat{U}\) is unitary.

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Key Concepts

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

Unitary Operator
In quantum mechanics, unitary operators play a crucial role because they represent transformations that preserve the total probability within a quantum system. This is essential, as probabilities must always sum to one. A unitary operator, denoted here as \( \hat{U} \), satisfies the condition that its adjoint (or Hermitian conjugate) \( \hat{U}^{\dagger} \) multiplied by itself is equal to the identity operator:
  • \( \hat{U}^{\dagger} \hat{U} = 1 \)
To understand why an operator like \( \hat{U} = \exp(i \hat{Q}) \) is unitary, we need to prove that this condition holds true. In the exercise, it is shown that by calculating \( \hat{U}^{\dagger} \) as \( \exp(-i \hat{Q}) \) and then verifying that \( \hat{U}^{\dagger} \hat{U} = \exp(-i \hat{Q}) \exp(i \hat{Q}) = 1 \), we indeed confirm its unitarity. Unitary operators are central in quantum mechanics for representing reversible evolution and symmetry transformations.
Eigenvalues
Eigenvalues are fundamental to understanding the behavior of quantum systems in relation to operators. When a quantum system is measured, it collapses to one of its eigenstates, and the outcome of this measurement corresponds to an eigenvalue of the operator associated with the observable being measured.
  • An eigenvalue equation is expressed as: \( \hat{A} | \psi \rangle = \lambda | \psi \rangle \), where \( \lambda \) is an eigenvalue.
  • For Hermitian operators, these eigenvalues are real numbers.
This property makes Hermitian operators particularly valuable in quantum mechanics as they are used to describe physical observables like energy and momentum. Since our operator \( \hat{Q} \) in the exercise is Hermitian, the eigenvalues are guaranteed to be real, ensuring that measurements of the corresponding observable are also real, aligning with physical reality.
Quantum Mechanics
Quantum mechanics is the branch of physics providing a mathematical framework for understanding the behavior of subatomic particles, where classical mechanics is insufficient. It revolves around concepts like wave functions, superpositions, and entanglement.
  • Central principles include the uncertainty principle, indicating inherent uncertainties in knowing pairs of quantities.
  • The Schr枚dinger equation, which describes how the quantum state of a physical system changes over time.
  • Operators like Hermitian and unitary operators represent physical observables and transformations in quantum systems.
The exercise involves Hermitian and unitary operators, both of which are essential in quantum theory. Hermitian operators connect with observables, while unitary operators describe system evolution without loss of information. These concepts highlight how quantum mechanics provides insights into the inherently probabilistic nature of microscopic phenomena.

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Most popular questions from this chapter

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.

Consider the free particle in one dimension: \(\hat{H}=\hat{p}^{2} / 2 m .\) This Hamiltonian has both translational symmetry and inversion symmetry. (a) Show that translations and inversion don't commute. (b) Because of the translational symmetry we know that the eigenstates of \(\hat{H}\) can be chosen to be simultaneous eigenstates of momentum, namely \(f_{p}(x)\) (Equation 3.32 ). Show that the parity operator turns \(f_{p}(x)\) into \(f_{-p}(x) ;\) these two states must therefore have the same energy. (c) Alternatively, because of the inversion symmetry we know that the eigenstates of \(\hat{H}\) can be chosen to be simultaneous eigenstates of parity, namely \(\frac{1}{\sqrt{\pi \hbar}} \cos \left(\frac{p x}{\hbar}\right)\) and \(\frac{1}{\sqrt{\pi \hbar}} \sin \left(\frac{p x}{\hbar}\right).\) Show that the translation operator mixes these two states together; they therefore must be degenerate. Note: Both parity and translational invariance are required to explain the degeneracy in the free-particle spectrum. Without parity, there is no reason for \(f_{p}(x)\) and \(f_{-p}(x)\) to have the same energy (I mean no reason based on symmetries discussed thus far ...obviously you can plug them in to the timeindependent Schr枚dinger equation and show it's true).

Consider a free particle of mass \(m .\) Show that the position and momentum operators in the Heisenberg picture are given by \(\hat{x}_{H}(t)=\hat{x}_{H}(0)+\frac{1}{m} \hat{p}_{H}(0) t\) \(\hat{p}_{H}(t)=\hat{p}_{H}(0).\) Comment on the relationship between these equations and the classical equations of motion. Hint: you will first need to evaluate the commutator \(\left[\hat{x}, \hat{H}^{n}\right] ;\) this will allow you to evaluate the commutator \([\hat{x}, \hat{U}].\)

Consider a time-independent Hamiltonian for a particle moving in one dimension that has stationary states \(\psi_{n}(x)\) with energies \(E_{n}\) (a) Show that the solution to the time-dependent Schr枚dinger equation can be written \(\Psi(x, t)=\hat{U}(t) \Psi(x, 0)=\int K\left(x, x^{\prime}, t\right) \Psi\left(x^{\prime}, 0\right) d x^{\prime}\) where \(K\left(x, x^{\prime}, t\right),\) known as the propagator, is \(K\left(x, x^{\prime}, t\right)=\sum_{n} \psi_{n}^{*}\left(x^{\prime}\right) e^{-i E_{n} t / \hbar} \psi_{n}(x)\) Here \(\left|K\left(x, x^{\prime}, t\right)\right|^{2}\) is the probability for a quantum mechanical particle to travel from position \(x^{\prime}\) to position \(x\) in time \(t.\) (b) Find \(K\) for a particle of mass \(m\) in a simple harmonic oscillator potential of frequency \(\omega .\) You will need the identity $$\frac{1}{\sqrt{1-z^{2}}} \exp \left[-\frac{\xi^{2}+\eta^{2}-2 \xi \eta z}{1-z^{2}}\right]=e^{-\xi^{2}} e^{-\eta^{2}} \sum_{n=0}^{\infty} \frac{z^{n}}{2^{n} n !} H_{n}(\xi) H_{n}(\eta)$$ (c) Find \(\Psi(x, t)\) if the particle from part (a) is initially in the state \(^{40}\) \(\Psi(x, 0)=\left(\frac{2 a}{\pi}\right)^{1 / 4} e^{-a\left(x-x_{0}\right)^{2}}\) Compare your answer with Problem \(2.49 .\) Note: Problem 2.49 is a special case with \(a=m \omega / 2 \hbar.\) (d) Find \(K\) for a free particle of mass \(m\). In this case the stationary states are continuous, not discrete, and one must make the replacement \(\sum_{n} \rightarrow \int_{-\infty}^{\infty} d p\) in Equation 6.79. (e) Find \(\Psi(x, t)\) for a free particle that starts out in the state \(\Psi(x, 0)=\left(\frac{2 a}{\pi}\right)^{1 / 4} e^{-a x^{2}}\) Compare your answer with Problem 2.21.

(a) Under parity, a "true" scalar operator does not change: \(\hat{\Pi}^{\dagger} \hat{f} \hat{\Pi}=\hat{f}\) whereas a pseudoscalar changes sign. Show therefore that \([\hat{\Pi}, \hat{f}]=0\) for a "true" scalar, whereas \(\\{\hat{\Pi}, \hat{f}\\}=0\) for a pseudoscalar. Note: the anti- commutator of two operators \(\hat{A}\) and \(\hat{B}\) is defined as $$ \\{\hat{A}, \hat{B}\\} \equiv \hat{A} \hat{B}+\hat{B} \hat{A} $$ (b) Similarly, a "true" vector changes sign \(\hat{\Pi}^{\dagger} \hat{\mathbf{V}} \hat{\Pi}=-\hat{\mathbf{V}}\) whereas a pseudovector is unchanged. Show therefore that \(\\{\hat{\Pi}, \hat{\mathbf{V}}\\}=\mathbf{0}\) for a "true" vector and \([\hat{\Pi}, \hat{\mathbf{V}}]=\mathbf{0}\) for a pseudovector.
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