/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 4 Using the rules of bra-ket algeb... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 1: Problem 4

Using the rules of bra-ket algebra, prove or evaluate the following: (a) \(\operatorname{tr}(X Y)=\operatorname{tr}(Y X),\) where \(X\) and \(Y\) are operators. (b) \((X Y)^{\dagger}=Y^{\prime} X^{\dagger},\) where \(X\) and \(Y\) are operators. (c) \(\exp [i f(A)]=?\) in ket-bra form, where \(A\) is a Hermitian operator whose eigenvalues are known. (d) \(\sum_{a^{\prime}} \psi_{a^{\prime}}^{*}\left(\mathbf{x}^{\prime}\right) \psi_{a^{\prime}}\left(\mathbf{x}^{\prime \prime}\right),\) where \(\psi_{a^{\prime}}\left(\mathbf{x}^{\prime}\right)=\left\langle\mathbf{x}^{\prime} | a^{\prime}\right\rangle\).

Short Answer

Expert verified
(a) True by trace property; (b) follows Hermitian conjugate rule; (c) is spectral decomposition; (d) simplifies to identity resolution.

Step by step solution

01

Prove Part (a) - Cyclic property of the trace

The trace of a matrix or operator is defined as the sum of its diagonal elements, specifically, \( \operatorname{tr}(X) = \sum_i \langle i | X | i \rangle \), where \( |i\rangle \) are the basis kets. For two operators \(X\) and \(Y\), we want to show that \( \operatorname{tr}(XY) = \operatorname{tr}(YX) \). Using the cyclic property of the trace, which states \( \operatorname{tr}(ABC) = \operatorname{tr}(BCA) = \operatorname{tr}(CAB) \) for any operators \(A, B, C\), we can re-arrange the product inside the trace:\[\operatorname{tr}(XY) = \operatorname{tr}(YX).\] This property holds because the trace is invariant to cyclic permutations due to the summation over a complete orthonormal basis, so the products of operators commute within the trace.
02

Prove Part (b) - Hermitian conjugate of a product

The Hermitian conjugate (also known as the adjoint) of an operator \( O \) is denoted as \( O^{\dagger} \). For any two operators \( X \) and \( Y \), the Hermitian conjugate of their product is given as \[ (XY)^{\dagger} = Y^{\dagger}X^{\dagger} \].This property follows from the definition of the Hermitian conjugate: for any product of operators, \[ (AB)^{\dagger} = B^{\dagger}A^{\dagger}. \]This is due to the rule that the order of operators is reversed when taking the Hermitian conjugate.
03

Evaluate Part (c) - Exponential of a Hermitian operator

Given a Hermitian operator \( A \) with eigenvalues known, denoted as \( \lambda_i \) for eigenstate \( |i\rangle \), the exponential of this operator can be expressed in terms of its spectral decomposition:\[ \exp[i f(A)] = \sum_i \exp[i f(\lambda_i)] |i\rangle \langle i|. \]This results from the fact that any function of an operator, especially a Hermitian one, can be expanded using its eigenvectors and eigenvalues. The term \( \exp[i f(\lambda_i)] \) corresponds to the action of the function applied to the eigenvalues, multiplying back the projection operator \( |i\rangle \langle i| \).
04

Simplifying Part (d) - Summation over basis

The expression \( \sum_{a'} \psi_{a'}^{*}(\mathbf{x'}) \psi_{a'}(\mathbf{x''}) \) with \( \psi_{a'}(\mathbf{x'}) = \langle \mathbf{x'} | a' \rangle \) implies we are summing over a complete set of states \( |a'\rangle \). In the context of bra-ket notation, the completeness relation states \( \sum_{a'} |a'\rangle \langle a'| = I \), the identity operator. Hence, this expression can be written as \[\sum_{a'} \langle \mathbf{x'} | a' \rangle \langle a' | \mathbf{x''} \rangle = \langle \mathbf{x'} | \mathbf{x''} \rangle.\] This represents the resolution of identity over the basis states, revealing the inner product between position states \( |\mathbf{x'}\rangle \) and \( |\mathbf{x''}\rangle \).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Bra-ket Notation
Bra-ket notation, also known as Dirac notation, is a compact form used extensively in quantum mechanics to describe states and operations. It represents states as "kets" and their duals as "bras." For example, a state vector is written as \(| \psi \rangle\) and its dual is \(\langle \psi |\). This notation is incredibly handy when expressing the inner product, which translates to \(\langle \phi | \psi \rangle\) for states \(|\phi\rangle\) and \(|\psi\rangle\). It symbolizes the "overlap" or probability amplitude between two states.

Let's break down the usage:
  • A "bra" is \(\langle \phi |\) representing a row vector.
  • A "ket" is \(| \psi \rangle\) representing a column vector.
  • The inner product is the matrix product of a bra and a ket.
  • Outer products, like \(| \phi \rangle \langle \psi |\), represent operators.
Bra-ket notation simplifies expressing physical concepts such as the evolution of quantum states through operators, resolution of identity, and even more complex constructs in quantum mechanics.

It is a foundation stone that supports the entire quantum mechanics framework, coupling elegantly with Hermitian operators and eigenstates.
Hermitian Operators
Hermitian operators play a crucial role in quantum mechanics. They are operators \(A\) for which \(A^{\dagger} = A\), meaning their Hermitian conjugate equals the operator itself. This property gives them real eigenvalues, making them ideal for representing observable quantities like energy or momentum.

To understand why Hermitian operators are significant, consider:
  • They ensure that measurements are real numbers, matching physical observations (no imaginary parts).
  • They have orthogonal eigenvectors, which form a complete basis set.
  • They are associated with conserved quantities in quantum systems due to unitary evolution.
Additionally, the spectral theorem states that any Hermitian operator can be decomposed into its eigenvalues and eigenvectors: \[A = \sum_i \lambda_i |i\rangle \langle i| \]where \( \lambda_i \) are the eigenvalues, and \(|i\rangle\) are the corresponding eigenstates. This provides a powerful tool for manipulating and analyzing complex quantum mechanical operators.
Matrix Trace
The trace of a matrix or operator \(X\), denoted \( \operatorname{tr}(X) \), is the sum of its diagonal elements. It's an essential concept in linear algebra and quantum mechanics due to its mathematical properties, such as invariance under change of basis.

In simpler terms, the trace is:
  • The sum of eigenvalues of the operator, repeated according to their algebraic multiplicity.
  • Invariant to cyclic permutations (\( \operatorname{tr}(XY) = \operatorname{tr}(YX) \)), a useful property utilized in proofs.
  • Useful in quantum mechanics for partial tracing, reducing a state representation to its relevant subsystem.
Let's see what this means in quantum mechanics. Consider an operator \(X\) acting on a state within our system. The trace is the sum over a complete basis \(|i\rangle\): \[ \operatorname{tr}(X) = \sum_i \langle i | X | i \rangle \]This concept applies significantly in the context of various proofs and calculations, efficiently summarizing the intrinsic properties of operators in their operational form.
Eigenvalues
Eigenvalues are intrinsic numbers associated with operators that provide insight into their properties. For an operator \(A\), an eigenvalue \(\lambda\) satisfies the equation:\[A \mathbf{v} = \lambda \mathbf{v}\]where \(\mathbf{v}\) is the corresponding eigenvector.

In quantum mechanics, eigenvalues have special physical meanings:
  • They are often related to measurable quantities, like energy levels in quantum systems.
  • They indicate the "allowed" values that might be observed upon a measurement.
  • Hermitian operators have real eigenvalues, connecting them closely with observables.
Understanding eigenvalues involves understanding how operators act on quantum states and change them (or leave them unchanged), correlating the spectral properties with measurable outcomes in real physical situations. They unravel layers in quantum systems from energy spectra to state transformations and serve as the backbone of mathematical physics regarding operator analysis.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A beam of spin \(\frac{1}{2}\) atoms goes through a series of Stern-Gerlach-type measurements as follows: (a) The first measurement accepts \(s_{z}=h / 2\) atoms and rejects \(s_{z}=-h / 2\) atoms. (b) The second measurement accepts \(s_{n}=\hbar / 2\) atoms and rejects \(s_{n}=-\hbar / 2\) atoms, where \(s_{n}\) is the eigenvalue of the operator \(\mathbf{S} \cdot \hat{\mathbf{n}},\) with \(\hat{\mathbf{n}}\) making an angle \(\beta\) in the \(x z\) -plane with respect to the \(z\) -axis. (c) The third measurement accepts \(s_{z}=-\hbar / 2\) atoms and rejects \(s_{z}=\hbar / 2\) atoms. What is the intensity of the final \(s_{z}=-\hbar / 2\) beam when the \(s_{z}=\hbar / 2\) beam surviving the first measurement is normalized to unity? How must we orient the second measuring apparatus if we are to maximize the intensity of the final \(s_{z}=-\hbar / 2\) beam?

Let \(A\) and \(B\) be observables. Suppose the simultancous eigenkets of \(A\) and \(B\) \(\left\\{\left|a^{\prime}, b^{\prime}\right\rangle\right\\}\) form a complete orthonormal set of base kets. Can we always conclude that \\[[A, B]=0 ?\\] If your answer is yes, prove the assertion. If your answer is no, give a counterexample.

Estimate the rough order of magnitude of the length of time that an ice pick can be balanced on its point if the only limitation is that set by the Heisenberg uncertainty principle. Assume that the point is sharp and that the point and the surface on which it rests are hard. You may make approximations that do not alter the general order of magnitude of the result. Assume reasonable values for the dimensions and weight of the ice pick. Obtain an approximate numerical result and express it in seconds.

Two Hermitian operators anticommute: \\[\\{A, B\\}=A B+B A=0\\] Is it possible to have a simultaneous (that is, common) eigenket of \(A\) and \(B\) ? Prove or illustrate your assertion.

(a) Consider two kets \(|\alpha\rangle\) and \(|\beta\rangle .\) Suppose \(\left\langle a^{\prime} | \alpha\right\rangle,\left\langle a^{\prime \prime} | \alpha\right\rangle, \ldots\) and \(\left\langle a^{\prime} | \beta\right\rangle\) \(\left\langle a^{\prime \prime} | \beta\right\rangle, \ldots\) are all known, where \(\left|a^{\prime}\right\rangle,\left|a^{\prime \prime}\right\rangle, \ldots\) form a complete set of base kets. Find the matrix representation of the operator \(|\alpha\rangle\langle\beta|\) in that basis. (b) We now consider a spin \(\frac{1}{2}\) system and let \(| \alpha\) ) and \(|\beta\rangle\) be \(| s_{z}=h / 2\) ) and \(\left|s_{x}=h / 2\right\rangle,\) respectively. Write down explicitly the square matrix that corresponds to \(|\alpha\rangle\langle\beta|\) in the usual \((s_{z}\) diagonal) basis..
See all solutions

Recommended explanations on Physics Textbooks

Magnetism and Electromagnetic Induction

Read Explanation

Measurements

Read Explanation

Fluids

Read Explanation

Kinematics Physics

Read Explanation

Electricity and Magnetism

Read Explanation

Physics of Motion

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.