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Chapter 1: Problem 17

The operator \(e^{A}\) has a meaning if it is expanded as a power scrics: \(\mathrm{e}^{A}=\Sigma_{n}(1 / n !) A^{n} .\) Show that if \(|a\rangle\) is an eigenstate of \(A\) with eigenvalue \(a,\) then it is also an eigenstate of \(\mathrm{e}^{A} .\) Find the latter's eigenvalue.

Short Answer

Expert verified
Yes, if \(|a\rangle\) is an eigenstate of \(A\) with eigenvalue \(a\), then it is also an eigenstate of \(\mathrm{e}^{A}\). The eigenvalue of the latter operator will be \(e^{a}\).

Step by step solution

01

Understand the Power Series of an Exponentiated Operator

The exponential of an operator \(A\), denoted as \(e^{A}\), can be represented as a power series \(\mathrm{e}^{A}=\Sigma_{n}(1 / n !) A^{n}\) This is a direct usage of the power series expansion of \(e^{x}\) where \(x\) has been replaced by the operator \(A\) and each power of \(A\) corresponds to repeated applications of the operator.
02

Recognizing the Eigenvector of the Operator

Given the eigenstate \(|a\rangle\) and its corresponding eigenvalue \(a\) of operator \(A\), \(A\) acts on \(|a\rangle\) to give \(a |a\rangle\), i.e., \(A|a\rangle = a|a\rangle\).
03

Apply the Eigenstate to the Power Series

We need to apply the eigenstate \(|a\rangle\) to the power series of \(e^{A}\) and see how it transforms. Given that \(|a\rangle\) is an eigenstate of \(A\), exponentiation of \(A\) applied to \(|a\rangle\) is expressed as \(e^{A}|a\rangle = \Sigma_n (1/n!) (A^n|a\rangle)\) = \(\Sigma_n (1/n!)(a^n |a\rangle)\) by taking into account the eigen equation whose result is \(a^n |a\rangle\) for \(A^n|a\rangle\). The summation then reduces to \(e^{a}|a\rangle\), the power series expansion of \(e^{a}\) where all instances of \(x\) have been replaced with \(a\). This implies that \(|a\rangle\) is also an eigenstate of \(e^{A}\).
04

Find the Eigenvalue of the Exponentiated Operator

As shown in step 3, \(e^{A}|a\rangle = e^{a}|a\rangle\). This implies that the eigenvalue for operator \(e^{A}\) corresponding to eigenstate \(|a\rangle\) is \(e^{a}\).

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Most popular questions from this chapter

Find the operator for position \(x\) if the operator for momentum \(p\) is taken to be \((\hbar / 2 m)^{1 / 2}(A+B),\) with \([A, B]=1\) and all other commutators zero. Hint. Write \(x=a A+b B\) and find one set of solutions for \(a\) and \(b\)

Evaluate the expectation values of the operators \(p_{x}\) and \(p_{x}^{2}\) for a particle with wavefunction \((2 / L)^{1 / 2} \sin (\pi x / L)\) in the range 0 to \(L\)

(a) Show that \(\mathrm{e}^{A} \mathrm{e}^{B}=\mathrm{e}^{A+B}\) only if \([A, B]=0 .\) (b) If \([A, B] \neq 0\) but \([A,[A, B]]=[B,[A, B]]=0,\) show that \(\mathrm{e}^{A} \mathrm{e}^{B}=\) \(\mathrm{e}^{A+B} \mathrm{e}^{f},\) where \(f\) is a simple function of \([A, B] .\) Hint. This is another example of the differences between operators \((q-\text { numbers })\) and ordinary numbers (c-numbers). The simplest approach is to expand the exponentials and to collect and compare terms on both sides of the equality. Note that \(\mathrm{e}^{A} \mathrm{e}^{B}\) will give terms like \(2 A B\) while \(\mathrm{e}^{A+B}\) will give \(A B+B A .\) Be careful with order.

Show that if the Schrödinger equation had the form of a true wave equation, then the integrated probability would be time dependent. Hint. A wave equation has \(\kappa \partial^{2} / \partial t^{2}\) in place of it a chere \(\kappa\) is a constant with the appropriate dimensions (what are they?). Solve the time component of the separable equation and investigate the behaviour of \(\int \Psi^{*} \Psi d \tau\)

Are the linear combinations \(2 x-y-z, 2 y-x-z\) \(2 z-x-y\) linearly independent?
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