Why Are Eigenvalues of Unitary Operators Pure Phases?

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SUMMARY

All eigenvalues of a unitary operator are pure phases, as demonstrated through the relationship between eigenfunctions and their corresponding eigenvalues. Specifically, if \( U \) is a unitary operator, then \( U^\dagger U = I \), leading to the conclusion that \( |\lambda| = 1 \) for eigenvalues \( \lambda \). Additionally, it is established that if \( M \) is a Hermitian operator, then \( e^{iM} \) is a unitary operator, satisfying the condition \( UU^\dagger = I \).

PREREQUISITES
  • Understanding of unitary operators and their properties
  • Knowledge of Hermitian operators and their significance in quantum mechanics
  • Familiarity with eigenvalues and eigenfunctions in linear algebra
  • Basic concepts of matrix exponentiation, particularly \( e^{A} \)
NEXT STEPS
  • Study the properties of Hermitian operators in quantum mechanics
  • Learn about the spectral theorem and its implications for unitary operators
  • Explore the concept of matrix exponentiation in greater depth
  • Investigate the applications of unitary operators in quantum computing
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Students of quantum mechanics, mathematicians focusing on linear algebra, and physicists interested in the properties of unitary operators and their applications in quantum theory.

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Homework Statement



We only briefly mentioned this in class and now its on our problem set...

Show that all eigenvalues i of a Unitary operator are pure phases.
Suppose M is a Hermitian operator. Show that e^iM is a Unitary operator.

Homework Equations





The Attempt at a Solution



Uf = λf where is is an eigenfunction, U dagger = U inverse
multiply by either maybe...
 
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Uf = λf

denote your inner product (a,b)

a.
(f, U^\dagger U f) = ?
calculate this two ways (in terms of λ and λ*)
(f, U^\dagger U f ) = (Uf, Uf) = ?
(f, U^\dagger U f ) = (f, f) [since U^\dagger = U^{-1} ]
so what does this say about λ and λ*?

b.
The second part follows from

1. \left( e^A \right)^\dagger<br /> = \left( 1 + A + \frac{1}{2!}A^2 + \cdots \right)^\dagger<br /> = \left( 1 + A^\dagger + \frac{1}{2!}(A^\dagger)^2 + \cdots \right)<br /> = e^{(A^\dagger)}.

2. For commuting matrices (operators)
e^A e^B = e^{(A+B)} .

now you need to show show U = e^(iM) satisfies UU^(dagger) = 1.

can you fill in the rest?
 

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