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Show that a hermitian operator is positive iff its eigenvalues are positive.

Short Answer

Expert verified
A hermitian operator \( H \) is positive if and only if all its eigenvalues are positive. This is proven by showing that if \( H \) has only positive eigenvalues, it is a positive operator, and if \( H \) is a positive hermitian operator, all its eigenvalues are positive.

Step by step solution

01

Define Terms

First, clarify the meanings of the important terms in this problem. A hermitian operator \( H \) is defined as \( H = H^† \), which means it equals its own hermitian conjugate. A positive operator \( H \) has the property that \( (Ψ|H|Ψ) ≥ 0 \) for any wave function \( Ψ \). An eigenvalue is a scalar \( λ \) for which there exists a non-zero wave function \( Ψ \) such that \( H|Ψ) = λ|Ψ) \).
02

Showing a hermitian operator is positive if its eigenvalues are positive

Assume \( H \) has only positive eigenvalues \( λ \). Because \( H \) is hermitian, it is guaranteed to have a complete set of orthogonal eigenfunctions. Assume \( |Ψ) is an arbitrary state vector that can be written as a superposition of these eigenfunctions. Then if we operate with \( H \) on \( |Ψ) \), the result \( (Ψ|H|Ψ) \) is a sum of positive terms (since \( λ \) is positive), and hence the operator \( H \) is positive.
03

Showing if a hermitian operator is positive, its eigenvalues are positive

Now, take \( H \) to be a positive operator. For any of its eigenvectors \( |u), H|u) = λ|u) \). Taking the inner product of both sides of this equation with \( |u) gives \( (u|H|u) = λ(u|u) \). Because \( (u|u) > 0 \) and \( (u|H|u) ≥ 0 \) as \( H \) is positive, it is guaranteed that \( λ ≥ 0 \). Therefore, all eigenvalues of \( H \) are positive.

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Key Concepts

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

Eigenvalues
Eigenvalues are important in understanding how operators work in quantum mechanics. They are scalar values, typically represented by the symbol \( \lambda \). These scalars help in determining how operators affect vectors in a given space. To find an eigenvalue, we look for non-zero wave functions \( |\Psi\rangle \) such that when an operator \( H \) acts on \(|\Psi\rangle\), the result is simply the wave function scaled by the eigenvalue \( \lambda \), expressed mathematically as:
  • \( H|\Psi\rangle = \lambda|\Psi\rangle \)
If \( \lambda \) is positive for every \( |\Psi\rangle \), then after applying the operator, the original wave function remains in the same direction, just scaled. If complex or negative \( \lambda \) would be allowed, the nature of the operation could be changing the wave function's direction, introducing more complexities.
Positive operator
A positive operator is a special type of operator with a significant property that simplifies many mathematical processes in quantum mechanics. Specifically, for a positive operator \( H \), the inner product with any vector \( |\Psi\rangle \) results in a non-negative value:
  • \( (\Psi|H|\Psi) \geq 0 \)
This condition ensures that the outcomes of this operation do not diverge into negative values, meaning that every influence by this operator shifts probability amplitudes in a consistent and predictable manner. This property is crucial for ensuring that certain physical quantities (like energy) remain positive, and it correlates directly with the eigenvalues being positive, as shown by:
  • Positive eigenvalues result in positive operations critically reflecting stability and physical realism in many systems.
Hermitian conjugate
The Hermitian conjugate of an operator, often denoted as \( H^† \), plays a crucial role in quantum mechanics as it relates closely to measurable quantities. An operator \( H \) is Hermitian if it equals its own Hermitian conjugate, i.e., \( H = H^† \). This characteristic implies that:
  • All eigenvalues of Hermitian operators are real numbers.
  • They have a complete set of orthogonal eigenvectors.
These properties make Hermitian operators fundamental in defining observables in quantum physics, such as position, momentum, and energy, which must yield real, measurable values. The connection with the Hermitian conjugate ensures that the mathematical operations on quantum states closely align with physically realizable observations, thus solidifying its importance in both theoretical and applied physics.

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Most popular questions from this chapter

Find the unitary matrices that diagonalize the following hermitian matrices: $$\begin{array}{c}\mathrm{A}_{1}=\left(\begin{array}{cc}2 & -1+i \\\\-1-i & -1\end{array}\right), \quad \mathrm{A}_{2}=\left(\begin{array}{cc}3 & i \\\\-i & 3\end{array}\right), \quad \mathrm{A}_{3}=\left(\begin{array}{cc}1 & -i \\\i & 0\end{array}\right) \\\\\mathrm{B}_{1}=\left(\begin{array}{ccc} 1 & -1 & -i \\\\-1 & 0 & i \\\i & -i & -1\end{array}\right), \quad \mathrm{B}_{2}=\left(\begin{array}{ccc}2 & 0 & i \\ 0 & -1 & -i \\\\-i & i & 0\end{array}\right) .\end{array}$$

Show that a subspace \(\mathcal{M}\) of an inner product space \(\mathcal{V}\) is invariant under the linear operator \(\mathbf{A}\) if and only if \(\mathcal{M}^{\perp}\) is invariant under \(\mathbf{A}^{\dagger}\).

Show that (a) the coefficient of \(\lambda^{N}\) in the characteristic polynomial of any linear operator is \((-1)^{N}\), where \(N=\operatorname{dim} \mathcal{V}\), and (b) the constant in the characteristic polynomial of an operator is its determinant.

What are the spectral decompositions of \(\mathbf{A}^{\dagger}, \mathbf{A}^{-1}\), and \(\mathbf{A A}^{\dagger}\) for an invertible normal operator \(\mathbf{A}\) ?

Find the polar decomposition of the following matrices: $$\mathrm{A}=\left(\begin{array}{cc}2 i & 0 \\\\\sqrt{7} & 3 \end{array}\right), \quad \mathrm{B}=\left(\begin{array}{cc}41 & -12 i \\ 12 i & 34\end{array}\right), \quad \mathrm{C}=\left(\begin{array}{ccc}1 & 0 & 1 \\\0 & 1 & -i \\ 1 & i & 0\end{array}\right)$$

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